COMP90038
Algorithms and Complexity
Lecture 6: Recursion
(with thanks to Harald Søndergaard)
DMD 8.17 (Level 8, Doug McDonell Bldg)
http://people.eng.unimelb.edu.au/tobym @tobycmurray
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Recursion
We’ve already seen some examples
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A very natural approach when the data structure is
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recursive (e.g. lists, trees)
But also examples of naturally recursive array
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processing algorithms
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Next week we’ll express depth first graph traversal recursively (the natural way); later we’ll meet other examples of recursion too
Example: Factorial
n!: we can use recursion (left) or iteration (right)
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F(5) =F(4)⋅5
= (F(3) ⋅ 4) ⋅ 5
= ((F(2) ⋅ 3) ⋅ 4) ⋅ 5
= (((F(1) ⋅ 2) ⋅ 3) ⋅ 4) ⋅ 5
= ((((F(0) ⋅ 1) ⋅ 2) ⋅ 3) ⋅ 4) ⋅ 5 = ((((1 ⋅ 1) ⋅ 2) ⋅ 3) ⋅ 4) ⋅ 5
= 120
n: 420315
result: 61521200
Iterative version normally preferred since it is constant space
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Complexity is exponential in n
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Example: Fibonacci Numbers
Follows the mathematical definition of Fibonacci numbers very closely.
Easy to understand
But performs lots of redundant computation
To generate the nth number of sequence: 1 1 2 3 5
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8 13 21 34 55 …
Basic operation: addition
Fibonacci Again
Of course we only need to remember the latest two
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items. Recursive version: left; iterative version: right
Initial call: Fib(n, 1, 0) Time complexity of both
solutions is linear in n
(There is a cleverer, still recursive, way which is O(log n).)
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Tracing Recursive Fibonacci
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Initial call: Fib(5, 1, 0)
Fib(5,1,0) = Fib(4,1,1) = Fib(3,2,1) = Fib(2,3,2)
= Fib(1,5,3)
= Fib(0,8,5) =8
Fibonacci Sequence: 1 12 3 5 8 …
Tower of Hanoi
Init Aux Fin
Move n disks from Init to Fin. A larger disk can never be placed on top of a smaller one.
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Tower of Hanoi: Recursive Solution
Init Aux Fin Move n-1 disks from Init to Aux.
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Tower of Hanoi: Recursive Solution
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Init Aux Fin Move n-1 disks from Init to Aux.
Then move the nth disk to Fin.
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Tower of Hanoi: Recursive Solution
Init Aux Fin
Move n-1 disks from Init to Aux. Then move the nth disk to Fin. Then move the n-1 disks from Aux to Fin.
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Tower of Hanoi: Recursive Solution
Init Aux Fin
Move n-1 disks from Init to Aux. Then move the nth disk to Fin. Then move the n-1 disks from Aux to Fin.
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Tower Of Hanoi: Recursive Algorithm
Init Aux Fin
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Tracing Tower of Hanoi Recursive Algorithm
http://vornlocher.de/tower.html
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A Challenge:
Coin Change Problem
There are 6 different kinds of Australian coin
In cents, their values are: 5, 10, 20, 50, 100, 200
In how many different ways can I produce a handful of coins adding up to $4?
This is not an easy problem!
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Key to solving it is to find a way to break it down
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into simpler sub-problems
Coin Change Problem: Decomposition
$4 made from
Does the bag contain at least one $2 coin? Yes No
+
made from
$2 $4
made from
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Coin Change Problem: Decomposition
The number of ways of making $4 is therefore:
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1 x the number of ways of making $2
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the number of ways of making $4 without using a $2 coin
Coin Change Problem: Partial Algorithm
function WAYS(amount, denominations) // … base cases ….
d ← selectLargest(denominations)
return WAYS(amount − d, denominations) +
WAYS(amount, denominations \ {d})
For example:
Ways(400, {5,10,20,50,100,200}) = Ways(200, {5,10,20,50,100,200}) + Ways(400, {5,10,20,50,100})
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Coin Change Problem: Base Cases
Each time we recurse, we decrease either:
amount (by subtracting some quantity from it), or
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demonisations (by removing an item from the set) •
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Consider each of these separately.
amount base cases:
amount = 0: there is one way (using no coins)
amount < 0: there are no ways to make this denominations = ∅ (and amount > 0):
there are no ways to make this amount Copyright University of Melbourne 2016, provided under Creative Commons Attribution License
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•
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Coin Change Problem: Full Recursive Algorithm
function WAYS(amount, denominations) if amount = 0 then
return 1
if amount < 0 then
return 0
if denominations = ∅ then
return 0
d ← selectLargest(denominations)
return WAYS(amount − d, denominations) +
WAYS(amount, denominations \ {d})
Initial call: WAYS(amount, {5, 10, 20, 50, 100, 200}).
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Recursive Solution and its Complexity
Although our recursive algorithm is short and elegant, it is not the most efficient way of solving the problem.
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More efficient solutions can be developed using memoing or dynamic programming—more about that later (around Week 10).
Its running time grows exponentially as you grow
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the input amount.
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Next Time...
Graphs, trees, graph traversal and allied
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algorithms.