266 Solutions to Exercises synunetric. Finally, if U '" v and v'" w, then there is a walk U = UO, Ub ... ,Uk = v from U to v and a walk v = vO,vb . .. ,Ve = w from v to w. Butthenu = UO,Ub . .. ,Uk = V = VO,V1, . .. ,Ve = w is a walk from U to w, proving U '" w and establishing transitivity. 19. No. Let 9 be as shown and let circuit C be ABC D A. After removal of the edges of C, the remaining graph 91 consists of the 3-cycles AF EA and DHGD. This is not connected. F G 20. Let A be a vertex in 91 and B be a vertex in the circuit C 1. Since 9 is connected, there is a walk in 9 from A to B. If the edge AU1 is in C 1, then A is a vertex in both 91 and C 1 and we are done. Otherwise, the edge AU1 is not in C1, so AU1 is in 91 and hence U1 is in 91. Repeat this argument. Eventually, either we find a vertex Ui in both 91 and C 1 or else none of the edges along the above path is in C 1. In this case, B is in 91 as well as C 1 and again, we are done. 21. We are given that there is an isomorphism
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