CS代考程序代写 database AI flex computational biology chain prolog algorithm DNA data structure ER interpreter Excel scheme Algorithms

34 Algorithms Fermat’s little theorem If p is prime, then for every 1 ≤ a < p, ap−1 ≡ 1 (mod p). Proof. Let S be the nonzero integers modulo p; that is, S = {1, 2, . . . , p − 1}. Here’s the crucial observation: the effect of multiplying these numbers by a (modulo p) is simply to permute them. For instance, here’s a picture of the case a = 3, p = 7: 11 22 33 44 55 66 Let’s carry this example a bit further. From the picture, we can conclude {1,2,...,6} = {3·1mod7, 3·2mod7,..., 3·6mod7}. Multiplying all the numbers in each representation then gives 6! ≡ 36 ·6! (mod 7), and dividing by6!weget36 ≡1 (mod7),exactlytheresultwewantedinthecasea=3,p=7. Now let’s generalize this argument to other values of a and p, with S = {1, 2, . . . , p − 1}. We’ll prove that when the elements of S are multiplied by a modulo p, the resulting numbers are all distinct and nonzero. And since they lie in the range [1, p − 1], they must simply be a permutation of S. The numbers a · i mod p are distinct because if a · i ≡ a · j (mod p), then dividing both sides byagivesi≡j (modp). Theyarenonzerobecausea·i≡0similarlyimpliesi≡0. (Andwe can divide by a, because by assumption it is nonzero and therefore relatively prime to p.) We now have two ways to write set S: S = {1,2,...,p−1} = {a·1modp, a·2modp,..., a·(p−1)modp}. We can multiply together its elements in each of these representations to get (p−1)!≡ap−1·(p−1)! (modp). Dividing by (p − 1)! (which we can do because it is relatively prime to p, since p is assumed prime) then gives the theorem. This theorem suggests a “factorless” test for determining whether a number N is prime: S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani Figure 1.7 An algorithm for testing primality. function primality(N) Input: Positive integer N Output: yes/no Pick a positive integer a < N at random if aN−1 ≡ 1 (mod N): 35 return yes else: return no Pick some a Pass Fail Fermat’s test “prime” “composite” IsaN−1 ≡1modN? The problem is that Fermat’s theorem is not an if-and-only-if condition; it doesn’t say what happens when N is not prime, so in these cases the preceding diagram is questionable. In fact, it is possible for a composite number N to pass Fermat’s test (that is, aN−1 ≡ 1 mod N) for certain choices of a. For instance, 341 = 11 · 31 is not prime, and yet 2340 ≡ 1 mod 341. Nonetheless, we might hope that for composite N, most values of a will fail the test. This is indeed true, in a sense we will shortly make precise, and motivates the algorithm of Figure 1.7: rather than fixing an arbitrary value of a in advance, we should choose it randomly from {1, . . . , N − 1}. In analyzing the behavior of this algorithm, we first need to get a minor bad case out of the way. It turns out that certain extremely rare composite numbers N, called Carmichael num- bers, pass Fermat’s test for all a relatively prime to N. On such numbers our algorithm will fail; but they are pathologically rare, and we will later see how to deal with them (page 38), so let’s ignore these numbers for the time being. In a Carmichael-free universe, our algorithm works well. Any prime number N will of course pass Fermat’s test and produce the right answer. On the other hand, any non- Carmichael composite number N must fail Fermat’s test for some value of a; and as we will now show, this implies immediately that N fails Fermat’s test for at least half the possible values of a! Lemma If aN−1 ̸≡ 1 mod N for some a relatively prime to N, then it must hold for at least half the choices of a < N. Proof. FixsomevalueofaforwhichaN−1̸≡1modN.Thekeyistonoticethateveryelement b < N that passes Fermat’s test with respect to N (that is, bN−1 ≡ 1 mod N) has a twin, a · b, that fails the test: (a·b)N−1 ≡aN−1 ·bN−1 ≡aN−1 ̸≡1modN. 36 Algorithms Moreover, all these elements a · b, for fixed a but different choices of b, are distinct, for the same reason a · i ̸≡ a · j in the proof of Fermat’s test: just divide by a. The set {1,2,...,N − 1} b Pass a·b Fail The one-to-one function b 􏰚→ a · b shows that at least as many elements fail the test as pass it. Hey, that was group theory! For any integer N, the set of all numbers mod N that are relatively prime to N constitute what mathematicians call a group: • There is a multiplication operation defined on this set. • The set contains a neutral element (namely 1: any number multiplied by this remains unchanged). • All elements have a well-defined inverse. This particular group is called the multiplicative group of N, usually denoted Z∗N. Group theory is a very well developed branch of mathematics. One of its key concepts is that a group can contain a subgroup—a subset that is a group in and of itself. And an important fact about a subgroup is that its size must divide the size of the whole group. ConsidernowthesetB={b:bN−1 ≡1modN}.Itisnothardtoseethatitisasubgroup of Z∗N (just check that B is closed under multiplication and inverses). Thus the size of B must divide that of Z∗N . Which means that if B doesn’t contain all of Z∗N , the next largest size it can have is |Z∗N |/2. We are ignoring Carmichael numbers, so we can now assert IfN isprime,thenaN−1 ≡1modN foralla 10 where the exponentiation can be performed using fewer multiplications, by some other method.
1.17. Consider the problem of computing xy for given integers x and y: we want the whole answer, not modulo a third integer. We know two algorithms for doing this: the iterative algorithm which performs y − 1 multiplications by x; and the recursive algorithm based on the binary expansion of y.
Compare the time requirements of these two algorithms, assuming that the time to multiply an n-bit number by an m-bit number is O(mn).
1.18. Compute gcd(210, 588) two different ways: by finding the factorization of each number, and by using Euclid’s algorithm.
1.19. The Fibonacci numbers F0, F1, . . . are given by the recurrence Fn+1 = Fn + Fn−1, F0 = 0, F1 = 1. Show that for any n ≥ 1, gcd(Fn+1, Fn) = 1.
1.20. Find the inverse of: 20 mod 79, 3 mod 62, 21 mod 91, 5 mod 23.
1.21. How many integers modulo 113 have inverses? (Note: 113 = 1331.)
1.22. Prove or disprove: If a has an inverse modulo b, then b has an inverse modulo a.
1.23. Show that if a has a multiplicative inverse modulo N , then this inverse is unique (modulo N ).
1.24. If p is prime, how many elements of {0, 1, . . . , pn − 1} have an inverse modulo pn?
1.25. Calculate 2125 mod 127 using any method you choose. (Hint: 127 is prime.)
1.26. What is the least significant decimal digit of 171717 ? (Hint: For distinct primes p, q, and any a ̸≡ 0 (mod pq), we proved the formula a(p−1)(q−1) ≡ 1 (mod pq) in Section 1.4.2.)
1.27. Consider an RSA key set with p = 17,q = 23,N = 391, and e = 3 (as in Figure 1.9). What value of d should be used for the secret key? What is the encryption of the message M = 41?
1.28. In an RSA cryptosystem, p = 7 and q = 11 (as in Figure 1.9). Find appropriate exponents d and e.
1.29. Let [m] denote the set {0, 1, . . . , m − 1}. For each of the following families of hash functions, say whether or not it is universal, and determine how many random bits are needed to choose a function from the family.
(a) H = {ha1,a2 : a1,a2 ∈ [m]}, where m is a fixed prime and ha1,a2(x1,x2)=a1x1 +a2x2 modm.
Notice that each of these functions has signature h of integers in [m] to a single integer in [m].
a1 ,a2
: [m]2 → [m], that is, it maps a pair
(b) H is as before, except that now m = 2k is some fixed power of 2. (c) H isthesetofallfunctionsf :[m]→[m−1].

50 Algorithms
1.30. The grade-school algorithm for multiplying two n-bit binary numbers x and y consists of adding together n copies of x, each appropriately left-shifted. Each copy, when shifted, is at most 2n bits long.
In this problem, we will examine a scheme for adding n binary numbers, each m bits long, using a circuit or a parallel architecture. The main parameter of interest in this question is therefore the depth of the circuit or the longest path from the input to the output of the circuit. This determines the total time taken for computing the function.
To add two m-bit binary numbers naively, we must wait for the carry bit from position i − 1 before we can figure out the ith bit of the answer. This leads to a circuit of depth O(m). However carry lookahead circuits (see wikipedia.com if you want to know more about this) can add in O(log m) depth.
(a) Assuming you have carry lookahead circuits for addition, show how to add n numbers each m bits long using a circuit of depth O((log n)(log m)).
(b) Whenaddingthreem-bitbinarynumbersx+y+z,thereisatrickwecanusetoparallelize the process. Instead of carrying out the addition completely, we can re-express the result as the sum of just two binary numbers r + s, such that the ith bits of r and s can be computed independently of the other bits. Show how this can be done. (Hint: One of the numbers represents carry bits.)
(c) Show how to use the trick from the previous part to design a circuit of depth O(log n) for multiplying two n-bit numbers.
1.31. ConsidertheproblemofcomputingN!=1·2·3···N.
(a) If N is an n-bit number, how many bits long is N !, approximately (in Θ(·) form)?
(b) Give an algorithm to compute N ! and analyze its running time.
1.32. A positive integer N is a power if it is of the form qk, where q, k are positive integers and k > 1.
(a) Give an efficient algorithm that takes as input a number N and determines whether it is a square, that is, whether it can be written as q2 for some positive integer q. What is the running time of your algorithm?
(b) Show that if N = qk (with N, q, and k all positive integers), then either k ≤ logN or N = 1.
(c) GiveanefficientalgorithmfordeterminingwhetherapositiveintegerNisapower.Analyze
its running time.
1.33. Give an efficient algorithm to compute the least common multiple of two n-bit numbers x and y, that is, the smallest number divisible by both x and y. What is the running time of your algorithm as a function of n?
1.34. On page 38, we claimed that since about a 1/n fraction of n-bit numbers are prime, on average it is sufficient to draw O(n) random n-bit numbers before hitting a prime. We now justify this rigorously.
Suppose a particular coin has a probability p of coming up heads. How many times must you toss it, on aver􏰐age, before it comes up heads? (Hint: Method 1: start by showing that the correct expression is ∞i=1 i(1 − p)i−1p. Method 2: if E is the average number of coin tosses, show that E = 1 + (1 − p)E.)
1.35. Wilson’s theorem says that a number N is prime if and only if (N−1)!≡−1 (modN).

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 51
(a) If p is prime, then we know every number 1 ≤ x < p is invertible modulo p. Which of these numbers are their own inverse? (b) By pairing up multiplicative inverses, show that (p − 1)! ≡ −1 (mod p) for prime p. (c) Show that if N is not prime, then (N − 1)! ̸≡ −1 (mod N ). (Hint: Consider d = gcd(N, (N − 1)!).) (d) UnlikeFermat’sLittletheorem,Wilson’stheoremisanif-and-only-ifconditionforprimality. Why can’t we immediately base a primality test on this rule? 1.36. Square roots. In this problem, we’ll see that it is easy to compute square roots modulo a prime p with p ≡ 3 (mod 4). (a) Suppose p ≡ 3 (mod 4). Show that (p + 1)/4 is an integer. (b) Wesayxisasquarerootofamodulopifa≡x2 (modp).Showthatifp≡3(mod4)andif a has a square root modulo p, then a(p+1)/4 is such a square root. 1.37. The Chinese remainder theorem. (a) Make a table with three columns. The first column is all numbers from 0 to 14. The second is the residues of these numbers modulo 3; the third column is the residues modulo 5. What do you observe? (b) Prove that if p and q are distinct primes, then for every pair (j,k) with 0 ≤ j < p and 0 ≤ k < q, there is a unique integer 0 ≤ i < pq such that i ≡ j mod p and i ≡ k mod q. (Hint: Prove that no two different i’s in this range can have the same (j, k), and then count.) (c) Inthisone-to-onecorrespondencebetweenintegersandpairs,itiseasytogofromito(j,k). Prove that the following formula takes you the other way: i = {j·q·(q−1 modp)+k·p·(p−1 modq)}modpq. (d) Can you generalize parts (b) and (c) to more than two primes? 1.38. To see if a number, say 562437487, is divisible by 3, you just add up the digits of its decimal representation,andseeiftheresultisdivisibleby3. (5+6+2+4+3+7+4+8+7=46,soit is not divisible by 3.) To see if the same number is divisible by 11, you can do this: subdivide the number into pairs of digits, from the right-hand end (87, 74, 43, 62, 5), add these numbers, and see if the sum is divisible by 11 (if it’s too big, repeat). How about 37? To see if the number is divisible by 37, subdivide it into triples from the end (487, 437, 562) add these up, and see if the sum is divisible by 37. This is true for any prime p other than 2 and 5. That is, for any prime p ̸= 2, 5, there is an integer r such that in order to see if p divides a decimal number n, we break n into r-tuples of decimal digits (starting from the right-hand end), add up these r-tuples, and check if the sum is divisible by p. (a) Whatisthesmallestsuchrforp=13?Forp=17? (b) Showthatrisadivisorofp−1. 1.39. Give a polynomial-time algorithm for computing abc mod p, given a, b, c, and prime p. 52 Algorithms 1.40. Show that if x is a nontrivial square root of 1 modulo N, that is, if x2 ≡ 1modN but x ̸≡ ±1 mod N, then N must be composite. (For instance, 42 ≡ 1 mod 15 but 4 ̸≡ ±1 mod 15; thus 4 is a nontrivial square root of 1 modulo 15.) 1.41. Quadratic residues. Fix a positive integer N. We say that a is a quadratic residue modulo N if there exists x such that a ≡ x2 mod N. (a) Let N be an odd prime and a be a non-zero quadratic residue modulo N. Show that there areexactlytwovaluesin{0,1,...,N−1}satisfyingx2 ≡amodN. (b) ShowthatifNisanoddprime,thereareexactly(N+1)/2quadraticresiduesin{0,1,...,N− 1}. (c) Give an example of positive integers a and N such that x2 ≡ a mod N has more than two solutions in {0,1,...,N −1}. 1.42. Suppose that instead of using a composite N = pq in the RSA cryptosystem (Figure 1.9), we simply use a prime modulus p. As in RSA, we would have an encryption exponent e, and the encryption of a message m mod p would be me mod p. Prove that this new cryptosystem is not secure, by giving an efficient algorithm to decrypt: that is, an algorithm that given p, e, and me mod p as input, computes m mod p. Justify the correctness and analyze the running time of your decryption algorithm. 1.43. In the RSA cryptosystem, Alice’s public key (N,e) is available to everyone. Suppose that her private key d is compromised and becomes known to Eve. Show that if e = 3 (a common choice) then Eve can efficiently factor N. 1.44. Alice and her three friends are all users of the RSA cryptosystem. Her friends have public keys (Ni, ei = 3), i = 1, 2, 3, where as always, Ni = piqi for randomly chosen n-bit primes pi, qi. Show that if Alice sends the same n-bit message M (encrypted using RSA) to each of her friends, then anyone who intercepts all three encrypted messages will be able to efficiently recover M . (Hint: It helps to have solved problem 1.37 first.) 1.45. RSA and digital signatures. Recall that in the RSA public-key cryptosystem, each user has a public key P = (N, e) and a secret key d. In a digital signature scheme, there are two algorithms, sign and verify. The sign procedure takes a message and a secret key, then outputs a signa- ture σ. The verify procedure takes a public key (N, e), a signature σ, and a message M , then returns “true” if σ could have been created by sign (when called with message M and the secret key corresponding to the public key (N, e)); “false” otherwise. (a) Why would we want digital signatures? (b) An RSA signature consists of sign(M, d) = M d (mod N ), where d is a secret key and N is part of the public key. Show that anyone who knows the public key (N,e) can perform verify((N,e),Md,M), i.e., they can check that a signature really was created by the pri- vate key. Give an implementation and prove its correctness. (c) Generate your own RSA modulus N = pq, public key e, and private key d (you don’t need to use a computer). Pick p and q so you have a 4-digit modulus and work by hand. Now sign your name using the private exponent of this RSA modulus. To do this you will need to specify some one-to-one mapping from strings to integers in [0, N − 1]. Specify any mapping you like. Give the mapping from your name to numbers m1,m2,...mk, then sign the first number by giving the value md1 (mod N), and finally show that (md1)e = m1 (mod N). (d) Alice wants to write a message that looks like it was digitally signed by Bob. She notices that Bob’s public RSA key is (17, 391). To what exponent should she raise her message? S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 53 1.46. Digital signatures, continued. Consider the signature scheme of Exercise 1.45. (a) Signinginvolvesdecryption,andisthereforerisky.ShowthatifBobagreestosignanything he is asked to, Eve can take advantage of this and decrypt any message sent by Alice to Bob. (b) Suppose that Bob is more careful, and refuses to sign messages if their signatures look suspiciously like text. (We assume that a randomly chosen message—that is, a random number in the range {1, . . . , N − 1}—is very unlikely to look like text.) Describe a way in which Eve can nevertheless still decrypt messages from Alice to Bob, by getting Bob to sign messages whose signatures look random. 54 Algorithms Chapter 2 Divide-and-conquer algorithms The divide-and-conquer strategy solves a problem by: 1. Breaking it into subproblems that are themselves smaller instances of the same type of problem 2. Recursively solving these subproblems 3. Appropriately combining their answers The real work is done piecemeal, in three different places: in the partitioning of problems into subproblems; at the very tail end of the recursion, when the subproblems are so small that they are solved outright; and in the gluing together of partial answers. These are held together and coordinated by the algorithm’s core recursive structure. As an introductory example, we’ll see how this technique yields a new algorithm for multi- plying numbers, one that is much more efficient than the method we all learned in elementary school! 2.1 Multiplication The mathematician Carl Friedrich Gauss (1777–1855) once noticed that although the product of two complex numbers (a+bi)(c+di) = ac−bd+(bc+ad)i seems to involve four real-number multiplications, it can in fact be done with just three: ac, bd, and (a + b)(c + d), since bc+ad = (a+b)(c+d)−ac−bd. In our big-O way of thinking, reducing the number of multiplications from four to three seems wasted ingenuity. But this modest improvement becomes very significant when applied recur- sively. 55 56 Algorithms Let’s move away from complex numbers and see how this helps with regular multiplica- tion. Suppose x and y are two n-bit integers, and assume for convenience that n is a power of 2 (the more general case is hardly any different). As a first step toward multiplying x and y, split each of them into their left and right halves, which are n/2 bits long: x= =2n/2xL +xR y= =2n/2yL +yR. For instance, if x = 101101102 (the subscript 2 means “binary”) then xL = 10112, xR = 01102, and x = 10112 × 24 + 01102. The product of x and y can then be rewritten as xy = (2n/2xL + xR)(2n/2yL + yR) = 2n xLyL + 2n/2 (xLyR + xRyL) + xRyR. We will compute xy via the expression on the right. The additions take linear time, as do the multiplications by powers of 2 (which are merely left-shifts). The significant operations are the four n/2-bit multiplications, xLyL, xLyR, xRyL, xRyR; these we can handle by four recursive calls. Thus our method for multiplying n-bit numbers starts by making recursive calls to multiply these four pairs of n/2-bit numbers (four subproblems of half the size), and then evaluates the preceding expression in O(n) time. Writing T(n) for the overall running time on n-bit inputs, we get the recurrence relation T(n) = 4T(n/2)+O(n). We will soon see general strategies for solving such equations. In the meantime, this particu- lar one works out to O(n2), the same running time as the traditional grade-school multiplica- tion technique. So we have a radically new algorithm, but we haven’t yet made any progress in efficiency. How can our method be sped up? This is where Gauss’s trick comes to mind. Although the expression for xy seems to de- mandfourn/2-bitmultiplications,asbeforejustthreewilldo: xLyL,xRyR,and(xL+xR)(yL+yR), sincexLyR+xRyL =(xL+xR)(yL+yR)−xLyL−xRyR.Theresultingalgorithm,showninFigure2.1, has an improved running time of1 T(n) = 3T(n/2)+O(n). The point is that now the constant factor improvement, from 4 to 3, occurs at every level of the recursion, and this compounding effect leads to a dramatically lower time bound of O(n1.59). This running time can be derived by looking at the algorithm’s pattern of recursive calls, which form a tree structure, as in Figure 2.2. Let’s try to understand the shape of this tree. At each successive level of recursion the subproblems get halved in size. At the (log2 n)th level, 1Actually, the recurrence should read T(n) ≤ 3T(n/2 + 1) + O(n) since the numbers (xL + xR) and (yL + yR) could be n/2 + 1 bits long. The one we’re using is simpler to deal with and can be seen to imply exactly the same big-O running time. xL xR yL yR S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 57 Figure 2.1 A divide-and-conquer algorithm for integer multiplication. function multiply(x, y) Input: Positive integers x and y, in binary Output: Their product n = max(size of x, size of y) if n = 1: return xy xL, xR = leftmost ⌈n/2⌉, rightmost ⌊n/2⌋ bits of x yL, yR = leftmost ⌈n/2⌉, rightmost ⌊n/2⌋ bits of y P1 =multiply(xL,yL) P2 =multiply(xR,yR) P3 =multiply(xL +xR,yL +yR) return P1 ×2n +(P3 −P1 −P2)×2n/2 +P2 the subproblems get down to size 1, and so the recursion ends. Therefore, the height of the tree is log2 n. The branching factor is 3—each problem recursively produces three smaller ones—with the result that at depth k in the tree there are 3k subproblems, each of size n/2k. For each subproblem, a linear amount of work is done in identifying further subproblems and combining their answers. Therefore the total time spent at depth k in the tree is k 􏰄n􏰅 􏰆3􏰇k 3 ×O 2k = 2 ×O(n). At the very top level, when k = 0, this works out to O(n). At the bottom, when k = log2 n, it is O(3log2 n), which can be rewritten as O(nlog2 3) (do you see why?). Between these two endpoints, the work done increases geometrically from O(n) to O(nlog2 3), by a factor of 3/2 per level. The sum of any increasing geometric series is, within a constant factor, simply the last term of the series: such is the rapidity of the increase (Exercise 0.2). Therefore the overall running time is O(nlog2 3), which is about O(n1.59). In the absence of Gauss’s trick, the recursion tree would have the same height, but the branching factor would be 4. There would be 4log2 n = n2 leaves, and therefore the running time would be at least this much. In divide-and-conquer algorithms, the number of subprob- lems translates into the branching factor of the recursion tree; small changes in this coefficient can have a big impact on running time. A practical note: it generally does not make sense to recurse all the way down to 1 bit. For most processors, 16- or 32-bit multiplication is a single operation, so by the time the numbers get into this range they should be handed over to the built-in procedure. Finally, the eternal question: Can we do better? It turns out that even faster algorithms for multiplying numbers exist, based on another important divide-and-conquer algorithm: the fast Fourier transform, to be explained in Section 2.6. 58 Algorithms Figure 2.2 Divide-and-conquer integer multiplication. (a) Each problem is divided into three subproblems. (b) The levels of recursion. (a) (b) 1011 × 0110 0010 × 0011 1101 × 1001 10110010 × 01100011 Size n Size n/2 Size n/4 . ··· ··· log n levels 2 2 2 2 1 11 1 11 1 11 1 11 2.2 Recurrence relations Divide-and-conquer algorithms often follow a generic pattern: they tackle a problem of size n by recursively solving, say, a subproblems of size n/b and then combining these answers in O(nd) time, for some a, b, d > 0 (in the multiplication algorithm, a = 3, b = 2, and d = 1). Their running time can therefore be captured by the equation T (n) = aT (⌈n/b⌉) + O(nd). We next derive a closed-form solution to this general recurrence so that we no longer have to solve it explicitly in each new instance.
Master theorem2 If T(n) = aT(⌈n/b⌉) + O(nd) for some constants a > 0, b > 1, and d ≥ 0, 2There are even more general results of this type, but we will not be needing them.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani
Figure 2.3 Each problem of size n is divided into a subproblems of size n/b.
Size n Size n/b Size n/b2
Size 1
59
Branching factor a
Depth
logb n
Width alogb n = nlogb a
then
 O(nd) if d > logb a T(n) = O(ndlogn) ifd=logba
O(nlogb a) if d < logb a . This single theorem tells us the running times of most of the divide-and-conquer procedures we are likely to use. Proof. To prove the claim, let’s start by assuming for the sake of convenience that n is a power of b. This will not influence the final bound in any important way—after all, n is at most a multiplicative factor of b away from some power of b (Exercise 2.2)—and it will allow us to ignore the rounding effect in ⌈n/b⌉. Next, notice that the size of the subproblems decreases by a factor of b with each level of recursion, and therefore reaches the base case after logb n levels. This is the height of the recursion tree. Its branching factor is a, so the kth level of the tree is made up of ak subproblems, each of size n/bk (Figure 2.3). The total work done at this level is ak×O􏰄n􏰅d = O(nd)×􏰄a􏰅k. bk bd As k goes from 0 (the root) to logb n (the leaves), these numbers form a geometric series with 60 Algorithms ratio a/bd. Finding the sum of such a series in big-O notation is easy (Exercise 0.2), and comes down to three cases. 1. The ratio is less than 1. Then the series is decreasing, and its sum is just given by its first term, O(nd). 2. The ratio is greater than 1. The series is increasing and its sum is given by its last term, O(nlogb a): nd􏰄a􏰅logbn = nd􏰆 alogbn 􏰇 = alogbn = a(logan)(logba) = nlogba. bd (blogb n)d 3. The ratio is exactly 1. In this case all O(log n) terms of the series are equal to O(nd). These cases translate directly into the three contingencies in the theorem statement. Binary search The ultimate divide-and-conquer algorithm is, of course, binary search: to find a key k in a large file containing keys z[0, 1, . . . , n − 1] in sorted order, we first compare k with z[n/2], and depending on the result we recurse either on the first half of the file, z[0, . . . , n/2 − 1], or on the second half, z[n/2, . . . , n − 1]. The recurrence now is T (n) = T (⌈n/2⌉) + O(1), which is the case a = 1, b = 2, d = 0. Plugging into our master theorem we get the familiar solution: a running time of just O(log n). 2.3 Mergesort The problem of sorting a list of numbers lends itself immediately to a divide-and-conquer strategy: split the list into two halves, recursively sort each half, and then merge the two sorted sublists. function mergesort(a[1 . . . n]) Input: Anarrayofnumbersa[1...n] Output: A sorted version of this array if n > 1:
return merge(mergesort(a[1 . . . ⌊n/2⌋]), mergesort(a[⌊n/2⌋ + 1 . . . n]))
else: return a
The correctness of this algorithm is self-evident, as long as a correct merge subroutine is specified. If we are given two sorted arrays x[1 . . . k] and y[1 . . . l], how do we efficiently merge them into a single sorted array z[1 . . . k + l]? Well, the very first element of z is either x[1] or y[1], whichever is smaller. The rest of z[·] can then be constructed recursively.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani
Figure 2.4 The sequence of merge operations in mergesort. Input:
61
10
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13
1
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10
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.
function merge(x[1 . . . k], y[1 . . . l]) if k = 0: return y[1…l]
if l = 0: return x[1…k]
if x[1] ≤ y[1]:
return x[1]◦merge(x[2…k],y[1…l]) else:
return y[1]◦merge(x[1…k],y[2…l])
Here ◦ denotes concatenation. This merge procedure does a constant amount of work per recursive call (provided the required array space is allocated in advance), for a total running time of O(k + l). Thus merge’s are linear, and the overall time taken by mergesort is
T (n) = 2T (n/2) + O(n),
or O(n log n).
Looking back at the mergesort algorithm, we see that all the real work is done in merg- ing, which doesn’t start until the recursion gets down to singleton arrays. The singletons are merged in pairs, to yield arrays with two elements. Then pairs of these 2-tuples are merged, producing 4-tuples, and so on. Figure 2.4 shows an example.
This viewpoint also suggests how mergesort might be made iterative. At any given mo- ment, there is a set of “active” arrays—initially, the singletons—which are merged in pairs to give the next batch of active arrays. These arrays can be organized in a queue, and processed by repeatedly removing two arrays from the front of the queue, merging them, and putting the result at the end of the queue.

62 Algorithms In the following pseudocode, the primitive operation inject adds an element to the end
of the queue while eject removes and returns the element at the front of the queue. function iterative-mergesort(a[1 . . . n])
Input: elementsa1,a2,…,an tobesorted
Q = [ ] (empty queue) for i=1 to n:
inject(Q, [ai ]) while |Q| > 1:
inject(Q, merge(eject(Q), eject(Q))) return eject(Q)

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 63 An n log n lower bound for sorting
Sorting algorithms can be depicted as trees. The one in the following figure sorts an array of three elements, a1, a2, a3. It starts by comparing a1 to a2 and, if the first is larger, compares it with a3; otherwise it compares a2 and a3. And so on. Eventually we end up at a leaf, and this leaf is labeled with the true order of the three elements as a permutation of 1, 2, 3. For example, if a2 < a1 < a3, we get the leaf labeled “2 1 3.” No a1 0. Therearemanywaystosee this. The easiest is to notice that n! ≥ (n/2)(n/2) because n! = 1 · 2 · · · · · n contains at least n/2 factors larger than n/2; and to then take logs of both sides. Another is to recall Stirling’s formula 􏰕
Either way, we have established that any comparison tree that sorts n elements must make, in the worst case, Ω(n log n) comparisons, and hence mergesort is optimal!
Well, there is some fine print: this neat argument applies only to algorithms that use comparisons. Is it conceivable that there are alternative sorting strategies, perhaps using sophisticated numerical manipulations, that work in linear time? The answer is yes, under certain exceptional circumstances: the canonical such example is when the elements to be sorted are integers that lie in a small range (Exercise 2.20).
􏰆 1􏰇 n −n n!≈ π 2n+3 ·n ·e .
312
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123

64 Algorithms 2.4 Medians
The median of a list of numbers is its 50th percentile: half the numbers are bigger than it, and half are smaller. For instance, the median of [45, 1, 10, 30, 25] is 25, since this is the middle element when the numbers are arranged in order. If the list has even length, there are two choices for what the middle element could be, in which case we pick the smaller of the two,
say.The purpose of the median is to summarize a set of numbers by a single, typical value. The mean, or average, is also very commonly used for this, but the median is in a sense more typical of the data: it is always one of the data values, unlike the mean, and it is less sensitive to outliers. For instance, the median of a list of a hundred 1’s is (rightly) 1, as is the mean. However, if just one of these numbers gets accidentally corrupted to 10,000, the mean shoots up above 100, while the median is unaffected.
Computing the median of n numbers is easy: just sort them. The drawback is that this takes O(n log n) time, whereas we would ideally like something linear. We have reason to be hopeful, because sorting is doing far more work than we really need—we just want the middle element and don’t care about the relative ordering of the rest of them.
When looking for a recursive solution, it is paradoxically often easier to work with a more general version of the problem—for the simple reason that this gives a more powerful step to recurse upon. In our case, the generalization we will consider is selection.
SELECTION
Input: A list of numbers S; an integer k Output: The kth smallest element of S
For instance, if k = 1, the minimum of S is sought, whereas if k = ⌊|S|/2⌋, it is the median. A randomized divide-and-conquer algorithm for selection
Here’s a divide-and-conquer approach to selection. For any number v, imagine splitting list S into three categories: elements smaller than v, those equal to v (there might be duplicates), and those greater than v. Call these SL, Sv, and SR respectively. For instance, if the array
S:
is split on v = 5, the three subarrays generated are
SL : Sv : SR :
2
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8
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The search can instantly be narrowed down to one of these sublists. If we want, say, the eighth-smallest element of S, we know it must be the third-smallest element of SR since |SL| + |Sv| = 5. That is, selection(S,8) = selection(SR,3). More generally, by checking k against the sizes of the subarrays, we can quickly determine which of them holds the desired element:
 selection(SL, k) if k ≤ |SL|
selection(S, k) = v if |SL| < k ≤ |SL| + |Sv| selection(SR,k−|SL|−|Sv|) ifk>|SL|+|Sv|.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 65
The three sublists SL,Sv, and SR can be computed from S in linear time; in fact, this compu- tation can even be done in place, that is, without allocating new memory (Exercise 2.15). We then recurse on the appropriate sublist. The effect of the split is thus to shrink the number of elements from |S| to at most max{|SL|, |SR|}.
Our divide-and-conquer algorithm for selection is now fully specified, except for the crucial detail of how to choose v. It should be picked quickly, and it should shrink the array substan- tially, the ideal situation being |SL|,|SR| ≈ 1|S|. If we could always guarantee this situation, we would get a running time of 2
T (n) = T (n/2) + O(n),
which is linear as desired. But this requires picking v to be the median, which is our ultimate
goal! Instead, we follow a much simpler alternative: we pick v randomly from S. Efficiency analysis
Naturally, the running time of our algorithm depends on the random choices of v. It is possible that due to persistent bad luck we keep picking v to be the largest element of the array (or the smallest element), and thereby shrink the array by only one element each time. In the earlier example, we might first pick v = 36, then v = 21, and so on. This worst-case scenario would force our selection algorithm to perform
n + (n − 1) + (n − 2) + · · · + n = Θ(n2) 2
operations (when computing the median), but it is extremely unlikely to occur. Equally un- likely is the best possible case we discussed before, in which each randomly chosen v just happens to split the array perfectly in half, resulting in a running time of O(n). Where, in this spectrum from O(n) to Θ(n2), does the average running time lie? Fortunately, it lies very close to the best-case time.
To distinguish between lucky and unlucky choices of v, we will call v good if it lies within the 25th to 75th percentile of the array that it is chosen from. We like these choices of v because they ensure that the sublists SL and SR have size at most three-fourths that of S (do you see why?), so that the array shrinks substantially. Fortunately, good v’s are abundant: half the elements of any list must fall between the 25th to 75th percentile!
Given that a randomly chosen v has a 50% chance of being good, how many v’s do we need to pick on average before getting a good one? Here’s a more familiar reformulation (see also Exercise 1.34):
Lemma On average a fair coin needs to be tossed two times before a “heads” is seen.
Proof. Let E be the expected number of tosses before a heads is seen. We certainly need at
least one toss, and if it’s heads, we’re done. If it’s tails (which occurs with probability 1/2), we
need to repeat. Hence E = 1 + 1 E, which works out to E = 2. 2

66 Algorithms Therefore, after two split operations on average, the array will shrink to at most three-
fourths of its size. Letting T (n) be the expected running time on an array of size n, we get T (n) ≤ T (3n/4) + O(n).
This follows by taking expected values of both sides of the following statement: Time taken on an array of size n
≤ (time taken on an array of size 3n/4) + (time to reduce array size to ≤ 3n/4),
and, for the right-hand side, using the familiar property that the expectation of the sum is the sum of the expectations.
From this recurrence we conclude that T (n) = O(n): on any input, our algorithm returns the correct answer after a linear number of steps, on the average.
The Unix sort command
Comparing the algorithms for sorting and median-finding we notice that, beyond the com- mon divide-and-conquer philosophy and structure, they are exact opposites. Mergesort splits the array in two in the most convenient way (first half, second half), without any regard to the magnitudes of the elements in each half; but then it works hard to put the sorted sub- arrays together. In contrast, the median algorithm is careful about its splitting (smaller numbers first, then the larger ones), but its work ends with the recursive call.
Quicksort is a sorting algorithm that splits the array in exactly the same way as the me- dian algorithm; and once the subarrays are sorted, by two recursive calls, there is nothing more to do. Its worst-case performance is Θ(n2), like that of median-finding. But it can be proved (Exercise 2.24) that its average case is O(n log n); furthermore, empirically it outper- forms other sorting algorithms. This has made quicksort a favorite in many applications— for instance, it is the basis of the code by which really enormous files are sorted.
2.5 Matrix multiplication
Theproductoftwon×nmatricesXandY is􏰑athirdn×nmatrixZ=XY,with(i,j)thentry n
Zij = XikYkj. k=1
To make it more visual, Zij is the dot product of the ith row of X with the jth column of Y : j
i
×= XYZ
(i,j)

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 67
In general, XY is not the same as Y X; matrix multiplication is not commutative.
The preceding formula implies an O(n3) algorithm for matrix multiplication: there are n2 entries to be computed, and each takes O(n) time. For quite a while, this was widely believed to be the best running time possible, and it was even proved that in certain models of com- putation no algorithm could do better. It was therefore a source of great excitement when in 1969, the German mathematician Volker Strassen announced a significantly more efficient
algorithm, based upon divide-and-conquer.
Matrix multiplication is particularly easy to break into subproblems, because it can be
performed blockwise. To see what this means, carve X into four n/2 × n/2 blocks, and also Y : X = 􏰈A B􏰉, Y = 􏰈E F􏰉.
CD GH
Then their product can be expressed in terms of these blocks and is exactly as if the blocks were single elements (Exercise 2.11).
XY = 􏰈A B􏰉􏰈E F􏰉 = 􏰈AE+BG AF+BH􏰉 C D G H CE+DG CF+DH
We now have a divide-and-conquer strategy: to compute the size-n product XY , recursively compute eight size-n/2 products AE, BG, AF, BH, CE, DG, CF, DH, and then do a few O(n2)- time additions. The total running time is described by the recurrence relation
T (n) = 8T (n/2) + O(n2).
This comes out to an unimpressive O(n3), the same as for the default algorithm. But the efficiency can be further improved, and as with integer multiplication, the key is some clever algebra. It turns out XY can be computed from just seven n/2 × n/2 subproblems, via a decomposition so tricky and intricate that one wonders how Strassen was ever able to discover it!
where
XY = 􏰈P5+P4−P2+P6 P1+P2 􏰉 P3 +P4 P1 +P5 −P3 −P7
P1= P2 = P3 = P4 =
A(F−H) (A+B)H (C+D)E D(G−E)
P5 = (A+D)(E+H) P6 = (B−D)(G+H) P7 = (A−C)(E+F)
The new running time is
which by the master theorem works out to O(nlog2 7) ≈ O(n2.81).
T (n) =
7T (n/2) + O(n2),

68 Algorithms 2.6 The fast Fourier transform
We have so far seen how divide-and-conquer gives fast algorithms for multiplying integers and matrices; our next target is polynomials. The product of two degree-d polynomials is a polynomial of degree 2d, for example:
(1+2x+3×2)·(2+x+4×2)=2+5x+12×2 +11×3 +12×4.
More generally, if A(x) = a0 + a1x + · · · + adxd and B(x) = b0 + b1x + · · · + bdxd, their product
C(x)=A(x)·B(x)=c0 +c1x+···+c2dx2d hascoefficients
􏰑k i=0
(for i > d, take ai and bi to be zero). Computing ck from this formula takes O(k) steps, and finding all 2d + 1 coefficients would therefore seem to require Θ(d2) time. Can we possibly multiply polynomials faster than this?
The solution we will develop, the fast Fourier transform, has revolutionized—indeed, defined—the field of signal processing (see the following box). Because of its huge impor- tance, and its wealth of insights from different fields of study, we will approach it a little more leisurely than usual. The reader who wants just the core algorithm can skip directly to Section 2.6.4.
ck = a0bk +a1bk−1 +···+akb0 =
aibk−i

