# HW12_Solutions (1).pdf - Homework 12 Solutions December 8...

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Homework 12 Solutions December 8, 2016 Section 10.4 10. An actor is in the same connected component as Kevin Bacon if there is a path from that person to Bacon. This means that the actor was in a movie with someone who was in a movie with someone who . . . who was in a movie with Kevin Bacon. This includes Kevin Bacon, all actors who appeared in a movie with Kevin Bacon, all actors who appeared in movies with those people, and so on. 16. The given conditions imply that there is a path from u to v , a path from v to u , a path from v to w, and a path from w to v. Concatenating the first and third of these paths gives a path from u to w , and concatenating the fourth and second of these paths gives a path from w to u . Therefore u and w are mutually reachable. 18. Let a, b, c, . . . , z be the directed path. Since z and a are in the same strongly connected component, there is a directed path from z to a. This path appended to the given path gives us a circuit. We can reach any vertex on the original path from any other vertex on that path by going around this circuit. 20. The graph G has a simple closed path containing exactly the vertices of degree 3, namely u 1 u 2 u 6 u 5 u 1 . The graph H has no simple closed path containing exactly the vertices of degree 3 . Therefore the two graphs are not isomorphic. 24. (a) Adjacent vertices are in different parts, so every path between them must have odd length. There- fore there are no paths of length 2. (b) A path of length 3 is specified by choosing a vertex in one part for the second vertex in the path and a vertex in the other part for the third vertex in the path (the first and fourth vertices are the given adjacent vertices). Therefore there are 3 · 3 = 9 paths. (c) As in part (a), the answer is 0. (d) This is similar to part (b); therefore the answer is 3 4 = 81. 28. We show this by induction on n . For n = 1 there is nothing to prove. Now assume the inductive hypothesis, and let G be a connected graph with n + 1 vertices and fewer than n edges, where n 1 . Since the sum of the degrees of the vertices of G is equal to 2 times the number of edges, we know that the sum of the degrees is less than 2 n , which is less than 2( n + 1). Therefore some vertex has degree less than 2. Since G is connected, this vertex is not isolated, so it must have degree 1. Remove this vertex and its edge. Clearly the result is still connected, and it has n vertices and fewer than n - 1 edges, contradicting the inductive hypothesis. Therefore the statement holds for G , and the proof is complete. 30. Let v be a vertex of odd degree, and let H be the component of G containing v . Then H is a graph itself, so it has an even number of vertices of odd degree. In particular, there is another vertex w in H with odd degree. By definition of connectivity, there is a path from v to w .

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• Fall '19
• Graph Theory, Glossary of graph theory, Hamiltonian path, Graph connectivity

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