The exercises
School of Computing and Information Systems COMP90038 Algorithms and Complexity
Tutorial Week 11 Sample Answers
74. Use the dynamic-programming algorithm developed in Lecture 18 to solve this instance of the coin-row problem: 20, 50, 20, 5, 10, 20, 5.
Answer: We build the table S of optimal values as follows:
i: 01234567 C[i]: − 20 50 20 5 10 20 5 S[i]: 0 20 50 50 55 60 75 75
The optimal selection uses the coins at indices 2, 4, and 6.
75. In Week 12 we will meet the concept of problem reduction. This question prepares you for that. First, when we talk about the length of a path in an un-weighted directed acyclic graph (dag), we mean the number of edges in the path. (You could also consider the un-weighted graph weighted, will all edges having weight 1.)
Show how to reduce the coin-row problem to the problem of finding a longest path in a dag. That is, give an algorithm that transforms any coin-row instance into a longest-path-in-dag instance in such as way that a solution to the latter provides a solution to the former.
Hint: If there are n coins, use n + 1 nodes; let an edge with weight i correspond to picking a coin with value i.
Answer: Assume we have n coins c1, . . . , cn. We generate a weighted dag with n + 1 nodes C0, C1, . . . , Cn. The dag has edges as follows:
• n − 1 edges (C0, Cn), (C1, Cn), . . . , (Cn−2, Cn), each with weight cn.
• n − 2 edges (C0, Cn−1), (C1, Cn−1), . . . , (Cn−3, Cn−1), each with weight cn−1. • and so on, down to two edges (C0,C3) and (C1,C3), each with weight c3.
• one edge (C0,C2) with weight c2, and
• one edge (C0, C1) with weight c1.
Any path in the generated dag corresponds to a legal selection of coins, and the sum of the weights along a given path is exactly the sum of the coins chosen.
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76. Consider the problem of finding the length of a “longest” path in a weighted, not necessarily connected, dag. We assume that all weights are positive, and that a “longest” path is a path whose edge weights add up to the maximal possible value. For example, for the following graph, the longest path is of length 15:
6
3b d4
a42fg9h 1c e1
3
Use a dynamic programming approach to the problem of finding longest path in a weighted dag.
Answer: This is easy if we process the nodes in topologically sorted order. For each node t we want to find its longest distance from a source, and to store these distances in an array L. That is, for each t we want to calculate
max({0} ∪ {L[u] + weight[u, t] | (u, t) ∈ E})
So:
T ← TopSort(⟨V, E⟩) — List of nodes sorted topologically for each t ∈ T (in topological order) do
L[t] ← 0 foreachu∈V do
if (u,t) ∈ E then
if L[u] + weight[u, t] > L[t] then
L[t] ← L[u] + weight[u, t]
max ← 0 foreachu∈V do
if L[u] > max then max ← L[u]
return max
For the sample graph, DFS-based topsort yields the sequence g, h, a, c, b, d, e, f. The “longest
path” table L gets filled as follows:
t: abcdefgh
L[t] :
0
0
1
5
11
13
15
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9
77. Design a dynamic programming algorithm for the version of the knapsack problem in which there are unlimited numbers of copies of each item. That is, we are given items I1, . . . , In that have values v1,…,vn and weights w1,…,wn as usual, but each item Ii can be selected several times. Hint: This actually makes the knapsack problem a bit easier, as there is only one parameter (namely the remaining capacity w) in the recurrence relation.
Answer: Assume the items I1,…,In have values v1,…,vn and weights w1,…,wn. Let V (w) denote the optimal value we can achieve given capacity w. With capacity w we are in a position to select any item Ii which weighs no more than w. And if we pick item Ii then the best value we can achieve is vi + V (w − wi). As we want to maximise the value for capacity w, we have the recurrence
V(w)=max{vi +V(w−wi)|1≤i≤n∧wi ≤w} That leads to this table-filling approach:
forw←1toW do
V[w]←max({0}∪{vi +V(w−wi)|1≤i≤n∧wi ≤w})
return V [W ]
As an example, consider the case W = 10, and three items I1, I2, and I3, with weights 4, 5 and 3, respectively, and values 11, 12, and 7, respectively. The table V is filled from left to right, as follows:
w: 12345678910 V[w]: 0 0 7 11 12 14 18 22 23 25
Hence the optimal bag is [I1,I3,I3] for a total value of 25.
78. Work through Warshall’s algorithm to find the transitive closure of the binary relation given
by this table (or directed graph):
abcd a0011 b0010 c1000 d0000
bcad
Answer: We run down the columns from left to right, stopping when we meet a 1. This first happens when we are in row 3, column 1. At that point, ‘or’ row 1 onto row 3 (and so on):
abcd abcd abcd a0011 a1011 a1011 b0010⇒b0010⇒b1011 c1011 c1011 c1011 d0000 d0000 d0000
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