chap23-solutions - is a light edge crossing the cut .S; V N...

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Selected Solutions for Chapter 23: Minimum Spanning Trees Solution to Exercise 23.1-1 Theorem 23.1 shows this. Let A be the empty set and S be any set containing u but not ± . Solution to Exercise 23.1-4 A triangle whose edge weights are all equal is a graph in which every edge is a light edge crossing some cut. But the triangle is cyclic, so it is not a minimum spanning tree. Solution to Exercise 23.1-6 Suppose that for every cut of G , there is a unique light edge crossing the cut. Let us consider two minimum spanning trees, T and T 0 , of G . We will show that every edge of T is also in T 0 , which means that T and T 0 are the same tree and hence there is a unique minimum spanning tree. Consider any edge .u; ±/ 2 T . If we remove .u; ±/ from T , then T becomes disconnected, resulting in a cut .S; V N S/ . The edge .u; ±/
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Unformatted text preview: is a light edge crossing the cut .S; V N S/ (by Exercise 23.1-3). Now consider the edge .x; y/ 2 T that crosses .S; V N S/ . It, too, is a light edge crossing this cut. Since the light edge crossing .S; V N S/ is unique, the edges .u; ±/ and .x; y/ are the same edge. Thus, .u;±/ 2 T . Since we chose .u; ±/ arbitrarily, every edge in T is also in T . Here’s a counterexample for the converse: x y z 1 1 23-2 Selected Solutions for Chapter 23: Minimum Spanning Trees Here, the graph is its own minimum spanning tree, and so the minimum spanning tree is unique. Consider the cut . f x g ; f y; ´ g / . Both of the edges .x; y/ and .x; ´/ are light edges crossing the cut, and they are both light edges....
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chap23-solutions - is a light edge crossing the cut .S; V N...

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