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1 Professor Kindred Math 104, Graph Theory Homework 2 Solutions February 7, 2013 Introduction to Graph Theory, West Section 1.2: 26, 38, 42 Section 1.3: 14, 18 Section 2.1: 26, 29, 30 DO NOT RE-DISTRIBUTE THIS SOLUTION FILE Prove that a graph G is bipartite if and only if every subgraph H of G has an independent set consisting of at least half of V(H). ( ) Let G be a bipartite graph. By def n, G has a bipartition X, Y such that X and Y are independent sets (though not necessarily maximum independent sets). Without loss of generality, assume X Y. Then so X contains at least half of V(G). X = 1 2 ( X + X ) 1 2 ( X + Y ) = 1 2 V(G), Since every subgraph of a bipartite graph is bipartite, the above argument applies to all subgraphs of a bipartite graph. ( ) For the converse, we prove the contrapositive statement, so suppose that G is not bipartite. By the characterization of bipartite graphs, G contains an odd cycle H. This subgraph H has no independent set containing at least half its vertices, because every set consisting of at least half the vertices in an odd cycle must have two consecutive vertices on the cycle Prove that every n-vertex graph with at least n edges contains a cycle. Proof 1: Suppose, to the contrary, that G is an n-vertex graph with at least n edges that does not contain a cycle. By Theorem , every edge of G is a cut-edge, and this remains true as edges are deleted from G (since removing an edge of G cannot create cycles). Thus, deleting all edges produces at least n + 1 components, but G has only n vertices. Therefore, it must be that G contains a cycle. 1

2 Proof 2: We prove the statement by induction on n. Base case: The only simple graph on 3 vertices with 3 edges is K 3, which clearly contains a cycle (of length 3). Induction hypothesis: Suppose any graph on n vertices with at least n edges, where n 3, has a cycle. Let G be a graph on n + 1 vertices with at least n + 1 edges. We consider two cases. Case 1: G has a vertex v of degree at most 1. Then G v has n vertices and at least n edges, so by the induction hypothesis, G v must have a cycle, and this cycle also appears in G. Case 2: All vertices in G are of degree at least 2. Then, by Lemma , it follows that G must contain a cycle. In both cases, we conclude that G has a cycle, and so the result holds by induction Let G be a connected (simple) graph that does not have P 4 or C 4 as an induced subgraph. Prove that G has a vertex adjacent to all other vertices. 1 Assume G is a connected graph that does not have P 4 or C 4 as an induced subgraph. Let v be a vertex of maximum degree. Assume, for sake of contradiction, that there exists a vertex z not adjacent to v. Since G is connected, there exists a v, z-path, so consider a shortest one and call it P. x y z v w dashed line -- edge does not exist open circle -- vertex may or may not exist Suppose P is v, x, y,..., z, where y may be z. (See above figure.) Since d(v) d(x), there must be a neighbor w of v such that x w. If not, then d(x) d(v) + 1 = (G) + 1, which is impossible. Also, since P is a shortest v, z-path, we know that v y. Then G[{w, v, x, y}] is either a path of length 4 (if w y) or a cycle of length 4 (if w y). It must be that v is adjacent to all other vertices. 1 In the textbook, the problem statement has a hint look it up if you need it! 2

