Assignment # 4 Selected Solutions

Transcription

1 Assignment # 4 Selected Solutions Problem Let G be a connected graph which is not a tree (did you notice this is redundant?) and let C be a cycle in G. Prove that the complement of any spanning tree of G contains at least one edge of C. Solution 1 Direct proof We know that and edge of G is a bridge if and only if it is not in any cycle. But since C is a cycle in G, none of its edges is a bridge. Pick one of them, e, then removing any edge e of C does not disconnect G. Then G e still has a spanning tree T. Now by definition of the complement, T contains exactly those edges that are not in T. Hence, e T. Solution 2 Proof by contradiction Let C be a cycle and T a spanning tree such that no edge of C belongs to T. Then by the definition of the complement, all edges of C belong to T. Hence, T cannot be a tree, since a tree is acyclic, a contradiction. 1

2 2 Problem 2.3.4a Let G be an edge of a connected graph G. bridge if and only if it is in every spanning tree of G. Prove that e is a Solution 1 Proof by contrapositive both ways This is a nice proof, because both directions are proved by contraposition. Hence, we are in fact proving by direct proofs the statement e is not a bridge if and only if there is a spanning tree of G not containing e. So let s prove if e is in every spanning tree of G, then e is a bridge. Assume that e is not a bridge ( q). We want to show p, that is, there is a spanning tree of G not containing e. Because e is not a bridge, removing it from G does not make the graph disconnected. Because G e is connected, it contains a spanning tree T. Apparently, e / T. (What were we proving, a necessary or sufficient condition?) Now the other direction (necessary or sufficient?:) if e is a bridge, then e is in every spanning tree of G. Suppose there is a spanning tree S not containing e. Adding e to S joins two vertices, that were in S connected by a path, thus e now belongs to a cycle in S + e and therefore in G. By Theorem 2.7, it is not a bridge. This can also be rephrased as a proof by contradiction. Now p is e is in every spanning tree of G and q is e is a bridge. We assume p q, that is, e is in every spanning tree of G but e is not a bridge. In fact, this would only be really rephrased contraposition, because we would have (p q) p. Solution 2 Direct proof of the necessary condition Let s finally confirm that if e is a bridge, then e is in every spanning tree of G is the necessary condition. Let e be a bridge of G. Then by definition of a bridge, G e is disconnected. By Theorem 2.12 (H has a spanning tree iff H is connected) G e has no spanning tree. In other words, there is no spanning tree without e and hence e is in every spanning tree. (Notice that if you just state the argument that G e has no spanning tree without citing the theorem, your argument is not supported and therefore not complete.) Or we can say: Let e be a bridge of G. Then by definition of a bridge, G e is disconnected. Hence, every spanning subgraph of G e must be disconnected. Then no spanning subgraph of G e is a spanning tree. Because G was connected and had a spanning tree, more precisely spanning trees S 1,..., S k where k 1, e must be in every S i. Solution 2 Direct proof of the sufficient condition We are proving if e is in every spanning tree of G then e is a bridge. Because e is in every spanning tree, then it is in at least one, and G is connected. Also because it is in every spanning tree, then there is no spanning tree in G e, hence G e is disconnected. So it follows from the definition that e is a bridge.

3 3 Problem Prove that if G has exactly one spanning tree, then G is a tree. Solution 1 Proof by contraposition Suppose that G is not a tree. Then it is either not connected, or contains a cycle. In the former case, it has no spanning tree. In the latter case, if it is connected, it contains a cycle C. No edge of C can be a bridge, so removing any edge e of C still keeps the graph G e connected. Find a spanning tree T of G e. Now the graph H = T + e contains a cycle C. Remove from H any edge of C other than e, say f. Resulting graph H f is still connected, since f belonged to C and hence was not a bridge. Now T = H f is a tree (why? can you prove it?) containing all vertices of G. We know that T T and hence we have more than one spanning tree. In both cases, we have p. Solution 2 Proof by contradiction Suppose that G has exactly one spanning tree and is not a tree itself. Then it contains cycle(s). You cannot use the argument from Solution 1 here yet! If you remove e from a cycle C, you cannot argue that if you start by removing another edge f from C, you eventually obtain different spanning tree. You do not know whether after removing e you will not have to remove f later to get rid of another cycle C. Keep removing edges and always check whether you got a tree. If yes, then call it T and put back the last removed edge e. This creates a cycle that contains another edge f. Remove f to obtain T. Apparently, T T, which we wanted to show.

4 4 Problem Prove that a vertex v of a tree T is a cut vertex iff deg(v) > 1. Solution 1 Proof by contrapositive both ways First we prove that if a vertex v of a tree T is a cut vertex then deg(v) > 1. If deg(v) = 0, then T = K 1 and there is no cut-vertex at all. So suppose that deg(v) = 1 and call its only neighbor u. Because T is connected, then there is a (unique) (u, w)-path for any vertex w u in T. Moreover, if w v, then the path does not contain edge uv, since v is of degree one. Now if we remove v, there is still the same (u, w)-path in T v as before, and hence T v is still connected and v is not a cut-vertex. Now we prove that if v is a vertex of a tree T and deg(v) > 2, then v is a cut-vertex. Hence we assume that v is not a cut-vertex, which means that T v is connected. We want to show that the statement v is a vertex of a tree T and deg(v) > 2 is not true. To show that, it is enough to show that either deg(v) 1 or T is not a tree. If deg(v) 1, we are done. Hence we suppose that v is of degree at least two in T. Then v has at least two neighbors, say u and w. We supposed that T v is connected, and therefore there is a (u, w)-path in T v. But this path together with the edges uv and wv creates a cycle, hence T is not a tree. Solution 2 Proof by a counting argument both ways First we prove that if a vertex v of a tree T is a cut vertex then deg(v) > 1. Suppose T has n vertices. Because v is a cut vertex, then T v has k components, M 1, M 2,..., M k, where k 2. We proved in the previous assignment that this graph has (n 1) k edges (because we now have only n 1 vertices in T v), and we know that k 2 and hence (n 1) k n 3. At the same time, there were exactly n 1 edges in T. Therefore, by removing v we must have removed at least 2 edges and deg(v) 2. Or even simpler, we have a tree T with n vertices and n 1 edges, and a vertex v such that deg(v) > 1. Removing v from T, we remove at least two edges, so our new graph T = T v is still acyclic, has n 1 vertices, and at most (n 1) 2 = n 3 < n 2 edges. Having n 1 vertices and less than n 2 edges, T cannot be a tree, so v is a cut vertex. Now we use a contraposition and prove that if v is a vertex of a tree T and deg(v) = 1, then v is not a cut-vertex. Suppose T has n vertices. Remove v from T to obtain a graph T. By removing v, we have removed exactly one edge, as deg(v) = 1. Because by removing a vertex we do not create a new cycle, T is still acyclic. T had n vertices and n 1 edges and T has n = n 1 vertices and n 2 = n 1 edges. Because it is acyclic, has n vertices and n 1 edges, by a Theorem in the book it is a tree and hence connected. Thus, v was not a cut-vertex.

5 Solution 3 Proof of sufficient condition by contradiction Let T have n vertices and deg(v) 2. Suppose that T v is still connected. Since T v has n 1 edges and is still acyclic, it is again a tree and has n 2 edges. But T had n 1 edges and we have removed at least 2 of them, hence can have at most n 3 edges, which is a contradiction. 5