Math350 hw2solutions
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This document contains solutions to problems in graph theory and combinatorics from a course assignment. It addresses questions about bipartite graphs, matchings, connectivity, and more. The problems cover topics like showing a graph is bipartite based on independent sets, determining the size of maximum matchings, existence of matchings covering subsets of vertices, and conditions for when edges lie on common cycles. Counterexamples are provided for some statements about linking and joining vertices and subsets.
Math350 hw2solutions
- 1. MATH 350: Graph Theory and Combinatorics. Fall 2012. Assignment #2: Bipartite graphs, matchings and connectivity. 1. Show that a graph G is bipartite if and only if for every subgraph H of G, there is a subset X ⊆ V (H) with |X| ≥ 1 |V (H)| so that no two 2 members of X are adjacent in H. Solution: If G is bipartite and H is a subgraph of G then either A ∩ V (H) or B ∩ V (H) satisfies the requirement. If G is not bipartite then it contains an odd cycle H as a subgraph and the size of a maximum independent set in H is 1 (|V (H)| − 1). 2 Let G be a loopless graph in which every vertex has degree ≥ 1. Let X be the largest matching in G, and let Y be the smallest set of edges of G so that every vertex of G is incident with ≥ 1 edge in Y . Show that |X| + |Y | = |V (G)|. 2. Solution: The solution of this problem was presented in class as Theorem 11.2. 3. Let G be a bipartite graph with bipartition (A, B) in which every vertex has degree ≥ 1. Assume that for every edge of G with ends a ∈ A and b ∈ B we have deg(a) ≥ deg(b). Show that there exists a matching in G covering A. Solution: Suppose not. By Hall’s theorem there exists X ⊂ A with < |X| neighbors in B. Choose such a set X with |X| minimum. Let Y ⊂ B denote the set of vertices adjacent to any of the vertices in X. Then there exists a matching M consisting of edges joining vertices of Y to vertices of X of size |Y |. Indeed, otherwise, by Hall’s theorem, there exists Y ′ ⊂ Y so that the set X ′ of vertices in X adjacent to any of the vertices of Y ′ satisfies |X ′ | < |Y ′ |. It follows that |X − X ′ | < |Y − Y ′ | and, as the vertices of X − X ′ have no neighbors in Y ′ , we deduce that X − X ′ contradicts the minimality of X.
- 2. Let F denote the set of edges joining X and Y . Then ∑ ∑ |F | = deg(a) > deg(a) ≥ a∈A = ∑ e=ab∈M a∈A,b∈B deg(b) ≥ |F |, e=ab∈M a∈A,b∈B a contradiction. Thus G contains a matching covering A, as desired. Given integers n ≥ m ≥ k ≥ 0, determine the maximum possible number of edges in a simple bipartite graph G with bipartition (A, B), with |A| = n, |B| = m and no matching of size k. 4. Solution: If G has no matching of size k then by K¨nig’s theorem it o contains a set X with |X| ≤ k−1 so that every edge has an end in X. Every vertex in X is incident with at most n edges. Therefore, |E(G)| ≤ (k −1)n. One can have a graph with these many edges satisfying all the criteria by having exactly k − 1 vertices of B with non-zero degree, each joined to all the vertices of A. 5. Let G be a connected graph in which every vertex has degree three. Show that if G has no cut-edge then every two edges of G lie on a common cycle. Solution: Note that G is loopless, as otherwise it would contain a cutedge. Consider e1 , e2 ∈ E(G) and let xi , yi be the ends of ei for i = 1, 2. If there exist two vertex-disjoint paths from x1 , y1 to x2 , y2 then these paths together with e1 and e2 form the required cycle. Otherwise, by Menger’s theorem, there exists a separation (A, B) of order 1 with x1 , y1 ∈ A, x2 , y2 ∈ B. Let {v} = A ∩ B. Let u1 , u2 , u3 be the other ends of edges incident to v. (These three vertices are not necessarily distinct.) Without loss of generality, u1 ∈ A, u2 , u3 ∈ B. Then the edge joining u1 and v is a cut-edge, a contradiction. 6. a) Distinct u, v ∈ V (G) are k-linked if there are k paths P1 , ..., Pk of G from u to v so that E(Pi ∩ Pj ) = ∅ (1 ≤ i < j ≤ k). Suppose u, v, w
- 3. X Z X,Y Y,Z Figure 1: Counterexample for Problem 6b). are distinct and u, v are k-linked, and so are v, w. Does it follow that u, w are k-linked? Solution: Yes. By Theorem 9.4 if u and w are not k-linked then there exists X ⊆ V (G) with u in X, w ̸∈ X and |δ(X)| < k. By symmetry, we may assume v ∈ X. Then the opposite direction of Theorem 9.4 implies that v and w are not k-linked. b) Subsets X, Y ⊆ V (G) are k-joined if |X| = |Y | = k and there are k paths P1 , ..., Pk of G from X to Y so that V (Pi ∩ Pj ) = ∅ (1 ≤ i < j ≤ k). Suppose X, Y, Z ⊆ V (G) and X, Y are k-joined, and so are Y, Z. Does it follow that X, Z are k-joined? Solution: No. See Figure 1 for an example with k = 2.