3 Prove that every simple graph with at least two vertices has two vertices of equal degree. Is the conclusion true for loopless graphs (i.e., graph may have multiple edges)? Let G be a graph on n 2 vertices. For any vertex v, d(v) is an integer with 0 d(v) n 1. Furthermore, if there exists a vertex of degree 0 (an isolated vertex), then there can be no vertex of degree n 1, since such a vertex would have to be adjacent to the isolated vertex. Thus, either the vertex degrees are in the set {0, 1,..., n 2} or {1, 2,..., n 1}, both of which are of size n 1. Since we have n vertices to which we need to assign n 1 different values for degrees, by the pigeonhole principle, there must be two vertices with the same degree. The conclusion is NOT TRUE for loopless graphs. To see this, consider the following graph in which no two vertices have equal degrees (degrees are 1, 2, 3) For k 2, prove that a k-regular bipartite graph has no cut edge. Suppose G is a k-regular bipartite graph, where k 2. Assume BWOC that G has a cut edge uv. Let X, Y be a bipartition of G, and WLOG, assume that u X and v Y. Let H be the component of G uv containing u. All vertices in H have the same degree as they did in G (degree k) except for vertex u which has degree k 1. If S = X V(H) and T = Y V(H), then S, T is a bipartition of H, and we have k S 1 = (k 1) + k( S 1) = d H (w) = E(H) = d H (w) = k T. w S w T We know that s = S and t = T are integers, so ks 1 = kt = k (s t) = 1 }{{} Z since k 2 is an integer. Thus, it must be that G has no cut edge For n 3, let G be an n-vertex graph such that every graph obtained by deleting one vertex is a tree. Determine E(G), and use this to determine G itself. Suppose a graph G = (V, E) has n vertices, and assume G v is a tree for all v V. Then it follows that for any vertex v, E(G v) = V(G v) 1 = (n 1) 1 = n 2. 3

4 Furthermore, every edge uw of G appears in exactly n 2 of the subgraphs {G v : v V} (all except G u and G w). Thus, We conclude that E(G) = n. (n 2) E(G) = E(G v) = (n 2) = n(n 2). v V v V By problem , since G has n edges, we know G must contain a cycle. Because G v contains no cycles, any cycle in G must contain v, implying that every cycle in G contains every vertex. Thus, we know G has a spanning cycle and since a spanning cycle contains n edges, G has no edges beyond those in the spanning cycle. Thus, we conclude that G is C n Every tree is bipartite. Prove that every tree has a leaf in its larger partite set (in both if they have equal size). Let X and Y be the parts of a bipartition of V(T) for a tree T with X Y. Suppose there are k leaves in X. Note that because T is a tree, T is connected, so there are no isolated vertices in T, i.e., vertices that are not leaves have degree at least 2. Then 2 X k = k + 2( X k) d(v) v X = E(T) since every edge of T has exactly one endpoint in X = V(T) 1 because T is a tree = X + Y 1 X + X 1 since X Y = 2 X 1. Solving for k, we find that k 1. Thus, T has at least one leaf in its larger part Let T be a tree in which all vertices adjacent to leaves have degree at least 3. Prove that T has some pair of leaves with a common neighbor. We give four proofs below (though only one is required). Let T be a tree in which all vertices adjacent to leaves have degree at least 3. extremality Consider a longest path P in T. Let u be an endpoint of this path, and let v be u s neighbor in P. Since u is a leaf (P is a longest path, T is acyclic), by our previous assumption, d(v) 3. Since v has only two neighbors in P, there exists a vertex w not in P that is adjacent to v. If w has a neighbor x = v, then replacing (u, v) with (x, w, v) in P yields a longer path than P (x cannot be a vertex of P already since T is acyclic), so such a vertex x cannot exist. Thus, d(w) = 1, and u and w are a pair of leaves with a common neighbor. 4

5 contradiction Suppose, BWOC, that all leaves have different neighbors. Deleting all leaves reduces degree of each neighbor of a leaf by 1. All neighbors had degree at least 3 to begin with, so after deletion of leaves they have degree at least 2. The resulting graph is still a tree but has no leaves. counting Suppose T has k leaves, and let p be the number of vertices that are adjacent to leaves. (We claim that p < k.) Using the degree-sum formula, we have 2n 2 = 2 E(T) = Solving for p, we obtain d(v) k + 3p + 2(n k p) = k + 2n + p. v V(T) p k 2 < k. Thus, there are more leaves than neighbors of leaves, so by the pigeonhole principle, it must be that there exists a pair of leaves with a common neighbor. induction Proof omitted, but such a proof is possible. 5