Discrete Mathematics with Graph Theory (3rd Edition) 358

Discrete Mathematics with Graph Theory (3rd Edition) 358 -...

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356 Solutions to Review Exercises 2. (a) This is JC3,3. (b) There are twelve Hamiltonian cycles. By checking all cases, we see that there are two pairs of solutions: W4 Ws W4WS W4 WIW2W3W4W5W6WI (or its reverse) and Ws Ws W3W3 W3 WI W2W3W6W5W4WI (or its reverse). Ws W4 WSW4 W4W2 W4 W2 WI WI WI WI WI WI WI WI WI WI WI WI 3. (a) No. A tree with nine vertices has eight edges, and hence the sum of the degrees of its vertices should be 16. Here it is 18. (b) No. As in (a), the sum of the degrees of the vertices must be 16. Two vertices of degree 5 would leave a total of only 6 for the sum of the degrees of the remaining seven vertices. This forces some vertex to have degree 0, contradicting the fact that a tree is connected. 4. (i. ---+ ii.) Assume T has an Eulerian trail. By Theorem 10.1.5, T must have exactly two vertices of odd degree. But Corollary 12.1.7 says T has at least two vertices of degree 1. It follows that T has exactly two vertices of degree 1. (ii. ---+
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Unformatted text preview: iii.) Assume T has exactly two vertices of degree 1. Since T has n -1 edges, the sum of the degrees of its vertices is 2(n -1) = 2n - 2. This means that the remaining n - 2 vertices must have degrees totalling 2n - 4 = 2(n - 2). But each of these has degree at least 2, and it follows that each must have degree exactly 2. (iii. ---+ i.) Assume T has n - 2 vertices of degree 2. Corollary 12.1.7 tells us that the other two vertices have degree 1. Since T is connected and has exactly two vertices of odd degree, Theorem 10.1.5 guarantees an Eulerian trail. 5. (a) 9 is connected because there is a path between any two vertices. If it contained a cycle, there would be two paths between any vertices of this cycle, contrary to fact. So 9 is acyclic and connected, hence a tree. (b) A tree with n vertices has n-l edges and I: degv = 211. Here then, 33(1)+6(3) +(n-39)7 = 2( n -1). This gives 5n = 220, so n = 44 and the number of vertices of degree 7 is 44 - 39 = 5....
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