1.5 Problems and solutions

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1 15 PROBLEMS AND SOLUTIONS Problems and solutions Homework 1: 01 Show by induction that n n 2 = n(n + 1)(2n + 1) 01 Show by induction that n n 2 = n(n + 1)(2n + 1) We ve already seen that the first two cases are valid Suppose that the statement is true in case k Then we have We would like to show that the truth of this statement implies the truth of the equivalent statement with k replaced by k + 1: k+1 (k + 1)(k + 2)(2k + 3) To show this, let s start with what we are given Now add (k + 1) 2 to both sides (k + 1) 2 + (k + 1) 2 + The left side is k+1 j 2

2 12 CHAPTER 1 INTRODUCTION while the right side is (k + 1) 2 + ( k + + k(2k + 1) ) ( k + + 2k 2 + k ) ( (k + 2)(2k + 3) ) (k + 1)(k + 2)(2k + 3) = Thus we ve shown that if the formula is true in case n = k, then it is true in case n = k + 1 But since it is true in the case n = 1, it must be true for every positive integer n 121 If G is a graph of order N, what is the maximum number of edges in G For a complete graph, with every edge present, we get ( N 2) = N(N 1)/2 edges 122 Prove that for any graph G of order at least 2, the degree sequence has at least one pair of repeated entries If the graph has no edges, but at least 2 vertices, then 0 is repeated in the degree sequence If the graph has at least one edge, but at most one isolated vertex, discard the isolated vertex and consider the remaining graph The degree sequence has one entry for each of the N vertices The minimum degree is 1, while the maximum degree is N 1 Since there are at most N 1 distinct degrees, but there are N entries, at least one must appear twice 124 Is it true that a finite graph having exactly two vertices of odd degree must contain a path from one to the other? If not, then there are two components of the graph with a single vertex of odd degree Treating such a component as a graph, it has a single vertex of odd degree This violates Theorem Let G be a FINITE graph where δ(g) k (a) Prove that G has a path of length at least k Recall that the length of a path is the number of edges it contains Try a proof by induction on k If k = 1 there is an edge, and this is the needed path Suppose K 2, the result is true for k < K, and δ(g) = K By the induction hypothesis G has a path P of length at least K 1 If the length of P is at least K we re done, so suppose the length is K 1 and the sequence of vertices is v 1,,v K The vertex v 1 has at least K adjacent

3 15 PROBLEMS AND SOLUTIONS 13 vertices, so there is a vertex w which is adjacent to v 1, but is not on the path P Thus w, v 1,,v K is a path in G with K edges (b) If K 2, prove that G has a cycle of length at least K + 1 Let P = v 0, v 1,,v J be a path of maximum length in G The argument of part (a) shows that J K, and that every vertex w k adjacent to v 0 is on the path P Suppose, starting at v 0, the vertices w k appear in the order w 1,,w L Then extend the path v 0,,w L to a cycle by adding the edge w L v 0 The resulting cycle has at least K + 1 vertices, and thus has length at least K Prove that every closed odd walk in a graph contains an odd cycle First consider the shortest possible closed odd walk, which has 3 edges It must have 3 distinct vertices, and then a 3-cycle Proceed by induction Let the walk contain vertices v 1,,v N and edges v n v n+1 along with v N v 1 If this walk is not a cycle already, there must be a least index j such that v j = v k for some k with j < k N Construct two new walks, w 1 with vertices v 1,,v j, v k+1,,v N and w 2 with vertices v j+1,,v k If k j is even, then w 1 is a shorter odd walk, otherwise w 2 is a shorter odd walk Eventually we reach an odd cycle 128 Let P 1 and P 2 be two paths of maximum length in a connected graph G Prove that P 1 and P 2 have a common vertex If not, let P 3 be the path of minimal length connecting a vertex v on P 1 to a vertex w on P 2 Since the path has minimal length it contains no other vertices from P 1 or P 2 (or you could shrink the path) One of the paths joining an end of P 1 to an end of P 2 through P 3 now has greater length 129 Let G be a graph of order N that is not connected What is the maximum size of G? Start with a complete graph, with size N(N 1)/2 Remove the N 1 edges incident on a single vertex The resulting graph has (N 1)[N/2 1] edges If G is not connected, we may divide it into two subgraphs with no connecting edges Let the orders be K and N K The maximum number of edges for the two graphs (if complete) is N(N 1)/2 K(N K) Now the minimum (by calculus) of x(n x) is at an endpoint, x = 1, (N 1), so the best case is to use K = 1 as above 1210 Let G be a graph of order N and size strictly less than N 1 Prove that G is not connected

4 14 CHAPTER 1 INTRODUCTION We give two proofs A connected graph G can be constructed sequentially in the following way Start with any vertex v 0, and define the graph G 0 to have the single vertex v 0 and no edges Given G n with vertex set V n and edge set E n, define G n+1 by finding an edge vw which is not in G n, but with at least one vertex v in G n Define V n+1 = V n w and E n+1 = E n vw Continue this process until no more edges can be added Since G is connected, every edge of G will be added in this process At each step we added one edge, and at most one vertex Thus if G has order N and is connected, we find that the number of edges is at least N 1 The second proof is by induction The first case has N = 2, with zero edges This graph is not connected Suppose the result is true for graphs of order less than N, and let G have order N and strictly less than N 1 edges By Theorem 11 of the text, the sum of the degrees is less than 2(N 1) Thus some vertex v has degree less than two If v has degree zero we re done If v has degree 1 we can remove v and its one incident edge and get a disconnected graph G \ v with components G 1 and G 2 by the induction hypothesis Adding back v and its one adjacent edge cannot make the graph connected, since there is still no path from G 1 to G Prove that an edge e is a bridge of G if and only if e lies on no cycle of G Suppose that e = v 1 v 2 is on a cycle, with vertices v 1, v 2,,v K Any walk containing e can be replaced by a walk which uses v 1, v K,,v 2 instead, so if e is on a cycle, its removal will not change the number of connected components, that is e is not a bridge Suppose that e = v 1 v 2 is on no cycle Then v 1 and v 2, which are in the same component of G, are in different components of G e, since any path between them in G e allows us to build a cycle containing e Thus e is a bridge 1214 Prove that every 2-connected graph G of order N contains at least one cycle Since G is 2-connected, it must have δ(g) 2 By 125 (b) there is a cycle 121 Let G be a graph of order N (a) Show that if δ(g) (N 1)/2, then G is connected, but (b) there are graphs with δ(g) (N 2)/2 which are not connected (a) We use an argument similar to that from problem 9 above If G is not connected, we may divide it into two subgraphs G 1 and G 2

5 15 PROBLEMS AND SOLUTIONS 15 with no connecting edges and orders K and N K For vertices v in G 1 we have deg(v) K 1 and for vertices w in G 2 we have deg(w) N K 1 Suppose we know that δ(g) (N 1)/2 Then That is (N 1)/2 deg(v) K 1, (N 1)/2 deg(w) N K 1 K (N 1)/2 + 1, K N 1 (N 1)/2 = (N 1)/2 There is no such K (b) Suppose N is even Divide the vertice into two sets of size N/2, and let G 1 and G 2 be the complete graphs on N/2 vertices The combined graph G is disconnected, but the minimum degree is δ(g) = N/ Determine whether K 4 is a subgraph of K 4,4 K 4 has odd cycles, but since K 4,4 is bipartite it can t contain any odd cycles by Theorem List all of the unlabeled connected subgraphs of C 34 Any of the paths P n for n = 1,,33

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