11

1.5. PROBLEMS AND SOLUTIONS

1.5

Problems and solutions

Homework 1: 0.1 Show by induction that n X

j 2 = 1 + 22 + · · · + n2 =

j=1

n(n + 1)(2n + 1) . 6

0.1 Show by induction that n X

j 2 = 1 + 22 + · · · + n2 =

j=1

n(n + 1)(2n + 1) . 6

We’ve already seen that the first two cases are valid. Suppose that the statement is true in case k. Then we have k X

j2 =

j=1

k(k + 1)(2k + 1) . 6

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 X

j2 =

j=1

(k + 1)(k + 2)(2k + 3) . 6

To show this, let’s start with what we are given. k X

j2 =

j=1

k(k + 1)(2k + 1) . 6

Now add (k + 1)2 to both sides. 2

(k + 1) +

k X

j 2 = (k + 1)2 +

j=1

The left side is

k+1 X j=1

j2

k(k + 1)(2k + 1) . 6

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CHAPTER 1. INTRODUCTION

while the right side is 6(k + 1)2 k(k + 1)(2k + 1) + 6 6  6k + 6 + k(2k + 1)   6k + 6 + 2k 2 + k  = (k + 1) = (k + 1) 6 6  (k + 2)(2k + 3)  (k + 1)(k + 2)(2k + 3) = (k + 1) = . 6 6 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. 1.2.1 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 N2 = N(N − 1)/2 edges. 1.2.2 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. 1.2.4 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 1.1. 1.2.5 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 v1 , . . . , vK . The vertex v1 has at least K adjacent

1.5. PROBLEMS AND SOLUTIONS

13

vertices, so there is a vertex w which is adjacent to v1 , but is not on the path P . Thus w, v1 , . . . , vK 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 = v0 , v1 , . . . , vJ be a path of maximum length in G. The argument of part (a) shows that J ≥ K, and that every vertex wk adjacent to v0 is on the path P . Suppose, starting at v0 , the vertices wk appear in the order w1 , . . . , wL . Then extend the path v0 , . . . , wL to a cycle by adding the edge wL v0 . The resulting cycle has at least K + 1 vertices, and thus has length at least K + 1. 1.2.6 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 v1 , . . . , vN and edges vn vn+1 along with vN v1 . If this walk is not a cycle already, there must be a least index j such that vj = vk for some k with j < k ≤ N. Construct two new walks, w1 with vertices v1 , . . . , vj , vk+1 , . . . , vN and w2 with vertices vj+1 , . . . , vk . If k − j is even, then w1 is a shorter odd walk, otherwise w2 is a shorter odd walk. Eventually we reach an odd cycle. 1.2.8 Let P1 and P2 be two paths of maximum length in a connected graph G. Prove that P1 and P2 have a common vertex. If not, let P3 be the path of minimal length connecting a vertex v on P1 to a vertex w on P2 . Since the path has minimal length it contains no other vertices from P1 or P2 (or you could shrink the path). One of the paths joining an end of P1 to an end of P2 through P3 now has greater length. 1.2.9 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. 1.2.10 Let G be a graph of order N and size strictly less than N −1. Prove that G is not connected.

14

CHAPTER 1. INTRODUCTION

We give two proofs. A connected graph G can be constructed sequentially in the following way. Start with any vertex v0 , and define the graph G0 to have the single vertex v0 and no edges. Given Gn with vertex set Vn and edge set En , define Gn+1 by finding an edge vw which is not in Gn , but with at least one vertex v in Gn . Define Vn+1 = Vn ∪ w and En+1 = En ∪ 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 1.1 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 G1 and G2 by the induction hypothesis. Adding back v and its one adjacent edge cannot make the graph connected, since there is still no path from G1 to G2 . 1.2.11 Prove that an edge e is a bridge of G if and only if e lies on no cycle of G. Suppose that e = v1 v2 is on a cycle, with vertices v1 , v2 , . . . , vK . Any walk containing e can be replaced by a walk which uses v1 , vK , . . . , v2 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 = v1 v2 is on no cycle. Then v1 and v2 , 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. 1.2.14 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 1.2.5 (b) there is a cycle. 1.2.16 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 G1 and G2

1.5. PROBLEMS AND SOLUTIONS

15

with no connecting edges and orders K and N − K. For vertices v in G1 we have deg(v) ≤ K − 1 and for vertices w in G2 we have deg(w) ≤ N − K − 1. Suppose we know that δ(G) ≥ (N − 1)/2. Then (N − 1)/2 ≤ deg(v) ≤ K − 1,

(N − 1)/2 ≤ deg(w) ≤ N − K − 1.

That is 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 G1 and G2 be the complete graphs on N/2 vertices. The combined graph G is disconnected, but the minimum degree is δ(G) = N/2 − 1. 1.3.3 Determine whether K4 is a subgraph of K4,4 . K4 has odd cycles, but since K4,4 is bipartite it can’t contain any odd cycles by Theorem 1.3. 1.3.5 List all of the unlabeled connected subgraphs of C34 . Any of the paths Pn for n = 1, . . . , 33.