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 69 Why multiply polynomials?
For one thing, it turns out that the fastest algorithms we have for multiplying integers rely heavily on polynomial multiplication; after all, polynomials and binary integers are quite similar—just replace the variable x by the base 2, and watch out for carries. But perhaps more importantly, multiplying polynomials is crucial for signal processing.
A signal is any quantity that is a function of time (as in Figure (a)) or of position. It might, for instance, capture a human voice by measuring fluctuations in air pressure close to the speaker’s mouth, or alternatively, the pattern of stars in the night sky, by measuring brightness as a function of angle.
a(t) a(t) δ(t)
ttt
(a) (b) (c)
In order to extract information from a signal, we need to first digitize it by sampling (Figure (b))—and, then, to put it through a system that will transform it in some way. The output is called the response of the system:
signal −→ −→ response
An important class of systems are those that are linear—the response to the sum of two signals is just the sum of their individual responses—and time invariant—shifting the input signal by time t produces the same output, also shifted by t. Any system with these prop- erties is completely characterized by its response to the simplest possible input signal: the unit impulse δ(t), consisting solely of a “jerk” at t = 0 (Figure (c)). To see this, first consider the close relative δ(t − i), a shifted impulse in which the jerk occurs at time i. Any signal a(t) can be expressed as a linear combination of these, letting δ(t − i) pick out its behavior at time i,
T􏰑−1 i=0
(if the signal consists of T samples). By linearity, the system response to input a(t) is deter- mined by the responses to the various δ(t − i). And by time invariance, these are in turn just shifted copies of the impulse response b(t), the response to δ(t).
In other words, the output of the system at time k is
􏰑k
a(i)b(k−i), exactly the formula for polynomial multiplication!
a(t) =
a(i)δ(t − i)
c(k) =
SYSTEM
i=0
nomlkjihfgTledcb`aTf_^]\[ZXYUSRQPONMTUTSKLIJVWGHEFCDAB ?@ >=􏰜?􏰜=
<; :9􏰜<􏰜;􏰜:􏰜9 .- ,+ )* ('􏰜-􏰜+􏰜(􏰜' 􏰵􏰴 􏰳􏰲 􏰰􏰱 􏰯􏰮 􏰭􏰬􏰜􏰳􏰜􏰲􏰜􏰮􏰜􏰭􏰜􏰬 87 65 34 21 0/􏰜6􏰜5􏰜3􏰜2􏰜1􏰜0􏰜/ %& $# "!􏰜%􏰜$􏰜# 􏰷􏰶􏰜!􏰜􏰶 􏰪􏰫􏰪 􏰨􏰩􏰜􏰪􏰜􏰪 􏰞􏰟 􏰛􏰝􏰜􏰞􏰜􏰛 pq 􏰺􏰹􏰸􏰜􏰜􏰺􏰜􏰹􏰜􏰸 􏰫􏰦􏰧􏰤􏰥􏰢􏰣􏰡􏰠􏰜􏰢􏰜􏰡􏰜􏰠 70 Algorithms 2.6.1 An alternative representation of polynomials To arrive at a fast algorithm for polynomial multiplication we take inspiration from an impor- tant property of polynomials. Fact A degree-d polynomial is uniquely characterized by its values at any d + 1 distinct points. A familiar instance of this is that “any two points determine a line.” We will later see why the more general statement is true (page 76), but for the time being it gives us an alternative representation of polynomials. Fix any distinct points x0, . . . , xd. We can specify a degree-d polynomial A(x) = a0 + a1x + · · · + adxd by either one of the following: 1. Itscoefficientsa0,a1,...,ad 2. ThevaluesA(x0),A(x1),...,A(xd) Of these two representations, the second is the more attractive for polynomial multiplication. Since the product C(x) has degree 2d, it is completely determined by its value at any 2d + 1 points. And its value at any given point z is easy enough to figure out, just A(z) times B(z). Thus polynomial multiplication takes linear time in the value representation. The problem is that we expect the input polynomials, and also their product, to be specified by coefficients. So we need to first translate from coefficients to values—which is just a matter of evaluating the polynomial at the chosen points—then multiply in the value representation, and finally translate back to coefficients, a process called interpolation. Evaluation Interpolation Figure 2.5 presents the resulting algorithm. The equivalence of the two polynomial representations makes it clear that this high-level approach is correct, but how efficient is it? Certainly the selection step and the n multiplica- tions are no trouble at all, just linear time.3 But (leaving aside interpolation, about which we know even less) how about evaluation? Evaluating a polynomial of degree d ≤ n at a single point takes O(n) steps (Exercise 2.29), and so the baseline for n points is Θ(n2). We’ll now see that the fast Fourier transform (FFT) does it in just O(n log n) time, for a particularly clever choice of x0, . . . , xn−1 in which the computations required by the individual points overlap with one another and can be shared. 3In a typical setting for polynomial multiplication, the coefficients of the polynomials are real numbers and, moreover, are small enough that basic arithmetic operations (adding and multiplying) take unit time. We will assume this to be the case without any great loss of generality; in particular, the time bounds we obtain are easily adjustable to situations with larger numbers. Coefficient representation a0,a1,...,ad Value representation A(x0), A(x1), . . . , A(xd) S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 71 Figure 2.5 Polynomial multiplication Input: Coefficients of two polynomials, A(x) and B(x), of degree d Output: Their product C = A · B Selection Picksomepoints x0,x1,...,xn−1,where n≥2d+1 Evaluation Compute A(x0), A(x1), . . . , A(xn−1) and B(x0), B(x1), . . . , B(xn−1) Multiplication Compute C(xk) = A(xk)B(xk) for all k = 0, . . . , n − 1 Interpolation Recover C(x)=c0 +c1x+···+c2dx2d 2.6.2 Evaluation by divide-and-conquer Here’s an idea for how to pick the n points at which to evaluate a polynomial A(x) of degree ≤ n − 1. If we choose them to be positive-negative pairs, that is, ±x0, ±x1, . . . , ±xn/2−1, then the computations required for each A(xi) and A(−xi) overlap a lot, because the even powers of xi coincide with those of −xi. To investigate this, we need to split A(x) into its odd and even powers, for instance 3+4x+6x2 +2x3 +x4 +10x5 = (3+6x2 +x4)+x(4+2x2 +10x4). Notice that the terms in parentheses are polynomials in x2. More generally, A(x) = Ae(x2)+xAo(x2), where Ae(·), with the even-numbered coefficients, and Ao(·), with the odd-numbered coeffi- cients, are polynomials of degree ≤ n/2 − 1 (assume for convenience that n is even). Given paired points ±xi, the calculations needed for A(xi) can be recycled toward computing A(−xi): A(xi) = Ae(x2i ) + xiAo(x2i ) A(−xi) = Ae(x2i ) − xiAo(x2i ). In other words, evaluating A(x) at n paired points ±x0,...,±xn/2−1 reduces to evaluating Ae(x) and Ao(x) (which each have half the degree of A(x)) at just n/2 points, x20, . . . , x2n/2−1. 72 Algorithms Evaluate: Equivalently, evaluate: A(x) degree ≤ n − 1 at: +x0 −x0 +x1 −x1 · · · +xn/2−1 −xn/2−1 Ae(x) and Ao(x) degree ≤ n/2 − 1 at: x20 x21 · · · x2n/2−1 The original problem of size n is in this way recast as two subproblems of size n/2, followed by some linear-time arithmetic. If we could recurse, we would get a divide-and-conquer pro- cedure with running time T(n) = 2T(n/2)+O(n), which is O(n log n), exactly what we want. But we have a problem: The plus-minus trick only works at the top level of the recur- sion. To recurse at the next level, we need the n/2 evaluation points x20, x21, . . . , x2n/2−1 to be themselves plus-minus pairs. But how can a square be negative? The task seems impossible! Unless, of course, we use complex numbers. Fine, but which complex numbers? To figure this out, let us “reverse engineer” the process. At the very bottom of the recursion, we have a single point. This point might as well be 1, in which case the level above it must consist of its square roots, ±√1 = ±1. . +1 −1 +i −i +1 −1 +1 The next level up then has ±√+1 = ±1 as well as the complex numbers ±√−1 = ±i, where i is the imaginary unit. By continuing in this manner, we eventually reach the initial set of n points. Perhaps you have already guessed what they are: the complex nth roots of unity, that is, the n complex solutions to the equation zn = 1. Figure 2.6 is a pictorial review of some basic facts about complex numbers. The third panel of this figure introduces the nth roots of unity: the complex numbers 1, ω, ω2, . . . , ωn−1, where ω = e2πi/n. If n is even, 1. The nth roots are plus-minus paired, ωn/2+j = −ωj. S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 73 2. Squaring them produces the (n/2)nd roots of unity. Therefore, if we start with these numbers for some n that is a power of 2, then at successive levels of recursion we will have the (n/2k)th roots of unity, for k = 0, 1, 2, 3, . . .. All these sets of numbers are plus-minus paired, and so our divide-and-conquer, as shown in the last panel, works perfectly. The resulting algorithm is the fast Fourier transform (Figure 2.7). 74 Algorithms Figure 2.6 The complex roots of unity are ideal for our divide-and-conquer scheme. Imaginary b r θ a Real The complex plane z = a + bi is plotted at position (a, b). Polar coordinates: rewrite as z = r(cos θ + i sin θ) = reiθ , 5 + 5i (5√2, π/4) denoted (r, θ). •lengthr=√a2 +b2. • angle θ ∈ [0,2π): cosθ = a/r,sinθ = b/r. • θ can always be reduced modulo 2π. Examples: Number −1 i Polar coords (1, π) (1, π/2) (r1, θ1) (r2, θ2) (r1r2,θ1 +θ2) Multiplying is easy in polar coordinates Multiply the lengths and add the angles: (r1, θ1) × (r2, θ2) = (r1r2, θ1 + θ2). For any z = (r, θ), • −z = (r, θ + π) since −1 = (1, π). • If z is on the unit circle (i.e., r = 1), then zn = (1,nθ). 4π n Angle 2π n 2π +π n The nth complex roots of unity Solutions to the equation zn = 1. By the multiplication rule: solutions are z = (1, θ), for θ a multiple of 2π/n (shown here for n = 16). For even n: • These numbers are plus-minus paired: −(1, θ) = (1, θ+π). • Their squares are the (n/2)nd roots of unity, shown here with boxes around them. Divide-and-conquer step Evaluate A(x) at nth roots of unity Paired (n is a power of 2) Divide and conquer Evaluate Ae(x), Ao(x) at (n/2)nd roots Still paired S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 75 Figure 2.7 The fast Fourier transform (polynomial formulation) (A, ω) Input: Coefficient representation of a polynomial A(x) ofdegree ≤n−1,where n isapowerof 2 ω, an nth root of unity Output: ValuerepresentationA(ω0),...,A(ωn−1) if ω = 1: return A(1) express A(x) in the form Ae(x2) + xAo(x2) call FFT(Ae, ω2) to evaluate Ae at even powers of ω call FFT(Ao, ω2) to evaluate Ao at even powers of ω for j = 0 to n − 1: compute A(ωj) = Ae(ω2j) + ωjAo(ω2j) return A(ω0), . . . , A(ωn−1) 2.6.3 Interpolation Let’s take stock of where we are. We first developed a high-level scheme for multiplying polynomials (Figure 2.5), based on the observation that polynomials can be represented in two ways, in terms of their coefficients or in terms of their values at a selected set of points. Evaluation Interpolation The value representation makes it trivial to multiply polynomials, but we cannot ignore the coefficient representation since it is the form in which the input and output of our overall algorithm are specified. So we designed the FFT, a way to move from coefficients to values in time just O(n log n), when the points {xi} are complex nth roots of unity (1, ω, ω2, . . . , ωn−1). ⟨values⟩ = FFT(⟨coefficients⟩,ω). The last remaining piece of the puzzle is the inverse operation, interpolation. It will turn out, function FFT amazingly, that ⟨coefficients⟩ = 1 FFT(⟨values⟩, ω−1). n Coefficient representation a0,a1,...,an−1 Interpolation is thus solved in the most simple and elegant way we could possibly have hoped for—using the same FFT algorithm, but called with ω−1 in place of ω! This might seem like a miraculous coincidence, but it will make a lot more sense when we recast our polynomial oper- ations in the language of linear algebra. Meanwhile, our O(n log n) polynomial multiplication algorithm (Figure 2.5) is now fully specified. Value representation A(x0), A(x1), . . . , A(xn−1) 76 Algorithms A matrix reformulation To get a clearer view of interpolation, let’s explicitly set down the relationship between our two representations for a polynomial A(x) of degree ≤ n − 1. They are both vectors of n numbers, and one is a linear transformation of the other:  A ( x 0 )   1 x 0 x 20 · · · x n − 1   a 0   A(x )  1 x x2 ··· 11111  . . .  =  . . .   . . .  . A(xn−1) 1 xn−1 x2 ··· xn−1 an−1 n−1 n−1 Call the matrix in the middle M. Its specialized format—a Vandermonde matrix—gives it many remarkable properties, of which the following is particularly relevant to us. If x0, . . . , xn−1 are distinct numbers, then M is invertible. The existence of M −1 allows us to invert the preceding matrix equation so as to express coef- ficients in terms of values. In brief, Evaluation is multiplication by M , while interpolation is multiplication by M −1 . This reformulation of our polynomial operations reveals their essential nature more clearly. Among other things, it finally justifies an assumption we have been making throughout, that A(x) is uniquely characterized by its values at any n points—in fact, we now have an explicit formula that will give us the coefficients of A(x) in this situation. Vandermonde matrices also have the distinction of being quicker to invert than more general matrices, in O(n2) time in- stead of O(n3). However, using this for interpolation would still not be fast enough for us, so once again we turn to our special choice of points—the complex roots of unity. Interpolation resolved In linear algebra terms, the FFT multiplies an arbitrary n-dimensional vector—which we have been calling the coefficient representation—by the n × n matrix 1 M (ω) =  ω2(n−1)  ←− ω2  . 1 1 1 1 ··· ω ω2 ··· ω2 ω4 ··· . 1 ←−rowforω0=1 ωn−1  ←− ω ω(n−1)j  ←− ωj  .  . n 1 1 ω(n−1) ω2(n−1) · · · ω(n−1)(n−1) ←− ωn−1 ωj ω2j ··· where ω is a complex nth root of unity, and n is a power of 2. Notice how simple this matrix is to describe: its (j,k)th entry (starting row- and column-count at zero) is ωjk. Multiplication by M = Mn(ω) maps the kth coordinate axis (the vector with all zeros except for a 1 at position k) onto the kth column of M. Now here’s the crucial observation, which we’ll 0 xn−1 a  S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 77 Figure 2.8 The FFT takes points in the standard coordinate system, whose axes are shown here as x1,x2,x3, and rotates them into the Fourier basis, whose axes are the columns of Mn(ω), shown here as f1, f2, f3. For instance, points in direction x1 get mapped into direction f1. x2 x3 f1 f3 f2 x1 FFT prove shortly: the columns of M are orthogonal (at right angles) to each other. Therefore they can be thought of as the axes of an alternative coordinate system, which is often called the Fourier basis. The effect of multiplying a vector by M is to rotate it from the standard basis, with the usual set of axes, into the Fourier basis, which is defined by the columns of M (Figure 2.8). The FFT is thus a change of basis, a rigid rotation. The inverse of M is the opposite rotation, from the Fourier basis back into the standard basis. When we write out the orthogonality condition precisely, we will be able to read off this inverse transformation with ease: Inversion formula Mn(ω)−1 = 1 Mn(ω−1). n But ω−1 is also an nth root of unity, and so interpolation—or equivalently, multiplication by Mn(ω)−1—is itself just an FFT operation, but with ω replaced by ω−1. Now let’s get into the details. Take ω to be e2πi/n for convenience, and think of the columns of M as vectors in Cn. Recall that the angle between two vectors u = (u0, . . . , un−1) and v = (v0, . . . , vn−1) in Cn is just a scaling factor times their inner product u·v∗ = u0v0∗ +u1v1∗ +···+un−1vn∗−1, where z∗ denotes the complex conjugate4 of z. This quantity is maximized when the vectors lie in the same direction and is zero when the vectors are orthogonal to each other. The fundamental observation we need is the following. Lemma The columns of matrix M are orthogonal to each other. Proof. Take the inner product of any columns j and k of matrix M , 1 + ωj−k + ω2(j−k) + · · · + ω(n−1)(j−k). 4The complex conjugate of a complex number z = reiθ is z∗ = re−iθ. The complex conjugate of a vector (or matrix) is obtained by taking the complex conjugates of all its entries. 78 Algorithms This is a geometric series with first term 1, last term ω(n−1)(j−k), and ratio ω(j−k). Therefore it evaluates to (1 − ωn(j−k))/(1 − ω(j−k)), which is 0—except when j = k, in which case all terms are 1 and the sum is n. The orthogonality property can be summarized in the single equation MM∗ =nI, since (MM∗)ij is the inner product of the ith and jth columns of M (do you see why?). This immediately implies M−1 = (1/n)M∗: we have an inversion formula! But is it the same for- mula we earlier claimed? Let’s see—the (j, k)th entry of M ∗ is the complex conjugate of the corresponding entry of M, in other words ω−jk. Whereupon M∗ = Mn(ω−1), and we’re done. And now we can finally step back and view the whole affair geometrically. The task we need to perform, polynomial multiplication, is a lot easier in the Fourier basis than in the standard basis. Therefore, we first rotate vectors into the Fourier basis (evaluation), then perform the task (multiplication), and finally rotate back (interpolation). The initial vectors are coefficient representations, while their rotated counterparts are value representations. To efficiently switch between these, back and forth, is the province of the FFT. 2.6.4 A closer look at the fast Fourier transform Now that our efficient scheme for polynomial multiplication is fully realized, let’s hone in more closely on the core subroutine that makes it all possible, the fast Fourier transform. The definitive FFT algorithm The FFT takes as input a vector a = (a0, . . . , an−1) and a complex number ω whose powers 1, ω, ω2, . . . , ωn−1 are the complex nth roots of unity. It multiplies vector a by the n × n matrix Mn(ω), which has (j,k)th entry (starting row- and column-count at zero) ωjk. The potential for using divide-and-conquer in this matrix-vector multiplication becomes apparent when M ’s columns are segregated into evens and odds: k Mn(ω) Column 2k 2k + 1 Column 2k 2k + 1 a0 a0 a2 a2 . Row j . ω2jk ωj · ω2jk ω2jk ωj · ω2jk ω2jk −ωj · ω2jk j an−2 ωjk a0 a1 a2 a3 an−2 a4.=j a1= a1 . a3 a3 . . j + n/2 . an−1 an−1 a Even Odd an−1 columns columns In the second step, we have simplified entries in the bottom half of the matrix using ω n/2 = −1 and ωn = 1. Notice that the top left n/2 × n/2 submatrix is Mn/2(ω2), as is the one on the S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 79 Figure 2.9 The fast Fourier transform (a, ω) Input: An array a = (a0,a1,...,an−1), for n a power of 2 A primitive nth root of unity, ω Output: Mn(ω) a ifω=1: returna (s0,s1,...,sn/2−1) = FFT((a0,a2,...,an−2),ω2) (s′0, s′1, . . . , s′n/2−1) = FFT((a1, a3, . . . , an−1), ω2) for j = 0 to n/2 − 1: r j = s j + ω j s ′j rj+n/2 =sj −ωjs′j return (r0, r1, . . . , rn−1) bottom left. And the top and bottom right submatrices are almost the same as Mn/2(ω2), but with their jth rows multiplied through by ωj and −ωj, respectively. Therefore the final product is the vector function FFT Row j j + n/2 . 2 j . +ω a0 an−2 a0 an−2 a a1 a Mn/2 Mn/2 . 3 . an−1 a1 an−1 Mn/2 a .2j . −ω Mn/2 a . 3 . In short, the product of Mn(ω) with vector (a0, . . . , an−1), a size-n problem, can be expressed in terms of two size-n/2 problems: the product of Mn/2(ω2) with (a0,a2,...,an−2) and with (a1,a3,...,an−1). This divide-and-conquer strategy leads to the definitive FFT algorithm of Figure 2.9, whose running time is T (n) = 2T (n/2) + O(n) = O(n log n). The fast Fourier transform unraveled Throughout all our discussions so far, the fast Fourier transform has remained tightly co- cooned within a divide-and-conquer formalism. To fully expose its structure, we now unravel the recursion. The divide-and-conquer step of the FFT can be drawn as a very simple circuit. Here is how a problem of size n is reduced to two subproblems of size n/2 (for clarity, one pair of outputs (j, j + n/2) is singled out): 80 Algorithms FFTn (input: a0,...,an−1, output: r0,...,rn−1) a0 a an−2 a1 a an−1 j .2 . rj FFTn/2 .3 r . j + n/2 j+n/2 FFTn/2 We’re using a particular shorthand: the edges are wires carrying complex numbers from left to right. A weight of j means “multiply the number on this wire by ωj.” And when two wires come into a junction from the left, the numbers they are carrying get added up. So the two outputs depicted are executing the commands rj = sj+ωjs′j rj+n/2 = sj−ωjs′j from the FFT algorithm (Figure 2.9), via a pattern of wires known as a butterfly: . Unraveling the FFT circuit completely for n = 8 elements, we get Figure 10.4. Notice the following. 1. For n inputs there are log2 n levels, each with n nodes, for a total of n log n operations. 2. The inputs are arranged in a peculiar order: 0, 4, 2, 6, 1, 5, 3, 7. Why? Recall that at the top level of recursion, we first bring up the even coefficients of the input and then move on to the odd ones. Then at the next level, the even coefficients of this first group (which therefore are multiples of 4, or equivalently, have zero as their two least significant bits) are brought up, and so on. To put it otherwise, the inputs are arranged by increasing last bit of the binary representation of their index, resolving ties by looking at the nextmoresignificantbit(s). Theresultingorderinbinary,000,100,010,110,001,101,011,111, is the same as the natural one, 000, 001, 010, 011, 100, 101, 110, 111 except the bits are mirrored! 3. There is a unique path between each input aj and each output A(ωk). This path is most easily described using the binary representations of j and k (shown in Figure 10.4 for convenience). There are two edges out of each node, one going up (the 0-edge) and one going down (the 1-edge). To get to A(ωk) from any input node, simply follow the edges specified in the bit representation of k, starting from the rightmost bit. (Can you similarly specify the path in the reverse direction?) 4. On the path between aj and A(ωk), the labels add up to jk mod 8. Since ω8 = 1, this means that the contribution of input aj to output A(ωk) is ajωjk, and therefore the circuit computes correctly the values of polynomial A(x). S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 81 5. And finally, notice that the FFT circuit is a natural for parallel computation and direct implementation in hardware. Figure 2.10 The fast Fourier transform circuit. a0 000 100 010 110 001 101 011 111 4 A(ω0) 000 A(ω1) 001 4 A(ω2) 010 a4 a2 2 a6 a1 2 1 a5 5 2 a3 6 3 a7 A(ω3) 011 4 A(ω4) 100 A(ω5) 101 4 A(ω6) 110 46 4 467 A(ω7) 111 /. 􏰫􏰬 􏰩􏰪 -, 􏰭􏰮 􏰧􏰨 *+ 􏰯􏰰 􏰦􏰥 () 􏰲􏰱 􏰤􏰣 &' 􏰳􏰴 􏰢􏰡 %$ 􏰶􏰵 􏰟􏰠 #" 􏰷􏰸 􏰞􏰝 ! 􏰺􏰹 􏰛􏰜 82 Algorithms The slow spread of a fast algorithm In 1963, during a meeting of President Kennedy’s scientific advisors, John Tukey, a math- ematician from Princeton, explained to IBM’s Dick Garwin a fast method for computing Fourier transforms. Garwin listened carefully, because he was at the time working on ways to detect nuclear explosions from seismographic data, and Fourier transforms were the bot- tleneck of his method. When he went back to IBM, he asked John Cooley to implement Tukey’s algorithm; they decided that a paper should be published so that the idea could not be patented. Tukey was not very keen to write a paper on the subject, so Cooley took the initiative. And this is how one of the most famous and most cited scientific papers was published in 1965, co-authored by Cooley and Tukey. The reason Tukey was reluctant to publish the FFT was not secretiveness or pursuit of profit via patents. He just felt that this was a simple observation that was probably already known. This was typical of the period: back then (and for some time later) algorithms were considered second-class mathematical objects, devoid of depth and elegance, and unworthy of serious attention. But Tukey was right about one thing: it was later discovered that British engineers had used the FFT for hand calculations during the late 1930s. And—to end this chapter with the same great mathematician who started it—a paper by Gauss in the early 1800s on (what else?) interpolation contained essentially the same idea in it! Gauss’s paper had remained a secret for so long because it was protected by an old-fashioned cryptographic technique: like most scientific papers of its era, it was written in Latin. S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 83 Exercises 2.1. Use the divide-and-conquer integer multiplication algorithm to multiply the two binary integers 10011011 and 10111010. 2.2. Show that for any positive integers n and any base b, there must some power of b lying in the range [n, bn]. 2.3. Section 2.2 describes a method for solving recurrence relations which is based on analyzing the recursion tree and deriving a formula for the work done at each level. Another (closely related) method is to expand out the recurrence a few times, until a pattern emerges. For instance, let’s start with the familiar T (n) = 2T (n/2) + O(n). Think of O(n) as being ≤ cn for some constant c, so: T (n) ≤ 2T (n/2) + cn. By repeatedly applying this rule, we can bound T (n) in terms of T (n/2), then T (n/4), then T (n/8), and so on, at each step getting closer to the value of T (·) we do know, namely T (1) = O(1). T(n) ≤ 2T(n/2)+cn ≤ 2[2T(n/4)+cn/2]+cn = 4T(n/4)+2cn ≤ 4[2T(n/8)+cn/4]+2cn = 8T(n/8)+3cn ≤ 8[2T (n/16) + cn/8] + 3cn = 16T (n/16) + 4cn . A pattern is emerging... the general term is T(n) ≤ 2kT(n/2k) + kcn. Plugging in k = log2 n, we get T (n) ≤ nT (1) + cn log2 n = O(n log n). (a) Do the same thing for the recurrence T (n) = 3T (n/2) + O(n). What is the general kth term in this case? And what value of k should be plugged in to get the answer? (b) Now try the recurrence T (n) = T (n − 1) + O(1), a case which is not covered by the master theorem. Can you solve this too? 2.4. Suppose you are choosing between the following three algorithms: • Algorithm A solves problems by dividing them into five subproblems of half the size, recur- sively solving each subproblem, and then combining the solutions in linear time. • Algorithm B solves problems of size n by recursively solving two subproblems of size n − 1 and then combining the solutions in constant time. • Algorithm C solves problems of size n by dividing them into nine subproblems of size n/3, recursively solving each subproblem, and then combining the solutions in O(n2) time. What are the running times of each of these algorithms (in big-O notation), and which would you choose? 2.5. Solve the following recurrence relations and give a Θ bound for each of them. (a) T(n) = 2T(n/3) + 1 (b) T(n) = 5T(n/4) + n 84 Algorithms (c) T(n) = 7T(n/7) + n (d) T(n) = 9T(n/3) + n2 (e) T(n) = 8T(n/2) + n3 (f) T(n)=49T(n/25)+n3/2logn (g) T(n)=T(n−1)+2 (h) T(n)=T(n−1)+nc,wherec≥1isaconstant (i) T(n)=T(n−1)+cn,wherec>1issomeconstant
(j) T(n)=2T(n−1)+1
n)+1
(k) T(n)=T(√
2.6. A linear, time-invariant system has the following impulse response:
b(t)
1/t0
t0 (a) Describe in words the effect of this system.
(b) What is the corresponding polynomial?
t
2.7. What is the sum of the nth roots of unity? What is their product if n is odd? If n is even?
2.8. Practice with the fast Fourier transform.
(a) What is the FFT of (1, 0, 0, 0)? What is the appropriate value of ω in this case? And of which sequence is (1, 0, 0, 0) the FFT?
(b) Repeatfor(1,0,1,−1).
2.9. Practice with polynomial multiplication by FFT.
(a) Suppose that you want to multiply the two polynomials x + 1 and x2 + 1 using the FFT. Choose an appropriate power of two, find the FFT of the two sequences, multiply the results componentwise, and compute the inverse FFT to get the final result.
(b) Repeat for the pair of polynomials 1 + x + 2×2 and 2 + 3x.
2.10. Find the unique polynomial of degree 4 that takes on values p(1) = 2, p(2) = 1, p(3) = 0, p(4) = 4,
and p(5) = 0. Write your answer in the coefficient representation.
2.11. In justifying our matrix multiplication algorithm (Section 2.5), we claimed the following block-
wise property: if X and Y are n × n matrices, and
X = 􏰈A B􏰉, Y = 􏰈E F􏰉. CD GH
/. -, *+ () ‘& %$ “#( ! 􏰹􏰺%$ 􏰸􏰷 􏰵􏰶 􏰴􏰳 􏰷 􏰴􏰳
􏰮􏰭 􏰬􏰫 􏰩􏰪 􏰧􏰨 􏰥􏰦 􏰤􏰣 􏰰􏰯 􏰢􏰡 􏰠􏰟 􏰞􏰝 􏰛􏰜 􏰲􏰱

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 85 where A, B, C, D, E, F, G, and H are n/2×n/2 submatrices, then the product XY can be
expressed in terms of these blocks:
XY = 􏰈A B􏰉􏰈E F􏰉 = 􏰈AE+BG AF+BH􏰉 C D G H CE+DG CF+DH
Prove this property.
2.12. How many lines, as a function of n (in Θ(·) form), does the following program print? Write a
recurrence and solve it. You may assume n is a power of 2.
function f(n)
if n > 1:
print_line(‘‘still going’’) f(n/2)
f(n/2)
2.13. A binary tree is full if all of its vertices have either zero or two children. Let Bn denote the number of full binary trees with n vertices.
(a) By drawing out all full binary trees with 3, 5, or 7 vertices, determine the exact values of B3, B5, and B7. Why have we left out even numbers of vertices, like B4?
(b) For general n, derive a recurrence relation for Bn. (c) Show by induction that Bn is Ω(2n).
2.14. You are given an array of n elements, and you notice that some of the elements are duplicates; that is, they appear more than once in the array. Show how to remove all duplicates from the array in time O(n log n).
2.15. In our median-finding algorithm (Section 2.4), a basic primitive is the split operation, which takes as input an array S and a value v and then divides S into three sets: the elements less than v, the elements equal to v, and the elements greater than v. Show how to implement this split operation in place, that is, without allocating new memory.
2.16. You are given an infinite array A[·] in which the first n cells contain integers in sorted order and the rest of the cells are filled with ∞. You are not given the value of n. Describe an algorithm that takes an integer x as input and finds a position in the array containing x, if such a position exists, in O(log n) time. (If you are disturbed by the fact that the array A has infinite length, assume instead that it is of length n, but that you don’t know this length, and that the implementation of the array data type in your programming language returns the error message ∞ whenever elements A[i] with i > n are accessed.)
2.17. Given a sorted array of distinct integers A[1, . . . , n], you want to find out whether there is an index i for which A[i] = i. Give a divide-and-conquer algorithm that runs in time O(log n).
2.18. Consider the task of searching a sorted array A[1 . . . n] for a given element x: a task we usually perform by binary search in time O(log n). Show that any algorithm that accesses the array only via comparisons (that is, by asking questions of the form “is A[i] ≤ z?”), must take Ω(log n) steps.
2.19. A k-way merge operation. Suppose you have k sorted arrays, each with n elements, and you want to combine them into a single sorted array of kn elements.

86 Algorithms
(a) Here’s one strategy: Using the merge procedure from Section 2.3, merge the first two ar- rays, then merge in the third, then merge in the fourth, and so on. What is the time complexity of this algorithm, in terms of k and n?
(b) Give a more efficient solution to this problem, using divide-and-conquer. 2.20. Showthatanyarrayofintegersx[1…n]canbesortedinO(n+M)time,where
M = maxxi−minxi. ii
For small M , this is linear time: why doesn’t the Ω(n log n) lower bound apply in this case?
2.21. Mean and median. One of the most basic tasks in statistics is to summarize a set of observations
{x1 , x2 , . . . , xn } ⊆ R by a single number. Two popular choices for this summary statistic are:
• The median, which we’ll call μ1
• The mean, which we’ll call μ2
(a) Show that the median is the value of μ that minimizes the function
􏰑|xi −μ|. i
You can assume for simplicity that n is odd. (Hint: Show that for any μ ̸= μ1, the function decreases if you move μ either slightly to the left or slightly to the right.)
(b) Show that the mean is the value of μ that minimizes the function 􏰑(xi − μ)2.
i
One way to do this is by calculus. Another method is to prove that for any μ ∈ R, 􏰑(xi −μ)2 = 􏰑(xi −μ2)2 +n(μ−μ2)2.
ii
Notice how the function for μ2 penalizes points that are far from μ much more heavily than the function for μ1. Thus μ2 tries much harder to be close to all the observations. This might sound like a good thing at some level, but it is statistically undesirable because just a few outliers can severely throw off the estimate of μ2. It is therefore sometimes said that μ1 is a more robust estimator than μ2. Worse than either of them, however, is μ∞, the value of μ that minimizes the function
max|xi −μ|. i
(c) Show that μ∞ can be computed in O(n) time (assuming the numbers xi are small enough that basic arithmetic operations on them take unit time).
2.22. You are given two sorted lists of size m and n. Give an O(log m + log n) time algorithm for computing the kth smallest element in the union of the two lists.
2.23. An array A[1…n] is said to have a majority element if more than half of its entries are the same. Given an array, the task is to design an efficient algorithm to tell whether the array has a majority element, and, if so, to find that element. The elements of the array are not necessarily from some ordered domain like the integers, and so there can be no comparisons of the form “is A[i] > A[j]?”. (Think of the array elements as GIF files, say.) However you can answer questions of the form: “is A[i] = A[j]?” in constant time.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 87
(a) Show how to solve this problem in O(n log n) time. (Hint: Split the array A into two arrays A1 and A2 of half the size. Does knowing the majority elements of A1 and A2 help you figure out the majority element of A? If so, you can use a divide-and-conquer approach.)
(b) Can you give a linear-time algorithm? (Hint: Here’s another divide-and-conquer approach: • Pair up the elements of A arbitrarily, to get n/2 pairs
• Look at each pair: if the two elements are different, discard both of them; if they are
the same, keep just one of them
Show that after this procedure there are at most n/2 elements left, and that they have a
majority element if and only if A does.)
2.24. On page 66 there is a high-level description of the quicksort algorithm.
(a) Write down the pseudocode for quicksort.
(b) Show that its worst-case running time on an array of size n is Θ(n2).
(c) Show that its expected running time satisfies the recurrence relation
1 n􏰑− 1
T(n)≤O(n)+ n (T(i)+T(n−i)).
i=1
Then, show that the solution to this recurrence is O(n log n).
2.25. In Section 2.1 we described an algorithm that multiplies two n-bit binary integers x and y in time na, where a = log2 3. Call this procedure fastmultiply(x, y).
(a) We want to convert the decimal integer 10n (a 1 followed by n zeros) into binary. Here is the algorithm (assume n is a power of 2):
function pwr2bin(n)
if n = 1: return 10102 else:
z =???
return fastmultiply(z, z)
Fill in the missing details. Then give a recurrence relation for the running time of the algorithm, and solve the recurrence.
(b) Next, we want to convert any decimal integer x with n digits (where n is a power of 2) into binary. The algorithm is the following:
function dec2bin(x)
if n = 1: return binary[x] else:
split x into two decimal numbers xL, xR with n/2 digits each return ???
Here binary[·] is a vector that contains the binary representation of all one-digit integers. That is, binary[0] = 02, binary[1] = 12, up to binary[9] = 10012. Assume that a lookup in binary takes O(1) time.
Fill in the missing details. Once again, give a recurrence for the running time of the algo- rithm, and solve it.

88 Algorithms
2.26. Professor F. Lake tells his class that it is asymptotically faster to square an n-bit integer than to
multiply two n-bit integers. Should they believe him?
2.27. The square of a matrix A is its product with itself, AA.
(a) Show that five multiplications are sufficient to compute the square of a 2 × 2 matrix.
(b) What is wrong with the following algorithm for computing the square of an n × n matrix?
“Use a divide-and-conquer approach as in Strassen’s algorithm, except that in- stead of getting 7 subproblems of size n/2, we now get 5 subproblems of size n/2 thanks to part (a). Using the same analysis as in Strassen’s algorithm, we can conclude that the algorithm runs in time O(nlog2 5).”
(c) In fact, squaring matrices is no easier than matrix multiplication. In this part, you will show that if n × n matrices can be squared in time S(n) = O(nc), then any two n × n matrices can be multiplied in time O(nc).
i. Given two n × n matrices A and B, show that the matrix AB + BA can be computed in time 3S(n) + O(n2).
ii. Giventwon×nmatricesXandY,definethe2n×2nmatricesAandBasfollows: A=􏰈X 0􏰉andB=􏰈0 Y􏰉.
WhatisAB+BA,intermsofX andY?
iii. Using (i) and (ii), argue that the product XY can be computed in time 3S(2n) + O(n2).
Conclude that matrix multiplication takes time O(nc).
2.28. The Hadamard matrices H0 , H1 , H2 , . . . are defined as follows:
• H0 isthe1×1matrix􏰻1􏰼
• Fork>0,Hk isthe2k ×2k matrix
Hk=􏰈Hk−1 Hk−1
Hk−1 􏰉 −Hk−1
00 00
Show that if v is a column vector of length n = 2k, then the matrix-vector product Hkv can be calculated using O(n log n) operations. Assume that all the numbers involved are small enough that basic arithmetic operations like addition and multiplication take unit time.
2.29. Suppose we want to evaluate the polynomial p(x) = a0 + a1x + a2x2 + · · · + anxn at point x.
(a) Show that the following simple routine, known as Horner’s rule, does the job and leaves the
answer in z.
z = an
for i=n−1 downto 0:
z = zx + ai
(b) How many additions and multiplications does this routine use, as a function of n? Can you
find a polynomial for which an alternative method is substantially better?
2.30. This problem illustrates how to do the Fourier Transform (FT) in modular arithmetic, for exam- ple, modulo 7.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 89
(a) There is a number ω such that all the powers ω, ω2, . . . , ω6 are distinct (modulo 7). Find this ω, and show that ω + ω2 + · · · + ω6 = 0. (Interestingly, for any prime modulus there is such a number.)
(b) UsingthematrixformoftheFT,producethetransformofthesequence(0,1,1,1,5,2)mod- ulo 7; that is, multiply this vector by the matrix M6(ω), for the value of ω you found earlier. In the matrix multiplication, all calculations should be performed modulo 7.
(c) Write down the matrix necessary to perform the inverse FT. Show that multiplying by this matrix returns the original sequence. (Again all arithmetic should be performed modulo 7.)
(d) Now show how to multiply the polynomials x2 + x + 1 and x3 + 2x − 1 using the FT modulo 7.
2.31. In Section 1.2.3, we studied Euclid’s algorithm for computing the greatest common divisor (gcd) of two positive integers: the largest integer which divides them both. Here we will look at an alternative algorithm based on divide-and-conquer.
(a) Show that the following rule is true.
 2gcd(a/2,b/2) gcd(a, b) = gcd(a, b/2)
(c) How does the efficiency of your algorithm compare to Euclid’s algorithm if a and b are n-bit integers? (In particular, since n might be large you cannot assume that basic arithmetic operations like addition take constant time.)
2.32. Inthisproblemwewilldevelopadivide-and-conqueralgorithmforthefollowinggeometrictask.
CLOSEST PAIR
Input: A set of points in the plane, {p1 = (x1,y1),p2 = (x2,y2),…,pn = (xn,yn)}
Output: The closest pair of po􏰔ints: that is, the pair pi ̸= pj for which the distance between pi and pj , that is,
gcd((a − b)/2, b)
(b) Give an efficient divide-and-conquer algorithm for greatest common divisor.
ifa,bareeven
if a is odd, b is even if a, b are odd
(xi −xj)2 +(yi −yj)2,
For simplicity, assume that n is a power of two, and that all the x-coordinates xi are distinct, as
are the y-coordinates.
Here’s a high-level overview of the algorithm:
• Find a value x for which exactly half the points have xi < x, and half have xi > x. On this basis, split the points into two groups, L and R.
• Recursively find the closest pair in L and in R. Say these pairs are pL, qL ∈ L and pR, qR ∈ R, with distances dL and dR respectively. Let d be the smaller of these two distances.
• It remains to be seen whether there is a point in L and a point in R that are less than distance d apart from each other. To this end, discard all points with xi < x − d or xi > x + d and sort the remaining points by y-coordinate.
• Now, go through this sorted list, and for each point, compute its distance to the seven sub- sequent points in the list. Let pM , qM be the closest pair found in this way.
is minimized.

90
•
(a) (b) (c)
(d)
Algorithms
Theanswerisoneofthethreepairs{pL,qL},{pR,qR},{pM,qM},whicheverisclosest.
In order to prove the correctness of this algorithm, start by showing the following property:
any square of size d × d in the plane contains at most four points of L.
Now show that the algorithm is correct. The only case which needs careful consideration is
when the closest pair is split between L and R.
Write down the pseudocode for the algorithm, and show that its running time is given by
the recurrence:
T (n) = 2T (n/2) + O(n log n).
Show that the solution to this recurrence is O(n log2 n). Can you bring the running time down to O(n log n)?

Chapter 3
Decompositions of graphs
3.1 Why graphs?
A wide range of problems can be expressed with clarity and precision in the concise pictorial language of graphs. For instance, consider the task of coloring a political map. What is the minimum number of colors needed, with the obvious restriction that neighboring countries should have different colors? One of the difficulties in attacking this problem is that the map itself, even a stripped-down version like Figure 3.1(a), is usually cluttered with irrelevant information: intricate boundaries, border posts where three or more countries meet, open seas, and meandering rivers. Such distractions are absent from the mathematical object of Figure 3.1(b), a graph with one vertex for each country (1 is Brazil, 11 is Argentina) and edges between neighbors. It contains exactly the information needed for coloring, and nothing more. The precise goal is now to assign a color to each vertex so that no edge has endpoints of the same color.
Graph coloring is not the exclusive domain of map designers. Suppose a university needs to schedule examinations for all its classes and wants to use the fewest time slots possible. The only constraint is that two exams cannot be scheduled concurrently if some student will be taking both of them. To express this problem as a graph, use one vertex for each exam and put an edge between two vertices if there is a conflict, that is, if there is somebody taking both endpoint exams. Think of each time slot as having its own color. Then, assigning time slots is exactly the same as coloring this graph!
Some basic operations on graphs arise with such frequency, and in such a diversity of con- texts, that a lot of effort has gone into finding efficient procedures for them. This chapter is devoted to some of the most fundamental of these algorithms—those that uncover the basic connectivity structure of a graph.
Formally, a graph is specified by a set of vertices (also called nodes) V and by edges E between select pairs of vertices. In the map example, V = {1,2,3,…,13} and E includes, among many other edges, {1, 2}, {9, 11}, and {7, 13}. Here an edge between x and y specifically means “x shares a border with y.” This is a symmetric relation—it implies also that y shares a border with x—and we denote it using set notation, e = {x, y}. Such edges are undirected
91

92
Figure 3.1 (a) A map and (b) its graph.
(a) (b)
6 8
7
13
and are part of an undirected graph.
Sometimes graphs depict relations that do not have this reciprocity, in which case it is
necessary to use edges with directions on them. There can be directed edges e from x to y (written e = (x, y)), or from y to x (written (y, x)), or both. A particularly enormous example of a directed graph is the graph of all links in the World Wide Web. It has a vertex for each site on the Internet, and a directed edge (u, v) whenever site u has a link to site v: in total, billions of nodes and edges! Understanding even the most basic connectivity properties of the Web is of great economic and social interest. Although the size of this problem is daunting, we will soon see that a lot of valuable information about the structure of a graph can, happily, be determined in just linear time.
3.1.1 How is a graph represented?
We can represent a graph by an adjacency matrix; if there are n = |V | vertices v1, . . . , vn, this is an n × n array whose (i, j)th􏰌entry is
1 ifthereisanedgefromvi tovj 0 otherwise.
For undirected graphs, the matrix is symmetric since an edge {u, v} can be taken in either direction.
The biggest convenience of this format is that the presence of a particular edge can be checked in constant time, with just one memory access. On the other hand the matrix takes
Algorithms
54
32
1
9
10
12 11
aij =

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 93
up O(n2) space, which is wasteful if the graph does not have very many edges.
An alternative representation, with size proportional to the number of edges, is the adja- cency list. It consists of |V | linked lists, one per vertex. The linked list for vertex u holds the names of vertices to which u has an outgoing edge—that is, vertices v for which (u,v) ∈ E. Therefore, each edge appears in exactly one of the linked lists if the graph is directed or two of the lists if the graph is undirected. Either way, the total size of the data structure is O(|E|). Checking for a particular edge (u,v) is no longer constant time, because it requires sifting through u’s adjacency list. But it is easy to iterate through all neighbors of a vertex (by run- ning down the corresponding linked list), and, as we shall soon see, this turns out to be a very useful operation in graph algorithms. Again, for undirected graphs, this representation has a
symmetry of sorts: v is in u’s adjacency list if and only if u is in v’s adjacency list. How big is your graph?
Which of the two representations, adjacency matrix or adjacency list, is better? Well, it de- pends on the relationship between |V |, the number of nodes in the graph, and |E|, the num- ber of edges. |E| can be as small as |V | (if it gets much smaller, then the graph degenerates— for example, has isolated vertices), or as large as |V |2 (when all possible edges are present). When |E| is close to the upper limit of this range, we call the graph dense. At the other extreme, if |E| is close to |V |, the graph is sparse. As we shall see in this chapter and the next two chapters, exactly where |E| lies in this range is usually a crucial factor in selecting the right graph algorithm.
Or, for that matter, in selecting the graph representation. If it is the World Wide Web graph that we wish to store in computer memory, we should think twice before using an adjacency matrix: at the time of writing, search engines know of about eight billion vertices of this graph, and hence the adjacency matrix would take up dozens of millions of terabits. Again at the time we write these lines, it is not clear that there is enough computer memory in the whole world to achieve this. (And waiting a few years until there is enough memory is unwise: the Web will grow too and will probably grow faster.)
With adjacency lists, representing the World Wide Web becomes feasible: there are only a few dozen billion hyperlinks in the Web, and each will occupy a few bytes in the adjacency list. You can carry a device that stores the result, a terabyte or two, in your pocket (it may soon fit in your earring, but by that time the Web will have grown too).
The reason why adjacency lists are so much more effective in the case of the World Wide Web is that the Web is very sparse: the average Web page has hyperlinks to only about half a dozen other pages, out of the billions of possibilities.
3.2 Depth-first search in undirected graphs
3.2.1 Exploring mazes
Depth-first search is a surprisingly versatile linear-time procedure that reveals a wealth of information about a graph. The most basic question it addresses is,

94 Algorithms Figure 3.2 Exploring a graph is rather like navigating a maze.
DABE GHCFIJ
L
What parts of the graph are reachable from a given vertex?
To understand this task, try putting yourself in the position of a computer that has just been given a new graph, say in the form of an adjacency list. This representation offers just one basic operation: finding the neighbors of a vertex. With only this primitive, the reachability problem is rather like exploring a labyrinth (Figure 3.2). You start walking from a fixed place and whenever you arrive at any junction (vertex) there are a variety of passages (edges) you can follow. A careless choice of passages might lead you around in circles or might cause you to overlook some accessible part of the maze. Clearly, you need to record some intermediate information during exploration.
This classic challenge has amused people for centuries. Everybody knows that all you need to explore a labyrinth is a ball of string and a piece of chalk. The chalk prevents looping, by marking the junctions you have already visited. The string always takes you back to the starting place, enabling you to return to passages that you previously saw but did not yet investigate.
How can we simulate these two primitives, chalk and string, on a computer? The chalk marks are easy: for each vertex, maintain a Boolean variable indicating whether it has been visited already. As for the ball of string, the correct cyberanalog is a stack. After all, the exact role of the string is to offer two primitive operations—unwind to get to a new junction (the stack equivalent is to push the new vertex) and rewind to return to the previous junction (pop the stack).
Instead of explicitly maintaining a stack, we will do so implicitly via recursion (which is implemented using a stack of activation records). The resulting algorithm is shown in Figure 3.3.1 The previsit and postvisit procedures are optional, meant for performing operations on a vertex when it is first discovered and also when it is being left for the last time. We will soon see some creative uses for them.
1As with many of our graph algorithms, this one applies to both undirected and directed graphs. In such cases, we adopt the directed notation for edges, (x, y). If the graph is undirected, then each of its edges should be thought of as existing in both directions: (x, y) and (y, x).
L K
F B
E
H
J
CG
IAD
K

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 95 Figure 3.3 Finding all nodes reachable from a particular node.
procedure explore(G, v)
Input: G = (V,E) is a graph; v ∈ V
Output: visited(u) is set to true for all nodes u reachable from v
visited(v) = true previsit(v)
for each edge (v, u) ∈ E:
if not visited(u): explore(u) postvisit(v)
More immediately, we need to confirm that explore always works correctly. It certainly does not venture too far, because it only moves from nodes to their neighbors and can therefore never jump to a region that is not reachable from v. But does it find all vertices reachable from v? Well, if there is some u that it misses, choose any path from v to u, and look at the last vertex on that path that the procedure actually visited. Call this node z, and let w be the node immediately after it on the same path.
vzwu
So z was visited but w was not. This is a contradiction: while the explore procedure was at node z, it would have noticed w and moved on to it.
Incidentally, this pattern of reasoning arises often in the study of graphs and is in essence a streamlined induction. A more formal inductive proof would start by framing a hypothesis, such as “for any k ≥ 0, all nodes within k hops from v get visited.” The base case is as usual trivial, since v is certainly visited. And the general case—showing that if all nodes k hops away are visited, then so are all nodes k + 1 hops away—is precisely the same point we just argued.
Figure 3.4 shows the result of running explore on our earlier example graph, starting at node A, and breaking ties in alphabetical order whenever there is a choice of nodes to visit. The solid edges are those that were actually traversed, each of which was elicited by a call to explore and led to the discovery of a new vertex. For instance, while B was being visited, the edge B − E was noticed and, since E was as yet unknown, was traversed via a call to explore(E). These solid edges form a tree (a connected graph with no cycles) and are therefore called tree edges. The dotted edges were ignored because they led back to familiar terrain, to vertices previously visited. They are called back edges.

96 Algorithms Figure 3.4 The result of explore(A) on the graph of Figure 3.2.
A
BD EFG ICH
J
for all v ∈ V :
if not visited(v): explore(v)
3.2.2 Depth-first search
The explore procedure visits only the portion of the graph reachable from its starting point. To examine the rest of the graph, we need to restart the procedure elsewhere, at some vertex that has not yet been visited. The algorithm of Figure 3.5, called depth-first search (DFS), does this repeatedly until the entire graph has been traversed.
The first step in analyzing the running time of DFS is to observe that each vertex is explore’d just once, thanks to the visited array (the chalk marks). During the exploration of a vertex, there are the following steps:
1. Some fixed amount of work—marking the spot as visited, and the pre/postvisit.
2. A loop in which adjacent edges are scanned, to see if they lead somewhere new.
This loop takes a different amount of time for each vertex, so let’s consider all vertices to- gether. The total work done in step 1 is then O(|V |). In step 2, over the course of the entire DFS, each edge {x, y} ∈ E is examined exactly twice, once during explore(x) and once dur- ing explore(y). The overall time for step 2 is therefore O(|E|) and so the depth-first search
Figure 3.5 Depth-first search. procedure dfs(G)
for all v ∈ V : visited(v) = false

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 97 Figure 3.6 (a) A 12-node graph. (b) DFS search forest.
(a)
(b)
1,10 11,22 23,24 ACF
E 4,9 12,21 D I 5,8 13,20 H
J 14,17G L
ABCD EFGH IJKL
B 2,3
has a running time of O(|V | + |E|), linear in the size of its input. This is as efficient as we could possibly hope for, since it takes this long even just to read the adjacency list.
Figure 3.6 shows the outcome of depth-first search on a 12-node graph, once again break- ing ties alphabetically (ignore the pairs of numbers for the time being). The outer loop of DFS calls explore three times, on A, C , and finally F . As a result, there are three trees, each rooted at one of these starting points. Together they constitute a forest.
3.2.3 Connectivity in undirected graphs
An undirected graph is connected if there is a path between any pair of vertices. The graph of Figure 3.6 is not connected because, for instance, there is no path from A to K. However, it does have three disjoint connected regions, corresponding to the following sets of vertices:
{A,B,E,I,J} {C,D,G,H,K,L} {F}
These regions are called connected components: each of them is a subgraph that is internally connected but has no edges to the remaining vertices. When explore is started at a particular vertex, it identifies precisely the connected component containing that vertex. And each time the DFS outer loop calls explore, a new connected component is picked out.
Thus depth-first search is trivially adapted to check if a graph is connected and, more generally, to assign each node v an integer ccnum[v] identifying the connected component to which it belongs. All it takes is
procedure previsit(v) ccnum[v] = cc
where cc needs to be initialized to zero and to be incremented each time the DFS procedure calls explore.
6,7
18,19
K 15,16

98 Algorithms 3.2.4 Previsit and postvisit orderings
We have seen how depth-first search—a few unassuming lines of code—is able to uncover the connectivity structure of an undirected graph in just linear time. But it is far more versatile than this. In order to stretch it further, we will collect a little more information during the ex- ploration process: for each node, we will note down the times of two important events, the mo- ment of first discovery (corresponding to previsit) and that of final departure (postvisit). Figure 3.6 shows these numbers for our earlier example, in which there are 24 events. The fifth event is the discovery of I. The 21st event consists of leaving D behind for good.
One way to generate arrays pre and post with these numbers is to define a simple counter clock, initially set to 1, which gets updated as follows.
procedure previsit(v) pre[v] = clock
clock = clock + 1
procedure postvisit(v) post[v] = clock
clock = clock + 1
These timings will soon take on larger significance. Meanwhile, you might have noticed from Figure 3.4 that:
Property For any nodes u and v, the two intervals [pre(u), post(u)] and [pre(v), post(v)] are either disjoint or one is contained within the other.
Why? Because [pre(u), post(u)] is essentially the time during which vertex u was on the stack. The last-in, first-out behavior of a stack explains the rest.
3.3 Depth-first search in directed graphs
3.3.1 Types of edges
Our depth-first search algorithm can be run verbatim on directed graphs, taking care to tra- verse edges only in their prescribed directions. Figure 3.7 shows an example and the search tree that results when vertices are considered in lexicographic order.
In further analyzing the directed case, it helps to have terminology for important relation- ships between nodes of a tree. A is the root of the search tree; everything else is its descendant. Similarly, E has descendants F , G, and H , and conversely, is an ancestor of these three nodes. The family analogy is carried further: C is the parent of D, which is its child.
For undirected graphs we distinguished between tree edges and nontree edges. In the directed case, there is a slightly more elaborate taxonomy:

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani
Figure 3.7 DFS on a directed graph. BAC
99
A 1,16
B 2,11
C 12,15
D 13,14
E F D E 3,10
GH
4,7 F
5,6 G
H 8,9
Tree edges are actually part of the DFS forest.
Forward edges lead from a node to a nonchild descendant
in the DFS tree.
Back edges lead to an ancestor in the DFS tree.
Cross edges lead to neither descendant nor ancestor; they therefore lead to a node that has already been completely explored (that is, already postvisited).
C
DFS tree A
B
Cross D
Figure 3.7 has two forward edges, two back edges, and two cross edges. Can you spot them?
Ancestor and descendant relationships, as well as edge types, can be read off directly from pre and post numbers. Because of the depth-first exploration strategy, vertex u is an ancestor of vertex v exactly in those cases where u is discovered first and v is discovered during explore(u). This is to say pre(u) < pre(v) < post(v) < post(u), which we can depict pictorially as two nested intervals: uvvu The case of descendants is symmetric, since u is a descendant of v if and only if v is an an- cestor of u. And since edge categories are based entirely on ancestor-descendant relationships, Forward Back Tree 100 Algorithms it follows that they, too, can be read off from pre and post numbers. Here is a summary of the various possibilities for an edge (u, v): pre/post ordering for (u, v) uvvu vuuv vvuu Edge type Tree/forward Back Cross You can confirm each of these characterizations by consulting the diagram of edge types. Do you see why no other orderings are possible? 3.3.2 Directed acyclic graphs Acycleinadirectedgraphisacircularpathv0 →v1 →v2 →···→vk →v0. Figure3.7has quite a few of them, for example, B → E → F → B. A graph without cycles is acyclic. It turns out we can test for acyclicity in linear time, with a single depth-first search. Property A directed graph has a cycle if and only if its depth-first search reveals a back edge. Proof. One direction is quite easy: if (u, v) is a back edge, then there is a cycle consisting of this edge together with the path from v to u in the search tree. Conversely, if the graph has a cycle v0 → v1 → ··· → vk → v0, look at the first node on this cycle to be discovered (the node with the lowest pre number). Suppose it is vi. All the other vj on the cycle are reachable from it and will therefore be its descendants in the search tree. In particular, the edge vi−1 → vi (or vk → v0 if i = 0) leads from a node to its ancestor and is thus by definition a back edge. Directed acyclic graphs, or dags for short, come up all the time. They are good for modeling relations like causalities, hierarchies, and temporal dependencies. For example, suppose that you need to perform many tasks, but some of them cannot begin until certain others are completed (you have to wake up before you can get out of bed; you have to be out of bed, but not yet dressed, to take a shower; and so on). The question then is, what is a valid order in which to perform the tasks? Such constraints are conveniently represented by a directed graph in which each task is a node, and there is an edge from u to v if u is a precondition for v. In other words, before performing a task, all the tasks pointing to it must be completed. If this graph has a cycle, there is no hope: no ordering can possibly work. If on the other hand the graph is a dag, we would like if possible to linearize (or topologically sort) it, to order the vertices one after the other in such a way that each edge goes from an earlier vertex to a later vertex, so that all precedence constraints are satisfied. In Figure 3.8, for instance, one valid ordering is B, A, D, C, E, F . (Can you spot the other three?) S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 101 Figure 3.8 A directed acyclic graph with one source, two sinks, and four possible lineariza- tions. A CE BDF What types of dags can be linearized? Simple: All of them. And once again depth-first search tells us exactly how to do it: simply perform tasks in decreasing order of their post numbers. After all, the only edges (u, v) in a graph for which post(u) 1, define pk(v) = pk−1(p(v)) and p1(v) = p(v) (so pk(v) is the kth ancestor of v).
Each vertex v of the tree has an associated non-negative integer label l(v). Give a linear-time algorithm to update the labels of all the vertices in T according to the following rule: lnew(v) = l(pl(v)(v)).
3.21. Give a linear-time algorithm to find an odd-length cycle in a directed graph. (Hint: First solve this problem under the assumption that the graph is strongly connected.)

