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Homework Solutions – Unit 4: Chapter 23 CMPSC 465 Disclaimer: This is a draft of solutions that has been prepared by the TAs and the instructor makes no guarantees that solutions presented here contain the level of detail that would be expected on an exam. Any errors or explanations you find unclear should be reported to either of the TAs for corrections first. Exercise 23.1-2 Professor Sabatier conjectures the following converse of Theorem 23.1. Let G = (V, E) be a connected, undirected graph with a real-valued weight function w defined on E. Let A be a subset of E that is included in some minimum spanning tree for G, let (S, V – S) be any cut of G that respects A, and let (u, v) be a safe edge for A crossing (S, V – S). Then, (u, v) is a light edge for the cut. Show that the professor’s conjecture is incorrect by giving a counterexample. Solution: That is false. A safe edge (with respect to the set of edges A) is defined to be any edge such that when we add it to A, we still have a subset of some minimum spanning tree. This definition does not reference light edges in any way. In the situation from a counterexample, there may be multiple safe edges from S to V – S, not all of them have to be light. A counterexample is shown below. Let S = {a, b, c}, A = {(a, b), (b, c}, we can see that (a, d), (b, e), (c, f) are all safe edges for A crossing (S, V – S), but only (b, e) is a light edge for the cut.

Exercise 23.1-3 Show that if an edge (u, v) is contained in some minimum spanning tree, then it is a light edge crossing some cut of the graph. Solution: Suppose (u, v) T, a minimal spanning tree. Let A = {T – (u, v)}. A contains two trees: A = Tu Tv. Tu is the tree in which vertex u appears, and Tv is the tree in which vertex v appears. Moreover, Vu Vv = V, where Vu contains the vertices of Tu and Vv contains the vertices of Tv. That is, (Vu, Vv) is a cut, which is crossed by (u, v) and which respects A. Any edge crossing the cut rejoins the two trees. If there is a crossing edge (x, y) with w(x, y) < w(u, v), then T’ = Tu Tv (x, y) is a spanning tree with weight:

w(T’) = w(Tu Tv) + w(x, y) as T’ = Tu Tv (x, y) = w(Tu Tv) + w(u, v) + [w(x, y) – w(u, v)] by adding 0, that is w(u, v) – w(u, v) = w(T) – [w(u, v) – w(x, y)] as T = Tu Tv (u, v) < w(T) This contradicts with that (u, v) is contained in some minimum spanning tree. So, w(u, v) < w(x, y) for all (x, y) crossing the cut (Vu, Vv). That is, (u, v) is a light edge crossing the cut. Exercise 23.2-1 Kruskal’s algorithm can return different spanning trees for the same input graph G, depending on how it breaks ties when the edges are sorted into order. Show that for each minimum spanning tree T of G, there is a way to sort the edges of G in Kruskal’s algorithm so that the algorithm returns T.

 a

 e  d  f

 b  c 1 1

3 1 5 5

4

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Solution: We would start by sorting the edges in of G in non-descending order. In addition, we would want that among the edges of same weight, the edges which are contained in T are placed in first positions. Exercise 23.2-8 Professor Borden proposes a new divide-and-conquer algorithm for computing minimum spanning trees, which goes as follows. Given a graph G = (V, E), partition the set V of vertices into two sets V1 and V2 such that |V1| and |V2| differ by at most 1. Let E1 be the set of edges that are incident only on vertices in V1, and let E2 be the set of edges that are incident only on vertices in V2. Recursively solve a minimum-spanning-tree problem on each of the two subgraphs G1 = (V1, E1) and G2 = (V2, E2). Finally, select the minimum-weight edge in E that crosses the cut (V1, V2), and use this edge to unite the resulting two minimum spanning trees into a single spanning tree. Either argue that the algorithm correctly computes a minimum spanning tree of G, or provide an example for which the algorithm fails. Solution: We argue that the algorithm fails. Consider the graph G below. We partition G into V1

and V2 as follows: V1

= {A, B}, V2 = {C,

D}. E1 = {(A, B)}. E2

= {(C, D)}. The set of edges that cross the cut is Ec = {(A, C), (B, D)}.

Now, we must recursively find the minimum spanning trees of G1

and G2. We can see that in this case, MST(G1) = G1 and

MST(G2) = G2. The minimum spanning trees of G1 and G2

are shown below on the left. The minimum weighted edge of the two edges across the cut is edge (A, C). So (A, C) is used to connect G1

and G2. This is the minimum spanning tree returned by Professor Borden’s algorithm. It is shown below and to the right.

We can see that the minimum-spanning tree returned by Professor Borden’s algorithm is not the minimum spanning tree of G, therefore, this algorithm fails.

 A

 D  C

 B 1 G1

2 3

8 G2

 A

 D  C

 B 1 G1

2 3

8 G2

 A

 D  C

 B 1

2 3

8