290 Solutions to Review Exercises (b) The answer is 1 since there is an edge from Vs to WI. (c) The answer is 8 since there are 8 walks oflength two from W2 to W4 (via any of VI, ... , Vs). (d) The answer is 12 for the same reason as (c). (e) The answer is 0 because there is no walk of length two from WI2 to V6. (f) The answer is (12)(8)(12) = 1152. (g) The answer is 0 because there is no walk of odd length from V5 to V7. 18. Yes it is. By Theorem 10.1.4, a connected graph is Eulerian if and only if the degree of every vertex is even and this can easily be determined from the adjacency matrix. Remember that the degree of vertex i is the number of l's in row (or column) i. 19. There are at least four l's in each row. So the degree of each vertex is d ~ 4 ~ ~. The graph has 7 vertices, so the result follows from Dirac's Theorem. 20. Let the vertices of g be VI, . .• , Vn. The number b ii is the number of walks of length 2 from Vi to Vi, that is, the number of edges incident with Vi, equivalently, deg
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