110 Algorithms
3.22. Give an efficient algorithm which takes as input a directed graph G = (V,E), and determines
whether or not there is a vertex s ∈ V from which all other vertices are reachable.
3.23. Give an efficient algorithm that takes as input a directed acyclic graph G = (V,E), and two
vertices s, t ∈ V , and outputs the number of different directed paths from s to t in G.
3.24. Give a linear-time algorithm for the following task.
Input: A directed acyclic graph G
Question: Does G contain a directed path that touches every vertex exactly once?
3.25. You are given a directed graph in which each node u ∈ V has an associated price pu which is a positive integer. Define the array cost as follows: for each u ∈ V ,
cost[u] = price of the cheapest node reachable from u (including u itself).
For instance, in the graph below (with prices shown for each vertex), the cost values of the
nodesA,B,C,D,E,F are2,1,4,1,4,5,respectively. 264
ACE
BDF 315
Your goal is to design an algorithm that fills in the entire cost array (i.e., for all vertices).
(a) Give a linear-time algorithm that works for directed acyclic graphs. (Hint: Handle the
vertices in a particular order.)
(b) Extend this to a linear-time algorithm that works for all directed graphs. (Hint: Recall the
“two-tiered” structure of directed graphs.)
3.26. An Eulerian tour in an undirected graph is a cycle that is allowed to pass through each vertex multiple times, but must use each edge exactly once.
This simple concept was used by Euler in 1736 to solve the famous Konigsberg bridge problem, which launched the field of graph theory. The city of Konigsberg (now called Kaliningrad, in western Russia) is the meeting point of two rivers with a small island in the middle. There are seven bridges across the rivers, and a popular recreational question of the time was to determine whether it is possible to perform a tour in which each bridge is crossed exactly once.
Euler formulated the relevant information as a graph with four nodes (denoting land masses) and seven edges (denoting bridges), as shown here.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani
111
Small island
Big island
Northern bank
Southern bank
Notice an unusual feature of this problem: multiple edges between certain pairs of nodes.
(a) Show that an undirected graph has an Eulerian tour if and only if all its vertices have even degree. Conclude that there is no Eulerian tour of the Konigsberg bridges.
(b) An Eulerian path is a path which uses each edge exactly once. Can you give a similar if-and-only-if characterization of which undirected graphs have Eulerian paths?
(c) Can you give an analog of part (a) for directed graphs?
3.27. Two paths in a graph are called edge-disjoint if they have no edges in common. Show that in any undirected graph, it is possible to pair up the vertices of odd degree and find paths between each such pair so that all these paths are edge-disjoint.
3.28. In the 2SAT problem, you are given a set of clauses, where each clause is the disjunction (OR) of two literals (a literal is a Boolean variable or the negation of a Boolean variable). You are looking for a way to assign a value true or false to each of the variables so that all clauses are satisfied – that is, there is at least one true literal in each clause. For example, here’s an instance of 2SAT:
(x1 ∨x2) ∧ (x1 ∨x3) ∧ (x1 ∨x2) ∧ (x3 ∨x4) ∧ (x1 ∨x4).
This instance has a satisfying assignment: set x1, x2, x3, and x4 to true, false, false, and
true, respectively.
(a) Are there other satisfying truth assignments of this 2SAT formula? If so, find them all.
(b) Give an instance of 2SAT with four variables, and with no satisfying assignment.
The purpose of this problem is to lead you to a way of solving 2SAT efficiently by reducing it to the problem of finding the strongly connected components of a directed graph. Given an instance I of 2SAT with n variables and m clauses, construct a directed graph GI = (V, E) as follows.
• GI has 2n nodes, one for each variable and its negation.
• GI has 2m edges: for each clause (α ∨ β) of I (where α, β are literals), GI has an edge from
from the negation of α to β, and one from the negation of β to α.
Note that the clause (α ∨ β) is equivalent to either of the implications α ⇒ β or β ⇒ α. In this
sense, GI records all implications in I.
(c) Carry out this construction for the instance of 2SAT given above, and for the instance you constructed in (b).

112 Algorithms
(d) Show that if GI has a strongly connected component containing both x and x for some
variable x, then I has no satisfying assignment.
(e) Now show the converse of (d): namely, that if none of GI ’s strongly connected components contain both a literal and its negation, then the instance I must be satisfiable. (Hint: As- sign values to the variables as follows: repeatedly pick a sink strongly connected component of GI . Assign value true to all literals in the sink, assign false to their negations, and delete all of these. Show that this ends up discovering a satisfying assignment.)
(f) Conclude that there is a linear-time algorithm for solving 2SAT.
3.29. LetSbeafiniteset.AbinaryrelationonSissimplyacollectionRoforderedpairs(x,y)∈S×S.
For instance, S might be a set of people, and each such pair (x, y) ∈ R might mean “x knows y.” An equivalence relation is a binary relation which satisfies three properties:
• Reflexivity: (x,x) ∈ R for all x ∈ S
• Symmetry: if (x,y) ∈ R then (y,x) ∈ R
• Transitivity: if (x,y) ∈ R and (y,z) ∈ R then (x,z) ∈ R
For instance, the binary relation “has the same birthday as” is an equivalence relation, whereas “is the father of” is not, since it violates all three properties.
Show that an equivalence relation partitions set S into disjoint groups S1, S2, . . . , Sk (in other words,S=S1∪S2∪···∪Sk andSi∩Sj =∅foralli̸=j)suchthat:
• Any two members of a group are related, that is, (x,y) ∈ R for any x,y ∈ Si, for any i.
• Members of different groups are not related, that is, for all i ̸= j, for all x ∈ Si and y ∈ Sj,
we have (x, y) ̸∈ R.
(Hint: Represent an equivalence relation by an undirected graph.)
3.30. Onpage102,wedefinedthebinaryrelation“connected”onthesetofverticesofadirectedgraph. Show that this is an equivalence relation (see Exercise 3.29), and conclude that it partitions the vertices into disjoint strongly connected components.
3.31. Biconnected components Let G = (V, E) be an undirected graph. For any two edges e, e′ ∈ E, we’ll say e ∼ e′ if either e = e′ or there is a (simple) cycle containing both e and e′.
(a) Show that ∼ is an equivalence relation (recall Exercise 3.29) on the edges.
The equivalence classes into which this relation partitions the edges are called the biconnected
components of G. A bridge is an edge which is in a biconnected component all by itself. A separating vertex is a vertex whose removal disconnects the graph.
(b) Partitiontheedgesofthegraphbelowintobiconnectedcomponents,andidentifythebridges and separating vertices.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 113 CFG
ABDEIH ONML
KJ
Not only do biconnected components partition the edges of the graph, they also almost partition the vertices in the following sense.
(c) Associate with each biconnected component all the vertices that are endpoints of its edges. Show that the vertices corresponding to two different biconnected components are either disjoint or intersect in a single separating vertex.
(d) Collapse each biconnected component into a single meta-node, and retain individual nodes for each separating vertex. (So there are edges between each component-node and its sep- arating vertices.) Show that the resulting graph is a tree.
DFS can be used to identify the biconnected components, bridges, and separating vertices of a graph in linear time.
(e) Show that the root of the DFS tree is a separating vertex if and only if it has more than one child in the tree.
(f) Show that a non-root vertex v of the DFS tree is a separating vertex if and only if it has a child v′ none of whose descendants (including itself) has a backedge to a proper ancestor of v.
(g) For each vertex u define: low(u) = min 􏰌 pre(u)
pre(w) where (v, w) is a backedge for some descendant v of u Show that the entire array of low values can be computed in linear time.
(h) Show how to compute all separating vertices, bridges, and biconnected components of a graph in linear time. (Hint: Use low to identify separating vertices, and run another DFS with an extra stack of edges to remove biconnected components one at a time.)

114 Algorithms

Chapter 4
Paths in graphs
4.1 Distances
Depth-first search readily identifies all the vertices of a graph that can be reached from a designated starting point. It also finds explicit paths to these vertices, summarized in its search tree (Figure 4.1). However, these paths might not be the most economical ones possi- ble. In the figure, vertex C is reachable from S by traversing just one edge, while the DFS tree shows a path of length 3. This chapter is about algorithms for finding shortest paths in graphs.
Path lengths allow us to talk quantitatively about the extent to which different vertices of a graph are separated from each other:
The distance between two nodes is the length of the shortest path between them.
To get a concrete feel for this notion, consider a physical realization of a graph that has a ball for each vertex and a piece of string for each edge. If you lift the ball for vertex s high enough, the other balls that get pulled up along with it are precisely the vertices reachable from s. And to find their distances from s, you need only measure how far below s they hang.
Figure 4.1 (a) A simple graph and (b) its depth-first search tree.
(a)
(b)
S
AD
BE
C
ESA
DCB
115

116
Figure 4.2 A physical model of a graph.
ESA
DCB
Algorithms
S
ACDE
B
In Figure 4.2 for example, vertex B is at distance 2 from S, and there are two shortest paths to it. When S is held up, the strings along each of these paths become taut. On the other hand, edge (D, E) plays no role in any shortest path and therefore remains slack.
4.2 Breadth-first search
In Figure 4.2, the lifting of s partitions the graph into layers: s itself, the nodes at distance 1 from it, the nodes at distance 2 from it, and so on. A convenient way to compute distances from s to the other vertices is to proceed layer by layer. Once we have picked out the nodes at distance 0, 1, 2, . . . , d, the ones at d + 1 are easily determined: they are precisely the as-yet- unseen nodes that are adjacent to the layer at distance d. This suggests an iterative algorithm in which two layers are active at any given time: some layer d, which has been fully identified, and d + 1, which is being discovered by scanning the neighbors of layer d.
Breadth-first search (BFS) directly implements this simple reasoning (Figure 4.3). Ini- tially the queue Q consists only of s, the one node at distance 0. And for each subsequent distance d = 1, 2, 3, . . ., there is a point in time at which Q contains all the nodes at distance d and nothing else. As these nodes are processed (ejected off the front of the queue), their as-yet-unseen neighbors are injected into the end of the queue.
Let’s try out this algorithm on our earlier example (Figure 4.1) to confirm that it does the right thing. If S is the starting point and the nodes are ordered alphabetically, they get visited in the sequence shown in Figure 4.4. The breadth-first search tree, on the right, contains the edges through which each node is initially discovered. Unlike the DFS tree we saw earlier, it has the property that all its paths from S are the shortest possible. It is therefore a shortest- path tree.
Correctness and efficiency
We have developed the basic intuition behind breadth-first search. In order to check that the algorithm works correctly, we need to make sure that it faithfully executes this intuition. What we expect, precisely, is that
For each d = 0,1,2,…, there is a moment at which (1) all nodes at distance ≤ d

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 117 Figure 4.3 Breadth-first search.
procedure
Input:
Output:
bfs(G, s)
Graph G = (V, E), directed or undirected; vertex s ∈ V For all vertices u reachable from s, dist(u) is set to the distance from s to u.
for all u ∈ V : dist(u) = ∞
dist(s) = 0
Q = [s] (queue containing just s) while Q is not empty:
u = eject(Q)
for all edges (u, v) ∈ E:
if dist(v) = ∞: inject(Q, v)
dist(v) = dist(u) + 1
from s have their distances correctly set; (2) all other nodes have their distances set to ∞; and (3) the queue contains exactly the nodes at distance d.
This has been phrased with an inductive argument in mind. We have already discussed both the base case and the inductive step. Can you fill in the details?
The overall running time of this algorithm is linear, O(|V | + |E|), for exactly the same reasons as depth-first search. Each vertex is put on the queue exactly once, when it is first en- countered, so there are 2 |V | queue operations. The rest of the work is done in the algorithm’s innermost loop. Over the course of execution, this loop looks at each edge once (in directed graphs) or twice (in undirected graphs), and therefore takes O(|E|) time.
Now that we have both BFS and DFS before us: how do their exploration styles compare? Depth-first search makes deep incursions into a graph, retreating only when it runs out of new nodes to visit. This strategy gives it the wonderful, subtle, and extremely useful properties we saw in the Chapter 3. But it also means that DFS can end up taking a long and convoluted route to a vertex that is actually very close by, as in Figure 4.1. Breadth-first search makes sure to visit vertices in increasing order of their distance from the starting point. This is a broader, shallower search, rather like the propagation of a wave upon water. And it is achieved using almost exactly the same code as DFS—but with a queue in place of a stack.
Also notice one stylistic difference from DFS: since we are only interested in distances from s, we do not restart the search in other connected components. Nodes not reachable from s are simply ignored.

118
Figure 4.4 The result of breadth-first search on the graph of Figure 4.1.
Algorithms
Order
of visitation
Queue contents after processing node
S A C D E B
[S]
[A C D E] [C D E B] [D E B] [E B] [B]
[]
S
ACDE
B
Figure 4.5 Edge lengths often matter.
95 Francisco
Sacramento
133
San
290
Reno 271
409
Los Angeles
4.3 Lengths on edges
Bakersfield 112
275
445
291
Las Vegas
Breadth-first search treats all edges as having the same length. This is rarely true in ap- plications where shortest paths are to be found. For instance, suppose you are driving from San Francisco to Las Vegas, and want to find the quickest route. Figure 4.5 shows the major highways you might conceivably use. Picking the right combination of them is a shortest-path problem in which the length of each edge (each stretch of highway) is important. For the re- mainder of this chapter, we will deal with this more general scenario, annotating every edge e ∈ E with a length le. If e = (u,v), we will sometimes also write l(u,v) or luv.
These le’s do not have to correspond to physical lengths. They could denote time (driving time between cities) or money (cost of taking a bus), or any other quantity that we would like to conserve. In fact, there are cases in which we need to use negative lengths, but we will briefly overlook this particular complication.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani
Figure 4.6 Breaking edges into unit-length pieces. 2
119
2BD A132 A
1
BD
CE
C4E
4.4
4.4.1
Dijkstra’s algorithm
An adaptation of breadth-first search
Breadth-first search finds shortest paths in any graph whose edges have unit length. Can we adapt it to a more general graph G = (V, E) whose edge lengths le are positive integers?
A more convenient graph
Here is a simple trick for converting G into something BFS can handle: break G’s long edges into unit-length pieces, by introducing “dummy” nodes. Figure 4.6 shows an example of this transformation. To construct the new graph G′,
For any edge e = (u,v) of E, replace it by le edges of length 1, by adding le −1 dummy nodes between u and v.
Graph G′ contains all the vertices V that interest us, and the distances between them are exactly the same as in G. Most importantly, the edges of G′ all have unit length. Therefore, we can compute distances in G by running BFS on G′.
Alarm clocks
If efficiency were not an issue, we could stop here. But when G has very long edges, the G′ it engenders is thickly populated with dummy nodes, and the BFS spends most of its time diligently computing distances to these nodes that we don’t care about at all.
To see this more concretely, consider the graphs G and G′ of Figure 4.7, and imagine that the BFS, started at node s of G′, advances by one unit of distance per minute. For the first 99 minutes it tediously progresses along S − A and S − B, an endless desert of dummy nodes. Is there some way we can snooze through these boring phases and have an alarm wake us up whenever something interesting is happening—specifically, whenever one of the real nodes (from the original graph G) is reached?
We do this by setting two alarms at the outset, one for node A, set to go off at time T = 100, and one for B, at time T = 200. These are estimated times of arrival, based upon the edges currently being traversed. We doze off and awake at T = 100 to find A has been discovered. At

120 Algorithms
this point, the estimated time of arrival for B is adjusted to T = 150 and we change its alarm accordingly.
More generally, at any given moment the breadth-first search is advancing along certain edges of G, and there is an alarm for every endpoint node toward which it is moving, set to go off at the estimated time of arrival at that node. Some of these might be overestimates be- cause BFS may later find shortcuts, as a result of future arrivals elsewhere. In the preceding example, a quicker route to B was revealed upon arrival at A. However, nothing interesting can possibly happen before an alarm goes off. The sounding of the next alarm must therefore signal the arrival of the wavefront to a real node u ∈ V by BFS. At that point, BFS might also start advancing along some new edges out of u, and alarms need to be set for their endpoints.
The following “alarm clock algorithm” faithfully simulates the execution of BFS on G′.
• Set an alarm clock for node s at time 0.
• Repeat until there are no more alarms:
Say the next alarm goes off at time T , for node u. Then:
– ThedistancefromstouisT. – For each neighbor v of u in G:
∗ Ifthereisnoalarmyetforv,setonefortimeT +l(u,v).
∗ If v’s alarm is set for later than T + l(u, v), then reset it to this earlier time.
Dijkstra’s algorithm. The alarm clock algorithm computes distances in any graph with positive integral edge lengths. It is almost ready for use, except that we need to somehow implement the system of alarms. The right data structure for this job is a priority queue (usually implemented via a heap), which maintains a set of elements (nodes) with associated numeric key values (alarm times) and supports the following operations:
Insert. Add a new element to the set.
Decrease-key. Accommodate the decrease in key value of a particular element.1
1The name decrease-key is standard but is a little misleading: the priority queue typically does not itself change key values. What this procedure really does is to notify the queue that a certain key value has been decreased.
Figure 4.7 BFS on G′ is mostly uneventful. The dotted lines show some early “wavefronts.” G: A G′: A
100
S 50 S
200
BB

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 121 Delete-min. Return the element with the smallest key, and remove it from the set.
Make-queue. Build a priority queue out of the given elements, with the given key values. (In many implementations, this is significantly faster than inserting the elements one by one.)
The first two let us set alarms, and the third tells us which alarm is next to go off. Putting this all together, we get Dijkstra’s algorithm (Figure 4.8).
In the code, dist(u) refers to the current alarm clock setting for node u. A value of ∞ means the alarm hasn’t so far been set. There is also a special array, prev, that holds one crucial piece of information for each node u: the identity of the node immediately before it on the shortest path from s to u. By following these back-pointers, we can easily reconstruct shortest paths, and so this array is a compact summary of all the paths found. A full example of the algorithm’s operation, along with the final shortest-path tree, is shown in Figure 4.9.
In summary, we can think of Dijkstra’s algorithm as just BFS, except it uses a priority queue instead of a regular queue, so as to prioritize nodes in a way that takes edge lengths into account. This viewpoint gives a concrete appreciation of how and why the algorithm works, but there is a more direct, more abstract derivation that doesn’t depend upon BFS at all. We now start from scratch with this complementary interpretation.
Figure 4.8 Dijkstra’s shortest-path algorithm.
procedure
Input:
Output:
dijkstra(G, l, s)
Graph G = (V, E), directed or undirected;
positive edge lengths {le : e ∈ E}; vertex s ∈ V
For all vertices u reachable from s, dist(u) is set to the distance from s to u.
for all u ∈ V : dist(u) = ∞
prev(u) = nil dist(s) = 0
H = makequeue (V ) (using dist-values as keys) while H is not empty:
u = deletemin(H)
for all edges (u, v) ∈ E:
if dist(v) > dist(u) + l(u, v): dist(v) = dist(u) + l(u, v) prev(v) = u decreasekey(H, v)

122 Algorithms Figure 4.9 A complete run of Dijkstra’s algorithm, with node A as the starting point. Also
shown are the associated dist values and the final shortest-path tree.
B2D 4
4
A13 1
A: 0 B: 4 C: 2
D: ∞ E: ∞
2
3 C5E
B2D 4
4 A131
A: 0 B: 3 C: 2
D: 6 E: 7
2
3 C5E
B2D 4
4
A13 1
A: 0 B: 3 C: 2
D: 5 E: 6
2
4
3 C5E
B2D
4
A: 0 B: 3 C: 2
D: 5 E: 6
A13 1 3
2
C5E
B2D
13
2
CE
A

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 123 Figure 4.10 Single-edge extensions of known shortest paths.
Known region
R
s
u
v
4.4.2 An alternative derivation
Here’s a plan for computing shortest paths: expand outward from the starting point s, steadily growing the region of the graph to which distances and shortest paths are known. This growth should be orderly, first incorporating the closest nodes and then moving on to those further away. More precisely, when the “known region” is some subset of vertices R that includes s, the next addition to it should be the node outside R that is closest to s. Let us call this node v; the question is: how do we identify it?
To answer, consider u, the node just before v in the shortest path from s to v: suv
Since we are assuming that all edge lengths are positive, u must be closer to s than v is. This means that u is in R—otherwise it would contradict v’s status as the closest node to s outside R. So, the shortest path from s to v is simply a known shortest path extended by a single edge.
But there will typically be many single-edge extensions of the currently known shortest paths (Figure 4.10); which of these identifies v? The answer is, the shortest of these extended paths. Because, if an even shorter single-edge-extended path existed, this would once more contradict v’s status as the node outside R closest to s. So, it’s easy to find v: it is the node outside R for which the smallest value of distance(s, u) + l(u, v) is attained, as u ranges over R. In other words, try all single-edge extensions of the currently known shortest paths, find the shortest such extended path, and proclaim its endpoint to be the next node of R.
We now have an algorithm for growing R by looking at extensions of the current set of shortest paths. Some extra efficiency comes from noticing that on any given iteration, the only new extensions are those involving the node most recently added to region R. All other extensions will have been assessed previously and do not need to be recomputed. In the following pseudocode, dist(v) is the length of the currently shortest single-edge-extended path leading to v; it is ∞ for nodes not adjacent to R.
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124
Algorithms
Initialize dist(s) to 0, other dist(·) values to ∞ R = { } (the ‘‘known region’’)
while R ̸= V :
Pick the node v ̸∈ R with smallest dist(·) Add v to R
for all edges (v, z) ∈ E:
if dist(z) > dist(v) + l(v, z): dist(z) = dist(v) + l(v, z)
Incorporating priority queue operations gives us back Dijkstra’s algorithm (Figure 4.8).
To justify this algorithm formally, we would use a proof by induction, as with breadth-first
search. Here’s an appropriate inductive hypothesis.
At the end of each iteration of the while loop, the following conditions hold: (1) there is a value d such that all nodes in R are at distance ≤ d from s and all nodes outside R are at distance ≥ d from s, and (2) for every node u, the value dist(u) is the length of the shortest path from s to u whose intermediate nodes are constrained to be in R (if no such path exists, the value is ∞).
The base case is straightforward (with d = 0), and the details of the inductive step can be filled in from the preceding discussion.
4.4.3 Running time
At the level of abstraction of Figure 4.8, Dijkstra’s algorithm is structurally identical to breadth-first search. However, it is slower because the priority queue primitives are com- putationally more demanding than the constant-time eject’s and inject’s of BFS. Since makequeue takes at most as long as |V | insert operations, we get a total of |V | deletemin and |V | + |E| insert/decreasekey operations. The time needed for these varies by imple- mentation; for instance, a binary heap gives an overall running time of O((|V | + |E|) log |V |).

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 125 Which heap is best?
The running time of Dijkstra’s algorithm depends heavily on the priority queue implemen- tation used. Here are the typical choices.
Implementation
deletemin
insert/
decreasekey
|V | × deletemin + (|V | + |E|) × insert
Array
Binary heap d-ary heap Fibonacci heap
O(|V |) O(log |V |)
O(dlog|V|) log d
O(log |V |)
O(1) O(log |V |)
O( log |V | ) log d
O(1) (amortized)
O(|V |2)
O((|V|+|E|)log|V|)
O((|V|·d+|E|)log|V|) log d
O(|V|log|V|+|E|)
So for instance, even a naive array implementation gives a respectable time complexity of O(|V |2), whereas with a binary heap we get O((|V | + |E|) log |V |). Which is preferable?
This depends on whether the graph is sparse (has few edges) or dense (has lots of them). For all graphs, |E| is less than |V |2. If it is Ω(|V |2), then clearly the array implementation is the faster. On the other hand, the binary heap becomes preferable as soon as |E| dips below |V|2/log|V|.
The d-ary heap is a generalization of the binary heap (which corresponds to d = 2) and leads to a running time that is a function of d. The optimal choice is d ≈ |E|/|V |; in other words, to optimize we must set the degree of the heap to be equal to the average degree of the graph. This works well for both sparse and dense graphs. For very sparse graphs, in which |E| = O(|V |), the running time is O(|V | log |V |), as good as with a binary heap. For dense graphs, |E| = Ω(|V |2) and the running time is O(|V |2), as good as with a linked list. Finally, for graphs with intermediate density |E| = |V |1+δ, the running time is O(|E|), linear!
The last line in the table gives running times using a sophisticated data structure called a Fibonacci heap. Although its efficiency is impressive, this data structure requires con- siderably more work to implement than the others, and this tends to dampen its appeal in practice. We will say little about it except to mention a curious feature of its time bounds. Its insert operations take varying amounts of time but are guaranteed to average O(1) over the course of the algorithm. In such situations (one of which we shall encounter in Chapter 5) we say that the amortized cost of heap insert’s is O(1).

126 Algorithms 4.5 Priority queue implementations
4.5.1 Array
The simplest implementation of a priority queue is as an unordered array of key values for all potential elements (the vertices of the graph, in the case of Dijkstra’s algorithm). Initially, these values are set to ∞.
An insert or decreasekey is fast, because it just involves adjusting a key value, an O(1) operation. To deletemin, on the other hand, requires a linear-time scan of the list.
4.5.2 Binary heap
Here elements are stored in a complete binary tree, namely, a binary tree in which each level is filled in from left to right, and must be full before the next level is started. In addition, a special ordering constraint is enforced: the key value of any node of the tree is less than or equal to that of its children. In particular, therefore, the root always contains the smallest element. See Figure 4.11(a) for an example.
To insert, place the new element at the bottom of the tree (in the first available position), and let it “bubble up.” That is, if it is smaller than its parent, swap the two and repeat (Figure 4.11(b)–(d)). The number of swaps is at most the height of the tree, which is ⌊log2 n⌋ when there are n elements. A decreasekey is similar, except that the element is already in the tree, so we let it bubble up from its current position.
To deletemin, return the root value. To then remove this element from the heap, take the last node in the tree (in the rightmost position in the bottom row) and place it at the root. Let it “sift down”: if it is bigger than either child, swap it with the smaller child and repeat (Figure 4.11(e)–(g)). Again this takes O(log n) time.
The regularity of a complete binary tree makes it easy to represent using an array. The tree nodes have a natural ordering: row by row, starting at the root and moving left to right within each row. If there are n nodes, this ordering specifies their positions 1, 2, . . . , n within the array. Moving up and down the tree is easily simulated on the array, using the fact that node number j has parent ⌊j/2⌋ and children 2j and 2j + 1 (Exercise 4.16).
4.5.3 d-ary heap
A d-ary heap is identical to a binary heap, except that nodes have d children instead of just two. This reduces the height of a tree with n elements to Θ(log d n) = Θ((log n)/(log d)). Inserts are therefore speeded up by a factor of Θ(log d). Deletemin operations, however, take a little longer, namely O(d logd n) (do you see why?).
The array representation of a binary heap is easily extended to the d-ary case. This time, node number j has parent ⌈(j − 1)/d⌉ and children {(j − 1)d + 2, . . . , min{n, (j − 1)d + d + 1}} (Exercise 4.16).

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 127
Figure 4.11 (a) A binary heap with 10 elements. Only the key values are shown. (b)–(d) The intermediate “bubble-up” steps in inserting an element with key 7. (e)–(g) The “sift-down” steps in a delete-min operation.
(a) 3 (b) 3 105 105
(c)
11 12 6 8 15 20 13
3
105 75
11 12 6 8 15 20 13 7
11 7 6 8 11 10 6 8
15 20 13 12 15 20 13 12 (f) 12
75 75
(d) 3
(e)
(g)
11 10 6 8 15 20 13 12
5 (h)
11 10 6 8
15 20 13
5
712 76
11 10 6 8 11 10 12 8 15 20 13 15 20 13

128 Algorithms Figure 4.12 Dijkstra’s algorithm will not work if there are negative edges.
4.6
4.6.1
3
S −2
4
A
B
Shortest paths in the presence of negative edges
Negative edges
Dijkstra’s algorithm works in part because the shortest path from the starting point s to any node v must pass exclusively through nodes that are closer than v. This no longer holds when edge lengths can be negative. In Figure 4.12, the shortest path from S to A passes through B, a node that is further away!
What needs to be changed in order to accommodate this new complication? To answer this, let’s take a particular high-level view of Dijkstra’s algorithm. A crucial invariant is that the dist values it maintains are always either overestimates or exactly correct. They start off at ∞, and the only way they ever change is by updating along an edge:
procedure update((u, v) ∈ E)
dist(v) = min{dist(v), dist(u) + l(u, v)}
This update operation is simply an expression of the fact that the distance to v cannot possibly be more than the distance to u, plus l(u, v). It has the following properties.
1. It gives the correct distance to v in the particular case where u is the second-last node in the shortest path to v, and dist(u) is correctly set.
2. It will never make dist(v) too small, and in this sense it is safe. For instance, a slew of extraneous update’s can’t hurt.
This operation is extremely useful: it is harmless, and if used carefully, will correctly set distances. In fact, Dijkstra’s algorithm can be thought of simply as a sequence of update’s. We know this particular sequence doesn’t work with negative edges, but is there some other sequence that does? To get a sense of the properties this sequence must possess, let’s pick a node t and look at the shortest path to it from s.
st u1 u2 u3 uk
This path can have at most |V | − 1 edges (do you see why?). If the sequence of updates per- formed includes (s, u1), (u1, u2), (u2, u3), . . . , (uk, t), in that order (though not necessarily con- secutively), then by the first property the distance to t will be correctly computed. It doesn’t
􏰥􏰦 􏰥 􏰣􏰤 􏰡􏰢 􏰟􏰠 􏰝􏰞 􏰝 􏰛􏰜

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 129 Figure 4.13 The Bellman-Ford algorithm for single-source shortest paths in general graphs.
procedure
Input:
Output:
shortest-paths(G, l, s)
Directed graph G = (V, E);
edge lengths {le : e ∈ E} with no negative cycles; vertex s ∈ V
For all vertices u reachable from s, dist(u) is set to the distance from s to u.
for all u ∈ V : dist(u) = ∞
prev(u) = nil
dist(s) = 0
repeat |V | − 1 times:
forall e∈E: update(e)
matter what other updates occur on these edges, or what happens in the rest of the graph, because updates are safe.
But still, if we don’t know all the shortest paths beforehand, how can we be sure to update the right edges in the right order? Here is an easy solution: simply update all the edges, |V | − 1 times! The resulting O(|V | · |E|) procedure is called the Bellman-Ford algorithm and is shown in Figure 4.13, with an example run in Figure 4.14.
A note about implementation: for many graphs, the maximum number of edges in any shortest path is substantially less than |V | − 1, with the result that fewer rounds of updates are needed. Therefore, it makes sense to add an extra check to the shortest-path algorithm, to make it terminate immediately after any round in which no update occurred.
4.6.2 Negative cycles
If the length of edge (E, B) in Figure 4.14 were changed to −4, the graph would have a negative cycle A → E → B → A. In such situations, it doesn’t make sense to even ask about shortest paths. There is a path of length 2 from A to E. But going round the cycle, there’s also a path of length 1, and going round multiple times, we find paths of lengths 0, −1, −2, and so on.
The shortest-path problem is ill-posed in graphs with negative cycles. As might be ex- pected, our algorithm from Section 4.6.1 works only in the absence of such cycles. But where did this assumption appear in the derivation of the algorithm? Well, it slipped in when we asserted the existence of a shortest path from s to t.
Fortunately, it is easy to automatically detect negative cycles and issue a warning. Such a cycle would allow us to endlessly apply rounds of update operations, reducing dist estimates every time. So instead of stopping after |V | − 1 iterations, perform one extra round. There is a negative cycle if and only if some dist value is reduced during this final round.

130 Algorithms Figure 4.14 The Bellman-Ford algorithm illustrated on a sample graph.
S 10 A 8
1
Node
S A B C D E F G
0
1
2
Iteration
3
4
5
6
7
0 10 ∞ ∞ ∞ ∞ ∞ 8
0 10 ∞ ∞ ∞ 12 9 8
0
5 10 ∞ ∞ 8 9 8
0 5 6 11 ∞ 7 9 8
0 5 5 7 14 7 9 8
0 5 5 6 10 7 9 8
0 5 5 6 9 7 9 8
G−4 B 21
0
∞ ∞ ∞ ∞
1
4.7 Shortest paths in dags
F −2 C ∞
−1 3 E −1 D
∞ ∞
There are two subclasses of graphs that automatically exclude the possibility of negative cy- cles: graphs without negative edges, and graphs without cycles. We already know how to efficiently handle the former. We will now see how the single-source shortest-path problem can be solved in just linear time on directed acyclic graphs.
As before, we need to perform a sequence of updates that includes every shortest path as a subsequence. The key source of efficiency is that
In any path of a dag, the vertices appear in increasing linearized order.
Therefore, it is enough to linearize (that is, topologically sort) the dag by depth-first search, and then visit the vertices in sorted order, updating the edges out of each. The algorithm is given in Figure 4.15.
Notice that our scheme doesn’t require edges to be positive. In particular, we can find longest paths in a dag by the same algorithm: just negate all edge lengths.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 131 Figure 4.15 A single-source shortest-path algorithm for directed acyclic graphs.
procedure
Input:
Output:
dag-shortest-paths(G, l, s)
DagG=(V,E);
edgelengths {le :e∈E};vertex s∈V
For all vertices u reachable from s, dist(u) is set to the distance from s to u.
for all u ∈ V : dist(u) = ∞
prev(u) = nil
dist(s) = 0
Linearize G
for each u ∈ V , in linearized order:
for all edges (u, v) ∈ E: update(u, v)

132 Algorithms
Exercises
4.1. Suppose Dijkstra’s algorithm is run on the following graph, starting at node A. A1B2C1D
4866214
E5F1G1H
(a) Draw a table showing the intermediate distance values of all the nodes at each iteration of
the algorithm.
(b) Show the final shortest-path tree.
4.2. Just like the previous problem, but this time with the Bellman-Ford algorithm.
A
4
7
B −2
G 1 H −2 −1I1 3E
4.3. Squares. Design and analyze an algorithm that takes as input an undirected graph G = (V, E) and determines whether G contains a simple cycle (that is, a cycle which doesn’t intersect itself) of length four. Its running time should be at most O(|V |3).
You may assume that the input graph is represented either as an adjacency matrix or with adjacency lists, whichever makes your algorithm simpler.
4.4. Here’saproposalforhowtofindthelengthoftheshortestcycleinanundirectedgraphwithunit edge lengths.
When a back edge, say (v, w), is encountered during a depth-first search, it forms a cycle with the tree edges from w to v. The length of the cycle is level[v] − level[w] + 1, where the level of a vertex is its distance in the DFS tree from the root vertex. This suggests the following algorithm:
• Do a depth-first search, keeping track of the level of each vertex.
• Each time a back edge is encountered, compute the cycle length and save it if it is
smaller than the shortest one previously seen.
Show that this strategy does not always work by providing a counterexample as well as a brief (one or two sentence) explanation.
−2
C2D
6
S
5
1
3
−4
F 6

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 133
4.5. Often there are multiple shortest paths between two nodes of a graph. Give a linear-time algo-
rithm for the following task.
Input: Undirected graph G = (V, E) with unit edge lengths; nodes u, v ∈ V .
Output: The number of distinct shortest paths from u to v.
4.6. Prove that for the array prev computed by Dijkstra’s algorithm, the edges {u, prev[u]} (for all
u ∈ V ) form a tree.
4.7. You are given a directed graph G = (V, E) with (possibly negative) weighted edges, along with a specific node s ∈ V and a tree T = (V, E′), E′ ⊆ E. Give an algorithm that checks whether T is a shortest-path tree for G with starting point s. Your algorithm should run in linear time.
4.8. Professor F. Lake suggests the following algorithm for finding the shortest path from node s to node t in a directed graph with some negative edges: add a large constant to each edge weight so that all the weights become positive, then run Dijkstra’s algorithm starting at node s, and return the shortest path found to node t.
Is this a valid method? Either prove that it works correctly, or give a counterexample.
4.9. Consideradirectedgraphinwhichtheonlynegativeedgesarethosethatleaves;allotheredges
are positive. Can Dijkstra’s algorithm, started at s, fail on such a graph? Prove your answer.
4.10. You are given a directed graph with (possibly negative) weighted edges, in which the shortest path between any two vertices is guaranteed to have at most k edges. Give an algorithm that finds the shortest path between two vertices u and v in O(k|E|) time.
4.11. Give an algorithm that takes as input a directed graph with positive edge lengths, and returns the length of the shortest cycle in the graph (if the graph is acyclic, it should say so). Your algorithm should take time at most O(|V |3).
4.12. Give an O(|V |2) algorithm for the following task.
Input: An undirected graph G = (V, E); edge lengths le > 0; an edge e ∈ E.
Output: The length of the shortest cycle containing edge e.
4.13. You are given a set of cities, along with the pattern of highways between them, in the form of an undirected graph G = (V, E). Each stretch of highway e ∈ E connects two of the cities, and you know its length in miles, le. You want to get from city s to city t. There’s one problem: your car can only hold enough gas to cover L miles. There are gas stations in each city, but not between cities. Therefore, you can only take a route if every one of its edges has length le ≤ L.
(a) Given the limitation on your car’s fuel tank capacity, show how to determine in linear time whether there is a feasible route from s to t.
(b) You are now planning to buy a new car, and you want to know the minimum fuel tank capacity that is needed to travel from s to t. Give an O((|V | + |E|) log |V |) algorithm to determine this.
4.14. You are given a strongly connected directed graph G = (V, E) with positive edge weights along with a particular node v0 ∈ V . Give an efficient algorithm for finding shortest paths between all pairs of nodes, with the one restriction that these paths must all pass through v0.
4.15. Shortest paths are not always unique: sometimes there are two or more different paths with the minimum possible length. Show how to solve the following problem in O((|V | + |E|) log |V |) time.

134 Algorithms
Input: An undirected graph G = (V, E); edge lengths le > 0; starting vertex s ∈ V .
Output: A Boolean array usp[·]: for each node u, the entry usp[u] should be true if and only if there is a unique shortest path from s to u. (Note: usp[s] = true.)

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 135 Figure 4.16 Operations on a binary heap.
procedure insert(h, x) bubbleup(h, x, |h| + 1)
procedure decreasekey(h, x) bubbleup(h, x, h−1(x))
(h)
function deletemin
if |h| = 0: return null
else:
x = h(1)
siftdown(h, h(|h|), 1) return x
function makeheap(S)
h = empty array of size |S| for x∈S:
h(|h| + 1) = x
for i = |S| downto 1:
siftdown(h, h(i), i) return h
procedure bubbleup(h, x, i)
(place element x in position i of h, and let it bubble up) p = ⌈i/2⌉
while i ̸= 1 and key(h(p)) > key(x):
h(i)=h(p); i=p; p=⌈i/2⌉ h(i) = x
procedure siftdown(h, x, i)
(place element x in position i of h, and let it sift down) c = minchild(h, i)
while c ̸= 0 and key(h(c)) < key(x): h(i) = h(c); i = c; c = minchild(h, i) h(i) = x (h, i) function minchild (return the index of the smallest child of h(i)) if 2i > |h|:
return 0 (no children) else:
return arg min{key(h(j)) : 2i ≤ j ≤ min{|h|, 2i + 1}}

136 Algorithms 4.16. Section 4.5.2 describes a way of storing a complete binary tree of n nodes in an array indexed by
1,2,…,n.
(a) Consider the node at position j of the array. Show that its parent is at position ⌊j/2⌋ and
its children are at 2j and 2j + 1 (if these numbers are ≤ n).
(b) What the corresponding indices when a complete d-ary tree is stored in an array?
Figure 4.16 shows pseudocode for a binary heap, modeled on an exposition by R.E. Tarjan.2 The heap is stored as an array h, which is assumed to support two constant-time operations:
• |h|, which returns the number of elements currently in the array; • h−1, which returns the position of an element within the array.
The latter can always be achieved by maintaining the values of h−1 as an auxiliary array.
(c) Show that the makeheap procedure takes O(n) time when called on a set of n elements. 􏰐What is the worst-case input? (Hint: Start by showing that the running time is at most
ni=1 log(n/i).)
(a) What needs to be changed to adapt this pseudocode to d-ary heaps?
4.17. Suppose we want to run Dijkstra’s algorithm on a graph whose edge weights are integers in the range 0, 1, . . . , W , where W is a relatively small number.
(a) Show how Dijkstra’s algorithm can be made to run in time O(W |V | + |E|).
(b) ShowanalternativeimplementationthattakestimejustO((|V|+|E|)logW).
4.18. In cases where there are several different shortest paths between two nodes (and edges have varying lengths), the most convenient of these paths is often the one with fewest edges. For instance, if nodes represent cities and edge lengths represent costs of flying between cities, there might be many ways to get from city s to city t which all have the same cost. The most convenient of these alternatives is the one which involves the fewest stopovers. Accordingly, for a specific starting node s, define
best[u] = minimum number of edges in a shortest path from s to u.
In the example below, the best values for nodes S, A, B, C, D, E, F are 0, 1, 1, 1, 2, 2, 3, respectively.
2A2 1D1 S4C3F
2B21E
Give an efficient algorithm for the following problem.
Input: Graph G = (V, E); positive edge lengths le; starting node s ∈ V .
Output: The values of best[u] should be set for all nodes u ∈ V .
2See: R. E. Tarjan, Data Structures and Network Algorithms, Society for Industrial and Applied Mathematics, 1983.
1

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 137 4.19. Generalized shortest-paths problem. In Internet routing, there are delays on lines but also, more
significantly, delays at routers. This motivates a generalized shortest-paths problem.
Suppose that in addition to having edge lengths {le : e ∈ E}, a graph also has vertex costs {cv : v ∈ V}. Now define the cost of a path to be the sum of its edge lengths, plus the costs of all vertices on the path (including the endpoints). Give an efficient algorithm for the following problem.
Input: A directed graph G = (V, E); positive edge lengths le and positive vertex costs cv; a starting vertex s ∈ V .
Output: An array cost[·] such that for every vertex u, cost[u] is the least cost of any path from s to u (i.e., the cost of the cheapest path), under the definition above.
Notice that cost[s] = cs.
4.20. There is a network of roads G = (V, E) connecting a set of cities V . Each road in E has an associated length le. There is a proposal to add one new road to this network, and there is a list E′ of pairs of cities between which the new road can be built. Each such potential road e′ ∈ E′ has an associated length. As a designer for the public works department you are asked to determine the road e′ ∈ E′ whose addition to the existing network G would result in the maximum decrease in the driving distance between two fixed cities s and t in the network. Give an efficient algorithm for solving this problem.
4.21. Shortest path algorithms can be applied in currency trading. Let c1, c2, . . . , cn be various cur- rencies; for instance, c1 might be dollars, c2 pounds, and c3 lire. For any two currencies ci and cj , there is an exchange rate ri,j ; this means that you can purchase ri,j units of currency cj in exchange for one unit of ci. These exchange rates satisfy the condition that ri,j · rj,i < 1, so that if you start with a unit of currency ci, change it into currency cj and then convert back to currency ci, you end up with less than one unit of currency ci (the difference is the cost of the transaction). (a) Give an efficient algorithm for the following problem: Given a set of exchange rates ri,j, and two currencies s and t, find the most advantageous sequence of currency exchanges for converting currency s into currency t. Toward this goal, you should represent the currencies and rates by a graph whose edge lengths are real numbers. The exchange rates are updated frequently, reflecting the demand and supply of the various currencies. Occasionally the exchange rates satisfy the following property: there is a sequence of currenciesci1,ci2,...,cik suchthatri1,i2 ·ri2,i3 ···rik−1,ik ·rik,i1 >1. Thismeansthatbystarting with a unit of currency ci1 and then successively converting it to currencies ci2 , ci3 , . . . , cik , and finally back to ci1 , you would end up with more than one unit of currency ci1 . Such anomalies last only a fraction of a minute on the currency exchange, but they provide an opportunity for risk-free profits.
(b) Give an efficient algorithm for detecting the presence of such an anomaly. Use the graph representation you found above.
4.22. The tramp steamer problem. You are the owner of a steamship that can ply between a group of port cities V . You make money at each port: a visit to city i earns you a profit of pi dollars. Meanwhile, the transportation cost from port i to port j is cij > 0. You want to find a cyclic route in which the ratio of profit to cost is maximized.

138
Algorithms
To this end, consider a directed graph G = (V,E) whose nodes are ports, and which has edges between each pair of ports. For any cycle C in this graph, the profit-to-cost ratio is
􏰐(i,j)∈C pj r(C) = 􏰐 c .
(i,j)∈C ij
Let r∗ be the maximum ratio achievable by a simple cycle. One way to determine r∗ is by binary
search: by first guessing some ratio r, and then testing whether it is too large or too small. Consider any positive r > 0. Give each edge (i, j) a weight of wij = rcij − pj .
(a) Show that if there is a cycle of negative weight, then r < r∗. (b) Show that if all cycles in the graph have strictly positive weight, then r > r∗.
(c) Giveanefficientalgorithmthattakesasinputadesiredaccuracyε>0andreturnsasimple cycle C for which r(C) ≥ r∗ − ε. Justify the correctness of your algorithm and analyze its running time in terms of |V |, ε, and R = max(i,j)∈E (pj /cij ).

Chapter 5
Greedy algorithms
A game like chess can be won only by thinking ahead: a player who is focused entirely on immediate advantage is easy to defeat. But in many other games, such as Scrabble, it is possible to do quite well by simply making whichever move seems best at the moment and not worrying too much about future consequences.
This sort of myopic behavior is easy and convenient, making it an attractive algorithmic strategy. Greedy algorithms build up a solution piece by piece, always choosing the next piece that offers the most obvious and immediate benefit. Although such an approach can be disastrous for some computational tasks, there are many for which it is optimal. Our first example is that of minimum spanning trees.
5.1 Minimum spanning trees
Suppose you are asked to network a collection of computers by linking selected pairs of them. This translates into a graph problem in which nodes are computers, undirected edges are potential links, and the goal is to pick enough of these edges that the nodes are connected. But this is not all; each link also has a maintenance cost, reflected in that edge’s weight. What is the cheapest possible network?
1 434245
A B4D6F
CE
One immediate observation is that the optimal set of edges cannot contain a cycle, because removing an edge from this cycle would reduce the cost without compromising connectivity:
Property 1 Removing a cycle edge cannot disconnect a graph.
So the solution must be connected and acyclic: undirected graphs of this kind are called trees. The particular tree we want is the one with minimum total weight, known as the minimum spanning tree. Here is its formal definition.
139

140
Algorithms
Input: An undirected graph G = (V, E); edge weights we. Output: A tree T = (V,E′), with E′ ⊆ E, that􏰑minimizes
weight(T) = we. e∈E ′
In the preceding example, the minimum spanning tree has a cost of 16:
1
A
CE 4
25 B4DF
However, this is not the only optimal solution. Can you spot another?
5.1.1 A greedy approach
Kruskal’s minimum spanning tree algorithm starts with the empty graph and then selects edges from E according to the following rule.
Repeatedly add the next lightest edge that doesn’t produce a cycle.
In other words, it constructs the tree edge by edge and, apart from taking care to avoid cycles, simply picks whichever edge is cheapest at the moment. This is a greedy algorithm: every decision it makes is the one with the most obvious immediate advantage.
Figure 5.1 shows an example. We start with an empty graph and then attempt to add edges in increasing order of weight (ties are broken arbitrarily):
B−C, C−D, B−D, C−F, D−F, E−F, A−D, A−B, C−E, A−C.
The first two succeed, but the third, B − D, would produce a cycle if added. So we ignore it
and move along. The final result is a tree with cost 14, the minimum possible.
The correctness of Kruskal’s method follows from a certain cut property, which is general enough to also justify a whole slew of other minimum spanning tree algorithms.
Figure 5.1 The minimum spanning tree found by Kruskal’s algorithm. A6C5EACE
5412
34 B2D4FBDF

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 141 Trees
A tree is an undirected graph that is connected and acyclic. Much of what makes trees so useful is the simplicity of their structure. For instance,
Property 2 A tree on n nodes has n − 1 edges.
This can be seen by building the tree one edge at a time, starting from an empty graph. Initially each of the n nodes is disconnected from the others, in a connected component by itself. As edges are added, these components merge. Since each edge unites two different components, exactly n − 1 edges are added by the time the tree is fully formed.
In a little more detail: When a particular edge {u, v} comes up, we can be sure that u and v lie in separate connected components, for otherwise there would already be a path between them and this edge would create a cycle. Adding the edge then merges these two components, thereby reducing the total number of connected components by one. Over the course of this incremental process, the number of components decreases from n to one, meaning that n − 1 edges must have been added along the way.
The converse is also true.
Property 3 Any connected, undirected graph G = (V, E) with |E| = |V | − 1 is a tree.
We just need to show that G is acyclic. One way to do this is to run the following iterative procedure on it: while the graph contains a cycle, remove one edge from this cycle. The process terminates with some graph G′ = (V, E′), E′ ⊆ E, which is acyclic and, by Property 1 (from page 139), is also connected. Therefore G′ is a tree, whereupon |E′| = |V | − 1 by Property 2. So E′ = E, no edges were removed, and G was acyclic to start with.
In other words, we can tell whether a connected graph is a tree just by counting how many edges it has. Here’s another characterization.
Property 4 An undirected graph is a tree if and only if there is a unique path between any pair of nodes.
In a tree, any two nodes can only have one path between them; for if there were two paths, the union of these paths would contain a cycle.
On the other hand, if a graph has a path between any two nodes, then it is connected. If these paths are unique, then the graph is also acyclic (since a cycle has two paths between any pair of nodes).

142 Algorithms Figure 5.2 T ∪ {e}. The addition of e (dotted) to T (solid lines) produces a cycle. This cycle
must contain at least one other edge, shown here as e′, across the cut (S, V − S). S V−S
e e′
5.1.2 The cut property
Say that in the process of building a minimum spanning tree (MST), we have already chosen some edges and are so far on the right track. Which edge should we add next? The following lemma gives us a lot of flexibility in our choice.
Cut property Suppose edges X are part of a minimum spanning tree of G = (V, E). Pick any subset of nodes S for which X does not cross between S and V − S, and let e be the lightest edge across this partition. Then X ∪ {e} is part of some MST.
A cut is any partition of the vertices into two groups, S and V − S. What this property says is that it is always safe to add the lightest edge across any cut (that is, between a vertex in S and one in V − S), provided X has no edges across the cut.
Let’s see why this holds. Edges X are part of some MST T ; if the new edge e also happens to be part of T , then there is nothing to prove. So assume e is not in T . We will construct a different MST T ′ containing X ∪ {e} by altering T slightly, changing just one of its edges.
Add edge e to T . Since T is connected, it already has a path between the endpoints of e, so adding e creates a cycle. This cycle must also have some other edge e′ across the cut (S, V − S) (Figure 8.3). If we now remove this edge, we are left with T ′ = T ∪ {e} − {e′}, which we will show to be a tree. T ′ is connected by Property 1, since e′ is a cycle edge. And it has the same number of edges as T ; so by Properties 2 and 3, it is also a tree.
Moreover, T ′ is a minimum spanning tree. Compare its weight to that of T : weight(T′) = weight(T) + w(e) − w(e′).
Both e and e′ cross between S and V − S, and e is specifically the lightest edge of this type. Therefore w(e) ≤ w(e′), and weight(T′) ≤ weight(T). Since T is an MST, it must be the case that weight(T ′ ) = weight(T ) and that T ′ is also an MST.
Figure 5.3 shows an example of the cut property. Which edge is e′?
􏰨􏰧
􏰝􏰞
􏰟􏰠 􏰟
􏰸􏰷 􏰷 􏰶􏰵 􏰵
􏰥􏰤􏰣􏰦
􏰪􏰩
􏰬􏰫
􏰨􏰧
􏰛􏰜
􏰢􏰡
􏰢􏰡
􏰮􏰭
􏰯􏰰
􏰴􏰳 􏰲􏰱

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 143
Figure 5.3 The cut property at work. (a) An undirected graph. (b) Set X has three edges, and is part of the MST T on the right. (c) If S = {A,B,C,D}, then one of the minimum-weight edgesacrossthecut(S,V −S)ise={D,E}. X∪{e}ispartofMSTT′,shownontheright.
(a)
(b)
(c)
Edges X:
The cut:
A1C3E 22231
B1D4F
ACE ACE
MST T:
BDF BDF
ACE
e MST T′: BDF
ACE BDF
S V−S 5.1.3 Kruskal’s algorithm
We are ready to justify Kruskal’s algorithm. At any given moment, the edges it has already chosen form a partial solution, a collection of connected components each of which has a tree structure. The next edge e to be added connects two of these components; call them T1 and T2. Since e is the lightest edge that doesn’t produce a cycle, it is certain to be the lightest edge between T1 and V − T1 and therefore satisfies the cut property.
Now we fill in some implementation details. At each stage, the algorithm chooses an edge to add to its current partial solution. To do so, it needs to test each candidate edge u − v to see whether the endpoints u and v lie in different components; otherwise the edge produces a cycle. And once an edge is chosen, the corresponding components need to be merged. What kind of data structure supports such operations?
We will model the algorithm’s state as a collection of disjoint sets, each of which contains the nodes of a particular component. Initially each node is in a component by itself:
makeset(x): create a singleton set containing just x.
We repeatedly test pairs of nodes to see if they belong to the same set.
find(x): to which set does x belong?

144 Algorithms Figure 5.4 Kruskal’s minimum spanning tree algorithm.
procedure kruskal(G, w)
Input: A connected undirected graph G = (V, E) with edge weights we Output: A minimum spanning tree defined by the edges X
for all u ∈ V : makeset(u)
X = {}
Sort the edges E by weight
for all edges {u, v} ∈ E, in increasing order of weight:
if find(u) ̸= find(v): add edge {u, v} to X union(u, v)
And whenever we add an edge, we are merging two components. union(x, y): merge the sets containing x and y.
The final algorithm is shown in Figure 5.4. It uses |V | makeset, 2|E| find, and |V | − 1 union operations.
5.1.4 A data structure for disjoint sets
Union by rank
One way to store a set is as a directed tree (Figure 5.5). Nodes of the tree are elements of the set, arranged in no particular order, and each has parent pointers that eventually lead up to the root of the tree. This root element is a convenient representative, or name, for the set. It is distinguished from the other elements by the fact that its parent pointer is a self-loop.
Figure 5.5 A directed-tree representation of two sets {B, E} and {A, C, D, F, G, H}. EH
B CDF
GA

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 145 In addition to a parent pointer π, each node also has a rank that, for the time being, should
be interpreted as the height of the subtree hanging from that node.
procedure makeset(x) π(x) = x
rank(x) = 0
function find (x)
while x̸=π(x): x=π(x) return x
As can be expected, makeset is a constant-time operation. On the other hand, find follows parent pointers to the root of the tree and therefore takes time proportional to the height of the tree. The tree actually gets built via the third operation, union, and so we must make sure that this procedure keeps trees shallow.
Merging two sets is easy: make the root of one point to the root of the other. But we have a choice here. If the representatives (roots) of the sets are rx and ry, do we make rx point to ry or the other way around? Since tree height is the main impediment to computational efficiency, a good strategy is to make the root of the shorter tree point to the root of the taller tree. This way, the overall height increases only if the two trees being merged are equally tall. Instead of explicitly computing heights of trees, we will use the rank numbers of their root nodes—which is why this scheme is called union by rank.
procedure union(x, y) rx =find(x)
ry =find(y)
if rx = ry: return if rank(rx) > rank(ry):
π(ry) = rx else:
π(rx) = ry
ifrank(rx)=rank(ry): rank(ry)=rank(ry)+1
See Figure 5.6 for an example.
By design, the rank of a node is exactly the height of the subtree rooted at that node. This means, for instance, that as you move up a path toward a root node, the rank values along the way are strictly increasing.
Property 1 For any x, rank(x) < rank(π(x)). A root node with rank k is created by the merger of two trees with roots of rank k − 1. It follows by induction (try it!) that Property 2 Any root node of rank k has at least 2k nodes in its tree. 146 Algorithms This extends to internal (nonroot) nodes as well: a node of rank k has at least 2k de- scendants. After all, any internal node was once a root, and neither its rank nor its set of descendants has changed since then. Moreover, different rank-k nodes cannot have common descendants, since by Property 1 any element has at most one ancestor of rank k. Which means Property 3 If there are n elements overall, there can be at most n/2k nodes of rank k. This last observation implies, crucially, that the maximum rank is log n. Therefore, all the trees have height ≤ log n, and this is an upper bound on the running time of find and union. Figure 5.6 A sequence of disjoint-set operations. Superscripts denote rank. Aftermakeset(A),makeset(B),...,makeset(G): A0 B0 C0 D0 E0 F0 G0 Afterunion(A,D),union(B,E),union(C,F): D1 E1 F1 G0 A0 B0 C0 After union(C, G),union(E, A): D2 F1 After union(B, G): 1 E A0 C0G0 B0 D2 0 A E1 F1 B0 C0 G0 S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 147 Path compression With the data structure as presented so far, the total time for Kruskal’s algorithm becomes O(|E| log |V |) for sorting the edges (remember, log |E| ≈ log |V |) plus another O(|E| log |V |) for the union and find operations that dominate the rest of the algorithm. So there seems to be little incentive to make our data structure any more efficient. But what if the edges are given to us sorted? Or if the weights are small (say, O(|E|)) so that sorting can be done in linear time? Then the data structure part becomes the bottleneck, and it is useful to think about improving its performance beyond log n per operation. As it turns out, the improved data structure is useful in many other applications. But how can we perform union’s and find’s faster than log n? The answer is, by being a little more careful to maintain our data structure in good shape. As any housekeeper knows, a little extra effort put into routine maintenance can pay off handsomely in the long run, by forestalling major calamities. We have in mind a particular maintenance operation for our union-find data structure, intended to keep the trees short— during each find, when a series of parent pointers is followed up to the root of a tree, we will change all these pointers so that they point directly to the root (Figure 5.7). This path compression heuristic only slightly increases the time needed for a find and is easy to code. (x) function find if x ̸= π(x) : π(x) = find(π(x)) return π(x) The benefit of this simple alteration is long-term rather than instantaneous and thus neces- sitates a particular kind of analysis: we need to look at sequences of find and union opera- tions, starting from an empty data structure, and determine the average time per operation. This amortized cost turns out to be just barely more than O(1), down from the earlier O(log n). Think of the data structure as having a “top level” consisting of the root nodes, and below it, the insides of the trees. There is a division of labor: find operations (with or without path compression) only touch the insides of trees, whereas union’s only look at the top level. Thus path compression has no effect on union operations and leaves the top level unchanged. We now know that the ranks of root nodes are unaltered, but what about nonroot nodes? The key point here is that once a node ceases to be a root, it never resurfaces, and its rank is forever fixed. Therefore the ranks of all nodes are unchanged by path compression, even though these numbers can no longer be interpreted as tree heights. In particular, properties 1–3 (from page 145) still hold. If there are n elements, their rank values can range from 0 to log n by Property 3. Let’s divide the nonzero part of this range into certain carefully chosen intervals, for reasons that will soon become clear: {1},{2},{3,4},{5,6,...,16},{17,18,...,216 =65536},{65537,65538,...,265536},... Each group is of the form {k + 1, k + 2, . . . , 2k }, where k is a power of 2. The number of groups is log∗ n, which is defined to be the number of successive log operations that need to be applied 148 Algorithms Figure 5.7 The effect of path compression: find(I) followed by find(K). A3 A3 B0 C1 E2 B0 C1 E2 F1 I0 0 1 1 0 −→ 0 1 0 0 DFGH DGHJ I0 J0 K0 −→ to n to bring it down to 1 (or below 1). For instance, log∗ 1000 = 4 since log log log log 1000 ≤ 1. In practice there will just be the first five of the intervals shown; more are needed only if n ≥ 265536, in other words never. In a sequence of find operations, some may take longer than others. We’ll bound the overall running time using some creative accounting. Specifically, we will give each node a certain amount of pocket money, such that the total money doled out is at most n log ∗ n dollars. We will then show that each find takes O(log∗ n) steps, plus some additional amount of time that can be “paid for” using the pocket money of the nodes involved—one dollar per unit of time. Thus the overall time for m find’s is O(m log∗ n) plus at most O(n log∗ n). In more detail, a node receives its allowance as soon as it ceases to be a root, at which point itsrankisfixed. Ifthisrankliesintheinterval{k+1,...,2k},thenodereceives2k dollars. By Property 3, the number of nodes with rank > k is bounded by
n+n+···≤n. 2k+1 2k+2 2k
Therefore the total money given to nodes in this particular interval is at most n dollars, and since there are log∗ n intervals, the total money disbursed to all nodes is ≤ n log∗ n.
3 A
K0
B0 C1 E2 F1 I0 K0 G1
D0 H0 J0

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 149
Now, the time taken by a specific find is simply the number of pointers followed. Consider the ascending rank values along this chain of nodes up to the root. Nodes x on the chain fall into two categories: either the rank of π(x) is in a higher interval than the rank of x, or else it lies in the same interval. There are at most log∗ n nodes of the first type (do you see why?), so the work done on them takes O(log∗ n) time. The remaining nodes—whose parents’ ranks are in the same interval as theirs—have to pay a dollar out of their pocket money for their processing time.
This only works if the initial allowance of each node x is enough to cover all of its payments in the sequence of find operations. Here’s the crucial observation: each time x pays a dollar, its parent changes to one of higher rank. Therefore, if x’s rank lies in the interval {k + 1,…,2k}, it has to pay at most 2k dollars before its parent’s rank is in a higher interval; whereupon it never has to pay again.

150 Algorithms A randomized algorithm for minimum cut
We have already seen that spanning trees and cuts are intimately related. Here is another connection. Let’s remove the last edge that Kruskal’s algorithm adds to the spanning tree; this breaks the tree into two components, thus defining a cut (S,S) in the graph. What can we say about this cut? Suppose the graph we were working with was unweighted, and that its edges were ordered uniformly at random for Kruskal’s algorithm to process them. Here is a remarkable fact: with probability at least 1/n2, (S,S) is the minimum cut in the graph, where the size of a cut (S, S) is the number of edges crossing between S and S. This means that repeating the process O(n2) times and outputting the smallest cut found yields the minimum cut in G with high probability: an O(mn2 log n) algorithm for unweighted minimum cuts. Some further tuning gives the O(n2 log n) minimum cut algorithm, invented by David Karger, which is the fastest known algorithm for this important problem.
So let us see why the cut found in each iteration is the minimum cut with probability at least 1/n2. At any stage of Kruskal’s algorithm, the vertex set V is partitioned into connected components. The only edges eligible to be added to the tree have their two endpoints in distinct components. The number of edges incident to each component must be at least C, the size of the minimum cut in G (since we could consider a cut that separated this component from the rest of the graph). So if there are k components in the graph, the number of eligible edges is at least kC/2 (each of the k components has at least C edges leading out of it, and we need to compensate for the double-counting of each edge). Since the edges were randomly ordered, the chance that the next eligible edge in the list is from the minimum cut is at most C/(kC/2) = 2/k. Thus, with probability at least 1 − 2/k = (k − 2)/k, the choice leaves the minimum cut intact. But now the chance that Kruskal’s algorithm leaves the minimum cut intact all the way up to the choice of the last spanning tree edge is at least
n−2·n−3·n−4···2·1= 1 . n n−1 n−2 4 3 n(n−1)
5.1.5 Prim’s algorithm
Let’s return to our discussion of minimum spanning tree algorithms. What the cut property tells us in most general terms is that any algorithm conforming to the following greedy schema is guaranteed to work.
X = { } (edges picked so far) repeat until |X| = |V | − 1:
pickaset S ⊂V forwhich X hasnoedgesbetween S and V −S let e ∈ E be the minimum-weight edge between S and V − S
X = X ∪ {e}
A popular alternative to Kruskal’s algorithm is Prim’s, in which the intermediate set of edges X always forms a subtree, and S is chosen to be the set of this tree’s vertices.
On each iteration, the subtree defined by X grows by one edge, namely, the lightest edge between a vertex in S and a vertex outside S (Figure 5.8). We can equivalently think of S as

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 151 Figure 5.8 Prim’s algorithm: the edges X form a tree, and S consists of its vertices.
S V−S
X
e
growing to include the vertex v ̸∈ S of smallest cost: cost(v) = min w(u, v).
u∈S
This is strongly reminiscent of Dijkstra’s algorithm, and in fact the pseudocode (Figure 5.9) is almost identical. The only difference is in the key values by which the priority queue is ordered. In Prim’s algorithm, the value of a node is the weight of the lightest incoming edge from set S, whereas in Dijkstra’s it is the length of an entire path to that node from the starting point. Nonetheless, the two algorithms are similar enough that they have the same running time, which depends on the particular priority queue implementation.
Figure 5.9 shows Prim’s algorithm at work, on a small six-node graph. Notice how the final MST is completely specified by the prev array.
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􏰢􏰡 􏰢􏰡
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􏰞􏰝
􏰟􏰠 􏰟 􏰴􏰳
􏰦􏰥􏰣􏰤
􏰲􏰱
􏰫􏰬
􏰨􏰧
􏰰􏰯 􏰯 􏰺􏰹
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􏰮􏰭
􏰹􏰺
􏰷􏰸 􏰷
#” #”
#” #”
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%$
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152 Algorithms Figure 5.9 Top: Prim’s minimum spanning tree algorithm. Below: An illustration of Prim’s
algorithm, starting at node A. Also shown are a table of cost/prev values, and the final MST.
procedure prim(G, w)
Input: A connected undirected graph G = (V, E) with edge weights we Output: A minimum spanning tree defined by the array prev
for all u ∈ V : cost(u) = ∞
prev(u) = nil
Pick any initial node u0 cost(u0) = 0
H = makequeue (V ) (priority queue, using cost-values as keys) while H is not empty:
v = deletemin(H) for each {v, z} ∈ E:
if cost(z) > w(v, z): cost(z) = w(v, z) prev(z) = v decreasekey(H, z)
A6C5EACE
5412
34 B2D4FBDF
Set S
A
B
C
D
E
F
{}
A
A,D A,D,B A,D,B,C A,D,B,C,F
0/nil
∞/nil 5/A 2/D
∞/nil 6/A 2/D 1/B
∞/nil 4/A
∞/nil ∞/nil ∞/nil ∞/nil 5/C 4/F
∞/nil ∞/nil 4/D 4/D 3/C

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 153 5.2 Huffman encoding
In the MP3 audio compression scheme, a sound signal is encoded in three steps.
1. It is digitized by sampling at regular intervals, yielding a sequence of real numbers s1, s2, . . . , sT . For instance, at a rate of 44,100 samples per second, a 50-minute symphony would correspond to T = 50 × 60 × 44,100 ≈ 130 million measurements.1
2. Each real-valued sample st is quantized: approximated by a nearby number from a finite set Γ. This set is carefully chosen to exploit human perceptual limitations, with theintentionthattheapproximatingsequenceisindistinguishablefroms1,s2,…,sT by the human ear.
3. The resulting string of length T over alphabet Γ is encoded in binary.
It is in the last step that Huffman encoding is used. To understand its role, let’s look at a toy example in which T is 130 million and the alphabet Γ consists of just four values, denoted by the symbols A, B, C, D. What is the most economical way to write this long string in binary? The obvious choice is to use 2 bits per symbol—say codeword 00 for A, 01 for B, 10 for C, and 11 for D—in which case 260 megabits are needed in total. Can there possibly be a better encoding than this?
In search of inspiration, we take a closer look at our particular sequence and find that the four symbols are not equally abundant.
Symbol Frequency
A 70 million
B 3 million
C 20 million
D 37 million
Is there some sort of variable-length encoding, in which just one bit is used for the frequently occurring symbol A, possibly at the expense of needing three or more bits for less common symbols?
A danger with having codewords of different lengths is that the resulting encoding may not be uniquely decipherable. For instance, if the codewords are {0, 01, 11, 001}, the decoding of strings like 001 is ambiguous. We will avoid this problem by insisting on the prefix-free property: no codeword can be a prefix of another codeword.
Any prefix-free encoding can be represented by a full binary tree—that is, a binary tree in which every node has either zero or two children—where the symbols are at the leaves, and where each codeword is generated by a path from root to leaf, interpreting left as 0 and right as 1 (Exercise 5.28). Figure 5.10 shows an example of such an encoding for the four symbols A, B, C, D. Decoding is unique: a string of bits is decrypted by starting at the root, reading the string from left to right to move downward, and, whenever a leaf is reached, outputting the corresponding symbol and returning to the root. It is a simple scheme and pays off nicely
1For stereo sound, two channels would be needed, doubling the number of samples.

154 Algorithms Figure 5.10 A prefix-free encoding. Frequencies are shown in square brackets.
01
[60]
D [37]
Symbol
Codeword
A B C D
0 100 101 11
A [70]
[23]
B [3] C [20]
for our toy example, where (under the codes of Figure 5.10) the total size of the binary string
drops to 213 megabits, a 17% improvement.
In general, how do we find the optimal coding tree, given the frequencies f1, f2, . . . , fn of n symbols? To make the problem precise, we want a tree whose leaves each correspond to a symbol and which minimizes the overall length of the encoding,
􏰑n i=1
(the number of bits required for a symbol is exactly its depth in the tree).
There is another way to write this cost function that is very helpful. Although we are only given frequencies for the leaves, we can define the frequency of any internal node to be the sum of the frequencies of its descendant leaves; this is, after all, the number of times the internal node is visited during encoding or decoding. During the encoding process, each time we move down the tree, one bit gets output for every nonroot node through which we pass. So
the total cost—the total number of bits which are output—can also be expressed thus:
The cost of a tree is the sum of the frequencies of all leaves and internal nodes, except the root.
The first formulation of the cost function tells us that the two symbols with the smallest frequencies must be at the bottom of the optimal tree, as children of the lowest internal node (this internal node has two children since the tree is full). Otherwise, swapping these two symbols with whatever is lowest in the tree would improve the encoding.
This suggests that we start constructing the tree greedily: find the two symbols with the smallest frequencies, say i and j, and make them children of a new node, which then has frequency fi + fj. To keep the notation simple, let’s just assume these are f1 and f2. By the second formulation of the cost function, any tree in which f1 and f2 are sibling-leaves has cost f1 +f2 plusthecostforatreewithn−1leavesoffrequencies(f1 +f2),f3,f4,…,fn:
cost of tree =
fi · (depth of ith symbol in tree)

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 155
f1 + f2
f5 f4f3
f1 f2
The latter problem is just a smaller version of the one we started with. So we pull f1 and f2 off the list of frequencies, insert (f1 + f2), and loop. The resulting algorithm can be described in terms of priority queue operations (as defined on page 120) and takes O(n log n) time if a binary heap (Section 4.5.2) is used.
procedure Huffman(f)
Input: An array f [1 · · · n] of frequencies Output: An encoding tree with n leaves
let H be a priority queue of integers, ordered by f fori=1ton: insert(H,i)
for k = n + 1 to 2n − 1:
i = deletemin(H), j = deletemin(H)
create a node numbered k with children i, j f[k] = f[i] + f[j]
insert(H, k)
Returning to our toy example: can you tell if the tree of Figure 5.10 is optimal?

156 Algorithms Entropy
The annual county horse race is bringing in three thoroughbreds who have never competed against one another. Excited, you study their past 200 races and summarize these as prob- ability distributions over four outcomes: first (“first place”), second, third, and other.
Outcome Aurora Whirlwind Phantasm first 0.15 0.30 0.20 second 0.10 0.05 0.30 third 0.70 0.25 0.30 other 0.05 0.40 0.20
Which horse is the most predictable? One quantitative approach to this question is to look at compressibility. Write down the history of each horse as a string of 200 values (first, second, third, other). The total number of bits needed to encode these track- record strings can then be computed using Huffman’s algorithm. This works out to 290 bits for Aurora, 380 for Whirlwind, and 420 for Phantasm (check it!). Aurora has the shortest encoding and is therefore in a strong sense the most predictable.
The inherent unpredictability, or randomness, of a probability distribution can be mea- sured by the extent to which it is possible to compress data drawn from that distribution.
more compressible ≡ less random ≡ more predictable
Suppose there are n possible outcomes, with probabilities p1, p2, . . . , pn. If a sequence of m values is drawn from the distribution, then the ith outcome will pop up roughly mpi times (if m is large). For simplicity, assume these are exactly the observed frequencies, and moreover that the pi’s are all powers of 2 (that is, of the form 1/2k). It can be 􏰐seen by induction (Exercise 5.19) that the number of bits needed to encode the sequence is ni=1 mpi log(1/pi). Thus the average number of bits needed to encode a single draw from the distribution is
􏰑n 1 pilogp .
i=1 i
This is the entropy of the distribution, a measure of how much randomness it contains.
For example, a fair coin has two outcomes, each with probability 1/2. So its entropy is 1 log2+ 1 log2 = 1.
This is natural enough: the coin flip contains one bit of randomness. But what if the coin is not fair, if it has a 3/4 chance of turning up heads? Then the entropy is
3log4+1log4 = 0.81. 434
A biased coin is more predictable than a fair coin, and thus has lower entropy. As the bias becomes more pronounced, the entropy drops toward zero.
22
We explore these notions further in Exercise 5.18 and 5.19.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 157 5.3 Horn formulas
In order to display human-level intelligence, a computer must be able to perform at least some modicum of logical reasoning. Horn formulas are a particular framework for doing this, for expressing logical facts and deriving conclusions.
The most primitive object in a Horn formula is a Boolean variable, taking value either true or false. For instance, variables x, y, and z might denote the following possibilities.
x ≡
y ≡
z ≡
the murder took place in the kitchen the butler is innocent
the colonel was asleep at 8 pm
A literal is either a variable x or its negation x (“NOT x”). In Horn formulas, knowledge about variables is represented by two kinds of clauses:
1. Implications,whoseleft-handsideisanANDofanynumberofpositiveliteralsandwhose right-hand side is a single positive literal. These express statements of the form “if the conditions on the left hold, then the one on the right must also be true.” For instance,
(z ∧ w) ⇒ u
might mean “if the colonel was asleep at 8 pm and the murder took place at 8 pm then the colonel is innocent.” A degenerate type of implication is the singleton “⇒ x,” meaning simply that x is true: “the murder definitely occurred in the kitchen.”
2. Pure negative clauses, consisting of an OR of any number of negative literals, as in
u ∨ v ∨ y)
(“they can’t all be innocent”).
Given a set of clauses of these two types, the goal is to determine whether there is a consis- tent explanation: an assignment of true/false values to the variables that satisfies all the clauses. This is also called a satisfying assignment.
The two kinds of clauses pull us in different directions. The implications tell us to set some of the variables to true, while the negative clauses encourage us to make them false. Our strategy for solving a Horn formula is this: We start with all variables false. We then proceed to set some of them to true, one by one, but very reluctantly, and only if we absolutely have to because an implication would otherwise be violated. Once we are done with this phase and all implications are satisfied, only then do we turn to the negative clauses and make sure they are all satisfied.
In other words, our algorithm for Horn clauses is the following greedy scheme (stingy is perhaps more descriptive):
Input: a Horn formula
Output: a satisfying assignment, if one exists
(

158
Algorithms
set all variables to false
while there is an implication that is not satisfied: set the right-hand variable of the implication to true
if all pure negative clauses are satisfied: return the assignment else: return ‘‘formula is not satisfiable’’
For instance, suppose the formula is
(w∧y∧z)⇒x, (x∧z)⇒w, x⇒y, ⇒x, (x∧y)⇒w, (w∨x∨y), (z).
We start with everything false and then notice that x must be true on account of the sin-
gleton implication ⇒ x. Then we see that y must also be true, because of x ⇒ y. And so
on. To see why the algorithm is correct, notice that if it returns an assignment, this assign- ment satisfies both the implications and the negative clauses, and so it is indeed a satisfying truth assignment of the input Horn formula. So we only have to convince ourselves that if the algorithm finds no satisfying assignment, then there really is none. This is so because our “stingy” rule maintains the following invariant:
If a certain set of variables is set to true, then they must be true in any satisfying assignment.
Hence, if the truth assignment found after the while loop does not satisfy the negative clauses, there can be no satisfying truth assignment.
Horn formulas lie at the heart of Prolog (“programming by logic”), a language in which you program by specifying desired properties of the output, using simple logical expressions. The workhorse of Prolog interpreters is our greedy satisfiability algorithm. Conveniently, it can be implemented in time linear in the length of the formula; do you see how (Exercise 5.32)?
5.4 Set cover
The dots in Figure 5.11 represent a collection of towns. This county is in its early stages of planning and is deciding where to put schools. There are only two constraints: each school should be in a town, and no one should have to travel more than 30 miles to reach one of them. What is the minimum number of schools needed?
This is a typical set cover problem. For each town x, let Sx be the set of towns within 30 miles of it. A school at x will essentially “cover” these other towns. The question is then, how many sets Sx must be picked in order to cover all the towns in the county?
SET COVER
Input: A set of elements B; sets S1,…,Sm ⊆ B

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 159 Figure 5.11 (a) Eleven towns. (b) Towns that are within 30 miles of each other.
(a)
c (b) c bb
dfdf ae ae
kgkg ih ih
jj
Output: A selection of the Si whose union is B. Cost: Number of sets picked.
(In our example, the elements of B are the towns.) This problem lends itself immediately to a greedy solution:
Repeat until all elements of B are covered:
Pick the set Si with the largest number of uncovered elements.
This is extremely natural and intuitive. Let’s see what it would do on our earlier example: It would first place a school at town a, since this covers the largest number of other towns. Thereafter, it would choose three more schools—c, j, and either f or g—for a total of four. However, there exists a solution with just three schools, at b, e, and i. The greedy scheme is not optimal!
But luckily, it isn’t too far from optimal.
Claim Suppose B contains n elements and that the optimal cover consists of k sets. Then the greedy algorithm will use at most k ln n sets.2
Let nt be the number of elements still not covered after t iterations of the greedy algorithm (so n0 = n). Since these remaining elements are covered by the optimal k sets, there must be some set with at least nt/k of them. Therefore, the greedy strategy will ensure that
nt+1 ≤ nt−nt = nt􏰆1−1􏰇, kk
which by repeated application implies nt ≤ n0(1 − 1/k)t. A more convenient bound can be obtained from the useful inequality
1−x≤e−x forallx,withequalityifandonlyifx=0, 2ln means “natural logarithm,” that is, to the base e.

160
which is most easily proved by a picture:
Algorithms
Thus
􏰆
1􏰇t 1 − k
e−x 1−x
1 0
n0(e−1/k)t
At t = k ln n, therefore, nt is strictly less than ne− ln n = 1, which means no elements remain to
be covered.
The ratio between the greedy algorithm’s solution and the optimal solution varies from input to input but is always less than ln n. And there are certain inputs for which the ratio is very close to ln n (Exercise 5.33). We call this maximum ratio the approximation factor of the greedy algorithm. There seems to be a lot of room for improvement, but in fact such hopes are unjustified: it turns out that under certain widely-held complexity assumptions (which will be clearer when we reach Chapter 8), there is provably no polynomial-time algorithm with a smaller approximation factor.
nt ≤ n0
< = ne−t/k. x S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 161 Exercises 5.1. Consider the following graph. A6B5C6D 12254 E1F3G3H 57 (a) What is the cost of its minimum spanning tree? (b) How many minimum spanning trees does it have? (c) Suppose Kruskal’s algorithm is run on this graph. In what order are the edges added to the MST? For each edge in this sequence, give a cut that justifies its addition. 5.2. Suppose we want to find the minimum spanning tree of the following graph. A1B2C3D 4866214 E5F1G1H (a) Run Prim’s algorithm; whenever there is a choice of nodes, always use alphabetic ordering (e.g., start from node A). Draw a table showing the intermediate values of the cost array. (b) Run Kruskal’s algorithm on the same graph. Show how the disjoint-sets data structure looks at every intermediate stage (including the structure of the directed trees), assuming path compression is used. 5.3. Design a linear-time algorithm for the following task. Input: A connected, undirected graph G. Question: Is there an edge you can remove from G while still leaving G connected? Can you reduce the running time of your algorithm to O(|V |)? 5.4. Show that if an undirected graph with n vertices has k connected components, then it has at least n − k edges. 5.5. Consider an undirected graph G = (V, E) with nonnegative edge weights we ≥ 0. Suppose that you have computed a minimum spanning tree of G, and that you have also computed shortest paths to all nodes from a particular node s ∈ V . Nowsupposeeachedgeweightisincreasedby1:thenewweightsarewe′ =we+1. (a) Does the minimum spanning tree change? Give an example where it changes or prove it cannot change. (b) Do the shortest paths change? Give an example where they change or prove they cannot change. 162 Algorithms 5.6. Let G = (V, E) be an undirected graph. Prove that if all its edge weights are distinct, then it has a unique minimum spanning tree. 5.7. Show how to find the maximum spanning tree of a graph, that is, the spanning tree of largest total weight. 5.8. Suppose you are given a weighted graph G = (V,E) with a distinguished vertex s and where all edge weights are positive and distinct. Is it possible for a tree of shortest paths from s and a minimum spanning tree in G to not share any edges? If so, give an example. If not, give a reason. 5.9. The following statements may or may not be correct. In each case, either prove it (if it is cor- rect) or give a counterexample (if it isn’t correct). Always assume that the graph G = (V, E) is undirected. Do not assume that edge weights are distinct unless this is specifically stated. (a) If graph G has more than |V | − 1 edges, and there is a unique heaviest edge, then this edge cannot be part of a minimum spanning tree. (b) If G has a cycle with a unique heaviest edge e, then e cannot be part of any MST. (c) Let e be any edge of minimum weight in G. Then e must be part of some MST. (d) If the lightest edge in a graph is unique, then it must be part of every MST. (e) If e is part of some MST of G, then it must be a lightest edge across some cut of G. (f) If G has a cycle with a unique lightest edge e, then e must be part of every MST. (g) The shortest-path tree computed by Dijkstra’s algorithm is necessarily an MST. (h) The shortest path between two nodes is necessarily part of some MST. (i) Prim’s algorithm works correctly when there are negative edges. (j) (For any r > 0, define an r-path to be a path whose edges all have weight < r.) If G contains an r-path from node s to t, then every MST of G must also contain an r-path from node s to node t. 5.10. Let T be an MST of graph G. Given a connected subgraph H of G, show that T ∩ H is contained in some MST of H. 5.11. Givethestateofthedisjoint-setsdatastructureafterthefollowingsequenceofoperations,start- ing from singleton sets {1}, . . . , {8}. Use path compression. In case of ties, always make the lower numbered root point to the higher numbered one. union(1, 2), union(3, 4), union(5, 6), union(7, 8), union(1, 4), union(6, 7), union(4, 5), find(1) 5.12. Suppose you implement the disjoint-sets data structure using union-by-rank but not path com- pression. Give a sequence of m union and find operations on n elements that take Ω(m log n) time. 5.13. AlongstringconsistsofthefourcharactersA,C,G,T;theyappearwithfrequency31%,20%,9%, and 40%, respectively. What is the Huffman encoding of these four characters? 5.14. Supposethesymbolsa,b,c,d,eoccurwithfrequencies1/2,1/4,1/8,1/16,1/16,respectively. (a) What is the Huffman encoding of the alphabet? (b) If this encoding is applied to a file consisting of 1,000,000 characters with the given frequen- cies, what is the length of the encoded file in bits? S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 163 5.15. WeuseHuffman’salgorithmtoobtainanencodingofalphabet{a,b,c}withfrequenciesfa,fb,fc. In each of the following cases, either give an example of frequencies (fa, fb, fc) that would yield the specified code, or explain why the code cannot possibly be obtained (no matter what the frequencies are). (a) Code:{0,10,11} (b) Code:{0,1,00} (c) Code:{10,01,00} 5.16. Prove the following two properties of the Huffman encoding scheme. (a) If some character occurs with frequency more than 2/5, then there is guaranteed to be a codeword of length 1. (b) If all characters occur with frequency less than 1/3, then there is guaranteed to be no codeword of length 1. 5.17. Under a Huffman encoding of n symbols with frequencies f1,f2,...,fn, what is the longest a codeword could possibly be? Give an example set of frequencies that would produce this case. 5.18. The following table gives the frequencies of the letters of the English language (including the blank for separating words) in a particular corpus. blank 18.3% r 4.8% y 1.6% e 10.2% d 3.5% p 1.6% t 7.7%l3.4%b1.3% a 6.8%c2.6%v0.9% o 5.9%u2.4%k0.6% i 5.8% m 2.1% j 0.2% n 5.5%w1.9%x0.2% s 5.1%f1.8%q0.1% h 4.9%g1.7%z0.1% (a) What is the optimum Huffman encoding of this alphabet? (b) What is the expected number of bits per letter? (c) Suppose now that we calculate the entropy of these frequencies 􏰑26 1 H = pi log pi i=0 (see the box in page 156). Would you expect it to be larger or smaller than your answer above? Explain. (d) Do you think that this is the limit of how much English text can be compressed? What features of the English language, besides letters and their frequencies, should a better compression scheme take into account? 5.19. Entropy.Consideradistributionovernpossibleoutcomes,withprobabilitiesp1,p2,...,pn. 164 Algorithms (a) Just for this part of the problem, assume that each pi is a power of 2 (that is, of the form 1/2k). Suppose a long sequence of m samples is drawn from the distribution and that for all 1 ≤ i ≤ n, the ith outcome occurs exactly mpi times in the sequence. Show that if Huffman encoding is applied to this sequence, the resulting encoding will have length 􏰑n 1 mpi log p . i=1 i (b) Now consider arbitrary distributions—that is, the probabilities pi are not restricted to pow- ers of 2. The most commonly used measure of the amount of randomness in the distribution is the entropy 􏰑n 1 pilogp . i=1 i For what distribution (over n outcomes) is the entropy the largest possible? The smallest possible? 5.20. Give a linear-time algorithm that takes as input a tree and determines whether it has a perfect matching: a set of edges that touches each node exactly once. A feedback edge set of an undirected graph G = (V , E ) is a subset of edges E ′ ⊆ E that intersects every cycle of the graph. Thus, removing the edges E′ will render the graph acyclic. Give an efficient algorithm for the following problem: Input: Undirected graph G = (V, E) with positive edge weights􏰐we Output: A feedback edge set E′ ⊆ E of minimum total weight e∈E′ we 5.21. Inthisproblem,wewilldevelopanewalgorithmforfindingminimumspanningtrees.Itisbased upon the following property: Pick any cycle in the graph, and let e be the heaviest edge in that cycle. Then there is a minimum spanning tree that does not contain e. (a) Prove this property carefully. (b) Here is the new MST algorithm. The input is some undirected graph G = (V, E) (in adja- cency list format) with edge weights {we}. sort the edges according to their weights for each edge e ∈ E, in decreasing order of we: if e ispartofacycleof G: G=G−e (thatis,remove e from G) return G Prove that this algorithm is correct. (c) On each iteration, the algorithm must check whether there is a cycle containing a specific edge e. Give a linear-time algorithm for this task, and justify its correctness. (d) What is the overall time taken by this algorithm, in terms of |E|? Explain your answer. S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 165 5.22. You are given a graph G = (V,E) with positive edge weights, and a minimum spanning tree T = (V , E ′ ) with respect to these weights; you may assume G and T are given as adjacency lists. Now suppose the weight of a particular edge e ∈ E is modified from w(e) to a new value wˆ(e). You wish to quickly update the minimum spanning tree T to reflect this change, without recomputing the entire tree from scratch. There are four cases. In each case give a linear-time algorithm for updating the tree. (a) e ̸∈ E′ and wˆ(e) > w(e). (b) e ̸∈ E′ and wˆ(e) < w(e). (c) e∈E′ andwˆ(e)w(e).
5.23. Sometimes we want light spanning trees with certain special properties. Here’s an example.
Input: Undirected graph G = (V, E); edge weights we; subset of vertices U ⊂ V Output: The lightest spanning tree in which the nodes of U are leaves (there might be other leaves in this tree as well).
(The answer isn’t necessarily a minimum spanning tree.)
Give an algorithm for this problem which runs in O(|E| log |V |) time. (Hint: When you remove
nodes U from the optimal solution, what is left?)
5.24. Abinarycounterofunspecifiedlengthsupportstwooperations:increment(whichincreasesits value by one) and reset (which sets its value back to zero). Show that, starting from an initially zero counter, any sequence of n increment and reset operations takes time O(n); that is, the amortized time per operation is O(1).
5.25. Here’s a problem that occurs in automatic program analysis. For a set of variables x1,…,xn, you are given some equality constraints, of the form “xi = xj ” and some disequality constraints, of the form “xi ̸= xj .” Is it possible to satisfy all of them?
For instance, the constraints
x1 =x2,x2 =x3,x3 =x4,x1 ̸=x4
cannot be satisfied. Give an efficient algorithm that takes as input m constraints over n variables
and decides whether the constraints can be satisfied.
5.26. Graphswithprescribeddegreesequences.Givenalistofnpositiveintegersd1,d2,…,dn,wewant to efficiently determine whether there exists an undirected graph G = (V, E) whose nodes have degrees precisely d1, d2, . . . , dn. That is, if V = {v1, . . . , vn}, then the degree of vi should be exactly di. We call (d1, . . . , dn) the degree sequence of G. This graph G should not contain self-loops (edges with both endpoints equal to the same node) or multiple edges between the same pair of nodes.
(a) Give an example of d1,d2,d3,d4 where all the di ≤ 3 and d1 + d2 + d3 + d4 is even, but for which no graph with degree sequence (d1, d2, d3, d4) exists.
(b) Suppose that d1 ≥ d2 ≥ ··· ≥ dn and that there exists a graph G = (V,E) with degree sequence (d1 , . . . , dn ). We want to show that there must exist a graph that has this degree sequence and where in addition the neighbors of v1 are v2, v3, . . . , vd1+1. The idea is to gradually transform G into a graph with the desired additional property.
i. Suppose the neighbors of v1 in G are not v2, v3, . . . , vd1+1. Show that there exists i < j ≤ n and u ∈ V such that {v1,vi},{u,vj} ∈/ E and {v1,vj},{u,vi} ∈ E. 166 Algorithms ii. Specify the changes you would make to G to obtain a new graph G′ = (V,E′) with the same degree sequence as G and where (v1, vi) ∈ E′. iii. Now show that there must be a graph with the given degree sequence but in which v1 has neighbors v2, v3, . . . , vd1+1. (c) Using the result from part (b), describe an algorithm that on input d1 , . . . , dn (not necessar- ily sorted) decides whether there exists a graph􏰐with this degree sequence. Your algorithm should run in time polynomial in n and in m = ni=1 di. 5.27. Alice wants to throw a party and is deciding whom to call. She has n people to choose from, and she has made up a list of which pairs of these people know each other. She wants to pick as many people as possible, subject to two constraints: at the party, each person should have at least five other people whom they know and five other people whom they don’t know. Give an efficient algorithm that takes as input the list of n people and the list of pairs who know each other and outputs the best choice of party invitees. Give the running time in terms of n. 5.28. A prefix-free encoding of a finite alphabet Γ assigns each symbol in Γ a binary codeword, such that no codeword is a prefix of another codeword. Show that such an encoding can be represented by a full binary tree in which each leaf corre- sponds to a unique element of Γ, whose codeword is generated by the path from the root to that leaf (interpreting a left branch as 0 and a right branch as 1). 5.29. Ternary Huffman. Trimedia Disks Inc. has developed “ternary” hard disks. Each cell on a disk can now store values 0, 1, or 2 (instead of just 0 or 1). To take advantage of this new technology, provide a modified Huffman algorithm for compressing sequences of characters from an alpha- bet of size n, where the characters occur with known frequencies f1,f2,...,fn. Your algorithm should encode each character with a variable-length codeword over the values 0, 1, 2 such that no codeword is a prefix of another codeword and so as to obtain the maximum possible compression. Prove that your algorithm is correct. 5.30. The basic intuition behind Huffman’s algorithm, that frequent blocks should have short en- codings and infrequent blocks should have long encodings, is also at work in English, where typical words like I, you, is, and, to, from, and so on are short, and rarely used words like velociraptor are longer. However, words like fire!, help!, and run! are short not because they are frequent, but perhaps because time is precious in situations where they are used. To make things theoretical, suppose we have a file composed of m different words, with frequen- cies f1, . . . , fm. Suppose also that for the ith word, the cost per bit of encoding is ci. Thus, if we find a prefix-free􏰐code where the ith word has a codeword of length li, then the total cost of the encoding will be i fi · ci · li. Show how to modify Huffman’s algorithm to find the prefix-free encoding of minimum total cost. 5.31. A server has n customers waiting to be served. The service time required by each customer is known in advance: it is ti minutes for customer i. So if, 􏰐for example, the customers are served in order of increasing i, then the ith customer has to wait ij=1 tj minutes. We wish to minimize the total waiting time T = 􏰑n (time spent waiting by customer i). i=1 Give an efficient algorithm for computing the optimal order in which to process the customers. S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 167 5.32. ShowhowtoimplementthestingyalgorithmforHornformulasatisfiability(Section5.3)intime that is linear in the length of the formula (the number of occurrences of literals in it). (Hint: Use a directed graph, with one node per variable, to represent the implications.) 5.33. Show that for any integer n that is a power of 2, there is an instance of the set cover problem (Section 5.4) with the following properties: i. There are n elements in the base set. ii. The optimal cover uses just two sets. iii. The greedy algorithm picks at least log n sets. Thus the approximation ratio we derived in the chapter is tight. 168 Algorithms Chapter 6 Dynamic programming In the preceding chapters we have seen some elegant design principles—such as divide-and- conquer, graph exploration, and greedy choice—that yield definitive algorithms for a variety of important computational tasks. The drawback of these tools is that they can only be used on very specific types of problems. We now turn to the two sledgehammers of the algorithms craft, dynamic programming and linear programming, techniques of very broad applicability that can be invoked when more specialized methods fail. Predictably, this generality often comes with a cost in efficiency. 6.1 Shortest paths in dags, revisited At the conclusion of our study of shortest paths (Chapter 4), we observed that the problem is especially easy in directed acyclic graphs (dags). Let’s recapitulate this case, because it lies at the heart of dynamic programming. The special distinguishing feature of a dag is that its nodes can be linearized; that is, they can be arranged on a line so that all edges go from left to right (Figure 6.1). To see why this helps with shortest paths, suppose we want to figure out distances from node S to the other nodes. For concreteness, let’s focus on node D. The only way to get to it is through its Figure 6.1 A dag and its linearization (topological ordering). A6B 3 2 1 S4 1ES2C4A6B1D1E 2C3D1 1 2 169 170 Algorithms predecessors, B or C; so to find the shortest path to D, we need only compare these two routes: dist(D) = min{dist(B) + 1,dist(C) + 3}. A similar relation can be written for every node. If we compute these dist values in the left-to-right order of Figure 6.1, we can always be sure that by the time we get to a node v, we already have all the information we need to compute dist(v). We are therefore able to compute all distances in a single pass: initialize all dist(·) values to ∞ dist(s) = 0 for each v ∈ V \{s}, in linearized order: dist(v) = min(u,v)∈E {dist(u) + l(u, v)} Notice that this algorithm is solving a collection of subproblems, {dist(u) : u ∈ V }. We start with the smallest of them, dist(s), since we immediately know its answer to be 0. We then proceed with progressively “larger” subproblems—distances to vertices that are further and further along in the linearization—where we are thinking of a subproblem as large if we need to have solved a lot of other subproblems before we can get to it. This is a very general technique. At each node, we compute some function of the values of the node’s predecessors. It so happens that our particular function is a minimum of sums, but we could just as well make it a maximum, in which case we would get longest paths in the dag. Or we could use a product instead of a sum inside the brackets, in which case we would end up computing the path with the smallest product of edge lengths. Dynamic programming is a very powerful algorithmic paradigm in which a problem is solved by identifying a collection of subproblems and tackling them one by one, smallest first, using the answers to small problems to help figure out larger ones, until the whole lot of them is solved. In dynamic programming we are not given a dag; the dag is implicit. Its nodes are the subproblems we define, and its edges are the dependencies between the subproblems: if to solve subproblem B we need the answer to subproblem A, then there is a (conceptual) edge from A to B. In this case, A is thought of as a smaller subproblem than B—and it will always be smaller, in an obvious sense. But it’s time we saw an example. 6.2 Longest increasing subsequences In the longest increasing subsequence problem, the input is a sequence of numbers a 1 , . . . , an . Asubsequenceisanysubsetofthesenumberstakeninorder,oftheformai1,ai2,...,aik where 1 ≤ i1 < i2 < ··· < ik ≤ n, and an increasing subsequence is one in which the numbers are getting strictly larger. The task is to find the increasing subsequence of greatest length. For instance, the longest increasing subsequence of 5, 2, 8, 6, 3, 6, 9, 7 is 2, 3, 6, 9: 52863697 S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 171 Figure 6.2 The dag of increasing subsequences. 52863697 In this example, the arrows denote transitions between consecutive elements of the opti- mal solution. More generally, to better understand the solution space, let’s create a graph of all permissible transitions: establish a node i for each element ai, and add directed edges (i, j) whenever it is possible for ai and aj to be consecutive elements in an increasing subsequence, that is, whenever i < j and ai < aj (Figure 6.2). Notice that (1) this graph G = (V,E) is a dag, since all edges (i,j) have i < j, and (2) there is a one-to-one correspondence between increasing subsequences and paths in this dag. Therefore, our goal is simply to find the longest path in the dag! Here is the algorithm: for j = 1,2,...,n: L(j) = 1 + max{L(i) : (i, j) ∈ E} return maxj L(j) L(j) is the length of the longest path—the longest increasing subsequence—ending at j (plus 1, since strictly speaking we need to count nodes on the path, not edges). By reasoning in the same way as we did for shortest paths, we see that any path to node j must pass through one of its predecessors, and therefore L(j) is 1 plus the maximum L(·) value of these predecessors. If there are no edges into j, we take the maximum over the empty set, zero. And the final answer is the largest L(j), since any ending position is allowed. This is dynamic programming. In order to solve our original problem, we have defined a collection of subproblems {L(j) : 1 ≤ j ≤ n} with the following key property that allows them to be solved in a single pass: (*) There is an ordering on the subproblems, and a relation that shows how to solve a subproblem given the answers to “smaller” subproblems, that is, subproblems that appear earlier in the ordering. In our case, each subproblem is solved using the relation L(j) = 1+max{L(i):(i,j)∈E}, 172 Algorithms an expression which involves only smaller subproblems. How long does this step take? It requires the predecessors of j to be known; for this the adjacency list of the reverse graph GR, constructible in linear time (recall Exercise 3.5), is handy. The computation of L(j) then takes time proportional to the indegree of j, giving an overall running time linear in |E|. This is at most O(n2), the maximum being when the input array is sorted in increasing order. Thus the dynamic programming solution is both simple and efficient. There is one last issue to be cleared up: the L-values only tell us the length of the optimal subsequence, so how do we recover the subsequence itself? This is easily managed with the same bookkeeping device we used for shortest paths in Chapter 4. While computing L(j), we should also note down prev(j), the next-to-last node on the longest path to j. The optimal subsequence can then be reconstructed by following these backpointers. S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 173 Recursion? No, thanks. Returning to our discussion of longest increasing subsequences: the formula for L(j) also suggests an alternative, recursive algorithm. Wouldn’t that be even simpler? Actually, recursion is a very bad idea: the resulting procedure would require exponential time! To see why, suppose that the dag contains edges (i, j) for all i < j—that is, the given sequence of numbers a1, a2, . . . , an is sorted. In that case, the formula for subproblem L(j) becomes L(j) = 1 + max{L(1), L(2), . . . , L(j − 1)}. The following figure unravels the recursion for L(5). Notice that the same subproblems get solved over and over again! L(1) L(2) L(3) L(4) L(1) L(1) L(2) L(1) L(2) L(3) L(5) L(1) L(1) L(1) L(2) L(1) For L(n) this tree has exponentially many nodes (can you bound it?), and so a recursive solution is disastrous. Then why did recursion work so well with divide-and-conquer? The key point is that in divide-and-conquer, a problem is expressed in terms of subproblems that are substantially smaller, say half the size. For instance, mergesort sorts an array of size n by recursively sorting two subarrays of size n/2. Because of this sharp drop in problem size, the full recursion tree has only logarithmic depth and a polynomial number of nodes. In contrast, in a typical dynamic programming formulation, a problem is reduced to subproblems that are only slightly smaller—for instance, L(j) relies on L(j − 1). Thus the full recursion tree generally has polynomial depth and an exponential number of nodes. However, it turns out that most of these nodes are repeats, that there are not too many distinct subproblems among them. Efficiency is therefore obtained by explicitly enumerating the distinct subproblems and solving them in the right order. 174 Algorithms Programming? The origin of the term dynamic programming has very little to do with writing code. It was first coined by Richard Bellman in the 1950s, a time when computer programming was an esoteric activity practiced by so few people as to not even merit a name. Back then programming meant “planning,” and “dynamic programming” was conceived to optimally plan multistage processes. The dag of Figure 6.2 can be thought of as describing the possible ways in which such a process can evolve: each node denotes a state, the leftmost node is the starting point, and the edges leaving a state represent possible actions, leading to different states in the next unit of time. The etymology of linear programming, the subject of Chapter 7, is similar. 6.3 Edit distance When a spell checker encounters a possible misspelling, it looks in its dictionary for other words that are close by. What is the appropriate notion of closeness in this case? A natural measure of the distance between two strings is the extent to which they can be aligned, or matched up. Technically, an alignment is simply a way of writing the strings one above the other. For instance, here are two possible alignments of SNOWY and SUNNY: S−NOWY −SNOW−Y SUNN−Y SUN− −NY Cost: 3 Cost: 5 The “−” indicates a “gap”; any number of these can be placed in either string. The cost of an alignment is the number of columns in which the letters differ. And the edit distance between two strings is the cost of their best possible alignment. Do you see that there is no better alignment of SNOWY and SUNNY than the one shown here with a cost of 3? Edit distance is so named because it can also be thought of as the minimum number of edits—insertions, deletions, and substitutions of characters—needed to transform the first string into the second. For instance, the alignment shown on the left corresponds to three edits: insert U, substitute O → N, and delete W. In general, there are so many possible alignments between two strings that it would be terribly inefficient to search through all of them for the best one. Instead we turn to dynamic programming. A dynamic programming solution When solving a problem by dynamic programming, the most crucial question is, What are the subproblems? As long as they are chosen so as to have the property (*) from page 171. it is an easy matter to write down the algorithm: iteratively solve one subproblem after the other, in order of increasing size. Our goal is to find the edit distance between two strings x[1···m] and y[1···n]. What is a good subproblem? Well, it should go part of the way toward solving the whole problem; so how S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani Figure 6.3 The subproblem E(7, 5). TIAL OMIAL about looking at the edit distance between some prefix of the first string, x[1 · · · i], and some prefix of the second, y[1 · · · j]? Call this subproblem E(i, j) (see Figure 6.3). Our final objective, then, is to compute E(m, n). For this to work, we need to somehow express E(i,j) in terms of smaller subproblems. Let’s see—what do we know about the best alignment between x[1 · · · i] and y[1 · · · j]? Well, its rightmost column can only be one of three things: x[i] or − or x[i] 175 EXPONEN POLYN − y[j] y[j] The first case incurs a cost of 1 for this particular column, and it remains to align x[1 · · · i − 1] with y[1 · · · j]. But this is exactly the subproblem E(i−1, j)! We seem to be getting somewhere. In the second case, also with cost 1, we still need to align x[1···i] with y[1···j − 1]. This is again another subproblem, E(i, j − 1). And in the final case, which either costs 1 (if x[i] ̸= y[j]) or 0 (if x[i] = y[j]), what’s left is the subproblem E(i − 1, j − 1). In short, we have expressed E(i, j) in terms of three smaller subproblems E(i − 1, j), E(i, j − 1), E(i − 1, j − 1). We have no idea which of them is the right one, so we need to try them all and pick the best: E(i,j) = min{1+E(i−1,j), 1+E(i,j−1), diff(i,j)+E(i−1,j−1)} where for convenience diff(i, j) is defined to be 0 if x[i] = y[j] and 1 otherwise. For instance, in computing the edit distance between EXPONENTIAL and POLYNOMIAL, subproblem E(4, 3) corresponds to the prefixes EXPO and POL. The rightmost column of their best alignment must be one of the following: −O or −L or OL Thus,E(4,3)=min{1+E(3,3), 1+E(4,2), 1+E(3,2)}. The answers to all the subproblems E(i, j) form a two-dimensional table, as in Figure 6.4. In what order should these subproblems be solved? Any order is fine, as long as E(i − 1, j), E(i, j − 1), and E(i − 1, j − 1) are handled before E(i, j). For instance, we could fill in the table one row at a time, from top row to bottom row, and moving left to right across each row. Or alternatively, we could fill it in column by column. Both methods would ensure that by the time we get around to computing a particular table entry, all the other entries we need are already filled in. 176 Algorithms Figure 6.4 (a) The table of subproblems. Entries E(i − 1, j − 1), E(i − 1, j), and E(i, j − 1) are needed to fill in E(i, j). (b) The final table of values found by dynamic programming. (a) i−1 i m (b) POLYNOMIAL E X P O N E N T I A L 0 1 2 3 4 5 6 7 8 9 10 1 1 2 3 4 5 6 7 8 9 10 2 2 2 3 4 5 6 7 8 9 10 3 2 3 3 4 5 6 7 8 9 10 43234556789 54334456789 65444556789 76555456789 87666556789 98777666678 10 9 8 8 8 7 7 7 7 6 7 11 10 9 8 9 8 8 8 8 7 6 j−1j n GOAL With both the subproblems and the ordering specified, we are almost done. There just remain the “base cases” of the dynamic programming, the very smallest subproblems. In the present situation, these are E(0, ·) and E(·, 0), both of which are easily solved. E(0, j) is the edit distance between the 0-length prefix of x, namely the empty string, and the first j letters of y: clearly, j. And similarly, E(i, 0) = i. At this point, the algorithm for edit distance basically writes itself. for i = 0,1,2,...,m: E(i,0) = i for j = 1,2,...,n: E(0,j) = j for i = 1,2,...,m: for j = 1,2,...,n: E(i, j) = min{E(i − 1, j) + 1, E(i, j − 1) + 1, E(i − 1, j − 1) + diff(i, j)} return E(m, n) This procedure fills in the table row by row, and left to right within each row. Each entry takes constant time to fill in, so the overall running time is just the size of the table, O(mn). And in our example, the edit distance turns out to be 6: EXPONEN−TIAL − −POLYNOMIAL S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 177 Figure 6.5 The underlying dag, and a path of length 6. POLYNOMIAL E X P O N E N T I A L Every dynamic program has an underlying dag structure: think of each node as representing a subproblem, and each edge as a precedence constraint on the order in which the subproblems can be tackled. Having nodes u1, . . . , uk point to v means “subproblem v can only be solved once the answers to u1, . . . , uk are known.” In our present edit distance application, the nodes of the underlying dag correspond to subproblems, or equivalently, to positions (i,j) in the table. Its edges are the precedence constraints, of the form (i − 1, j) → (i, j), (i, j − 1) → (i, j), and (i − 1, j − 1) → (i, j) (Figure 6.5). In fact, we can take things a little further and put weights on the edges so that the edit distances are given by shortest paths in the dag! To see this, set all edge lengths to 1, except for {(i − 1, j − 1) → (i, j) : x[i] = y[j]} (shown dotted in the figure), whose length is 0. The final answer is then simply the distance between nodes s = (0, 0) and t = (m, n). One possible shortest path is shown, the one that yields the alignment we found earlier. On this path, each move down is a deletion, each move right is an insertion, and each diagonal move is either a match or a substitution. By altering the weights on this dag, we can allow generalized forms of edit distance, in which insertions, deletions, and substitutions have different associated costs. The underlying dag 178 Algorithms Common subproblems Finding the right subproblem takes creativity and experimentation. But there are a few standard choices that seem to arise repeatedly in dynamic programming. i. The input is x1,x2,...,xn and a subproblem is x1,x2,...,xi. x7 x8 x9 x10 The number of subproblems is therefore linear. ii. The input is x1,...,xn, and y1,...,ym. A subproblem is x1, ...,xi and y1,...,yj. x7 x8 x9 x10 y6 y7 y8 The number of subproblems is O(mn). iii. The input is x1,...,xn and a subproblem is xi,xi+1,...,xj. x1 x2 x7 x8 x9 x10 The number of subproblems is O(n2). iv. The input is a rooted tree. A subproblem is a rooted subtree. If the tree has n nodes, how many subproblems are there? We’ve already encountered the first two cases, and the others are coming up shortly. x1 x2 x3 x4 x5 x6 x1 x2 x3 x4 x5 x6 y1 y2 y3 y4 y5 x3 x4 x5 x6 S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 179 180 Algorithms Of mice and men Our bodies are extraordinary machines: flexible in function, adaptive to new environments, and able to interact and reproduce. All these capabilities are specified by a program unique to each of us, a string that is 3 billion characters long over the alphabet {A, C, G, T }—our DNA. The DNA sequences of any two people differ by only about 0.1%. However, this still leaves 3 million positions on which they vary, more than enough to explain the vast range of human diversity. These differences are of great scientific and medical interest—for instance, they might help predict which people are prone to certain diseases. DNA is a vast and seemingly inscrutable program, but it can be broken down into smaller units that are more specific in their role, rather like subroutines. These are called genes. Computers have become a crucial tool in understanding the genes of humans and other organisms, to the extent that computational genomics is now a field in its own right. Here are examples of typical questions that arise. 1. When a new gene is discovered, one way to gain insight into its function is to find known genes that match it closely. This is particularly helpful in transferring knowl- edge from well-studied species, such as mice, to human beings. A basic primitive in this search problem is to define an efficiently computable notion of when two strings approximately match. The biology suggests a generalization of edit distance, and dynamic programming can be used to compute it. Then there’s the problem of searching through the vast thicket of known genes: the database GenBank already has a total length of over 1010, and this number is growing rapidly. The current method of choice is BLAST, a clever combination of algorithmic tricks and biological intuitions that has made it the most widely used software in com- putational biology. 2. Methods for sequencing DNA (that is, determining the string of characters that consti- tute it) typically only find fragments of 500–700 characters. Billions of these randomly scattered fragments can be generated, but how can they be assembled into a coherent DNA sequence? For one thing, the position of any one fragment in the final sequence is unknown and must be inferred by piecing together overlapping fragments. A showpiece of these efforts is the draft of human DNA completed in 2001 by two groups simultaneously: the publicly funded Human Genome Consortium and the pri- vate Celera Genomics. 3. When a particular gene has been sequenced in each of several species, can this infor- mation be used to reconstruct the evolutionary history of these species? We will explore these problems in the exercises at the end of this chapter. Dynamic pro- gramming has turned out to be an invaluable tool for some of them and for computational biology in general. S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 181 6.4 Knapsack During a robbery, a burglar finds much more loot than he had expected and has to decide what to take. His bag (or “knapsack”) will hold a total weight of at most W pounds. There are n items to pick from, of weight w1, . . . , wn and dollar value v1, . . . , vn. What’s the most valuable combination of items he can fit into his bag?1 For instance, take W = 10 and Item Weight Value 1 6 $30 2 3 $14 3 4 $16 4 2 $9 There are two versions of this problem. If there are unlimited quantities of each item avail- able, the optimal choice is to pick item 1 and two of item 4 (total: $48). On the other hand, if there is one of each item (the burglar has broken into an art gallery, say), then the optimal knapsack contains items 1 and 3 (total: $46). As we shall see in Chapter 8, neither version of this problem is likely to have a polynomial- time algorithm. However, using dynamic programming they can both be solved in O(nW) time, which is reasonable when W is small, but is not polynomial since the input size is proportional to log W rather than W . Knapsack with repetition Let’s start with the version that allows repetition. As always, the main question in dynamic programming is, what are the subproblems? In this case we can shrink the original problem in two ways: we can either look at smaller knapsack capacities w ≤ W, or we can look at fewer items (for instance, items 1, 2, . . . , j, for j ≤ n). It usually takes a little experimentation to figure out exactly what works. The first restriction calls for smaller capacities. Accordingly, define K(w) = maximum value achievable with a knapsack of capacity w. Can we express this in terms of smaller subproblems? Well, if the optimal solution to K(w) includes item i, then removing this item from the knapsack leaves an optimal solution to K(w − wi). In other words, K(w) is simply K(w − wi) + vi, for some i. We don’t know which i, so we need to try all possibilities. K(w)= max{K(w−wi)+vi}, i:wi ≤w where as usual our convention is that the maximum over an empty set is 0. We’re done! The algorithm now writes itself, and it is characteristically simple and elegant. 1If this application seems frivolous, replace “weight” with “CPU time” and “only W pounds can be taken” with “only W units of CPU time are available.” Or use “bandwidth” in place of “CPU time,” etc. The knapsack problem generalizes a wide variety of resource-constrained selection tasks. 182 Algorithms K(0) = 0 for w = 1 to W: K(w)=max{K(w−wi)+vi :wi ≤w} return K(W) This algorithm fills in a one-dimensional table of length W + 1, in left-to-right order. Each entry can take up to O(n) time to compute, so the overall running time is O(nW ). As always, there is an underlying dag. Try constructing it, and you will be rewarded with a startling insight: this particular variant of knapsack boils down to finding the longest path in a dag! Knapsack without repetition On to the second variant: what if repetitions are not allowed? Our earlier subproblems now become completely useless. For instance, knowing that the value K(w − wn) is very high doesn’t help us, because we don’t know whether or not item n already got used up in this partial solution. We must therefore refine our concept of a subproblem to carry additional information about the items being used. We add a second parameter, 0 ≤ j ≤ n: K(w, j) = maximum value achievable using a knapsack of capacity w and items 1, . . . , j. The answer we seek is K(W,n). How can we express a subproblem K(w, j) in terms of smaller subproblems? Quite simple: either item j is needed to achieve the optimal value, or it isn’t needed: K(w,j) = max{K(w − wj,j − 1) + vj,K(w,j − 1)}. (The first case is invoked only if wj ≤ w.) In other words, we can express K(w,j) in terms of subproblems K (·, j − 1). The algorithm then consists of filling out a two-dimensional table, with W + 1 rows and n + 1 columns. Each table entry takes just constant time, so even though the table is much larger than in the previous case, the running time remains the same, O(nW ). Here’s the code. Initialize all K(0, j) = 0 and all K(w, 0) = 0 for j = 1 to n: for w = 1 to W: ifwj>w: K(w,j)=K(w,j−1)
else: K(w,j)=max{K(w,j−1),K(w−wj,j−1)+vj}
return K(W,n)

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 183 Memoization
In dynamic programming, we write out a recursive formula that expresses large problems in terms of smaller ones and then use it to fill out a table of solution values in a bottom-up manner, from smallest subproblem to largest.
The formula also suggests a recursive algorithm, but we saw earlier that naive recursion can be terribly inefficient, because it solves the same subproblems over and over again. What about a more intelligent recursive implementation, one that remembers its previous invocations and thereby avoids repeating them?
On the knapsack problem (with repetitions), such an algorithm would use a hash table (recall Section 1.5) to store the values of K(·) that had already been computed. At each recursive call requesting some K(w), the algorithm would first check if the answer was already in the table and then would proceed to its calculation only if it wasn’t. This trick is called memoization:
A hash table, initially empty, holds values of K(w) indexed by w
function knapsack(w)
if w is in hash table: return K(w) K(w) = max{knapsack(w − wi) + vi : wi ≤ w} insert K(w) into hash table, with key w return K(w)
Since this algorithm never repeats a subproblem, its running time is O(nW), just like the dynamic program. However, the constant factor in this big-O notation is substantially larger because of the overhead of recursion.
In some cases, though, memoization pays off. Here’s why: dynamic programming au- tomatically solves every subproblem that could conceivably be needed, while memoization only ends up solving the ones that are actually used. For instance, suppose that W and all the weights wi are multiples of 100. Then a subproblem K(w) is useless if 100 does not divide w. The memoized recursive algorithm will never look at these extraneous table entries.

184 Algorithms
Figure 6.6 A × B × C × D = (A × (B × C)) × D. (a)
(b)
(c)
6.5
50×20 20×1 1×10
× A × (B × C)
10×100 50×20 20×10 10×100 (d)
××× ×× ABCDAB×CD
50×10 10×100
Chain matrix multiplication
D
(A × (B × C)) × D 50×100
Suppose that we want to multiply four matrices, A × B × C × D, of dimensions 50 × 20, 20 × 1, 1 × 10, and 10 × 100, respectively (Figure 6.6). This will involve iteratively multiplying two matrices at a time. Matrix multiplication is not commutative (in general, A×B ̸= B×A), but it is associative, which means for instance that A×(B×C) = (A×B)×C. Thus we can compute our product of four matrices in many different ways, depending on how we parenthesize it. Are some of these better than others?
Multiplying an m × n matrix by an n × p matrix takes mnp multiplications, to a good enough approximation. Using this formula, let’s compare several different ways of evaluating A × B × C × D:
Parenthesization Cost computation Cost
A×((B×C)×D) 20·1·10+20·10·100+50·20·100 120,200 (A × (B × C)) × D 20 · 1 · 10 + 50 · 20 · 10 + 50 · 10 · 100 60, 200 (A × B) × (C × D) 50 · 20 · 1 + 1 · 10 · 100 + 50 · 1 · 100 7, 000
As you can see, the order of multiplications makes a big difference in the final running time! Moreover, the natural greedy approach, to always perform the cheapest matrix multiplication available, leads to the second parenthesization shown here and is therefore a failure.
How do we determine the optimal order, if we want to compute A1 × A2 × · · · × An, where the Ai’s are matrices with dimensions m0 × m1, m1 × m2, . . . , mn−1 × mn, respectively? The first thing to notice is that a particular parenthesization can be represented very naturally by a binary tree in which the individual matrices correspond to the leaves, the root is the final

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 185 Figure6.7(a)((A×B)×C)×D; (b)A×((B×C)×D); (c)(A×(B×C))×D.
(a) (b) (c)
DAD CDA
ABBCBC
product, and interior nodes are intermediate products (Figure 6.7). The possible orders in which to do the multiplication correspond to the various full binary trees with n leaves, whose number is exponential in n (Exercise 2.13). We certainly cannot try each tree, and with brute force thus ruled out, we turn to dynamic programming.
The binary trees of Figure 6.7 are suggestive: for a tree to be optimal, its subtrees must also be optimal. What are the subproblems corresponding to the subtrees? They are products of the form Ai × Ai+1 × · · · × Aj . Let’s see if this works: for 1 ≤ i ≤ j ≤ n, define
C(i,j)=minimumcostofmultiplyingAi ×Ai+1 ×···×Aj.
The size of this subproblem is the number of matrix multiplications, |j − i|. The smallest subproblem is when i = j, in which case there’s nothing to multiply, so C(i,i) = 0. For j > i, consider the optimal subtree for C(i,j). The first branch in this subtree, the one at the top, will split the product in two pieces, of the form Ai × ··· × Ak and Ak+1 × ··· × Aj, for some k between i and j. The cost of the subtree is then the cost of these two partial products, plus the cost of combining them: C(i, k) + C(k + 1, j) + mi−1 · mk · mj. And we just need to find the splitting point k for which this is smallest:
C(i,j)= min{C(i,k)+C(k+1,j)+mi−1·mk·mj}. i≤k 1, we define C(S, 1) = ∞ since the path cannot both start and end at 1.
Now, let’s express C(S,j) in terms of smaller subproblems. We need to start at 1 and end at j; what should we pick as the second-to-last city? It has to be some i ∈ S, so the overall path length is the distance from 1 to i, namely, C(S − {j}, i), plus the length of the final edge,
dij . We must pick the best such i:
C(S,j)= min C(S−{j},i)+dij.
i∈S :i̸=j
The subproblems are ordered by |S|. Here’s the code.
C({1},1) = 0
for s=2 to n:
for all subsets S ⊆ {1,2,…,n} of size s and containing 1: C(S,1) = ∞
forall j ∈S,j ̸=1: C(S,j)=min{C(S−{j},i)+dij :i∈S,i̸=j}
return minj C({1,…,n},j)+dj1
There are at most 2n · n subproblems, and each one takes linear time to solve. The total
running time is therefore O(n22n).
6.7 Independent sets in trees
A subset of nodes S ⊂ V is an independent set of graph G = (V,E) if there are no edges between them. For instance, in Figure 6.10 the nodes {1, 5} form an independent set, but nodes {1, 4, 5} do not, because of the edge between 4 and 5. The largest independent set is {2, 3, 6}.
Like several other problems we have seen in this chapter (knapsack, traveling salesman), finding the largest independent set in a graph is believed to be intractable. However, when the graph happens to be a tree, the problem can be solved in linear time, using dynamic programming. And what are the appropriate subproblems? Already in the chain matrix multiplication problem we noticed that the layered structure of a tree provides a natural definition of a subproblem—as long as one node of the tree has been identified as a root.
So here’s the algorithm: Start by rooting the tree at any node r. Now, each node defines a subtree—the one hanging from it. This immediately suggests subproblems:
I(u) = size of largest independent set of subtree hanging from u.

190 Algorithms On time and memory
The amount of time it takes to run a dynamic programming algorithm is easy to discern from the dag of subproblems: in many cases it is just the total number of edges in the dag! All we are really doing is visiting the nodes in linearized order, examining each node’s inedges, and, most often, doing a constant amount of work per edge. By the end, each edge of the dag has been examined once.
But how much computer memory is required? There is no simple parameter of the dag characterizing this. It is certainly possible to do the job with an amount of memory propor- tional to the number of vertices (subproblems), but we can usually get away with much less. The reason is that the value of a particular subproblem only needs to be remembered until the larger subproblems depending on it have been solved. Thereafter, the memory it takes up can be released for reuse.
For example, in the Floyd-Warshall algorithm the value of dist(i, j, k) is not needed once the dist(·, ·, k + 1) values have been computed. Therefore, we only need two |V | × |V | arrays to store the dist values, one for odd values of k and one for even values: when computing dist(i, j, k), we overwrite dist(i, j, k − 2).
(And let us not forget that, as always in dynamic programming, we also need one more ar- ray, prev(i, j), storing the next to last vertex in the current shortest path from i to j, a value that must be updated with dist(i, j, k). We omit this mundane but crucial bookkeeping step from our dynamic programming algorithms.)
Can you see why the edit distance dag in Figure 6.5 only needs memory proportional to the length of the shorter string?
Our final goal is I(r).
Dynamic programming proceeds as always from smaller subproblems to larger ones, that
is to say, bottom-up in the rooted tree. Suppose we know the largest independent sets for all subtrees below a certain node u; in other words, suppose we know I(w) for all descendants w of u. How can we compute I(u)? Let’s split the computation into two cases: any independent set either includes u or it doesn’t (Figure 6.11).
I(u) = max1+ 􏰑 I(w), 􏰑 I(w). grandchildren w of u children w of u
If the independent set includes u, then we get one point for it, but we aren’t allowed to include the children of u—therefore we move on to the grandchildren. This is the first case in the formula. On the other hand, if we don’t include u, then we don’t get a point for it, but we can move on to its children.
The number of subproblems is exactly the number of vertices. With a little care, the running time can be made linear, O(|V | + |E|).

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 191 Figure 6.10 The largest independent set in this graph has size 3.
12 34
56
Figure 6.11 I(u) is the size of the largest independent set of the subtree rooted at u. Two cases: either u is in this independent set, or it isn’t.
r u
Exercises
6.1. A contiguous subsequence of a list S is a subsequence made up of consecutive elements of S. For instance, if S is
5, 15, −30, 10, −5, 40, 10,
then 15, −30, 10 is a contiguous subsequence but 5, 15, 40 is not. Give a linear-time algorithm for
the following task:
Input: A list of numbers, a1, a2, . . . , an.
Output: The contiguous subsequence of maximum sum (a subsequence of length zero has sum zero).
For the preceding example, the answer would be 10, −5, 40, 10, with a sum of 55.
(Hint: Foreachj∈{1,2,…,n},considercontiguoussubsequencesendingexactlyatpositionj.)
6.2. You are going on a long trip. You start on the road at mile post 0. Along the way there are n hotels, at mile posts a1 < a2 < · · · < an, where each ai is measured from the starting point. The only places you are allowed to stop are at these hotels, but you can choose which of the hotels you stop at. You must stop at the final hotel (at distance an), which is your destination. 192 Algorithms You’d ideally like to travel 200 miles a day, but this may not be possible (depending on the spacing of the hotels). If you travel x miles during a day, the penalty for that day is (200 − x)2. You want to plan your trip so as to minimize the total penalty—that is, the sum, over all travel days, of the daily penalties. Give an efficient algorithm that determines the optimal sequence of hotels at which to stop. 6.3. Yuckdonald’sisconsideringopeningaseriesofrestaurantsalongQuaintValleyHighway(QVH). The n possible locations are along a straight line, and the distances of these locations from the start of QVH are, in miles and in increasing order, m1, m2, . . . , mn. The constraints are as follows: • At each location, Yuckdonald’s may open at most one restaurant. The expected profit from opening a restaurant at location i is pi, where pi > 0 and i = 1,2,…,n.
• Any two restaurants should be at least k miles apart, where k is a positive integer.
Give an efficient algorithm to compute the maximum expected total profit subject to the given
constraints.
6.4. Youaregivenastringofncharacterss[1…n],whichyoubelievetobeacorruptedtextdocument in which all punctuation has vanished (so that it looks something like “itwasthebestoftimes…”). You wish to reconstruct the document using a dictionary, which is available in the form of a Boolean function dict(·): for any strin􏰌g w,
true if w is a valid word false otherwise.
(a) Give a dynamic programming algorithm that determines whether the string s[·] can be reconstituted as a sequence of valid words. The running time should be at most O(n2), assuming calls to dict take unit time.
(b) In the event that the string is valid, make your algorithm output the corresponding se- quence of words.
6.5. Pebbling a checkerboard. We are given a checkerboard which has 4 rows and n columns, and has an integer written in each square. We are also given a set of 2n pebbles, and we want to place some or all of these on the checkerboard (each pebble can be placed on exactly one square) so as to maximize the sum of the integers in the squares that are covered by pebbles. There is one constraint: for a placement of pebbles to be legal, no two of them can be on horizontally or vertically adjacent squares (diagonal adjacency is fine).
(a) Determine the number of legal patterns that can occur in any column (in isolation, ignoring the pebbles in adjacent columns) and describe these patterns.
Call two patterns compatible if they can be placed on adjacent columns to form a legal placement. Let us consider subproblems consisting of the first k columns 1 ≤ k ≤ n. Each subproblem can be assigned a type, which is the pattern occurring in the last column.
(b) Using the notions of compatibility and type, give an O(n)-time dynamic programming algo- rithm for computing an optimal placement.
6.6. Let us define a multiplication operation on three symbols a, b, c according to the following table; thus ab = b, ba = c, and so on. Notice that the multiplication operation defined by the table is neither associative nor commutative.
dict(w) =

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 193
Find an efficient algorithm that examines a string of these symbols, say bbbbac, and decides whether or not it is possible to parenthesize the string in such a way that the value of the resulting expression is a. For example, on input bbbbac your algorithm should return yes because ((b(bb))(ba))c = a.
6.7. A subsequence is palindromic if it is the same whether read left to right or right to left. For instance, the sequence
A,C,G,T,G,T,C,A,A,A,A,T,C,G
has many palindromic subsequences, including A, C, G, C, A and A, A, A, A (on the other hand, the subsequence A, C, T is not palindromic). Devise an algorithm that takes a sequence x[1 . . . n] and returns the (length of the) longest palindromic subsequence. Its running time should be O(n2).
6.8. Given two strings x = x1x2 ···xn and y = y1y2 ···ym, we wish to find the length of their longest commonsubstring,thatis,thelargestkforwhichthereareindicesiandjwithxixi+1···xi+k−1 = yjyj+1 ···yj+k−1. Show how to do this in time O(mn).
6.9. A certain string-processing language offers a primitive operation which splits a string into two pieces. Since this operation involves copying the original string, it takes n units of time for a string of length n, regardless of the location of the cut. Suppose, now, that you want to break a string into many pieces. The order in which the breaks are made can affect the total running time. For example, if you want to cut a 20-character string at positions 3 and 10, then making the first cut at position 3 incurs a total cost of 20 + 17 = 37, while doing position 10 first has a better cost of 20 + 10 = 30.
Give a dynamic programming algorithm that, given the locations of m cuts in a string of length n, finds the minimum cost of breaking the string into m + 1 pieces.
6.10. Counting heads. Given integers n and k, along with p1, . . . , pn ∈ [0, 1], you want to determine the probability of obtaining exactly k heads when n biased coins are tossed independently at random, where pi is the probability that the ith coin comes up heads. Give an O(n2) algorithm for this task.2 Assumeyoucanmultiplyandaddtwonumbersin[0,1]inO(1)time.
6.11. Given two strings x = x1x2 ···xn and y = y1y2 ···ym, we wish to find the length of their longest common subsequence, that is, the largest k for which there are indices i1 < i2 < · · · < ik and j1 0ontheedges.
We would like to send as much oil as possible from s to t without exceeding the capacities of any of the edges. A particular shipping scheme is called a flow and consists of a variable fe for each edge e of the network, satisfying the following two properties:
1. It doesn’t violate edge capacities: 0 ≤ fe ≤ ce for all e ∈ E.
2. For all nodes u except s and t, the amount of flow entering u equals the amount leaving
5
u:
In other words, flow is conserved.
􏰑 fwu = 􏰑 fuz. (w,u)∈E (u,z)∈E
The size of a flow is the total quantity sent from s to t and, by the conservation principle, is equal to the quantity leaving s:
size(f)= 􏰑 fsu. (s,u)∈E
In short, our goal is to assign values to {fe : e ∈ E} that will satisfy a set of linear constraints and maximize a linear objective function. But this is a linear program! The maximum-flow problem reduces to linear programming.
For example, for the network of Figure 7.4 the LP has 11 variables, one per edge. It seeks to maximize fsa + fsb + fsc subject to a total of 27 constraints: 11 for nonnegativity (such as fsa ≥ 0), 11 for capacity (such as fsa ≤ 3), and 5 for flow conservation (one for each node of the graph other than s and t, such as fsc + fdc = fce). Simplex would take no time at all to correctly solve the problem and to confirm that, in our example, a flow of 7 is in fact optimal.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 213
Figure 7.5 An illustration of the max-flow algorithm. (a) A toy network. (b) The first path chosen. (c) The second path chosen. (d) The final flow. (e) We could have chosen this path first. (f) In which case, we would have to allow this second path.
(a) (b)
a1a 1
s1tst
1
b
(c) (d)
1
a 1
1 sts0t
b1b
(f)
(e)
1
1
s1ts1t 11
bb
aa 1
7.2.3
A closer look at the algorithm
All we know so far of the simplex algorithm is the vague geometric intuition that it keeps making local moves on the surface of a convex feasible region, successively improving the objective function until it finally reaches the optimal solution. Once we have studied it in more detail (Section 7.6), we will be in a position to understand exactly how it handles flow LPs, which is useful as a source of inspiration for designing direct max-flow algorithms.
It turns out that in fact the behavior of simplex has an elementary interpretation: Start with zero flow.
Repeat: choose an appropriate path from s to t, and increase flow along the edges of this path as much as possible.
Figure 7.5(a)–(d) shows a small example in which simplex halts after two iterations. The final flow has size 2, which is easily seen to be optimal.

214 Algorithms
There is just one complication. What if we had initially chosen a different path, the one in Figure 7.5(e)? This gives only one unit of flow and yet seems to block all other paths. Simplex gets around this problem by also allowing paths to cancel existing flow. In this particular case, it would subsequently choose the path of Figure 7.5(f). Edge (b, a) of this path isn’t in the original network and has the effect of canceling flow previously assigned to edge (a, b).
To summarize, in each iteration simplex looks for an s − t path whose edges (u, v) can be of two types:
1. (u, v) is in the original network, and is not yet at full capacity.
2. The reverse edge (v, u) is in the original network, and there is some flow along it.
Ifthecurrentflowisf,theninthefirstcase,edge(u,v)canhandleuptocuv −fuv additional units of flow, and in the second case, upto fvu additional units (canceling all or part of the existing flow on (v,u)). These flow-increasing opportunities can be captured in a residual networkGf =(V,Ef),whichhasexactlythetwotypesofedgeslisted,withresidualcapacities cf: 􏰌
cuv−fuv if(u,v)∈Eandfuv0
Thus we can equivalently think of simplex as choosing an s − t path in the residual network. By simulating the behavior of simplex, we get a direct algorithm for solving max-flow. It proceeds in iterations, each time explicitly constructing Gf , finding a suitable s − t path in Gf by using, say, a linear-time breadth-first search, and halting if there is no longer any such
path along which flow can be increased.
Figure 7.6 illustrates the algorithm on our oil example.
7.2.4 A certificate of optimality
Now for a truly remarkable fact: not only does simplex correctly compute a maximum flow, but it also generates a short proof of the optimality of this flow!
Let’s see an example of what this means. Partition the nodes of the oil network (Figure 7.4) into two groups, L = {s,a,b} and R = {c,d,e,t}:
La2dR
3
4
c5e
Any oil transmitted must pass from L to R. Therefore, no flow can possibly exceed the total capacity of the edges from L to R, which is 7. But this means that the flow we found earlier, of size 7, must be optimal!
10 1 s3b11t
2
5

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 215
More generally, an (s, t)-cut partitions the vertices into two disjoint groups L and R such thatsisinLandtisinR. ItscapacityisthetotalcapacityoftheedgesfromLtoR,andas argued previously, is an upper bound on any flow:
Pick any flow f and any (s, t)-cut (L, R). Then size(f ) ≤ capacity(L, R).
Some cuts are large and give loose upper bounds—cut ({s, b, c}, {a, d, e, t}) has a capacity of 19. But there is also a cut of capacity 7, which is effectively a certificate of optimality of the maximum flow. This isn’t just a lucky property of our oil network; such a cut always exists.
Max-flow min-cut theorem The size of the maximum flow in a network equals the capacity of the smallest (s, t)-cut.
Moreover, our algorithm automatically finds this cut as a by-product!
Let’s see why this is true. Suppose f is the final flow when the algorithm terminates. We know that node t is no longer reachable from s in the residual network Gf . Let L be the nodes thatarereachablefromsinGf,andletR=V −Lbetherestofthenodes. Then(L,R)isa cut in the graph G:
LR
s
e
e′
= capacity(L, R).
t
We claim that
size(f )
To see this, observe that by the way L is defined, any edge going from L to R must be at full capacity (in the current flow f), and any edge from R to L must have zero flow. (So, in the figure, fe = ce and fe′ = 0.) Therefore the net flow across (L, R) is exactly the capacity of the cut.
7.2.5 Efficiency
Each iteration of our maximum-flow algorithm is efficient, requiring O(|E|) time if a depth- first or breadth-first search is used to find an s − t path. But how many iterations are there?
Suppose all edges in the original network have integer capacities ≤ C. Then an inductive argument shows that on each iteration of the algorithm, the flow is always an integer and increases by an integer amount. Therefore, since the maximum flow is at most C|E| (why?), it follows that the number of iterations is at most this much. But this is hardly a reassuring bound: what if C is in the millions?
We examine this issue further in Exercise 7.31. It turns out that it is indeed possible to construct bad examples in which the number of iterations is proportional to C, if s − t paths are not carefully chosen. However, if paths are chosen in a sensible manner—in particular, by

216 Algorithms
using a breadth-first search, which finds the path with the fewest edges—then the number of iterations is at most O(|V | · |E|), no matter what the capacities are. This latter bound gives an overall running time of O(|V | · |E|2) for maximum flow.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 217
Figure 7.6 The max-flow algorithm applied to the network of Figure 7.4. At each iteration, the current flow is shown on the left and the residual network on the right. The paths chosen are shown in bold.
Current flow Residual graph (a)ad a2d
2 sbts3b11t
4
3
10 1
5 cec5e
(b) a 1 d
1 12101 2
sb1ts3b11t
41 114
c1e c4e (c) a 2 d a 2 d
2423 c1e c14e
(d) a 2 d a 2 d
21101 2 sb11ts3b11t
35315 c4e c14e
a 1 d
2 sb11ts3b11t
21101 2
2

218
Figure 7.6 Continued
Current Flow
(e) a 2 d 2
Algorithms
Residual Graph
a 2 d
1
1
21 10 1 sb1ts3b11t
1
4545 c5e c5e
(f) a 2 d a 2 d 212 211012
s1b1 t s21b11t
4545 c5e c5e

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 219 Figure 7.7 An edge between two people means they like each other. Is it possible to pair
everyone up happily?
7.3 Bipartite matching
BOYS
Al Bob Chet Dan
GIRLS
Alice Beatrice Carol Danielle
Figure 7.7 shows a graph with four nodes on the left representing boys and four nodes on the right representing girls.1 There is an edge between a boy and girl if they like each other (for instance, Al likes all the girls). Is it possible to choose couples so that everyone has exactly one partner, and it is someone they like? In graph-theoretic jargon, is there a perfect matching?
This matchmaking game can be reduced to the maximum-flow problem, and thereby to linear programming! Create a new source node, s, with outgoing edges to all the boys; a new sink node, t, with incoming edges from all the girls; and direct all the edges in the original bipartite graph from boy to girl (Figure 7.8). Finally, give every edge a capacity of 1. Then there is a perfect matching if and only if this network has a flow whose size equals the number of couples. Can you find such a flow in the example?
Actually, the situation is slightly more complicated than just stated: what is easy to see is that the optimum integer-valued flow corresponds to the optimum matching. We would be at a bit of a loss interpreting a flow that ships 0.7 units along the edge Al–Carol, for instance!
1This kind of graph, in which the nodes can be partitioned into two groups such that all edges are between the groups, is called bipartite.
Figure 7.8 A matchmaking network. Each edge has a capacity of one.
Alice Bob Beatrice
Chet Carol
Al
s
t
Dan
Danielle

220 Algorithms
Fortunately, the maximum-flow problem has the following property: if all edge capacities are integers, then the optimal flow found by our algorithm is integral. We can see this directly from the algorithm, which in such cases would increment the flow by an integer amount on each iteration.
Hence integrality comes for free in the maximum-flow problem. Unfortunately, this is the exception rather than the rule: as we will see in Chapter 8, it is a very difficult problem to find the optimum solution (or for that matter, any solution) of a general linear program, if we also demand that the variables be integers.
7.4 Duality
We have seen that in networks, flows are smaller than cuts, but the maximum flow and mini- mum cut exactly coincide and each is therefore a certificate of the other’s optimality. Remark- able as this phenomenon is, we now generalize it from maximum flow to any problem that can be solved by linear programming! It turns out that every linear maximization problem has a dual minimization problem, and they relate to each other in much the same way as flows and cuts.
To understand what duality is about, recall our introductory LP with the two types of chocolate:
max x1 + 6×2 x1 ≤ 200
x2 ≤ 300 x1 + x2 ≤ 400
x1,x2 ≥0
Simplex declares the optimum solution to be (x1,x2) = (100,300), with objective value 1900. Can this answer be checked somehow? Let’s see: suppose we take the first inequality and add it to six times the second inequality. We get
x1+6×2 ≤2000.
This is interesting, because it tells us that it is impossible to achieve a profit of more than 2000. Can we add together some other combination of the LP constraints and bring this upper bound even closer to 1900? After a little experimentation, we find that multiplying the three inequalities by 0, 5, and 1, respectively, and adding them up yields
x1+6×2 ≤1900.
So 1900 must indeed be the best possible value! The multipliers (0, 5, 1) magically constitute a certificate of optimality! It is remarkable that such a certificate exists for this LP—and even if we knew there were one, how would we systematically go about finding it?

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 221 Let’s investigate the issue by describing what we expect of these three multipliers, call
them y1, y2, y3.
Multiplier
y1 y2 y3
Inequality
≤ 200 x2 ≤300 x1 + x2 ≤ 400
x1
To start with, these yi’s must be nonnegative, for otherwise they are unqualified to multiply inequalities (multiplying an inequality by a negative number would flip the ≤ to ≥). After the multiplication and addition steps, we get the bound:
(y1 + y3)x1 + (y2 + y3)x2 ≤ 200y1 + 300y2 + 400y3.
We want the left-hand side to look like our objective function x1 + 6×2 so that the right-hand side is an upper bound on the optimum solution. For this we need y1 + y3 to be 1 and y2 + y3 to be 6. Come to think of it, it would be fine if y1 + y3 were larger than 1—the resulting certificate would be all the more convincing. Thus, we get an upper bound
 y 1 , y 2 , y 3 ≥ 0  x1 +6×2 ≤ 200y1 +300y2 +400y3 if  y1 +y3 ≥1 .
y2 + y3 ≥ 6
We can easily find y’s that satisfy the inequalities on the right by simply making them large enough, for example (y1, y2, y3) = (5, 3, 6). But these particular multipliers would tell us that the optimum solution of the LP is at most 200 · 5 + 300 · 3 + 400 · 6 = 4300, a bound that is far too loose to be of interest. What we want is a bound that is as tight as possible, so we should minimize 200y1 + 300y2 + 400y3 subject to the preceding inequalities. And this is a new linear program!
Therefore, finding the set of multipliers that gives the best upper bound on our original LP is tantamount to solving a new LP:
min 200y1 + 300y2 + 400y3 y1 + y3 ≥ 1
y2 + y3 ≥ 6 y1,y2,y3 ≥0
By design, any feasible value of this dual LP is an upper bound on the original primal LP. So if we somehow find a pair of primal and dual feasible values that are equal, then they must both be optimal. Here is just such a pair:
Primal : (x1, x2) = (100, 300); Dual : (y1, y2, y3) = (0, 5, 1).
They both have value 1900, and therefore they certify each other’s optimality (Figure 7.9).
Amazingly, this is not just a lucky example, but a general phenomenon. To start with, the preceding construction—creating a multiplier for each primal constraint; writing a constraint

222 Algorithms Figure 7.9 By design, dual feasible values ≥ primal feasible values. The duality theorem
tells us that moreover their optima coincide.
Primal Dual Primal feasible opt opt
This duality gap is zero
Dual feasible
Objective value
Figure 7.10 A generic primal LP in matrix-vector form, and its dual.
Primal LP:
Dual LP:
max cTx Ax ≤ b x≥0
min yTb yT A ≥ cT y≥0
in the dual for every variable of the primal, in which the sum is required to be above the objective coefficient of the corresponding primal variable; and optimizing the sum of the mul- tipliers weighted by the primal right-hand sides—can be carried out for any LP, as shown in Figure 7.10, and in even greater generality in Figure 7.11. The second figure has one notewor- thy addition: if the primal has an equality constraint, then the corresponding multiplier (or dual variable) need not be nonnegative, because the validity of equations is preserved when multiplied by negative numbers. So, the multipliers of equations are unrestricted variables. Notice also the simple symmetry between the two LPs, in that the matrix A = (aij) defines one primal constraint with each of its rows, and one dual constraint with each of its columns.
By construction, any feasible solution of the dual is an upper bound on any feasible solution of the primal. But moreover, their optima coincide!
Duality theorem If a linear program has a bounded optimum, then so does its dual, and the two optimum values coincide.
When the primal is the LP that expresses the max-flow problem, it is possible to assign interpretations to the dual variables that show the dual to be none other than the minimum- cut problem (Exercise 7.25). The relation between flows and cuts is therefore just a specific instance of the duality theorem. And in fact, the proof of this theorem falls out of the simplex algorithm, in much the same way as the max-flow min-cut theorem fell out of the analysis of the max-flow algorithm.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 223
Figure 7.11 In the most general case of linear programming, we have a set I of inequalities and a set E of equalities (a total of m = |I| + |E| constraints) over n variables, of which a subset N are constrained to be nonnegative. The dual has m = |I| + |E| variables, of which only those corresponding to I have nonnegativity constraints.
Primal LP:
max c1x1 +···+cnxn ai1x1+···+ainxn≤bi fori∈I ai1x1+···+ainxn=bi fori∈E xj ≥0 forj∈N
Visualizing duality
Dual LP:
min b1y1 +···+bmym a1jy1+···+amjym≥cj forj∈N a1jy1+···+amjym=cj forj̸∈N yi ≥0 fori∈I
One can solve the shortest-path problem by the following “analog” device: Given a weighted undirected graph, build a physical model of it in which each edge is a string of length equal to the edge’s weight, and each node is a knot at which the appropriate endpoints of strings are tied together. Then to find the shortest path from s to t, just pull s away from t until the gadget is taut. It is intuitively clear that this finds the shortest path from s to t.
BSA
DCT
There is nothing remarkable or surprising about all this until we notice the following: the shortest-path problem is a minimization problem, right? Then why are we pulling s away from t, an act whose purpose is, obviously, maximization? Answer: By pulling s away from t we solve the dual of the shortest-path problem! This dual has a very simple form (Exercise 7.28), with one variable xu for each node u:
max xS − xT |xu−xv|≤wuv foralledges{u,v}
In words, the dual problem is to stretch s and t as far apart as possible, subject to the constraint that the endpoints of any edge {u,v} are separated by a distance of at most wuv.

224 Algorithms 7.5 Zero-sum games
We can represent various conflict situations in life by matrix games. For example, the school- yard rock-paper-scissors game is specified by the payoff matrix illustrated here. There are two players, called Row and Column, and they each pick a move from {r, p, s}. They then look up the matrix entry corresponding to their moves, and Column pays Row this amount. It is Row’s gain and Column’s loss.
G= 0−11 1 0 −1 −1 1 0
Now suppose the two of them play this game repeatedly. If Row always makes the same move, Column will quickly catch on and will always play the countermove, winning every time. Therefore Row should mix things up: we can model this by allowing Row to have a mixed strategy, in which on each turn she plays r with probability x1, p with probability x2, and s with probability x3. This strategy is specified by the vector x = (x1,x2,x3), positive numbers that add up to 1. Similarly, Column’s mixed strategy is some y = (y1, y2, y3).2
On any given round of the game, there is an xiyj chance that Row and Column will play the ith and jth moves, respectively. Therefore the expected (average) payoff is
􏰑Gij ·Prob[Rowplaysi,Columnplaysj] = 􏰑Gijxiyj. i,j i,j
Row wants to maximize this, while Column wants to minimize it. What payoffs can they hope to achieve in rock-paper-scissors? Well, suppose for instance that Row plays the “completely random” strategy x = (1/3, 1/3, 1/3). If Column plays r, then the average payoff (reading the first column of the game matrix) will be
1·0+1·1+1·−1 = 0. 333
This is also true if Column plays p, or s. And since the payoff of any mixed strategy (y1, y2, y3) is just a weighted average of the individual payoffs for playing r, p, and s, it must also be zero. This can be seen directly from the preceding formula,
􏰑Gijxiyj = 􏰑Gij ·1yj = 􏰑yj􏰎􏰑1Gij􏰏 = 􏰑yj ·0 = 0, i,j i,j3ji3j
where the second-to-last equality is the observation that every column of G adds up to zero. Thus by playing the “completely random” strategy, Row forces an expected payoff of zero, no matter what Column does. This means that Column cannot hope for a negative (expected)
2Also of interest are scenarios in which players alter their strategies from round to round, but these can get very complicated and are a vast subject unto themselves.
Column
rps
r p s
Row

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 225
payoff (remember that he wants the payoff to be as small as possible). But symmetrically, if Column plays the completely random strategy, he also forces an expected payoff of zero, and thus Row cannot hope for a positive (expected) payoff. In short, the best each player can do is to play completely randomly, with an expected payoff of zero. We have mathematically confirmed what you knew all along about rock-paper-scissors!
Let’s think about this in a slightly different way, by considering two scenarios: 1. First Row announces her strategy, and then Column picks his.
2. First Column announces his strategy, and then Row chooses hers.
We’ve seen that the average payoff is the same (zero) in either case if both parties play op- timally. But this might well be due to the high level of symmetry in rock-paper-scissors. In general games, we’d expect the first option to favor Column, since he knows Row’s strategy and can fully exploit it while choosing his own. Likewise, we’d expect the second option to favor Row. Amazingly, this is not the case: if both play optimally, then it doesn’t hurt a player to announce his or her strategy in advance! What’s more, this remarkable property is a con- sequence of—and in fact equivalent to—linear programming duality.
Let’s investigate this with a nonsymmetric game. Imagine a presidential election scenario in which there are two candidates for office, and the moves they make correspond to campaign issues on which they can focus (the initials stand for economy, society, morality, and tax cut). The payoff entries are millions of votes lost by Column.
G= 3−1 −2 1
Suppose Row announces that she will play the mixed strategy x = (1/2, 1/2). What should Column do? Move m will incur an expected loss of 1/2, while t will incur an expected loss of 0. The best response of Column is therefore the pure strategy y = (0, 1).
More generally, once Row’s strategy x = (x1,x2) is fixed, there is always a pure strategy that is optimal for Column: either move m, with payoff 3×1 − 2×2, or t, with payoff −x1 + x2, whichever is smaller. After all, any mixed strategy y is a weighted average of these two pure strategies and thus cannot beat the better of the two.
Therefore, if Row is forced to announce x before Column plays, she knows that his best response will achieve an expected payoff of min{3×1 − 2×2, −x1 + x2}. She should choose x defensively to maximize her payoff against this best response:
Pick (x1, x2) that maximizes min{3×1 − 2×2, −x1 + x2} 􏰘 􏰗􏰖 􏰙
payoff from Column’s best response to x
This choice of xi’s gives Row the best possible guarantee about her expected payoff. And we will now see that it can be found by an LP! The main trick is to notice that for fixed x1 and x2 the following are equivalent:
mt
e s

226
z = min{3×1 − 2×2, −x1 + x2}
And Row needs to choose x1 and x2 to maximize this z.
Algorithms
max z −3×1 +2×2 +z x1 − x2 + z
x1 + x2 x1,x2
≤0 ≤ 0 =1 ≥0
Symmetrically, if Column has to announce his strategy first, his best bet is to choose the mixed strategy y that minimizes his loss under Row’s best response, in other words,
Pick (y1, y2) that minimizes max{3y1 − y2, −2y1 + y2}
max z
z ≤ 3×1 − 2×2 z ≤ −x1 + x2
􏰘 􏰗􏰖 􏰙
In LP form, this is
outcome of Row’s best response to y
min w
−3y1 +y2 +w ≥0
2y1 −y2 +w ≥0 y1 + y2 = 1 y1,y2 ≥0
The crucial observation now is that these two LPs are dual to each other (see Figure 7.11)! Hence, they have the same optimum, call it V .
Let us summarize. By solving an LP, Row (the maximizer) can determine a strategy for herself that guarantees an expected outcome of at least V no matter what Column does. And by solving the dual LP, Column (the minimizer) can guarantee an expected outcome of at most V , no matter what Row does. It follows that this is the uniquely defined optimal play: a priori it wasn’t even certain that such a play existed. V is known as the value of the game. In our example, it is 1/7 and is realized when Row plays her optimum mixed strategy (3/7, 4/7) and Column plays his optimum mixed strategy (2/7, 5/7).
This example is easily generalized to arbitrary games and shows the existence of mixed strategies that are optimal for both players and achieve the same value—a fundamental result of game theory called the min-max theorem. It can be written in equation form as follows:
max min 􏰑Gijxiyj = min max 􏰑Gijxiyj. xy yx
i,j i,j
This is surprising, because the left-hand side, in which Row has to announce her strategy first, should presumably be better for Column than the right-hand side, in which he has to go first. Duality equalizes the two, as it did with maximum flows and minimum cuts.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani
Figure 7.12 A polyhedron defined by seven inequalities. x2
A
227
⃝2
max x1+6×2+13×3 x1 ≤ 200
⃝1
⃝7
BC x2≤300⃝2
⃝5
⃝4
x3
⃝3
⃝1
x1 + x2 + x3 ≤ 400 ⃝3
x2 + 3×3 ≤ 600 ⃝4
x≥0 ⃝ x1 15
x2≥0 ⃝6 ⃝6 x 3 ≥ 0 ⃝7
7.6 The simplex algorithm
The extraordinary power and expressiveness of linear programs would be little consolation if we did not have a way to solve them efficiently. This is the role of the simplex algorithm.
At a high level, the simplex algorithm takes a set of linear inequalities and a linear objec- tive function and finds the optimal feasible point by the following strategy:
let v be any vertex of the feasible region
while there is a neighbor v′ of v with better objective value:
set v = v′
In our 2D and 3D examples (Figure 7.1 and Figure 7.2), this was simple to visualize and made intuitive sense. But what if there are n variables, x1, . . . , xn?
Any setting of the xi’s can be represented by an n-tuple of real numbers and plotted in n-dimensional space. A linear equation involving the xi’s defines a hyperplane in this same space Rn, and the corresponding linear inequality defines a half-space, all points that are either precisely on the hyperplane or lie on one particular side of it. Finally, the feasible region of the linear program is specified by a set of inequalities and is therefore the intersection of the corresponding half-spaces, a convex polyhedron.
But what do the concepts of vertex and neighbor mean in this general context? 7.6.1 Vertices and neighbors in n-dimensional space
Figure 7.12 recalls an earlier example. Looking at it closely, we see that each vertex is the unique point at which some subset of hyperplanes meet. Vertex A, for instance, is the sole point at which constraints ⃝2 , ⃝3 , and ⃝7 are satisfied with equality. On the other hand, the

228 Algorithms
hyperplanes corresponding to inequalities ⃝4 and ⃝6 do not define a vertex, because their intersection is not just a single point but an entire line.
Let’s make this definition precise.
Pick a subset of the inequalities. If there is a unique point that satisfies them with
equality, and this point happens to be feasible, then it is a vertex.
How many equations are needed to uniquely identify a point? When there are n variables, we need at least n linear equations if we want a unique solution. On the other hand, having more than n equations is redundant: at least one of them can be rewritten as a linear combination of the others and can therefore be disregarded. In short,
Each vertex is specified by a set of n inequalities.3 A notion of neighbor now follows naturally.
Two vertices are neighbors if they have n − 1 defining inequalities in common. InFigure7.12,forinstance,verticesAandCsharethetwodefininginequalities{⃝3,⃝7}and
are thus neighbors.
7.6.2 The algorithm
On each iteration, simplex has two tasks:
1. Check whether the current vertex is optimal (and if so, halt). 2. Determine where to move next.
As we will see, both tasks are easy if the vertex happens to be at the origin. And if the vertex is elsewhere, we will transform the coordinate system to move it to the origin!
First let’s see why the origin is so convenient. Suppose we have some generic LP
max cT x Ax ≤ b x≥0
where x is the vector of variables, x = (x1, . . . , xn). Suppose the origin is feasible. Then it is certainly a vertex, since it is the unique point at which the n inequalities {x1 ≥ 0, . . . , xn ≥ 0} are tight. Now let’s solve our two tasks. Task 1:
The origin is optimal if and only if all ci ≤ 0.
3There is one tricky issue here. It is possible that the same vertex might be generated by different subsets ofinequalities.InFigure7.12,vertexBisgeneratedby{⃝2,⃝3,⃝4},butalsoby{⃝2,⃝4,⃝5}.Suchverticesare called degenerate and require special consideration. Let’s assume for the time being that they don’t exist, and we’ll return to them later.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 229
If all ci ≤ 0, then considering the constraints x ≥ 0, we can’t hope for a better objective value. Conversely, if some ci > 0, then the origin is not optimal, since we can increase the objective function by raising xi.
Thus, for task 2, we can move by increasing some xi for which ci > 0. How much can we increase it? Until we hit some other constraint. That is, we release the tight constraint xi ≥ 0 and increase xi until some other inequality, previously loose, now becomes tight. At that point, we again have exactly n tight inequalities, so we are at a new vertex.
For instance, suppose we’re dealing with the following linear program.
max 2×1 + 5×2 2×1−x2≤4 ⃝1 x1+2×2≤9 ⃝2
−x1+x2≤3 ⃝3 x1 ≥ 0 ⃝4 x2 ≥ 0 ⃝5
Simplex can be started at the origin, which is specified by constraints ⃝4 and ⃝5 . To move, we release the tight constraint x2 ≥ 0. As x2 is gradually increased, the first constraint it runs into is −x1 + x2 ≤ 3, and thus it has to stop at x2 = 3, at which point this new inequality is tight. The new vertex is thus given by ⃝3 and ⃝4 .
So we know what to do if we are at the origin. But what if our current vertex u is else- where? The trick is to transform u into the origin, by shifting the coordinate system from the usual (x1, . . . , xn) to the “local view” from u. These local coordinates consist of (appropriately scaled) distances y1, . . . , yn to the n hyperplanes (inequalities) that define and enclose u:
u
y2
x
Specifically, if one of these enclosing inequalities is ai · x ≤ bi, then the distance from a point x to that particular “wall” is
yi = bi − ai · x.
The n equations of this type, one per wall, define the yi’s as linear functions of the xi’s, and this relationship can be inverted to express the xi’s as a linear function of the yi’s. Thus we can rewrite the entire LP in terms of the y’s. This doesn’t fundamentally change it (for instance, the optimal value stays the same), but expresses it in a different coordinate frame. The revised “local” LP has the following three properties:
y1

230 Algorithms 1. It includes the inequalities y ≥ 0, which are simply the transformed versions of the
inequalities defining u.
2. u itself is the origin in y-space.
3. The cost function becomes max c + c ̃T y, where c is the value of the objective function uu
at u and c ̃ is a transformed cost vector.
In short, we are back to the situation we know how to handle! Figure 7.13 shows this algo-
rithm in action, continuing with our earlier example.
The simplex algorithm is now fully defined. It moves from vertex to neighboring vertex, stopping when the objective function is locally optimal, that is, when the coordinates of the local cost vector are all zero or negative. As we’ve just seen, a vertex with this property must also be globally optimal. On the other hand, if the current vertex is not locally optimal, then its local coordinate system includes some dimension along which the objective function can be improved, so we move along this direction—along this edge of the polyhedron—until we reach a neighboring vertex. By the nondegeneracy assumption (see footnote 3 in Section 7.6.1), this edge has nonzero length, and so we strictly improve the objective value. Thus the process must eventually halt.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 231 Figure 7.13 Simplex in action.
Initial LP:
max 2×1 + 5×2 2×1−x2≤4 ⃝1 x1+2×2≤9 ⃝2
−x1+x2≤3 ⃝3 x1 ≥ 0 ⃝4 x2 ≥ 0 ⃝5
Currentvertex:{⃝4,⃝5}(origin). Objective value: 0.
Move: increase x2.
⃝5 is released, ⃝3 becomes tight. Stop at x2 = 3.
Newvertex{⃝4,⃝3}haslocalcoordinates(y1,y2): y1 =x1, y2 =3+x1−x2
Rewritten LP:
max 15+7y1 −5y2 y1+y2≤7 ⃝1 3y1−2y2≤3 ⃝2 y2 ≥ 0 ⃝3 y1 ≥ 0 ⃝4 −y1+y2≤3 ⃝5
Currentvertex:{⃝4,⃝3}. Objective value: 15.
Move: increase y1.
⃝4 is released, ⃝2 becomes tight. Stop at y1 = 1.
Newvertex{⃝2,⃝3}haslocalcoordinates(z1,z2): z1 =3−3y1+2y2, z2=y2
Rewritten LP:
max 22 − 7 z1 − 1 z2 33
−1z1+5z2≤6 ⃝1 33
z1 ≥ 0 ⃝2
z2 ≥ 0 ⃝3 1z1−2z2≤1 ⃝4
1z1+1z2≤4 ⃝5 33
33
Currentvertex:{⃝2,⃝3}. Objective value: 22.
Optimal: all ci < 0. Solve ⃝2 , ⃝3 (in original LP) to get optimal solution (x1, x2) = (1, 4). Increase y1 {⃝3 ,⃝4} Increase x2 {⃝4 ,⃝5} {⃝2 ,⃝3} {⃝1 ,⃝5} {⃝1 ,⃝2} 232 Algorithms 7.6.3 Loose ends There are several important issues in the simplex algorithm that we haven’t yet mentioned. The starting vertex. How do we find a vertex at which to start simplex? In our 2D and 3D examples we always started at the origin, which worked because the linear programs happened to have inequalities with positive right-hand sides. In a general LP we won’t always be so fortunate. However, it turns out that finding a starting vertex can be reduced to an LP and solved by simplex! To see how this is done, start with any linear program in standard form (recall Sec- tion 7.1.4), since we know LPs can always be rewritten this way. min cTx suchthat Ax=bandx≥0. We first make sure that the right-hand sides of the equations are all nonnegative: if bi < 0, just multiply both sides of the ith equation by −1. Then we create a new LP as follows: • Create m new artificial variables z1, . . . , zm ≥ 0, where m is the number of equations. • Add zi to the left-hand side of the ith equation. • Lettheobjective,tobeminimized,bez1+z2+···+zm. For this new LP, it’s easy to come up with a starting vertex, namely, the one with zi = bi for all i and all other variables zero. Therefore we can solve it by simplex, to obtain the optimum solution. There are two cases. If the optimum value of z1 + · · · + zm is zero, then all zi’s obtained by simplex are zero, and hence from the optimum vertex of the new LP we get a starting feasible vertex of the original LP, just by ignoring the zi’s. We can at last start simplex! But what if the optimum objective turns out to be positive? Let us think. We tried to minimize the sum of the zi’s, but simplex decided that it cannot be zero. But this means that the original linear program is infeasible: it needs some nonzero zi’s to become feasible. This is how simplex discovers and reports that an LP is infeasible. Degeneracy. In the polyhedron of Figure 7.12 vertex B is degenerate. Geometrically, this means that it is the intersection of more than n = 3 faces of the polyhedron (in this case, ⃝2 , ⃝3 , ⃝4 , ⃝5 ). Algebraically, it means that if we choose any one of four sets of three inequal- ities({⃝2,⃝3,⃝4},{⃝2,⃝3,⃝5},{⃝2,⃝4,⃝5},and{⃝3,⃝4,⃝5})andsolvethecorrespondingsystem of three linear equations in three unknowns, we’ll get the same solution in all four cases: (0, 300, 100). This is a serious problem: simplex may return a suboptimal degenerate vertex simply because all its neighbors are identical to it and thus have no better objective. And if we modify simplex so that it detects degeneracy and continues to hop from vertex to vertex despite lack of any improvement in the cost, it may end up looping forever. One way to fix this is by a perturbation: change each bi by a tiny random amount to bi ± εi. This doesn’t change the essence of the LP since the εi’s are tiny, but it has the effect of differ- entiating between the solutions of the linear systems. To see why geometrically, imagine that S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 233 the four planes ⃝2 , ⃝3 , ⃝4 , ⃝5 were jolted a little. Wouldn’t vertex B split into two vertices, very close to one another? Unboundedness. In some cases an LP is unbounded, in that its objective function can be made arbitrarily large (or small, if it’s a minimization problem). If this is the case, simplex will discover it: in exploring the neighborhood of a vertex, it will notice that taking out an inequality and adding another leads to an underdetermined system of equations that has an infinity of solutions. And in fact (this is an easy test) the space of solutions contains a whole line across which the objective can become larger and larger, all the way to ∞. In this case simplex halts and complains. 7.6.4 The running time of simplex What is the running time of simplex, for a generic linear program max cTx suchthat Ax≤0andx≥0, where there are n variables and A contains m inequality constraints? Since it is an iterative algorithm that proceeds from vertex to vertex, let’s start by computing the time taken for a single iteration. Suppose the current vertex is u. By definition, it is the unique point at which n inequality constraints are satisfied with equality. Each of its neighbors shares n − 1 of these inequalities, so u can have at most n · m neighbors: choose which inequality to drop and which new one to add. A naive way to perform an iteration would be to check each potential neighbor to see whether it really is a vertex of the polyhedron and to determine its cost. Finding the cost is quick, just a dot product, but checking whether it is a true vertex involves solving a system of n equations in n unknowns (that is, satisfying the n chosen inequalities exactly) and checking whether the result is feasible. By Gaussian elimination (see the following box) this takes O(n3) time, giving an unappetizing running time of O(mn4) per iteration. Fortunately, there is a much better way, and this mn4 factor can be improved to mn, mak- ing simplex a practical algorithm. Recall our earlier discussion (Section 7.6.2) about the local view from vertex u. It turns out that the per-iteration overhead of rewriting the LP in terms of the current local coordinates is just O((m + n)n); this exploits the fact that the local view changes only slightly between iterations, in just one of its defining inequalities. Next, to select the best neighbor, we recall that the (local view of) the objective function is of the form “max cu +c ̃·y” where cu is the value of the objective function at u. This immediately identifies a promising direction to move: we pick any c ̃i > 0 (if there is none, then the current vertex is optimal and simplex halts). Since the rest of the LP has now been rewritten in terms of the y-coordinates, it is easy to determine how much yi can be increased before some other inequality is violated. (And if we can increase yi indefinitely, we know the LP is unbounded.)
It follows that the running time per iteration of simplex is just O(mn). But how many iterations could there be? Naturally, there can’t be more than 􏰀m+n􏰁, which is an upper bound
on the number of vertices. But this upper bound is exponential in n. And in fact, there are examples of LPs for which simplex does indeed take an exponential number of iterations. In
n

234 Algorithms
other words, simplex is an exponential-time algorithm. However, such exponential examples do not occur in practice, and it is this fact that makes simplex so valuable and so widely used.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 235 Gaussian elimination
Under our algebraic definition, merely writing down the coordinates of a vertex involves solving a system of linear equations. How is this done?
We are given a system of n linear equations in n unknowns, say n = 4 and
x1 − 2×3 x2 + x3
= 2
The high school method for solving such systems is to repeatedly apply the following rule:
if we add a multiple of one equation to another equation, the overall system of equations remains equivalent. For example, adding −1 times the first equation to the third one, we get the equivalent system
x1 − 2×3 = 2 x2 + x3 = 3 x2 + 2×3 − x4 = 2 x2 + 3×3 + x4 = 5
This transformation is clever in the following sense: it eliminates the variable x1 from the third equation, leaving just one equation with x1. In other words, ignoring the first equation, we have a system of three equations in three unknowns: we decreased n by 1! We can solve this smaller system to get x2, x3, x4, and then plug these into the first equation to get x1.
This suggests an algorithm—once more due to Gauss.
procedure gauss(E, X)
Input: A system E = {e1,…,en} of equations in n unknowns X = {x1,…,xn}:
e1 :a11x1 +a12x2 +···+a1nxn =b1; ···; en :an1x1 +an2x2 +···+annxn =bn Output: A solution of the system, if one exists
if all coefficients ai1 are zero:
halt with message ‘‘either infeasible or not linearly independent’’
if n = 1: return b1/a11
choose the coefficient ap1 of largest magnitude, and swap equations e1, ep for i=2 to n:
ei = ei 􏰐− (ai1/a11) · e1
(x2,…,xn) =gauss(E −{e1},X −{x1})
x1=(b1− j>1a1jxj)/a11 return (x1,…,xn)
(When choosing the equation to swap into first place, we pick the one with largest |ap1| for reasons of numerical accuracy; after all, we will be dividing by ap1.)
Gaussian elimination uses O(n2) arithmetic operations to reduce the problem size from n to n − 1, and thus uses O(n3) operations overall. To show that this is also a good estimate of the total running time, we need to argue that the numbers involved remain polynomi- ally bounded—for instance, that the solution (x1,…,xn) does not require too much more precision to write down than the original coefficients aij and bi. Do you see why this is true?
x1 + x2
= 3 − x4 = 4 x2 + 3×3 + x4 = 5

236 Algorithms Linear programming in polynomial time
Simplex is not a polynomial time algorithm. Certain rare kinds of linear programs cause it to go from one corner of the feasible region to a better corner and then to a still better one, and so on for an exponential number of steps. For a long time, linear programming was considered a paradox, a problem that can be solved in practice, but not in theory!
Then, in 1979, a young Soviet mathematician called Leonid Khachiyan came up with the ellipsoid algorithm, one that is very different from simplex, extremely simple in its conception (but sophisticated in its proof) and yet one that solves any linear program in polynomial time. Instead of chasing the solution from one corner of the polyhedron to the next, Khachiyan’s algorithm confines it to smaller and smaller ellipsoids (skewed high- dimensional balls). When this algorithm was announced, it became a kind of “mathematical Sputnik,” a splashy achievement that had the U.S. establishment worried, in the height of the Cold War, about the possible scientific superiority of the Soviet Union. The ellipsoid algorithm turned out to be an important theoretical advance, but did not compete well with simplex in practice. The paradox of linear programming deepened: A problem with two algorithms, one that is efficient in theory, and one that is efficient in practice!
A few years later Narendra Karmarkar, a graduate student at UC Berkeley, came up with a completely different idea, which led to another provably polynomial algorithm for linear programming. Karmarkar’s algorithm is known as the interior point method, because it does just that: it dashes to the optimum corner not by hopping from corner to corner on the surface of the polyhedron like simplex does, but by cutting a clever path in the interior of the polyhedron. And it does perform well in practice.
But perhaps the greatest advance in linear programming algorithms was not Khachiyan’s theoretical breakthrough or Karmarkar’s novel approach, but an unexpected consequence of the latter: the fierce competition between the two approaches, simplex and interior point, resulted in the development of very fast code for linear programming.
7.7 Postscript: circuit evaluation
The importance of linear programming stems from the astounding variety of problems that reduce to it and thereby bear witness to its expressive power. In a sense, this next one is the ultimate application.
We are given a Boolean circuit, that is, a dag of gates of the following types. • Input gates have indegree zero, with value true or false.
• AND gates and OR gates have indegree 2.
• NOT gates have indegree 1.
In addition, one of the gates is designated as the output. Here’s an example.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani
237
output
AND
NOT
AND
OR
OR
true
NOT
true
false
The CIRCUIT VALUE problem is the following: when the laws of Boolean logic are applied to the gates in topological order, does the output evaluate to true?
There is a simple, automatic way of translating this problem into a linear program. Create a variable xg for each gate g, with constraints 0 ≤ xg ≤ 1. Add additional constraints for each type of gate:
gate g g true false
ggg
OR
AND
h h′
xg ≤xh
xg ≤xh′
xg ≥ xh + xh′ − 1
NOT
h
xg =1−xh
xg=1 xg=0 h h′ xg ≥xh
xg ≥xh′ xg ≤ xh + xh′
These constraints force all the gates to take on exactly the right values—0 for false, and 1 for true. We don’t need to maximize or minimize anything, and we can read the answer off from the variable xo corresponding to the output gate.
This is a straightforward reduction to linear programming, from a problem that may not seem very interesting at first. However, the CIRCUIT VALUE problem is in a sense the most general problem solvable in polynomial time! After all, any algorithm will eventually run on a computer, and the computer is ultimately a Boolean combinational circuit implemented on a chip. If the algorithm runs in polynomial time, it can be rendered as a Boolean circuit con- sisting of polynomially many copies of the computer’s circuit, one per unit of time, with the values of the gates in one layer used to compute the values for the next. Hence, the fact that CIRCUIT VALUE reduces to linear programming means that all problems that can be solved in polynomial time do!

238 Algorithms In our next topic, NP-completeness, we shall see that many hard problems reduce, much
the same way, to integer programming, linear programming’s difficult twin.
Another parting thought: by what other means can the circuit evaluation problem be solved? Let’s think—a circuit is a dag. And what algorithmic technique is most appropriate for solving problems on dags? That’s right: dynamic programming! Together with linear programming, the world’s two most general algorithmic techniques.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 239 Exercises
7.1. Consider the following linear program.
maximize 5x+3y 5x − 2y ≥ 0
x+y≤7 x≤5 x≥0 y≥0
Plot the feasible region and identify the optimal solution.
7.2. DuckwheatisproducedinKansasandMexicoandconsumedinNewYorkandCalifornia.Kansas produces 15 shnupells of duckwheat and Mexico 8. Meanwhile, New York consumes 10 shnupells and California 13. The transportation costs per shnupell are $4 from Mexico to New York, $1 from Mexico to California, $2 from Kansas to New York, and $3 and from Kansas to California.
Write a linear program that decides the amounts of duckwheat (in shnupells and fractions of a shnupell) to be transported from each producer to each consumer, so as to minimize the overall transportation cost.
7.3. A cargo plane can carry a maximum weight of 100 tons and a maximum volume of 60 cubic meters. There are three materials to be transported, and the cargo company may choose to carry any amount of each, upto the maximum available limits given below.
• Material1hasdensity2tons/cubicmeter,maximumavailableamount40cubicmeters,and revenue $1,000 per cubic meter.
• Material 2 has density 1 ton/cubic meter, maximum available amount 30 cubic meters, and revenue $1,200 per cubic meter.
• Material3hasdensity3tons/cubicmeter,maximumavailableamount20cubicmeters,and revenue $12,000 per cubic meter.
Write a linear program that optimizes revenue within the constraints.
7.4. MoeisdecidinghowmuchRegularDuffbeerandhowmuchDuffStrongbeertoordereachweek. Regular Duff costs Moe $1 per pint and he sells it at $2 per pint; Duff Strong costs Moe $1.50 per pint and he sells it at $3 per pint. However, as part of a complicated marketing scam, the Duff company will only sell a pint of Duff Strong for each two pints or more of Regular Duff that Moe buys. Furthermore, due to past events that are better left untold, Duff will not sell Moe more than 3,000 pints per week. Moe knows that he can sell however much beer he has. Formulate a linear program for deciding how much Regular Duff and how much Duff Strong to buy, so as to maximize Moe’s profit. Solve the program geometrically.
7.5. The Canine Products company offers two dog foods, Frisky Pup and Husky Hound, that are made from a blend of cereal and meat. A package of Frisky Pup requires 1 pound of cereal and 1.5 pounds of meat, and sells for $7. A package of Husky Hound uses 2 pounds of cereal and 1 pound of meat, and sells for $6. Raw cereal costs $1 per pound and raw meat costs $2 per pound. It also costs $1.40 to package the Frisky Pup and $0.60 to package the Husky Hound. A total of 240,000 pounds of cereal and 180,000 pounds of meat are available each month. The only production bottleneck is that the factory can only package 110,000 bags of Frisky Pup per month. Needless to say, management would like to maximize profit.

240 Algorithms (a) Formulate the problem as a linear program in two variables.
(b) Graph the feasible region, give the coordinates of every vertex, and circle the vertex maxi- mizing profit. What is the maximum profit possible?
7.6. Give an example of a linear program in two variables whose feasible region is infinite, but such that there is an optimum solution of bounded cost.
7.7. Find necessary and sufficient conditions on the reals a and b under which the linear program
(a) Is infeasible. (b) Is unbounded.
max x + y ax + by ≤ 1 x, y ≥ 0
(c) Has a unique optimal solution.
7.8. You are given the following points in the plane:
(1,3), (2,5), (3,7), (5,11), (7,14), (8,15), (10,19).
You want to find a line ax + by = c that approximately passes through these points (no line is a perfect fit). Write a linear program (you don’t need to solve it) to find the line that minimizes the maximum absolute error,
max |axi + byi − c|. 1≤i≤7
7.9. Aquadraticprogrammingproblemseekstomaximizeaquadratricobjectivefunction(withterms like 3×21 or 5x1x2) subject to a set of linear constraints. Give an example of a quadratic program in two variables x1, x2 such that the feasible region is nonempty and bounded, and yet none of the vertices of this region optimize the (quadratic) objective.
7.10. For the following network, with edge capacities as shown, find the maximum flow from S to T , along with a matching cut.
A4D 6212
5
G 12
26T C5F4
10 S 1 B 20 E
10
7.11. Write the dual to the following linear program.
max x + y 2x + y ≤ 3 x + 3y ≤ 5
x, y ≥ 0 Find the optimal solutions to both primal and dual LPs.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 241
7.12. For the linear program
max x1 − 2×3 x1 − x2 ≤ 1 2×2 − x3 ≤ 1 x1,x2,x3 ≥0
prove that the solution (x1 , x2 , x3 ) = (3/2, 1/2, 0) is optimal.
7.13. Matching pennies. In this simple two-player game, the players (call them R and C) each choose an outcome, heads or tails. If both outcomes are equal, C gives a dollar to R; if the outcomes are different, R gives a dollar to C.
(a) Represent the payoffs by a 2 × 2 matrix.
(b) What is the value of this game, and what are the optimal strategies for the two players?
7.14. The pizza business in Little Town is split between two rivals, Tony and Joey. They are each investigating strategies to steal business away from the other. Joey is considering either lowering prices or cutting bigger slices. Tony is looking into starting up a line of gourmet pizzas, or offering outdoor seating, or giving free sodas at lunchtime. The effects of these various strategies are summarized in the following payoff matrix (entries are dozens of pizzas, Joey’s gain and Tony’s loss).
TONY Gourmet Seating JOEY Lower price +2 0 Bigger slices −1 −2
Free soda
−3 +1
For instance, if Joey reduces prices and Tony goes with the gourmet option, then Tony will lose 2 dozen pizzas worth of business to Joey.
What is the value of this game, and what are the optimal strategies for Tony and Joey?
7.15. Find the value of the game specified by the following payoff matrix.
0 0 −1 −1
0 1 −2 −1 −1 −1 1 1 −1 0 0 1
1 −2 0 −3 1 −1 −1 −1 0 −3 2 −1 0 −2 1 −1
(Hint: Considerthemixedstrategies(1/3,0,0,1/2,1/6,0,0,0)and(2/3,0,0,1/3).)
7.16. A salad is any combination of the following ingredients: (1) tomato, (2) lettuce, (3) spinach, (4) carrot, and (5) oil. Each salad must contain: (A) at least 15 grams of protein, (B) at least 2 and at most 6 grams of fat, (C) at least 4 grams of carbohydrates, (D) at most 100 milligrams of sodium. Furthermore, (E) you do not want your salad to be more than 50% greens by mass. The nutritional contents of these ingredients (per 100 grams) are

242
Algorithms
ingredient energy (kcal) tomato 21 lettuce 16 spinach 371 carrot 346 oil 884
protein fat (grams) (grams) 0.85 0.33 1.62 0.20 12.78 1.58 8.39 1.39 0.00 100.00
carbohydrate sodium (grams) (milligrams) 4.64 9.00 2.37 8.00 74.69 7.00 80.70 508.20 0.00 0.00
Find a linear programming applet on the Web and
calories under the nutritional constraints. Describe your linear programming formulation and the optimal solution (the quantity of each ingredient and the value). Cite the Web resources that you used.
use it to make the salad with the fewest 7.17. Consider the following network (the numbers are edge capacities).
A4C
9
7
2
ST
2 6
5 B3D
(a) Find the maximum flow f and a minimum cut.
(b) DrawtheresidualgraphGf (alongwithitsedgecapacities).Inthisresidualnetwork,mark
the vertices reachable from S and the vertices from which T is reachable.
(c) An edge of a network is called a bottleneck edge if increasing its capacity results in an
increase in the maximum flow. List all bottleneck edges in the above network.
(d) Give a very simple example (containing at most four nodes) of a network which has no bottleneck edges.
(e) Give an efficient algorithm to identify all bottleneck edges in a network. (Hint: Start by running the usual network flow algorithm, and then examine the residual graph.)
7.18. There are many common variations of the maximum flow problem. Here are four of them.
(a) There are many sources and many sinks, and we wish to maximize the total flow from all sources to all sinks.
(b) Each vertex also has a capacity on the maximum flow that can enter it.
(c) Each edge has not only a capacity, but also a lower bound on the flow it must carry.
(d) The outgoing flow from each node u is not the same as the incoming flow, but is smaller by a factor of (1 − εu), where εu is a loss coefficient associated with node u.
Each of these can be solved efficiently. Show this by reducing (a) and (b) to the original max-flow problem, and reducing (c) and (d) to linear programming.
7.19. Suppose someone presents you with a solution to a max-flow problem on some network. Give a linear time algorithm to determine whether the solution does indeed give a maximum flow.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 243
7.20. Consider the following generalization of the maximum flow problem.
You are given a directed network G = (V,E) with edge capacities {ce}. Instead of a single (s,t) pair, you are given multiple pairs (s1, t1), (s2, t2), . . . , (sk, tk), where the si are sources of G and the ti are sinks of G. You are also given k demands d1,…,dk. The goal is to find k flows f(1),…,f(k) with the following properties:
doesnotexceedthecapacityc . e
• The size of the total flow (the sum of the flows) is as large as possible. How would you solve this problem?
7.21. An edge of a flow network is called critical if decreasing the capacity of this edge results in a decrease in the maximum flow. Give an efficient algorithm that finds a critical edge in a network.
7.22. InaparticularnetworkG=(V,E)whoseedgeshaveintegercapacitiesce,wehavealreadyfound the maximum flow f from node s to node t. However, we now find out that one of the capacity values we used was wrong: for edge (u, v) we used cuv whereas it should have been cuv − 1. This is unfortunate because the flow f uses that particular edge at full capacity: fuv = cuv.
We could redo the flow computation from scratch, but there’s a faster way. Show how a new optimal flow can be computed in O(|V | + |E|) time.
7.23. A vertex cover of an undirected graph G = (V, E) is a subset of the vertices which touches every edge—that is, a subset S ⊂ V such that for each edge {u,v} ∈ E, one or both of u,v are in S.
Show that the problem of finding the minimum vertex cover in a bipartite graph reduces to max- imum flow. (Hint: Can you relate this problem to the minimum cut in an appropriate network?)
7.24. Direct bipartite matching. We’ve seen how to find a maximum matching in a bipartite graph via reduction to the maximum flow problem. We now develop a direct algorithm.
Let G = (V1 ∪V2, E) be a bipartite graph (so each edge has one endpoint in V1 and one endpoint in V2), and let M ∈ E be a matching in the graph (that is, a set of edges that don’t touch). A vertex is said to be covered by M if it is the endpoint of one of the edges in M. An alternating path is a path of odd length that starts and ends with a non-covered vertex, and whose edges alternate between M and E − M.
(a) In the bipartite graph below, a matching M is shown in bold. Find an alternating path.
E A
F B
G C
H D
I
• f(i) isavalidflowfromsi toti.
• Foreachedgee,thetotalflowfe +fe +···+fe
(1) (2) (k) • The size of each flow f(i) is at least the demand di.

244 Algorithms (b) Prove that a matching M is maximal if and only if there does not exist an alternating path
with respect to it.
(c) Design an algorithm that finds an alternating path in O(|V | + |E|) time using a variant of breadth-first search.
(d) Give a direct O(|V | · |E|) algorithm for finding a maximal matching in a bipartite graph. 7.25. The dual of maximum flow. Consider the following network with edge capacities.
A
S1T 3B1
(a) Write the problem of finding the maximum flow from S to T as a linear program.
(b) Write down the dual of this linear program. There should be a dual variable for each edge
of the network and for each vertex other than S, T .
Now we’ll solve the same problem in full generality. Recall the linear program for a general
maximum flow problem (Section 7.2).
(c) Write down the dual of this general flow LP, using a variable ye for each edge and xu for
each vertex u ̸= s, t.
(d) Show that any solution to the general dual LP must satisfy the following property: for any directed path from s to t in the network, the sum of the ye values along the path must be at least 1.
(e) What are the intuitive meanings of the dual variables? Show that any s − t cut in the network can be translated into a dual feasible solution whose cost is exactly the capacity of that cut.
7.26. In a satisfiable system of linear inequalities
a11x1 +···+a1nxn ≤ b1 .
am1x1 +···+amnxn ≤ bm
we describe the jth inequality as forced-equ􏰐al if it is satisfied with equality by every solution
x = (x1,…,xn) of the system. Equivalently, 􏰐ajixi ≤ bj is not forced-equal if there exists an x
1
2
that satisfies the whole system and such that i ajixi < bj. For example, in i x1+x2 ≤ 2 −x1−x2 ≤ −2 x1 ≤ 1 −x2 ≤ 0 S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 245 the first two inequalities are forced-equal, while the third and fourth are not. A solution x to the system is called characteristic if, for every inequality I that is not forced-equal, x satisfies I without equality. In the instance above, such a solution is (x1 , x2 ) = (−1, 3), for which x1 < 1 and −x2 <0whilex1 +x2 =2and−x1 −x2 =−2. (a) Show that any satisfiable system has a characteristic solution. (b) Given a satisfiable system of linear inequalities, show how to use linear programming to determine which inequalities are forced-equal, and to find a characteristic solution. 7.27. Show that the change-making problem (Exercise 6.17) can be formulated as an integer linear program. Can we solve this program as an LP, in the certainty that the solution will turn out to be integral (as in the case of bipartite matching)? Either prove it or give a counterexample. 7.28. A linear program for shortest path. Suppose we want to compute the shortest path from node s to node t in a directed graph with edge lengths le > 0.
(a) Show that this is equivalent to finding an s − t flow f that minimizes 􏰐e lefe subject to size(f) = 1. There are no capacity constraints.
(b) Write the shortest path problem as a linear program.
(c) Show that the dual LP can be written as
max xs − xt
xu −xv ≤luv forall(u,v)∈E
(d) An interpretation for the dual is given in the box on page 223. Why isn’t our dual LP identical to the one on that page?
7.29. Hollywood. A film producer is seeking actors and investors for his new movie. There are n available actors; actor i charges si dollars. For funding, there are m available investors. Investor
j will provide pj dollars, but only on the condition that certain actors Lj ⊆ {1, 2, . . . , n} are included in the cast (all of these actors Lj must be chosen in order to receive funding from investor j).
The producer’s profit is the sum of the payments from investors minus the payments to actors. The goal is to maximize this profit.
(a) Express this problem as an integer linear program in which the variables take on values {0, 1}.
(b) Now relax this to a linear program, and show that there must in fact be an integral optimal solution (as is the case, for example, with maximum flow and bipartite matching).
7.30. Hall’s theorem. Returning to the matchmaking scenario of Section 7.3, suppose we have a bipar- tite graph with boys on the left and an equal number of girls on the right. Hall’s theorem says that there is a perfect matching if and only if the following condition holds: any subset S of boys is connected to at least |S| girls.
Prove this theorem. (Hint: The max-flow min-cut theorem should be helpful.)
7.31. Consider the following simple network with edge capacities as shown.

246
Algorithms
A S1T
1000 B 1000
1000
1000
(a) Showthat,iftheFord-Fulkersonalgorithmisrunonthisgraph,acarelesschoiceofupdates might cause it to take 1000 iterations. Imagine if the capacities were a million instead of 1000!
We will now find a strategy for choosing paths under which the algorithm is guaranteed to ter- minate in a reasonable number of iterations.
Consider an arbitrary directed network (G = (V, E), s, t, {ce}) in which we want to find the max- imum flow. Assume for simplicity that all edge capacities are at least 1, and define the capacity of an s − t path to be the smallest capacity of its constituent edges. The fattest path from s to t is the path with the most capacity.
(b) Show that the fattest s − t path in a graph can be computed by a variant of Dijkstra’s algorithm.
(c) Show that the maximum flow in G is the sum of individual flows along at most |E| paths from s to t.
(d) Now show that if we always increase flow along the fattest path in the residual graph, then the Ford-Fulkerson algorithm will terminate in at most O(|E| log F ) iterations, where F is the size of the maximum flow. (Hint: It might help to recall the proof for the greedy set cover algorithm in Section 5.4.)
In fact, an even simpler rule—finding a path in the residual graph using breadth-first search— guarantees that at most O(|V | · |E|) iterations will be needed.

Chapter 8
NP-complete problems
8.1 Search problems
Over the past seven chapters we have developed algorithms for finding shortest paths and minimum spanning trees in graphs, matchings in bipartite graphs, maximum increasing sub- sequences, maximum flows in networks, and so on. All these algorithms are efficient, because in each case their time requirement grows as a polynomial function (such as n, n2, or n3) of the size of the input.
To better appreciate such efficient algorithms, consider the alternative: In all these prob- lems we are searching for a solution (path, tree, matching, etc.) from among an exponential population of possibilities. Indeed, n boys can be matched with n girls in n! different ways, a graph with n vertices has nn−2 spanning trees, and a typical graph has an exponential num- ber of paths from s to t. All these problems could in principle be solved in exponential time by checking through all candidate solutions, one by one. But an algorithm whose running time is 2n, or worse, is all but useless in practice (see the next box). The quest for efficient algorithms is about finding clever ways to bypass this process of exhaustive search, using clues from the input in order to dramatically narrow down the search space.
So far in this book we have seen the most brilliant successes of this quest, algorithmic tech- niques that defeat the specter of exponentiality: greedy algorithms, dynamic programming, linear programming (while divide-and-conquer typically yields faster algorithms for problems we can already solve in polynomial time). Now the time has come to meet the quest’s most embarrassing and persistent failures. We shall see some other “search problems,” in which again we are seeking a solution with particular properties among an exponential chaos of al- ternatives. But for these new problems no shortcut seems possible. The fastest algorithms we know for them are all exponential—not substantially better than an exhaustive search. We now introduce some important examples.
247

248 Algorithms The story of Sissa and Moore
According to the legend, the game of chess was invented by the Brahmin Sissa to amuse and teach his king. Asked by the grateful monarch what he wanted in return, the wise man requested that the king place one grain of rice in the first square of the chessboard, two in the second, four in the third, and so on, doubling the amount of rice up to the 64th square. The king agreed on the spot, and as a result he was the first person to learn the valuable—-albeit humbling—lesson of exponential growth. Sissa’s request amounted to 2 64 − 1 = 18,446,744,073,709,551,615 grains of rice, enough rice to pave all of India several times over!
All over nature, from colonies of bacteria to cells in a fetus, we see systems that grow exponentially—for a while. In 1798, the British philosopher T. Robert Malthus published an essay in which he predicted that the exponential growth (he called it “geometric growth”) of the human population would soon deplete linearly growing resources, an argument that influenced Charles Darwin deeply. Malthus knew the fundamental fact that an exponential sooner or later takes over any polynomial.
In 1965, computer chip pioneer Gordon E. Moore noticed that transistor density in chips had doubled every year in the early 1960s, and he predicted that this trend would continue. This prediction, moderated to a doubling every 18 months and extended to computer speed, is known as Moore’s law. It has held remarkably well for 40 years. And these are the two root causes of the explosion of information technology in the past decades: Moore’s law and efficient algorithms.
It would appear that Moore’s law provides a disincentive for developing polynomial al- gorithms. After all, if an algorithm is exponential, why not wait it out until Moore’s law makes it feasible? But in reality the exact opposite happens: Moore’s law is a huge incen- tive for developing efficient algorithms, because such algorithms are needed in order to take advantage of the exponential increase in computer speed.
Here is why. If, for example, an O(2n) algorithm for Boolean satisfiability (SAT) were given an hour to run, it would have solved instances with 25 variables back in 1975, 31 vari- ables on the faster computers available in 1985, 38 variables in 1995, and about 45 variables with today’s machines. Quite a bit of progress—except that each extra variable requires a year and a half’s wait, while the appetite of applications (many of which are, ironically, re- lated to computer design) grows much faster. In contrast, the size of the instances solved by an O(n) or O(n log n) algorithm would be multiplied by a factor of about 100 each decade. In the case of an O(n2) algorithm, the instance size solvable in a fixed time would be mul- tiplied by about 10 each decade. Even an O(n6) algorithm, polynomial yet unappetizing, would more than double the size of the instances solved each decade. When it comes to the growth of the size of problems we can attack with an algorithm, we have a reversal: expo- nential algorithms make polynomially slow progress, while polynomial algorithms advance exponentially fast! For Moore’s law to be reflected in the world we need efficient algorithms.
As Sissa and Malthus knew very well, exponential expansion cannot be sustained in- definitely in our finite world. Bacterial colonies run out of food; chips hit the atomic scale. Moore’s law will stop doubling the speed of our computers within a decade or two. And then progress will depend on algorithmic ingenuity—or otherwise perhaps on novel ideas such as quantum computation, explored in Chapter 10.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 249 Satisfiability
SATISFIABILITY, or SAT (recall Exercise 3.28 and Section 5.3), is a problem of great practical importance, with applications ranging from chip testing and computer design to image analy- sis and software engineering. It is also a canonical hard problem. Here’s what an instance of SAT looks like:
(x ∨ y ∨ z) (x ∨ y) (y ∨ z) (z ∨ x) (x ∨ y ∨ z).
This is a Boolean formula in conjunctive normal form (CNF). It is a collection of clauses (the parentheses), each consisting of the disjunction (logical or, denoted ∨) of several literals, where a literal is either a Boolean variable (such as x) or the negation of one (such as x). A satisfying truth assignment is an assignment of false or true to each variable so that every clause contains a literal whose value is true. The SAT problem is the following: given a Boolean formula in conjunctive normal form, either find a satisfying truth assignment or else report that none exists.
In the instance shown previously, setting all variables to true, for example, satisfies every clause except the last. Is there a truth assignment that satisfies all clauses?
With a little thought, it is not hard to argue that in this particular case no such truth assignment exists. (Hint: The three middle clauses constrain all three variables to have the same value.) But how do we decide this in general? Of course, we can always search through all truth assignments, one by one, but for formulas with n variables, the number of possible assignments is exponential, 2n.
SAT is a typical search problem. We are given an instance I (that is, some input data specifying the problem at hand, in this case a Boolean formula in conjunctive normal form), and we are asked to find a solution S (an object that meets a particular specification, in this case an assignment that satisfies each clause). If no such solution exists, we must say so.
More specifically, a search problem must have the property that any proposed solution S to an instance I can be quickly checked for correctness. What does this entail? For one thing, S must at least be concise (quick to read), with length polynomially bounded by that of I. This is clearly true in the case of SAT, for which S is an assignment to the variables. To formalize the notion of quick checking, we will say that there is a polynomial-time algorithm that takes as input I and S and decides whether or not S is a solution of I. For SAT, this is easy as it just involves checking whether the assignment specified by S indeed satisfies every clause in I.
Later in this chapter it will be useful to shift our vantage point and to think of this efficient algorithm for checking proposed solutions as defining the search problem. Thus:
A search problem is specified by an algorithm C that takes two inputs, an instance I and a proposed solution S, and runs in time polynomial in |I|. We say S is a solution to I if and only if C(I, S) = true.
Given the importance of the SAT search problem, researchers over the past 50 years have tried hard to find efficient ways to solve it, but without success. The fastest algorithms we have are still exponential on their worst-case inputs.
Yet, interestingly, there are two natural variants of SAT for which we do have good algo- rithms. If all clauses contain at most one positive literal, then the Boolean formula is called

250 Algorithms Figure 8.1 The optimal traveling salesman tour, shown in bold, has length 18.
23
4
4
3
1
5
3
3
6
2
a Horn formula, and a satisfying truth assignment, if one exists, can be found by the greedy algorithm of Section 5.3. Alternatively, if all clauses have only two literals, then graph the- ory comes into play, and SAT can be solved in linear time by finding the strongly connected components of a particular graph constructed from the instance (recall Exercise 3.28). In fact, in Chapter 9, we’ll see a different polynomial algorithm for this same special case, which is called 2SAT.
On the other hand, if we are just a little more permissive and allow clauses to contain three literals, then the resulting problem, known as 3SAT (an example of which we saw earlier), once again becomes hard to solve!
The traveling salesman problem
In the traveling salesman problem (TSP) we are given n vertices 1, . . . , n and all n(n − 1)/2 distances between them, as well as a budget b. We are asked to find a tour, a cycle that passes through every vertex exactly once, of total cost b or less—or to report that no such tour exists. That is, we seek a permutation τ (1), . . . , τ (n) of the vertices such that when they are toured in this order, the total distance covered is at most b:
dτ(1),τ(2) +dτ(2),τ(3) +···+dτ(n),τ(1) ≤ b.
See Figure 8.1 for an example (only some of the distances are shown; assume the rest are very large).
Notice how we have defined the TSP as a search problem: given an instance, find a tour within the budget (or report that none exists). But why are we expressing the traveling salesman problem in this way, when in reality it is an optimization problem, in which the shortest possible tour is sought? Why dress it up as something else?
For a good reason. Our plan in this chapter is to compare and relate problems. The framework of search problems is helpful in this regard, because it encompasses optimization problems like the TSP in addition to true search problems like SAT.
Turning an optimization problem into a search problem does not change its difficulty at all, because the two versions reduce to one another. Any algorithm that solves the optimization

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 251 TSP also readily solves the search problem: find the optimum tour and if it is within budget,
return it; if not, there is no solution.
Conversely, an algorithm for the search problem can also be used to solve the optimization problem. To see why, first suppose that we somehow knew the cost of the optimum tour; then we could find this tour by calling the algorithm for the search problem, using the optimum cost as the budget. Fine, but how do we find the optimum cost? Easy: By binary search! (See Exercise 8.1.)
Incidentally, there is a subtlety here: Why do we have to introduce a budget? Isn’t any optimization problem also a search problem in the sense that we are searching for a solution that has the property of being optimal? The catch is that the solution to a search problem should be easy to recognize, or as we put it earlier, polynomial-time checkable. Given a po- tential solution to the TSP, it is easy to check the properties “is a tour” (just check that each vertex is visited exactly once) and “has total length ≤ b.” But how could one check the property “is optimal”?
As with SAT, there are no known polynomial-time algorithms for the TSP, despite much effort by researchers over nearly a century. Of course, there is an exponential algorithm for solving it, by trying all (n − 1)! tours, and in Section 6.6 we saw a faster, yet still exponential, dynamic programming algorithm.
The minimum spanning tree (MST) problem, for which we do have efficient algorithms, provides a stark contrast here. To phrase it as a search problem, we are ag􏰐ain given a distance matrix and a bound b, and are asked to find a tree T with total weight (i,j)∈T dij ≤ b. The TSP can be thought of as a tough cousin of the MST problem, in which the tree is not allowed to branch and is therefore a path.1 This extra restriction on the structure of the tree results in a much harder problem.
Euler and Rudrata
In the summer of 1735 Leonhard Euler (pronounced “Oiler”), the famous Swiss mathemati- cian, was walking the bridges of the East Prussian town of Ko ̈nigsberg. After a while, he noticed in frustration that, no matter where he started his walk, no matter how cleverly he continued, it was impossible to cross each bridge exactly once. And from this silly ambition, the field of graph theory was born.
Euler identified at once the roots of the park’s deficiency. First, you turn the map of the park into a graph whose vertices are the four land masses (two islands, two banks) and whose edges are the seven bridges:
1Actually the TSP demands a cycle, but one can define an alternative version that seeks a path, and it is not hard to see that this is just as hard as the TSP itself.

252
Algorithms
Small island
Big island
Northern bank
Southern bank
This graph has multiple edges between two vertices—a feature we have not been allowing so far in this book, but one that is meaningful for this particular problem, since each bridge must be accounted for separately. We are looking for a path that goes through each edge exactly once (the path is allowed to repeat vertices). In other words, we are asking this question: When can a graph be drawn without lifting the pencil from the paper?
The answer discovered by Euler is simple, elegant, and intuitive: If and only if (a) the graph is connected and (b) every vertex, with the possible exception of two vertices (the start and final vertices of the walk), has even degree (Exercise 3.26). This is why Ko ̈nigsberg’s park was impossible to traverse: all four vertices have odd degree.
To put it in terms of our present concerns, let us define a search problem called EULER PATH: Given a graph, find a path that contains each edge exactly once. It follows from Euler’s observation, and a little more thinking, that this search problem can be solved in polynomial time.
Almost a millennium before Euler’s fateful summer in East Prussia, a Kashmiri poet named Rudrata had asked this question: Can one visit all the squares of the chessboard, without repeating any square, in one long walk that ends at the starting square and at each step makes a legal knight move? This is again a graph problem: the graph now has 64 ver- tices, and two squares are joined by an edge if a knight can go from one to the other in a single move (that is, if their coordinates differ by 2 in one dimension and by 1 in the other). See Figure 8.2 for the portion of the graph corresponding to the upper left corner of the board. Can you find a knight’s tour on your chessboard?
This is a different kind of search problem in graphs: we want a cycle that goes through all vertices (as opposed to all edges in Euler’s problem), without repeating any vertex. And there is no reason to stick to chessboards; this question can be asked of any graph. Let us define the RUDRATA CYCLE search problem to be the following: given a graph, find a cycle that visits each vertex exactly once—or report that no such cycle exists.2 This problem is ominously reminiscent of the TSP, and indeed no polynomial algorithm is known for it.
There are two differences between the definitions of the Euler and Rudrata problems. The first is that Euler’s problem visits all edges while Rudrata’s visits all vertices. But there is
2In the literature this problem is known as the Hamilton cycle problem, after the great Irish mathematician who rediscovered it in the 19th century.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 253 Figure 8.2 Knight’s moves on a corner of a chessboard.
also the issue that one of them demands a path while the other requires a cycle. Which of these differences accounts for the huge disparity in computational complexity between the two problems? It must be the first, because the second difference can be shown to be purely cosmetic. Indeed, define the RUDRATA PATH problem to be just like RUDRATA CYCLE, except that the goal is now to find a path that goes through each vertex exactly once. As we will soon see, there is a precise equivalence between the two versions of the Rudrata problem.
Cuts and bisections
A cut is a set of edges whose removal leaves a graph disconnected. It is often of interest to find small cuts, and the MINIMUM CUT problem is, given a graph and a budget b, to find a cut with at most b edges. For example, the smallest cut in Figure 8.3 is of size 3. This problem can be solved in polynomial time by n − 1 max-flow computations: give each edge a capacity of 1, and find the maximum flow between some fixed node and every single other node. The smallest such flow will correspond (via the max-flow min-cut theorem) to the smallest cut. Can you see why? We’ve also seen a very different, randomized algorithm for this problem (page 150).
In many graphs, such as the one in Figure 8.3, the smallest cut leaves just a singleton vertex on one side—it consists of all edges adjacent to this vertex. Far more interesting are small cuts that partition the vertices of the graph into nearly equal-sized sets. More precisely, the BALANCED CUT problem is this: given a graph with n vertices and a budget b, partition the vertices into two sets S and T such that |S|,|T| ≥ n/3 and such that there are at most b edges between S and T . Another hard problem.
Balanced cuts arise in a variety of important applications, such as clustering. Consider for example the problem of segmenting an image into its constituent components (say, an elephant standing in a grassy plain with a clear blue sky above). A good way of doing this is to create a graph with a node for each pixel of the image and to put an edge between nodes whose corresponding pixels are spatially close together and are also similar in color. A single

254 Algorithms Figure 8.3 What is the smallest cut in this graph?
object in the image (like the elephant, say) then corresponds to a set of highly connected vertices in the graph. A balanced cut is therefore likely to divide the pixels into two clusters without breaking apart any of the primary constituents of the image. The first cut might, for instance, separate the elephant on the one hand from the sky and from grass on the other. A further cut would then be needed to separate the sky from the grass.
Integer linear programming
Even though the simplex algorithm is not polynomial time, we mentioned in Chapter 7 that there is a different, polynomial algorithm for linear programming. Therefore, linear pro- gramming is efficiently solvable both in practice and in theory. But the situation changes completely if, in addition to specifying a linear objective function and linear inequalities, we also constrain the solution (the values for the variables) to be integer. This latter problem is called INTEGER LINEAR PROGRAMMING (ILP). Let’s see how we might formulate it as a search problem. We are given a set of linear inequalities Ax ≤ b, where A is an m × n matrix and b is an m-vector; an objective function specified by an n-vector c; and finally, a goal g (the counterpart of a budget in maximization problems). We want to find a nonnegative integer n-vector x such that Ax ≤ b and c · x ≥ g.
But there is a redundancy here: the last constraint c · x ≥ g is itself a linear inequality and can be absorbed into Ax ≤ b. So, we define ILP to be following search problem: given A and b, find a nonnegative integer vector x satisfying the inequalities Ax ≤ b, or report that none exists. Despite the many crucial applications of this problem, and intense interest by researchers, no efficient algorithm is known for it.
There is a particularly clean special case of ILP that is very hard in and of itself: the goal is to find a vector x of 0’s and 1’s satisfying Ax = 1, where A is an m×n matrix with 0−1 entries and 1 is the m-vector of all 1’s. It should be apparent from the reductions in Section 7.1.4 that this is indeed a special case of ILP. We call it ZERO-ONE EQUATIONS (ZOE).
We have now introduced a number of important search problems, some of which are fa- miliar from earlier chapters and for which there are efficient algorithms, and others which are different in small but crucial ways that make them very hard computational problems. To

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 255 Figure 8.4 A more elaborate matchmaking scenario. Each triple is shown as a triangular-
shaped node joining boy, girl, and pet.
Chet Alice
Bob Beatrice
Al Carol
Armadillo Bobcat Canary
complete our story we will introduce a few more hard problems, which will play a role later in the chapter, when we relate the computational difficulty of all these problems. The reader is invited to skip ahead to Section 8.2 and then return to the definitions of these problems as required.
Three-dimensional matching
Recall the BIPARTITE MATCHING problem: given a bipartite graph with n nodes on each side (the boys and the girls), find a set of n disjoint edges, or decide that no such set exists. In Section 7.3, we saw how to efficiently solve this problem by a reduction to maximum flow. However, there is an interesting generalization, called 3D MATCHING, for which no polyno- mial algorithm is known. In this new setting, there are n boys and n girls, but also n pets, and the compatibilities among them are specified by a set of triples, each containing a boy, a girl,andapet. Intuitively,atriple(b,g,p)meansthatboyb,girlg,andpetpgetalongwell together. We want to find n disjoint triples and thereby create n harmonious households.
Can you spot a solution in Figure 8.4?
Independent set, vertex cover, and clique
In the INDEPENDENT SET problem (recall Section 6.7) we are given a graph and an integer g, and the aim is to find g vertices that are independent, that is, no two of which have an edge between them. Can you find an independent set of three vertices in Figure 8.5? How about four vertices? We saw in Section 6.7 that this problem can be solved efficiently on trees, but for general graphs no polynomial algorithm is known.
There are many other search problems about graphs. In VERTEX COVER, for example, the input is a graph and a budget b, and the idea is to find b vertices that cover (touch) every edge. Can you cover all edges of Figure 8.5 with seven vertices? With six? (And do you see the

256 Algorithms Figure 8.5 What is the size of the largest independent set in this graph?
intimate connection to the INDEPENDENT SET problem?)
VERTEX COVER is a special case of SET COVER, which we encountered in Chapter 5. In
that problem, we are given a set E and several subsets of it, S1, . . . , Sm, along with a budget b. We are asked to select b of these subsets so that their union is E. VERTEX COVER is the special case in which E consists of the edges of a graph, and there is a subset Si for each vertex, containing the edges adjacent to that vertex. Can you see why 3D MATCHING is also a special case of SET COVER?
And finally there is the CLIQUE problem: given a graph and a goal g, find a set of g ver- tices such that all possible edges between them are present. What is the largest clique in Figure 8.5?
Longest path
We know the shortest-path problem can be solved very efficiently, but how about the LONGEST PATH problem? Here we are given a graph G with nonnegative edge weights and two distin- guished vertices s and t, along with a goal g. We are asked to find a path from s to t with total weight at least g. Naturally, to avoid trivial solutions we require that the path be simple, containing no repeated vertices.
No efficient algorithm is known for this problem (which sometimes also goes by the name of TAXICAB RIP-OFF).
Knapsack and subset sum
Recall the KNAPSACK problem (Section 6.4): we are given integer weights w1,…,wn and integer values v1, . . . , vn for n items. We are also given a weight capacity W and a goal g (the former is present in the original optimization problem, the latter is added to make it a search problem). We seek a set of items whose total weight is at most W and whose total value is at least g. As always, if no such set exists, we should say so.
In Section 6.4, we developed a dynamic programming scheme for KNAPSACK with running

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 257
time O(nW ), which we noted is exponential in the input size, since it involves W rather than log W . And we have the usual exhaustive algorithm as well, which looks at all subsets of items—all 2n of them. Is there a polynomial algorithm for KNAPSACK? Nobody knows of one.
But suppose that we are interested in the variant of the knapsack problem in which the integers are coded in unary—for instance, by writing I I I I I I I I I I I I for 12. This is admittedly an exponentially wasteful way to represent integers, but it does define a legitimate problem, which we could call UNARY KNAPSACK. It follows from our discussion that this somewhat artificial problem does have a polynomial algorithm.
A different variation: suppose now that each item’s value is equal to its weight (all given in binary), and to top it off, the goal g is the same as the capacity W . (To adapt the silly break-in story whereby we first introduced the knapsack problem, the items are all gold nuggets, and the burglar wants to fill his knapsack to the hilt.) This special case is tantamount to finding a subset of a given set of integers that adds up to exactly W. Since it is a special case of KNAPSACK, it cannot be any harder. But could it be polynomial? As it turns out, this problem, called SUBSET SUM, is also very hard.
At this point one could ask: If SUBSET SUM is a special case that happens to be as hard as the general KNAPSACK problem, why are we interested in it? The reason is simplicity. In the complicated calculus of reductions between search problems that we shall develop in this chapter, conceptually simple problems like SUBSET SUM and 3SAT are invaluable.
8.2 NP-complete problems
Hard problems, easy problems
In short, the world is full of search problems, some of which can be solved efficiently, while others seem to be very hard. This is depicted in the following table.
Hard problems (NP-complete)
Easy problems (in P)
3SAT
2SAT, HORN SAT
TRAVELING SALESMAN PROBLEM
MINIMUM SPANNING TREE
LONGEST PATH
SHORTEST PATH
3D MATCHING
BIPARTITE MATCHING
KNAPSACK
UNARY KNAPSACK
INDEPENDENT SET
INDEPENDENT SET on trees
INTEGER LINEAR PROGRAMMING
LINEAR PROGRAMMING
RUDRATA PATH
EULER PATH
BALANCED CUT
MINIMUM CUT
This table is worth contemplating. On the right we have problems that can be solved efficiently. On the left, we have a bunch of hard nuts that have escaped efficient solution over many decades or centuries.

258 Algorithms
The various problems on the right can be solved by algorithms that are specialized and diverse: dynamic programming, network flow, graph search, greedy. These problems are easy for a variety of different reasons.
In stark contrast, the problems on the left are all difficult for the same reason! At their core, they are all the same problem, just in different disguises! They are all equivalent: as we shall see in Section 8.3, each of them can be reduced to any of the others—and back.
P and NP
It’s time to introduce some important concepts. We know what a search problem is: its defin- ing characteristic is that any proposed solution can be quickly checked for correctness, in the sense that there is an efficient checking algorithm C that takes as input the given instance I (the data specifying the problem to be solved), as well as the proposed solution S, and outputs true if and only if S really is a solution to instance I. Moreover the running time of C(I, S) is bounded by a polynomial in |I|, the length of the instance. We denote the class of all search problems by NP.
We’ve seen many examples of NP search problems that are solvable in polynomial time. In such cases, there is an algorithm that takes as input an instance I and has a running time polynomial in |I|. If I has a solution, the algorithm returns such a solution; and if I has no solution, the algorithm correctly reports so. The class of all search problems that can be solved in polynomial time is denoted P. Hence, all the search problems on the right-hand side of the table are in P.
Why P and NP?
Okay, P must stand for “polynomial.” But why use the initials NP (the common chatroom abbreviation for “no problem”) to describe the class of search problems, some of which are terribly hard?
NP stands for “nondeterministic polynomial time,” a term going back to the roots of complexity theory. Intuitively, it means that a solution to any search problem can be found and verified in polynomial time by a special (and quite unrealistic) sort of algorithm, called a nondeterministic algorithm. Such an algorithm has the power of guessing correctly at every step.
Incidentally, the original definition of NP (and its most common usage to this day) was not as a class of search problems, but as a class of decision problems: algorithmic questions that can be answered by yes or no. Example: “Is there a truth assignment that satisfies this Boolean formula?” But this too reflects a historical reality: At the time the theory of NP- completeness was being developed, researchers in the theory of computation were interested in formal languages, a domain in which such decision problems are of central importance.
Are there search problems that cannot be solved in polynomial time? In other words, is P ̸= NP? Most algorithms researchers think so. It is hard to believe that exponential search can always be avoided, that a simple trick will crack all these hard problems, famously unsolved for decades and centuries. And there is a good reason for mathematicians to believe

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 259
that P ̸= NP—the task of finding a proof for a given mathematical assertion is a search problem and is therefore in NP (after all, when a formal proof of a mathematical statement is written out in excruciating detail, it can be checked mechanically, line by line, by an efficient algorithm). So if P = NP, there would be an efficient method to prove any theorem, thus eliminating the need for mathematicians! All in all, there are a variety of reasons why it is widely believed that P ̸= NP. However, proving this has turned out to be extremely difficult, one of the deepest and most important unsolved puzzles of mathematics.
Reductions, again
Even if we accept that P ̸= NP, what about the specific problems on the left side of the table? On the basis of what evidence do we believe that these particular problems have no efficient algorithm (besides, of course, the historical fact that many clever mathematicians and computer scientists have tried hard and failed to find any)? Such evidence is provided by reductions, which translate one search problem into another. What they demonstrate is that the problems on the left side of the table are all, in some sense, exactly the same problem, except that they are stated in different languages. What’s more, we will also use reductions to show that these problems are the hardest search problems in NP—if even one of them has a polynomial time algorithm, then every problem in NP has a polynomial time algorithm. Thus if we believe that P ̸= NP, then all these search problems are hard.
We defined reductions in Chapter 7 and saw many examples of them. Let’s now specialize this definition to search problems. A reduction from search problem A to search problem B is a polynomial-time algorithm f that transforms any instance I of A into an instance f(I) of B, together with another polynomial-time algorithm h that maps any solution S of f(I) back into a solution h(S) of I; see the following diagram. If f(I) has no solution, then neither does I. These two translation procedures f and h imply that any algorithm for B can be converted into an algorithm for A by bracketing it between f and h.
Algorithm for A
Instance
I
Solution h(S) of I
No solution to I
Instance f(I)
Solution S of f(I)
Algorithm for B
h
f
No solution to f(I)
And now we can finally define the class of the hardest search problems.
A search problem is NP-complete if all other search problems reduce to it.
This is a very strong requirement indeed. For a problem to be NP-complete, it must be useful in solving every search problem in the world! It is remarkable that such problems exist. But they do, and the first column of the table we saw earlier is filled with the most famous examples. In Section 8.3 we shall see how all these problems reduce to one another, and also why all other search problems reduce to them.

260 Algorithms Figure 8.6 The space NP of all search problems, assuming P ̸= NP.
NP− complete
Increasing difficulty
The two ways to use reductions
So far in this book the purpose of a reduction from a problem A to a problem B has been straightforward and honorable: We know how to solve B efficiently, and we want to use this knowledge to solve A. In this chapter, however, reductions from A to B serve a somewhat perverse goal: we know A is hard, and we use the reduction to prove that B is hard as well!
If we denote a reduction from A to B by
A −→ B
then we can say that difficulty flows in the direction of the arrow, while efficient algorithms move in the opposite direction. It is through this propagation of difficulty that we know NP-complete problems are hard: all other search problems reduce to them, and thus each NP-complete problem contains the complexity of all search problems. If even one NP-complete problem is in P, then P = NP.
Reductions also have the convenient property that they compose. If A −→ B and B −→ C, then A −→ C .
To see this, observe first of all that any reduction is completely specified by the pre- and postprocessing functions f and h (see the reduction diagram). If (fAB,hAB) and (fBC,hBC) define the reductions from A to B and from B to C, respectively, then a reduction from A to C is given by compositions of these functions: fBC ◦fAB maps an instance of A to an instance of C and hAB ◦ hBC sends a solution of C back to a solution of A.
This means that once we know a problem A is NP-complete, we can use it to prove that a new search problem B is also NP-complete, simply by reducing A to B. Such a reduction establishes that all problems in NP reduce to B, via A.
P

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 261 Factoring
One last point: we started off this book by introducing another famously hard search problem: FACTORING, the task of finding all prime factors of a given integer. But the difficulty of FACTORING is of a different nature than that of the other hard search problems we have just seen. For example, nobody believes that FACTORING is NP-complete. One major difference is that, in the case of FACTORING, the definition does not contain the now familiar clause “or report that none exists.” A number can always be factored into primes.
Another difference (possibly not completely unrelated) is this: as we shall see in Chap- ter 10, FACTORING succumbs to the power of quantum computation—while SAT, TSP and the other NP-complete problems do not seem to.

262
Figure 8.7 Reductions between search problems.
All of NP
SAT
3SAT
INDEPENDENT SET
VERTEX COVER CLIQUE
Algorithms
3D MATCHING
ZOE
SUBSET SUM ILP RUDRATA CYCLE
TSP
8.3 The reductions
We shall now see that the search problems of Section 8.1 can be reduced to one another as depicted in Figure 8.7. As a consequence, they are all NP-complete.
Before we tackle the specific reductions in the tree, let’s warm up by relating two versions of the Rudrata problem.
RUDRATA (s,t)-PATH−→RUDRATA CYCLE
Recall the RUDRATA CYCLE problem: given a graph, is there a cycle that passes through each vertex exactly once? We can also formulate the closely related RUDRATA (s, t)-PATH problem, in which two vertices s and t are specified, and we want a path starting at s and ending at t that goes through each vertex exactly once. Is it possible that RUDRATA CYCLE is easier than RUDRATA (s, t)-PATH? We will show by a reduction that the answer is no.
The reduction maps an instance (G = (V, E), s, t) of RUDRATA (s, t)-PATH into an instance G′ = (V ′, E′) of RUDRATA CYCLE as follows: G′ is simply G with an additional vertex x and two new edges {s, x} and {x, t}. For instance:
G G′ ss
t
xt

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 263 SoV′ =V ∪{x},andE′ =E∪{{s,x},{x,t}}. HowdowerecoveraRudrata(s,t)-pathinG
given any Rudrata cycle in G′? Easy, we just delete the edges {s, x} and {x, t} from the cycle. RUDRATA (s,t)-PATH
Instance:
G = (V, E) nodes s, t
Solution: path
No solution
Add node x and edges
{s, x}, {x, t}
G′ =(V′,E′)
RUDRATA CYCLE
Solution: cycle
Delete edges
{s, x}, {x, t}
No solution
To confirm the validity of this reduction, we have to show that it works in the case of either outcome depicted.
1. When the instance of RUDRATA CYCLE has a solution.
Since the new vertex x has only two neighbors, s and t, any Rudrata cycle in G′ must consec- utively traverse the edges {t, x} and {x, s}. The rest of the cycle then traverses every other vertex en route from s to t. Thus deleting the two edges {t, x} and {x, s} from the Rudrata cycle gives a Rudrata path from s to t in the original graph G.
2. When the instance of RUDRATA CYCLE does not have a solution.
In this case we must show that the original instance of RUDRATA (s,t)-PATH cannot have a solution either. It is usually easier to prove the contrapositive, that is, to show that if there is a Rudrata (s, t)-path in G, then there is also a Rudrata cycle in G′. But this is easy: just add the two edges {t, x} and {x, s} to the Rudrata path to close the cycle.
One last detail, crucial but typically easy to check, is that the pre- and postprocessing functions take time polynomial in the size of the instance (G, s, t).
It is also possible to go in the other direction and reduce RUDRATA CYCLE to RUDRATA (s,t)-PATH. Together, these reductions demonstrate that the two Rudrata variants are in essence the same problem—which is not too surprising, given that their descriptions are al- most the same. But most of the other reductions we will see are between pairs of problems that, on the face of it, look quite different. To show that they are essentially the same, our reductions will have to cleverly translate between them.
3SAT−→INDEPENDENT SET
One can hardly think of two more different problems. In 3SAT the input is a set of clauses, each with three or fewer literals, for example
x ∨ y ∨ z) (x ∨ y ∨ z) (x ∨ y ∨ z) (x ∨ y),
and the aim is to find a satisfying truth assignment. In INDEPENDENT SET the input is a graph and a number g, and the problem is to find a set of g pairwise non-adjacent vertices. We must somehow relate Boolean logic with graphs!
(

264 Algorithms Figure 8.8 The graph corresponding to (x ∨ y ∨ z) (x ∨ y ∨ z) (x ∨ y ∨ z) (x ∨ y).
yyyy xzxzxzx
Let us think. To form a satisfying truth assignment we must pick one literal from each clause and give it the value true. But our choices must be consistent: if we choose x in one clause, we cannot choose x in another. Any consistent choice of literals, one from each clause, specifies a truth assignment (variables for which neither literal has been chosen can take on either value).
So, let us represent a clause, say (x ∨ y ∨ z), by a triangle, with vertices labeled x, y, z. Why triangle? Because a triangle has its three vertices maximally connected, and thus forces us to pick only one of them for the independent set. Repeat this construction for all clauses—a clause with two literals will be represented simply by an edge joining the literals. (A clause with one literal is silly and can be removed in a preprocessing step, since the value of the variable is determined.) In the resulting graph, an independent set has to pick at most one literal from each group (clause). To force exactly one choice from each clause, take the goal g to be the number of clauses; in our example, g = 4.
All that is missing now is a way to prevent us from choosing opposite literals (that is, both x and x) in different clauses. But this is easy: put an edge between any two vertices that correspond to opposite literals. The resulting graph for our example is shown in Figure 8.8.
Let’s recap the construction. Given an instance I of 3SAT, we create an instance (G, g) of INDEPENDENT SET as follows.
• Graph G has a triangle for each clause (or just an edge, if the clause has two literals), with vertices labeled by the clause’s literals, and has additional edges between any two vertices that represent opposite literals.
• The goal g is set to the number of clauses.
Clearly, this construction takes polynomial time. However, recall that for a reduction we do not just need an efficient way to map instances of the first problem to instances of the second (the function f in the diagram on page 259), but also a way to reconstruct a solution to the first instance from any solution of the second (the function h). As always, there are two things to show.
1. Given an independent set S of g vertices in G, it is possible to efficiently recover a satis- fying truth assignment to I.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 265
For any variable x, the set S cannot contain vertices labeled both x and x, because any such pair of vertices is connected by an edge. So assign x a value of true if S contains a vertex labeled x, and a value of false if S contains a vertex labeled x (if S contains neither, then assign either value to x). Since S has g vertices, it must have one vertex per clause; this truth assignment satisfies those particular literals, and thus satisfies all clauses.
2. If graph G has no independent set of size g, then the Boolean formula I is unsatisfiable.
It is usually cleaner to prove the contrapositive, that if I has a satisfying assignment then G has an independent set of size g. This is easy: for each clause, pick any literal whose value under the satisfying assignment is true (there must be at least one such literal), and add the corresponding vertex to S. Do you see why set S must be independent?
SAT−→3SAT
This is an interesting and common kind of reduction, from a problem to a special case of itself. We want to show that the problem remains hard even if its inputs are restricted somehow—in the present case, even if all clauses are restricted to have ≤ 3 literals. Such reductions modify the given instance so as to get rid of the forbidden feature (clauses with ≥ 4 literals) while keeping the instance essentially the same, in that we can read off a solution to the original instance from any solution of the modified one.
Here’s the trick for reducing SAT to 3SAT: given an instance I of SAT, use exactly the same instance for 3SAT, except that any clause with more than three literals, (a1 ∨ a2 ∨ ··· ∨ ak) (where the ai’s are literals and k > 3), is replaced by a set of clauses,
(a1 ∨a2 ∨y1) (y1 ∨a3 ∨y2) (y2 ∨a4 ∨y3) ··· (yk−3 ∨ak−1 ∨ak),
where the yi’s are new variables. Call the resulting 3SAT instance I′. The conversion from I to I′ is clearly polynomial time.
Why does this reduction work? I′ is equivalent to I in terms of satisfiability, because for any assignment to the ai’s,
􏰌 (a1 ∨a2 ∨···∨ak) 􏰍  thereisasettingoftheyi’sforwhich  issatisfied ⇐⇒ (a1 ∨a2 ∨y1)(y1 ∨a3 ∨y2) ··· (yk−3 ∨ak−1 ∨ak) 
are all satisfied
To see this, first suppose that the clauses on the right are all satisfied. Then at least one of the literals a1, . . . , ak must be true—otherwise y1 would have to be true, which would in turn force y2 to be true, and so on, eventually falsifying the last clause. But this means (a1 ∨a2 ∨···∨ak)isalsosatisfied.
Conversely,if(a1 ∨a2 ∨···∨ak)issatisfied,thensomeai mustbetrue. Sety1,…,yi−2 to true and the rest to false. This ensures that the clauses on the right are all satisfied.
Thus, any instance of SAT can be transformed into an equivalent instance of 3SAT. In fact, 3SAT remains hard even under the further restriction that no variable appears in more than

266 Algorithms Figure 8.9 S is a vertex cover if and only if V − S is an independent set.
S
three clauses. To show this, we must somehow get rid of any variable that appears too many times.
Here’s the reduction from 3SAT to its constrained version. Suppose that in the 3SAT in- stance, variable x appears in k > 3 clauses. Then replace its first appearance by x1, its second appearance by x2, and so on, replacing each of its k appearances by a different new variable. Finally, add the clauses
x1 ∨x2)(x2 ∨x3) ··· (xk ∨x1). (
And repeat for every variable that appears more than three times.
It is easy to see that in the new formula no variable appears more than three times
(and in fact, no literal appears more than twice). Furthermore, the extra clauses involv- ing x1, x2, . . . , xk constrain these variables to have the same value; do you see why? Hence the original instance of 3SAT is satisfiable if and only if the constrained instance is satisfiable.
INDEPENDENT SET−→VERTEX COVER
Some reductions rely on ingenuity to relate two very different problems. Others simply record the fact that one problem is a thin disguise of another. To reduce INDEPENDENT SET to VERTEX COVER we just need to notice that a set of nodes S is a vertex cover of graph G = (V, E) (that is, S touches every edge in E) if and only if the remaining nodes, V − S, are an independent set of G (Figure 8.9).
Therefore, to solve an instance (G, g) of INDEPENDENT SET, simply look for a vertex cover of G with |V | − g nodes. If such a vertex cover exists, then take all nodes not in it. If no such vertex cover exists, then G cannot possibly have an independent set of size g.
INDEPENDENT SET−→CLIQUE
INDEPENDENT SET and CLIQUE are also easy to reduce to one another. Define the complement of a graph G = (V,E) to be G = (V,E), where E contains precisely those unordered pairs of vertices that are not in E. Then a set of nodes S is an independent set of G if and only if S is a clique of G. To paraphrase, these nodes have no edges between them in G if and only if they have all possible edges between them in G.
Therefore, we can reduce INDEPENDENT SET to CLIQUE by mapping an instance (G,g)

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 267 of INDEPENDENT SET to the corresponding instance (G, g) of CLIQUE; the solution to both is
identical.
3SAT−→3D MATCHING
Again, two very different problems. We must reduce 3SAT to the problem of finding, among a set of boy-girl-pet triples, a subset that contains each boy, each girl, and each pet exactly once. In short, we must design sets of boy-girl-pet triples that somehow behave like Boolean variables and gates!
Consider the following set of four triples, each represented by a triangular node joining a boy, girl, and pet:
p0
p1
b0 g0
g1 b1
p3
p2
Suppose that the two boys b0 and b1 and the two girls g0 and g1 are not involved in any other triples. (The four pets p0, . . . , p3 will of course belong to other triples as well; for otherwise the instance would trivially have no solution.) Then any matching must contain either the two triples (b0, g1, p0), (b1, g0, p2) or the two triples (b0, g0, p1), (b1, g1, p3), because these are the only ways in which these two boys and girls can find any match. Therefore, this “gadget” has two possible states: it behaves like a Boolean variable!
To then transform an instance of 3SAT to one of 3D MATCHING, we start by creating a copy of the preceding gadget for each variable x. Call the resulting nodes px1,bx0,gx1, and so on. The intended interpretation is that boy bx0 is matched with girl gx1 if x = true, and with girl gx0 if x = false.
Next we must create triples that somehow mimic clauses. For each clause, say c = (x∨y∨z), introduce a new boy bc and a new girl gc. They will be involved in three triples, one for each literal in the clause. And the pets in these triples must reflect the three ways whereby the clause can be satisfied: (1) x = true, (2) y = false, (3) z = true. For (1), we have the triple (bc, gc, px1), where px1 is the pet p1 in the gadget for x. Here is why we chose p1: if x = true, then bx0 is matched with gx1 and bx1 with gx0, and so pets px0 and px2 are taken. In which case bc and gc can be matched with px1. But if x = false, then px1 and px3 are taken, and so gc and bc cannot be accommodated this way. We do the same thing for the other two literals of the

268 Algorithms
clause, which yield triples involving bc and gc with either py0 or py2 (for the negated variable y) and with either pz1 or pz3 (for variable z).
We have to make sure that for every occurrence of a literal in a clause c there is a different pet to match with bc and gc. But this is easy: by an earlier reduction we can assume that no literal appears more than twice, and so each variable gadget has enough pets, two for negated occurrences and two for unnegated.
The reduction now seems complete: from any matching we can recover a satisfying truth assignment by simply looking at each variable gadget and seeing with which girl bx0 was matched. And from any satisfying truth assignment we can match the gadget corresponding to each variable x so that triples (bx0, gx1, px0) and (bx1, gx0, px2) are chosen if x = true and triples (bx0, gx0, px1) and (bx1, gx1, px3) are chosen if x = false; and for each clause c match bc and gc with the pet that corresponds to one of its satisfying literals.
But one last problem remains: in the matching defined at the end of the last paragraph, some pets may be left unmatched. In fact, if there are n variables and m clauses, then exactly 2n − m pets will be left unmatched (you can check that this number is sure to be positive, because we have at most three occurrences of every variable, and at least two literals in every clause). But this is easy to fix: Add 2n − m new boy-girl couples that are “generic animal- lovers,” and match them by triples with all the pets!
3D MATCHING−→ZOE
Recall that in ZOE we are given an m×n matrix A with 0−1 entries, and we must find a 0−1
vector x = (x1, . . . , xn) such that the m equations Ax = 1
are satisfied, where by 1 we denote the column vector of all 1’s. How can we express the 3D MATCHING problem in this framework?
ZOE and ILP are very useful problems precisely because they provide a format in which many combinatorial problems can be expressed. In such a formulation we think of the 0 − 1 variables as describing a solution, and we write equations expressing the constraints of the problem.
For example, here is how we express an instance of 3D MATCHING (m boys, m girls, m pets, and n boy-girl-pet triples) in the language of ZOE. We have 0 − 1 variables x1, . . . , xn, one per triple, where xi = 1 means that the ith triple is chosen for the matching, and xi = 0 means that it is not chosen.
Now all we have to do is write equations stating that the solution described by the xi’s is a legitimate matching. For each boy (or girl, or pet), suppose that the triples containing him (or her, or it) are those numbered j1, j2, . . . , jk; the appropriate equation is then
xj1 +xj2 +···+xjk =1,
which states that exactly one of these triples must be included in the matching. For example, here is the A matrix for an instance of 3D MATCHING we saw earlier.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani
269
Chet Alice
1 0 0 0 0 0 0 0 1 1 0 1 1 0 0 1 0 0 0 1
A=0 1 0 0 0 0 0 1 1 0 1 0 0 0 1
00110 01000
Bob Al
Beatrice Carol
Armadillo Bobcat Canary
The five columns of A correspond to the five triples, while the nine rows are for Al, Bob, Chet, Alice, Beatrice, Carol, Armadillo, Bobcat, and Canary, respectively.
It is straightforward to argue that solutions to the two instances translate back and forth.
ZOE−→SUBSET SUM
This is a reduction between two special cases of ILP: one with many equations but only 0 − 1 coefficients, and the other with a single equation but arbitrary integer coefficients. The reduction is based on a simple and time-honored idea: 0 − 1 vectors can encode numbers!
For example, given this instance of ZOE:
1 0 0 0
0 0 0 1 A =   0 1 1 0   ,
1000 0100
we are looking for a set of columns of A that, added together, make up the all-1’s vector. But if we think of the columns as binary integers (read from top to bottom), we are looking for a subset of the integers 18, 5, 4, 8 that add up to the binary integer 111112 = 31. And this is an instance of SUBSET SUM. The reduction is complete!
Except for one detail, the one that usually spoils the close connection between 0 − 1 vec- tors and binary integers: carry. Because of carry, 5-bit binary integers can add up to 31 (for example, 5 + 6 + 20 = 31 or, in binary, 001012 + 001102 + 101002 = 111112) even when the sum of the corresponding vectors is not (1, 1, 1, 1, 1). But this is easy to fix: Think of the column vectors not as integers in base 2, but as integers in base n + 1—one more than the number of columns. This way, since at most n integers are added, and all their digits are 0 and 1, there can be no carry, and our reduction works.
ZOE−→ILP
3SAT is a special case of SAT—or, SAT is a generalization of 3SAT. By special case we mean that the instances of 3SAT are a subset of the instances of SAT (in particular, the ones with no long clauses), and the definition of solution is the same in both problems (an assignment

270 Algorithms Figure 8.10 Rudrata cycle with paired edges: C = {(e1, e3), (e5, e6), (e4, e5), (e3, e7), (e3, e8)}.
e1 e2
e3 e8e7 e5
e4 e6
satisfying all clauses). Consequently, there is a reduction from 3SAT to SAT, in which the input undergoes no transformation, and the solution to the target instance is also kept unchanged. In other words, functions f and h from the reduction diagram (on page 259) are both the identity.
This sounds trivial enough, but it is a very useful and common way of establishing that a problem is NP-complete: Simply notice that it is a generalization of a known NP-complete problem. For example, the SET COVER problem is NP-complete because it is a generaliza- tion of VERTEX COVER (and also, incidentally, of 3D MATCHING). See Exercise 8.10 for more examples.
Often it takes a little work to establish that one problem is a special case of another. The reduction from ZOE to ILP is a case in point. In ILP we are looking for an integer vector x that satisfies Ax ≤ b, for given matrix A and vector b. To write an instance of ZOE in this precise form, we need to rewrite each equation of the ZOE instance as two inequalities (recall the transformations of Section 7.1.4), and to add for each variable xi the inequalities xi ≤ 1 and −xi ≤ 0.
ZOE−→RUDRATA CYCLE
In the RUDRATA CYCLE problem we seek a cycle in a graph that visits every vertex exactly once. We shall prove it NP-complete in two stages: first we will reduce ZOE to a generalization of RUDRATA CYCLE, called RUDRATA CYCLE WITH PAIRED EDGES, and then we shall see how to get rid of the extra features of that problem and reduce it to the plain RUDRATA CYCLE problem.
In an instance of RUDRATA CYCLE WITH PAIRED EDGES we are given a graph G = (V, E) and a set C ⊆ E × E of pairs of edges. We seek a cycle that (1) visits all vertices once, like a Rudrata cycle should, and (2) for every pair of edges (e,e′) in C, traverses either edge e or edge e′—exactly one of them. In the simple example of Figure 8.10 a solution is shown in bold. Notice that we allow two or more parallel edges between two nodes—a feature that doesn’t

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 271 Figure 8.11 Reducing ZOE to RUDRATA CYCLE WITH PAIRED EDGES.
equations
variables
make sense in most graph problems—since now the different copies of an edge can be paired with other copies of edges in ways that do make a difference.
Now for the reduction of ZOE to RUDRATA CYCLE WITH PAIRED EDGES. Given an instance of ZOE, Ax = 1 (where A is an m×n matrix with 0−1 entries, and thus describes m equations in n variables), the graph we construct has the very simple structure shown in Figure 8.11: a cycle that connects m+n collections of parallel edges. For each variable xi we have two parallel edges (corresponding to xi = 1 and xi = 0). And for each equation xj1 + · · · + xjk = 1 involving k variables we have k parallel edges, one for every variable appearing in the equation. This is the whole graph. Evidently, any Rudrata cycle in this graph must traverse the m + n collections of parallel edges one by one, choosing one edge from each collection. This way, the cycle “chooses” for each variable a value—0 or 1—and, for each equation, a variable appearing in it.
The whole reduction can’t be this simple, of course. The structure of the matrix A (and not just its dimensions) must be reflected somewhere, and there is one place left: the set C of pairs of edges such that exactly one edge in each pair is traversed. For every equation (recall there are m in total), and for every variable xi appearing in it, we add to C the pair (e,e′) where e is the edge corresponding to the appearance of xi in that particular equation (on the left-hand side of Figure 8.11), and e′ is the edge corresponding to the variable assignment xi = 0 (on the right side of the figure). This completes the construction.
Take any solution of this instance of RUDRATA CYCLE WITH PAIRED EDGES. As discussed before, it picks a value for each variable and a variable for every equation. We claim that the values thus chosen are a solution to the original instance of ZOE. If a variable xi has value 1, then the edge xi = 0 is not traversed, and thus all edges associated with xi on the equation

272 Algorithms
side must be traversed (since they are paired in C with the xi = 0 edge). So, in each equation exactly one of the variables appearing in it has value 1—which is the same as saying that all equations are satisfied. The other direction is straightforward as well: from a solution to the instance of ZOE one easily obtains an appropriate Rudrata cycle.
Getting Rid of the Edge Pairs. So far we have a reduction from ZOE to RUDRATA CYCLE WITH PAIRED EDGES; but we are really interested in RUDRATA CYCLE, which is a special case of the problem with paired edges: the one in which the set of pairs C is empty. To accomplish our goal, we need, as usual, to find a way of getting rid of the unwanted feature—in this case the edge pairs.
Consider the graph shown in Figure 8.12, and suppose that it is a part of a larger graph G in such a way that only the four endpoints a, b, c, d touch the rest of the graph. We claim that this graph has the following important property: in any Rudrata cycle of G the subgraph shown must be traversed in one of the two ways shown in bold in Figure 8.12(b) and (c). Here is why. Suppose that the cycle first enters the subgraph from vertex a continuing to f. Then it must continue to vertex g, because g has degree 2 and so it must be visited immediately after one of its adjacent nodes is visited—otherwise there is no way to include it in the cycle. Hence we must go on to node h, and here we seem to have a choice. We could continue on to j, or return to c. But if we take the second option, how are we going to visit the rest of the subgraph? (A Rudrata cycle must leave no vertex unvisited.) It is easy to see that this would be impossible, and so from h we have no choice but to continue to j and from there to visit the rest of the graph as shown in Figure 8.12(b). By symmetry, if the Rudrata cycle enters this subgraph at c, it must traverse it as in Figure 8.12(c). And these are the only two ways.
But this property tells us something important: this gadget behaves just like two edges {a, b} and {c, d} that are paired up in the RUDRATA CYCLE WITH PAIRED EDGES problem (see Figure 8.12(d)).
The rest of the reduction is now clear: to reduce RUDRATA CYCLE WITH PAIRED EDGES to RUDRATA CYCLE we go through the pairs in C one by one. To get rid of each pair ({a, b}, {c, d}) we replace the two edges with the gadget in Figure 8.12(a). For any other pair in C that involves {a, b}, we replace the edge {a, b} with the new edge {a, f }, where f is from the gadget: the traversal of {a,f} is from now on an indication that edge {a,b} in the old graph would be traversed. Similarly, {c, h} replaces {c, d}. After |C | such replacements (performed in polynomial time, since each replacement adds only 12 vertices to the graph) we are done, and the Rudrata cycles in the resulting graph will be in one-to-one correspondence with the Rudrata cycles in the original graph that conform to the constraints in C.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 273 Figure 8.12 A gadget for enforcing paired behavior.
(a)
(b)
(c)
(d)
aflmsb gknr chjpqd
ab
cd ab
cd
ab cd
C = {({a, b}, {c, d})}

274 Algorithms RUDRATA CYCLE−→TSP
Given a graph G = (V, E), construct the following instance of the TSP: the set of cities is the same as V , and the distance between cities u and v is 1 if {u, v} is an edge of G and 1 + α otherwise, for some α > 1 to be determined. The budget of the TSP instance is equal to the number of nodes, |V |.
It is easy to see that if G has a Rudrata cycle, then the same cycle is also a tour within the budget of the TSP instance; and that conversely, if G has no Rudrata cycle, then there is no solution: the cheapest possible TSP tour has cost at least n + α (it must use at least one edge of length 1 + α, and the total length of all n − 1 others is at least n − 1). Thus RUDRATA CYCLE reduces to TSP.
In this reduction, we introduced the parameter α because by varying it, we can obtain two interesting results. If α = 1, then all distances are either 1 or 2, and so this instance of the TSP satisfies the triangle inequality: if i, j, k are cities, then dij + djk ≥ dik (proof: a + b ≥ c holds for any numbers 1 ≤ a, b, c ≤ 2). This is a special case of the TSP which is of practical importance and which, as we shall see in Chapter 9, is in a certain sense easier, because it can be efficiently approximated.
If on the other hand α is large, then the resulting instance of the TSP may not satisfy the triangle inequality, but has another important property: either it has a solution of cost n or less, or all its solutions have cost at least n + α (which now can be arbitrarily larger than n). There can be nothing in between! As we shall see in Chapter 9, this important gap property implies that, unless P = NP, no approximation algorithm is possible.
ANY PROBLEM IN NP−→SAT
We have reduced SAT to the various search problems in Figure 8.7. Now we come full circle and argue that all these problems—and in fact all problems in NP—reduce to SAT.
In particular, we shall show that all problems in NP can be reduced to a generalization of SAT which we call CIRCUIT SAT. In CIRCUIT SAT we are given a (Boolean) circuit (see Figure 8.13, and recall Section 7.7), a dag whose vertices are gates of five different types:
• AND gates and OR gates have indegree 2.
• NOT gates have indegree 1.
• Known input gates have no incoming edges and are labeled false or true. • Unknown input gates have no incoming edges and are labeled “?”.
One of the sinks of the dag is designated as the output gate.
Given an assignment of values to the unknown inputs, we can evaluate the gates of the
circuit in topological order, using the rules of Boolean logic (such as false ∨ true = true), until we obtain the value at the output gate. This is the value of the circuit for the particular assignment to the inputs. For instance, the circuit in Figure8.13 evaluates to false under the assignment true, false, true (from left to right).
CIRCUIT SAT is then the following search problem: Given a circuit, find a truth assignment for the unknown inputs such that the output gate evaluates to true, or report that no such

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani
275
Figure 8.13 An instance of CIRCUIT SAT. output
AND
NOT
AND
OR
OR
AND
Gate g g
ggg
OR AND NOT
true ? ? ?
assignment exists. For example, if presented with the circuit in Figure 8.13 we could have returned the assignment (false,true,true) because, if we substitute these values to the unknown inputs (from left to right), the output becomes true.
CIRCUIT SAT is a generalization of SAT. To see why, notice that SAT asks for a satisfying truth assignment for a circuit that has this simple structure: a bunch of AND gates at the top join the clauses, and the result of this big AND is the output. Each clause is the OR of its literals. And each literal is either an unknown input gate or the NOT of one. There are no known input gates.
Going in the other direction, CIRCUIT SAT can also be reduced to SAT. Here is how we can rewrite any circuit in conjunctive normal form (the AND of clauses): for each gate g in the circuit we create a variable g, and we model the effect of the gate using a few clauses:
true
false
(g) (
g) h1 h2 h1 h2 h
(
(g ∨ h2)
(g ∨ h1) g∨h1 ∨h2)
(
g ∨ h1)
g ∨ h2) (
(g∨h1 ∨h2)
(g ∨ h)
(
g ∨ h)
(Do you see that these clauses do, in fact, force exactly the desired effect?) And to finish up, if g is the output gate, we force it to be true by adding the clause (g). The resulting instance

276 Algorithms
of SAT is equivalent to the given instance of CIRCUIT SAT: the satisfying truth assignments of this conjunctive normal form are in one-to-one correspondence with those of the circuit.
Now that we know CIRCUIT SAT reduces to SAT, we turn to our main job, showing that all search problems reduce to CIRCUIT SAT. So, suppose that A is a problem in NP. We must discover a reduction from A to CIRCUIT SAT. This sounds very difficult, because we know almost nothing about A!
All we know about A is that it is a search problem, so we must put this knowledge to work. The main feature of a search problem is that any solution to it can quickly be checked: there is an algorithm C that checks, given an instance I and a proposed solution S, whether or not S is a solution of I. Moreover, C makes this decision in time polynomial in the length of I (we can assume that S is itself encoded as a binary string, and we know that the length of this string is polynomial in the length of I).
Recall now our argument in Section 7.7 that any polynomial algorithm can be rendered as a circuit, whose input gates encode the input to the algorithm. Naturally, for any input length (number of input bits) the circuit will be scaled to the appropriate number of inputs, but the total number of gates of the circuit will be polynomial in the number of inputs. If the polynomial algorithm in question solves a problem that requires a yes or no answer (as is the situation with C: “Does S encode a solution to the instance encoded by I?”), then this answer is given at the output gate.
We conclude that, given any instance I of problem A, we can construct in polynomial time a circuit whose known inputs are the bits of I, and whose unknown inputs are the bits of S, such that the output is true if and only if the unknown inputs spell a solution S of I . In other words, the satisfying truth assignments to the unknown inputs of the circuit are in one-to-one correspondence with the solutions of instance I of A. The reduction is complete.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 277 Unsolvable problems
At least an NP-complete problem can be solved by some algorithm—the trouble is that this algorithm will be exponential. But it turns out there are perfectly decent computational problems for which no algorithms exist at all!
One famous problem of this sort is an arithmetical version of SAT. Given a polynomial equation in many variables, perhaps
x3yz + 2y4z2 − 7xy5z = 6,
are there integer values of x,y,z that satisfy it? There is no algorithm that solves this problem. No algorithm at all, polynomial, exponential, doubly exponential, or worse! Such problems are called unsolvable.
The first unsolvable problem was discovered in 1936 by Alan M. Turing, then a student of mathematics at Cambridge, England. ] When Turing came up with it, there were no computers or programming languages (in fact, it can be argued that these things came about later exactly because this brilliant thought occurred to Turing). But today we can state it in familiar terms.
Suppose that you are given a program in your favorite programming language, along with a particular input. Will the program ever terminate, once started on this input? This is a very reasonable question. Many of us would be ecstatic if we had an algorithm, call it terminates(p,x), that took as input a file containing a program p, and a file of data x, and after grinding away, finally told us whether or not p would ever stop if started on x.
But how would you go about writing the program terminates? (If you haven’t seen this before, it’s worth thinking about it for a while, to appreciate the difficulty of writing such an “universal infinite-loop detector.”)
Well, you can’t. Such an algorithm does not exist!
And here is the proof: Suppose we actually had such a program terminates(p,x). Then we could use it as a subroutine of the following evil program:
function paradox(z:file)
1: if terminates(z,z) goto 1
Notice what paradox does: it terminates if and only if program z does not terminate when given its own code as input.
You should smell trouble. What if we put this program in a file named paradox and we executed paradox(paradox)? Would this execution ever stop? Or not? Neither answer is possible. Since we arrived at this contradiction by assuming that there is an algorithm for telling whether programs terminate, we must conclude that this problem cannot be solved by any algorithm.
By the way, all this tells us something important about programming: It will never be automated, it will forever depend on discipline, ingenuity, and hackery. We now know that you can’t tell whether a program has an infinite loop. But can you tell if it has a buffer overrun? Do you see how to use the unsolvability of the “halting problem” to show that this, too, is unsolvable?

278 Algorithms Exercises
8.1. Optimization versus search. Recall the traveling salesman problem:
TSP
Input: A matrix of distances; a budget b
Output: A tour which passes through all the cities and has length ≤ b, if such a tour exists.
The optimization version of this problem asks directly for the shortest tour.
TSP-OPT
Input: A matrix of distances
Output: The shortest tour which passes through all the cities.
Show that if TSP can be solved in polynomial time, then so can TSP-OPT.
8.2. Search versus decision. Suppose you have a procedure which runs in polynomial time and tells you whether or not a graph has a Rudrata path. Show that you can use it to develop a polynomial- time algorithm for RUDRATA PATH (which returns the actual path, if it exists).
8.3. STINGY SAT is the following problem: given a set of clauses (each a disjunction of literals) and an integer k, find a satisfying assignment in which at most k variables are true, if such an assignment exists. Prove that STINGY SAT is NP-complete.
8.4. Consider the CLIQUE problem restricted to graphs in which every vertex has degree at most 3. Call this problem CLIQUE-3.
(a) Prove that CLIQUE-3 is in NP.
(b) What is wrong with the following proof of NP-completeness for CLIQUE-3?
We know that the CLIQUE problem in general graphs is NP-complete, so it is enough to present a reduction from CLIQUE-3 to CLIQUE. Given a graph G with vertices of degree ≤ 3, and a parameter g, the reduction leaves the graph and the parameter unchanged: clearly the output of the reduction is a possible input for the CLIQUE problem. Furthermore, the answer to both problems is identical. This proves the correctness of the reduction and, therefore, the NP-completeness of CLIQUE-3.
(c) It is true that the VERTEX COVER problem remains NP-complete even when restricted to graphs in which every vertex has degree at most 3. Call this problem VC-3. What is wrong with the following proof of NP-completeness for CLIQUE-3?
We present a reduction from VC-3 to CLIQUE-3. Given a graph G = (V, E) with node degrees bounded by 3, and a parameter b, we create an instance of CLIQUE-3 by leaving the graph unchanged and switching the parameter to |V | − b. Now, a subset C ⊆ V is a vertex cover in G if and only if the complementary set V − C is a clique in G. Therefore G has a vertex cover of size ≤ b if and only if it has a clique of size ≥ |V | − b. This proves the correctness of the reduction and, consequently, the NP-completeness of CLIQUE-3.
(d) Describe an O(|V |4) algorithm for CLIQUE-3.
8.5. Give a simple reduction from 3D MATCHING to SAT, and another from RUDRATA CYCLE to SAT. (Hint: In the latter case you may use variables xij whose intuitive meaning is “vertex i is the jth vertex of the Hamilton cycle”; you then need to write clauses that express the constraints of the problem.)

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 279
8.6. On page 266 we saw that 3SAT remains NP-complete even when restricted to formulas in which
each literal appears at most twice.
(a) Show that if each literal appears at most once, then the problem is solvable in polynomial time.
(b) Show that INDEPENDENT SET remains NP-complete even in the special case when all the nodes in the graph have degree at most 4.
8.7. Consideraspecialcaseof3SATinwhichallclauseshaveexactlythreeliterals,andeachvariable appears at most three times. Show that this problem can be solved in polynomial time. (Hint: create a bipartite graph with clauses on the left, variables on the right, and edges whenever a variable appears in a clause. Use Exercise 7.30 to show that this graph has a matching.)
8.8. In the EXACT 4SAT problem, the input is a set of clauses, each of which is a disjunction of exactly four literals, and such that each variable occurs at most once in each clause. The goal is to find a satisfying assignment, if one exists. Prove that EXACT 4SAT is NP-complete.
8.9. In the HITTING SET problem, we are given a family of sets {S1, S2, . . . , Sn} and a budget b, and we wish to find a set H of size ≤ b which intersects every Si, if such an H exists. In other words, wewantH∩Si ̸=∅foralli.
Show that HITTING SET is NP-complete.
8.10. Proving NP-completeness by generalization. For each of the problems below, prove that it is NP- complete by showing that it is a generalization of some NP-complete problem we have seen in this chapter.
(a) SUBGRAPH ISOMORPHISM: Given as input two undirected graphs G and H, determine whether G is a subgraph of H (that is, whether by deleting certain vertices and edges of H we obtain a graph that is, up to renaming of vertices, identical to G), and if so, return the corresponding mapping of V (G) into V (H).
(b) LONGEST PATH: Given a graph G and an integer g, find in G a simple path of length g.
(c) MAX SAT: Given a CNF formula and an integer g, find a truth assignment that satisfies at
least g clauses.
(d) DENSE SUBGRAPH: Given a graph and two integers a and b, find a set of a vertices of G
such that there are at least b edges between them.
(e) SPARSE SUBGRAPH: Given a graph and two integers a and b, find a set of a vertices of G
such that there are at most b edges between them.
(f) SET COVER. (This problem generalizes two known NP-complete problems.)
(g) RELIABLE NETWORK: We are given two n × n matrices, a distance matrix dij and a connec- tivity requirement matrix rij , as well as a budget b; we must find a graph G = ({1, 2, . . . , n}, E) such that (1) the total cost of all edges is b or less and (2) between any two distinct vertices i and j there are rij vertex-disjoint paths. (Hint: Suppose that all dij ’s are 1 or 2, b = n, and all rij ’s are 2. Which well known NP-complete problem is this?)
8.11. TherearemanyvariantsofRudrata’sproblem,dependingonwhetherthegraphisundirectedor directed, and whether a cycle or path is sought. Reduce the DIRECTED RUDRATA PATH problem to each of the following.
(a) The (undirected) RUDRATA PATH problem.

280 Algorithms (b) The undirected RUDRATA (s,t)-PATH problem, which is just like RUDRATA PATH except
that the endpoints of the path are specified in the input.
8.12. The k-SPANNING TREE problem is the following.
Input: An undirected graph G = (V, E)
Output: A spanning tree of G in which each node has degree ≤ k, if such a tree exists.
Show that for any k ≥ 2:
(a) k-SPANNING TREE is a search problem.
(b) k-SPANNING TREE is NP-complete. (Hint: Start with k = 2 and consider the relation between this problem and RUDRATA PATH.)
8.13. Determine which of the following problems are NP-complete and which are solvable in polyno- mial time. In each problem you are given an undirected graph G = (V, E), along with:
(a) A set of nodes L ⊆ V , and you must find a spanning tree such that its set of leaves includes the set L.
(b) A set of nodes L ⊆ V , and you must find a spanning tree such that its set of leaves is precisely the set L.
(c) A set of nodes L ⊆ V , and you must find a spanning tree such that its set of leaves is included in the set L.
(d) An integer k, and you must find a spanning tree with k or fewer leaves.
(e) An integer k, and you must find a spanning tree with k or more leaves.
(f) An integer k, and you must find a spanning tree with exactly k leaves.
(Hint: All the NP-completeness proofs are by generalization, except for one.)
8.14. Prove that the following problem is NP-complete: given an undirected graph G = (V, E) and an
integer k, return a clique of size k as well as an independent set of size k, provided both exist.
8.15. Show that the following problem is NP-complete.
MAXIMUM COMMON SUBGRAPH
Input: Two graphs G1 = (V1, E1) and G2 = (V2, E2); a budget b.
Output: Two set of nodes V1′ ⊆ V1 and V2′ ⊆ V2 whose deletion leaves at least b nodes in each graph, and makes the two graphs identical.
8.16. We are feeling experimental and want to create a new dish. There are various ingredients we can choose from and we’d like to use as many of them as possible, but some ingredients don’t go well with others. If there are n possible ingredients (numbered 1 to n), we write down an n × n matrix giving the discord between any pair of ingredients. This discord is a real number between 0.0 and 1.0, where 0.0 means “they go together perfectly” and 1.0 means “they really don’t go together.” Here’s an example matrix when there are five possible ingredients.
12345
1 2 3 4 5
0.0 0.4 0.2 0.9 1.0 0.4 0.0 0.1 1.0 0.2 0.2 0.1 0.0 0.8 0.5 0.9 1.0 0.8 0.0 0.2 1.0 0.2 0.5 0.2 0.0

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 281
In this case, ingredients 2 and 3 go together pretty well whereas 1 and 5 clash badly. Notice that this matrix is necessarily symmetric; and that the diagonal entries are always 0.0. Any set of ingredients incurs a penalty which is the sum of all discord values between pairs of ingredients. For instance, the set of ingredients {1, 3, 5} incurs a penalty of 0.2 + 1.0 + 0.5 = 1.7. We want this penalty to be small.
EXPERIMENTAL CUISINE
Input: n, the number of ingredients to choose from; D, the n×n “discord” matrix; some
number p ≥ 0
Output: The maximum number of ingredients we can choose with penalty ≤ p.
Show that if EXPERIMENTAL CUISINE is solvable in polynomial time, then so is 3SAT.
8.17. ShowthatforanyproblemΠinNP,thereisanalgorithmwhichsolvesΠintimeO(2p(n)),where
n is the size of the input instance and p(n) is a polynomial (which may depend on Π).
8.18. Show that if P = NP then the RSA cryptosystem (Section 1.4.2) can be broken in polynomial
time.
8.19. A kite is a graph on an even number of vertices, say 2n, in which n of the vertices form a clique and the remaining n vertices are connected in a “tail” that consists of a path joined to one of the vertices of the clique. Given a graph and a goal g, the KITE problem asks for a subgraph which is a kite and which contains 2g nodes. Prove that KITE is NP-complete.
8.20. In an undirected graph G = (V, E), we say D ⊆ V is a dominating set if every v ∈ V is either in D or adjacent to at least one member of D. In the DOMINATING SET problem, the input is a graph and a budget b, and the aim is to find a dominating set in the graph of size at most b, if one exists. Prove that this problem is NP-complete.
8.21. Sequencing by hybridization. One experimental procedure for identifying a new DNA sequence repeatedly probes it to determine which k-mers (substrings of length k) it contains. Based on these, the full sequence must then be reconstructed.
Let’s now formulate this as a combinatorial problem. For any string x (the DNA sequence), let Γ(x) denote the multiset of all of its k-mers. In particular, Γ(x) contains exactly |x| − k + 1 elements.
The reconstruction problem is now easy to state: given a multiset of k-length strings, find a string x such that Γ(x) is exactly this multiset.
(a) Show that the reconstruction problem reduces to RUDRATA PATH. (Hint: Construct a di- rected graph with one node for each k-mer, and with an edge from a to b if the last k − 1 characters of a match the first k − 1 characters of b.)
(b) But in fact, there is much better news. Show that the same problem also reduces to EULER PATH. (Hint: This time, use one directed edge for each k-mer.)
8.22. In task scheduling, it is common to use a graph representation with a node for each task and a directed edge from task i to task j if i is a precondition for j. This directed graph depicts the precedence constraints in the scheduling problem. Clearly, a schedule is possible if and only if the graph is acyclic; if it isn’t, we’d like to identify the smallest number of constraints that must be dropped so as to make it acyclic.
Given a directed graph G = (V, E), a subset E′ ⊆ E is called a feedback arc set if the removal of edges E′ renders G acyclic.

282 Algorithms FEEDBACK ARC SET (FAS): Given a directed graph G = (V, E) and a budget b, find a
feedback arc set of ≤ b edges, if one exists. (a) Show that FAS is in NP.
FAS can be shown to be NP-complete by a reduction from VERTEX COVER. Given an instance (G, b) of VERTEX COVER, where G is an undirected graph and we want a vertex cover of size ≤ b, we construct a instance (G′, b) of FAS as follows. If G = (V, E) has n vertices v1, . . . , vn, then make G′ = (V ′, E′) a directed graph with 2n vertices w1, w1′ , . . . , wn, wn′ , and n + 2|E| (directed) edges:
• (wi,wi′) for all i = 1,2,…,n.
• (wi′,wj) and (wj′ ,wi) for every (vi,vj) ∈ E.
(b) Show that if G contains a vertex cover of size b, then G′ contains a feedback arc set of size b.
(c) Show that if G′ contains a feedback arc set of size b, then G contains a vertex cover of size (at most) b. (Hint: given a feedback arc set of size b in G′, you may need to first modify it slightly to obtain another one which is of a more convenient form, but is of the same size or smaller. Then, argue that G must contain a vertex cover of the same size as the modified feedback arc set.)
8.23. In the NODE-DISJOINT PATHS problem, the input is an undirected graph in which some vertices have been specially marked: a certain number of “sources” s1, s2, . . . sk and an equal number of “destinations” t1 , t2 , . . . tk . The goal is to find k node-disjoint paths (that is, paths which have no nodes in common) where the ith path goes from si to ti. Show that this problem is NP-complete.
Here is a sequence of progressively stronger hints.
(a) Reduce from 3SAT.
(b) For a 3SAT formula with m clauses and n variables, use k = m+n sources and destinations. Introduce one source/destination pair (sx, tx) for each variable x, and one source/destination pair (sc, tc) for each clause c.
(c) For each 3SAT clause, introduce 6 new intermediate vertices, one for each literal occurring in that clause and one for its complement.
(d) Notice that if the path from sc to tc goes through some intermediate vertex representing, say, an occurrence of variable x, then no other path can go through that vertex. What vertex would you like the other path to be forced to go through instead?

Chapter 9
Coping with NP-completeness
You are the junior member of a seasoned project team. Your current task is to write code for solving a simple-looking problem involving graphs and numbers. What are you supposed to
do?
If you are very lucky, your problem will be among the half-dozen problems concerning
graphs with weights (shortest path, minimum spanning tree, maximum flow, etc.), that we have solved in this book. Even if this is the case, recognizing such a problem in its natural habitat—grungy and obscured by reality and context—requires practice and skill. It is more likely that you will need to reduce your problem to one of these lucky ones—or to solve it using dynamic programming or linear programming.
But chances are that nothing like this will happen. The world of search problems is a bleak landscape. There are a few spots of light—brilliant algorithmic ideas—each illuminating a small area around it (the problems that reduce to it; two of these areas, linear and dynamic programming, are in fact decently large). But the remaining vast expanse is pitch dark: NP- complete. What are you to do?
You can start by proving that your problem is actually NP-complete. Often a proof by generalization (recall the discussion on page 270 and Exercise 8.10) is all that you need; and sometimes a simple reduction from 3SAT or ZOE is not too difficult to find. This sounds like a theoretical exercise, but, if carried out successfully, it does bring some tangible rewards: now your status in the team has been elevated, you are no longer the kid who can’t do, and you have become the noble knight with the impossible quest.
But, unfortunately, a problem does not go away when proved NP-complete. The real ques- tion is, What do you do next?
This is the subject of the present chapter and also the inspiration for some of the most important modern research on algorithms and complexity. NP-completeness is not a death certificate—it is only the beginning of a fascinating adventure.
Your problem’s NP-completeness proof probably constructs graphs that are complicated and weird, very much unlike those that come up in your application. For example, even though SAT is NP-complete, satisfying assignments for HORN SAT (the instances of SAT that come up in logic programming) can be found efficiently (recall Section 5.3). Or, suppose the graphs that arise in your application are trees. In this case, many NP-complete problems,
283

284 Algorithms such as INDEPENDENT SET, can be solved in linear time by dynamic programming (recall
Section 6.7).
Unfortunately, this approach does not always work. For example, we know that 3SAT is NP-complete. And the INDEPENDENT SET problem, along with many other NP-complete problems, remains so even for planar graphs (graphs that can be drawn in the plane without crossing edges). Moreover, often you cannot neatly characterize the instances that come up in your application. Instead, you will have to rely on some form of intelligent exponential search—procedures such as backtracking and branch and bound which are exponential time in the worst-case, but, with the right design, could be very efficient on typical instances that come up in your application. We discuss these methods in Section 9.1.
Or you can develop an algorithm for your NP-complete optimization problem that falls short of the optimum but never by too much. For example, in Section 5.4 we saw that the greedy algorithm always produces a set cover that is no more than log n times the optimal set cover. An algorithm that achieves such a guarantee is called an approximation algorithm. As we will see in Section 9.2, such algorithms are known for many NP-complete optimization problems, and they are some of the most clever and sophisticated algorithms around. And the theory of NP-completeness can again be used as a guide in this endeavor, by showing that, for some problems, there are even severe limits to how well they can be approximated—unless of course P = NP.
Finally, there are heuristics, algorithms with no guarantees on either the running time or the degree of approximation. Heuristics rely on ingenuity, intuition, a good understanding of the application, meticulous experimentation, and often insights from physics or biology, to attack a problem. We see some common kinds in Section 9.3.
9.1 Intelligent exhaustive search
9.1.1 Backtracking
Backtracking is based on the observation that it is often possible to reject a solution by looking at just a small portion of it. For example, if an instance of SAT contains the clause (x1 ∨ x2), then all assignments with x1 = x2 = 0 (i.e., false) can be instantly eliminated. To put it differently, by quickly checking and discrediting this partial assignment, we are able to prune a quarter of the entire search space. A promising direction, but can it be systematically exploited?
Here’s how it is done. Consider the Boolean formula φ(w, x, y, z) specified by the set of clauses
(w∨x∨y∨z), (w∨x), (x∨y), (y∨z), (z∨w), (w∨z).
We will incrementally grow a tree of partial solutions. We start by branching on any one variable, say w:

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 285 Initial formula φ
w=0 w=1
Plugging w = 0 and w = 1 into φ, we find that no clause is immediately violated and thus neither of these two partial assignments can be eliminated outright. So we need to keep branching. We can expand either of the two available nodes, and on any variable of our choice. Let’s try this one:
Initial formula φ w=0 w=1
x=0 x=1
This time, we are in luck. The partial assignment w = 0, x = 1 violates the clause (w ∨ x) and can be terminated, thereby pruning a good chunk of the search space. We backtrack out of this cul-de-sac and continue our explorations at one of the two remaining active nodes.
In this manner, backtracking explores the space of assignments, growing the tree only at nodes where there is uncertainty about the outcome, and stopping if at any stage a satisfying assignment is encountered.
In the case of Boolean satisfiability, each node of the search tree can be described either by a partial assignment or by the clauses that remain when those values are plugged into the original formula. For instance, if w = 0 and x = 0 then any clause with w or x is instantly satisfied and any literal w or x is not satisfied and can be removed. What’s left is
Likewise, w = 0 and x = 1 leaves
(y ∨ z), (y), (y ∨ z). (), (y ∨ z),
with the “empty clause” ( ) ruling out satisfiability. Thus the nodes of the search tree, repre- senting partial assignments, are themselves SAT subproblems.
This alternative representation is helpful for making the two decisions that repeatedly arise: which subproblem to expand next, and which branching variable to use. Since the ben- efit of backtracking lies in its ability to eliminate portions of the search space, and since this happens only when an empty clause is encountered, it makes sense to choose the subproblem that contains the smallest clause and to then branch on a variable in that clause. If this clause

286 Algorithms Figure 9.1 Backtracking reveals that φ is not satisfiable.
(w ∨ x ∨ y ∨ z), (w ∨ x), (x ∨ y), (y ∨ z), (z ∨ w), (w ∨ z) w=0 w=1
(x ∨ y ∨ z), (x), (x ∨ y), (y ∨ z) x=0 x=1
(y ∨ z), (y), (y ∨ z) (), (y ∨ z)
y=0 y=1 (z), (z) ()
z=0 z=1 () ()
(x ∨ y), (y ∨ z), (z), (z) z=0 z=1
(x ∨ y), ()
(x ∨ y), (y), ()
happens to be a singleton, then at least one of the resulting branches will be terminated. (If there is a tie in choosing subproblems, one reasonable policy is to pick the one lowest in the tree, in the hope that it is close to a satisfying assignment.) See Figure 9.1 for the conclusion of our earlier example.
More abstractly, a backtracking algorithm requires a test that looks at a subproblem and quickly declares one of three outcomes:
1. Failure: the subproblem has no solution.
2. Success: a solution to the subproblem is found. 3. Uncertainty.
In the case of SAT, this test declares failure if there is an empty clause, success if there are no clauses, and uncertainty otherwise. The backtracking procedure then has the following format.
Start with some problem P0
Let S = {P0}, the set of active subproblems Repeat while S is nonempty:
choose a subproblem P ∈ S and remove it from S expand it into smaller subproblems P1, P2, . . . , Pk For each Pi:
(Pi) succeeds: halt and announce this solution If test
(Pi) fails: discard Pi
If test

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 287 Otherwise: add Pi to S
Announce that there is no solution
For SAT, the choose procedure picks a clause, and expand picks a variable within that clause. We have already discussed some reasonable ways of making such choices.
With the right test, expand, and choose routines, backtracking can be remarkably effec- tive in practice. The backtracking algorithm we showed for SAT is the basis of many successful satisfiability programs. Another sign of quality is this: if presented with a 2SAT instance, it will always find a satisfying assignment, if one exists, in polynomial time (Exercise 9.1)!
9.1.2 Branch-and-bound
The same principle can be generalized from search problems such as SAT to optimization problems. For concreteness, let’s say we have a minimization problem; maximization will follow the same pattern.
As before, we will deal with partial solutions, each of which represents a subproblem, namely, what is the (cost of the) best way to complete this solution? And as before, we need a basis for eliminating partial solutions, since there is no other source of efficiency in our method. To reject a subproblem, we must be certain that its cost exceeds that of some other solution we have already encountered. But its exact cost is unknown to us and is generally not efficiently computable. So instead we use a quick lower bound on this cost.
Start with some problem P0
Let S = {P0}, the set of active subproblems bestsofar = ∞
Repeat while S is nonempty:
choose a subproblem (partial solution) P ∈ S and remove it from S expand it into smaller subproblems P1, P2, . . . , Pk
For each Pi:
If Pi is a complete solution: update bestsofar
(Pi) < bestsofar: add Pi to S else if lowerbound return bestsofar Let’s see how this works for the traveling salesman problem on a graph G = (V, E) with edge lengths de > 0. A partial solution is a simple path a 􏰾 b passing through some vertices S ⊆ V , where S includes the endpoints a and b. We can denote such a partial solution by the tuple [a, S, b]—in fact, a will be fixed throughout the algorithm. The corresponding subproblem is to find the best completion of the tour, that is, the cheapest complementary path b 􏰾 a with intermediate nodes V −S. Notice that the initial problem is of the form [a, {a}, a] for any a ∈ V of our choosing.
At each step of the branch-and-bound algorithm, we extend a particular partial solution [a,S,b] by a single edge (b,x), where x ∈ V −S. There can be up to |V −S| ways to do this, and each of these branches leads to a subproblem of the form [a, S ∪ {x}, x].

288 Algorithms
How can we lower-bound the cost of completing a partial tour [a, S, b]? Many sophisticated methods have been developed for this, but let’s look at a rather simple one. The remainder of the tour consists of a path through V − S, plus edges from a and b to V − S. Therefore, its cost is at least the sum of the following:
1. The lightest edge from a to V − S.
2. The lightest edge from b to V − S.
3. The minimum spanning tree of V − S.
(Do you see why?) And this lower bound can be computed quickly by a minimum spanning tree algorithm. Figure 9.2 runs through an example: each node of the tree represents a partial tour (specifically, the path from the root to that node) that at some stage is considered by the branch-and-bound procedure. Notice how just 28 partial solutions are considered, instead of the 7! = 5,040 that would arise in a brute-force search.

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 289 Figure 9.2 (a) A graph and its optimal traveling salesman tour. (b) The branch-and-bound
search tree, explored left to right. Boxed numbers indicate lower bounds on cost. (a)
2F1E2 F1E
GDG1D 111111
HCHC
1A2B1
1A B1
(b)
A
10B F8 H
10C E EGCG 13 8
D H14 D F10 8B D 15
11EGG G C
10
1 11
5
10
8
8
8
14
12
14
14
inf
11F 11 G 11 H
Cost: 11
8D H 8 G
H Cost: 8
8
inf

290 Algorithms 9.2 Approximation algorithms
In an optimization problem we are given an instance I and are asked to find the optimum solution—the one with the maximum gain if we have a maximization problem like INDEPEN- DENT SET, or the minimum cost if we are dealing with a minimization problem such as the TSP. For every instance I, let us denote by OPT(I) the value (benefit or cost) of the optimum solution. It makes the math a little simpler (and is not too far from the truth) to assume that OPT(I) is always a positive integer.
We have already seen an example of a (famous) approximation algorithm in Section 5.4: the greedy scheme for SET COVER. For any instance I of size n, we showed that this greedy algorithm is guaranteed to quickly find a set cover of cardinality at most OPT(I)logn. This log n factor is known as the approximation guarantee of the algorithm.
More generally, consider any minimization problem. Suppose now that we have an algo- rithm A for our problem which, given an instance I, returns a solution with value A(I). The approximation ratio of algorithm A is defined to be
αA = max A(I) . I OPT(I)
In other words, αA measures by the factor by which the output of algorithm A exceeds the optimal solution, on the worst-case input. The approximation ratio can also be defined for maximization problems, such as INDEPENDENT SET, in the same way—except that to get a number larger than 1 we take the reciprocal.
So, when faced with an NP-complete optimization problem, a reasonable goal is to look for an approximation algorithm A whose αA is as small as possible. But this kind of guarantee might seem a little puzzling: How can we come close to the optimum if we cannot determine the optimum? Let’s look at a simple example.
9.2.1 Vertex cover
We already know the VERTEX COVER problem is NP-hard.
VERTEX COVER
Input: An undirected graph G = (V, E).
Output: A subset of the vertices S ⊆ V that touches every edge. Goal: Minimize |S|.
See Figure 9.3 for an example.
Since VERTEX COVER is a special case of SET COVER, we know from Chapter 5 that it can
be approximated within a factor of O(log n) by the greedy algorithm: repeatedly delete the vertex of highest degree and include it in the vertex cover. And there are graphs on which the greedy algorithm returns a vertex cover that is indeed log n times the optimum.
A better approximation algorithm for VERTEX COVER is based on the notion of a matching, a subset of edges that have no vertices in common (Figure 9.4). A matching is maximal if no

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 291 Figure 9.3 A graph whose optimal vertex cover, shown shaded, is of size 8.
Figure 9.4 (a) A matching, (b) its completion to a maximal matching, and (c) the resulting vertex cover.
(a) (b) (c)
more edges can be added to it. Maximal matchings will help us find good vertex covers, and moreover, they are easy to generate: repeatedly pick edges that are disjoint from the ones chosen already, until this is no longer possible.
What is the relationship between matchings and vertex covers? Here is the crucial fact: any vertex cover of a graph G must be at least as large as the number of edges in any matching in G; that is, any matching provides a lower bound on OPT. This is simply because each edge of the matching must be covered by one of its endpoints in any vertex cover! Finding such a lower bound is a key step in designing an approximation algorithm, because we must compare the quality of the solution found by our algorithm to OPT, which is NP-complete to compute.
One more observation completes the design of our approximation algorithm: let S be a set that contains both endpoints of each edge in a maximal matching M. Then S must be a vertex cover—if it isn’t, that is, if it doesn’t touch some edge e ∈ E, then M could not possibly be maximal since we could still add e to it. But our cover S has 2|M| vertices. And from the previous paragraph we know that any vertex cover must have size at least |M |. So we’re done.
Here’s the algorithm for VERTEX COVER. Find a maximal matching M ⊆ E

292
Algorithms
Return S = {all endpoints of edges in M }
This simple procedure always returns a vertex cover whose size is at most twice optimal!
In summary, even though we have no way of finding the best vertex cover, we can easily
find another structure, a maximal matching, with two key properties:
1. Its size gives us a lower bound on the optimal vertex cover.
2. It can be used to build a vertex cover, whose size can be related to that of the optimal cover using property 1.
Thus, this simple algorithm has an approximation ratio of αA ≤ 2. In fact, it is not hard to find examples on which it does make a 100% error; hence αA = 2.
9.2.2 Clustering
We turn next to a clustering problem, in which we have some data (text documents, say, or images, or speech samples) that we want to divide into groups. It is often useful to define “dis- tances” between these data points, numbers that capture how close or far they are from one another. Often the data are true points in some high-dimensional space and the distances are the usual Euclidean distance; in other cases, the distances are the result of some “similarity tests” to which we have subjected the data points. Assume that we have such distances and that they satisfy the usual metric properties:
1. d(x,y) ≥ 0 for all x,y.
2. d(x,y)=0ifandonlyifx=y.
3. d(x, y) = d(y, x).
4. (Triangle inequality) d(x, y) ≤ d(x, z) + d(z, y).
We would like to partition the data points into groups that are compact in the sense of having small diameter.
k-CLUSTER
Input: Points X = {x1, . . . , xn} with underlying distance metric d(·, ·); integer k. Output: A partition of the points into k clusters C1, . . . , Ck.
Goal: Minimize the diameter of the clusters,
max max d(xa , xb ). j xa ,xb ∈Cj
One way to visualize this task is to imagine n points in space, which are to be covered by k spheres of equal size. What is the smallest possible diameter of the spheres? Figure 9.5 shows an example.
This problem is NP-hard, but has a very simple approximation algorithm. The idea is to pick k of the data points as cluster centers and to then assign each of the remaining points to

S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 293 Figure 9.5 Some data points and the optimal k = 4 clusters.
Figure 9.6 (a) Four centers chosen by farthest-first traversal. (b) The resulting clusters.
(a)
2
(b)
4
1
3
the center closest to it, thus creating k clusters. The centers are picked one at a time, using an intuitive rule: always pick the next center to be as far as possible from the centers chosen so far (see Figure 9.6).
Pick any point μ1 ∈ X as the first cluster center for i=2 to k:
Let μi be the point in X that is farthest from μ1,…,μi−1
(i.e., that maximizes minj 0 produces an instance I(G, C) of the TSP such that:
(i) If G has a Rudrata path, then OPT(I(G, C)) = n, the number of vertices in G. (ii) If G has no Rudrata path, then OPT(I(G, C)) ≥ n + C.
This means that even an approximate solution to TSP would enable us to solve RUDRATA PATH! Let’s work out the details.
Consider an approximation algorithm A for TSP and let αA denote its approximation ratio. From any instance G of RUDRATA PATH, we will create an instance I(G,C) of TSP using the specific constant C = nαA. What happens when algorithm A is run on this TSP instance? In case (i), it must output a tour of length at most αAOPT(I(G,C)) = nαA, whereas in case (ii) it must output a tour of length at least OPT(I(G, C)) > nαA. Thus we can figure out whether G has a Rudrata path! Here is the resulting procedure:
Given any graph G:
compute I(G, C) (with C = n · αA) and run algorithm A on it if the resulting tour has length ≤ nαA:
conclude that G has a Rudrata path else: conclude that G has no Rudrata path
This tells us whether or not G has a Rudrata path; by calling the procedure a polynomial number of times, we can find the actual path (Exercise 8.2).
We’ve shown that if TSP has a polynomial-time approximation algorithm, then there is a polynomial algorithm for the NP-complete RUDRATA PATH problem. So, unless P = NP, there cannot exist an efficient approximation algorithm for the TSP.