Discrete Mathematics with Graph Theory (3rd Edition) 274

# Discrete Mathematics with Graph Theory (3rd Edition) 274 -...

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272 Solutions to Exercises (b) Yes, because the graph is Eulerian. An Eulerian circuit is ABCDOFGBOEA. 10. [BB] The result is obvious if n = 1 since in this case there are just two people who are friends. So we assume that n > 1. Consider the graph whose vertices correspond to people and where an edge between vertices v and u signifies that v and u are friends. The question asks us to prove that this graph is Hamiltonian. This is an immediate consequence of Dirac's Theorem since the graph in question has 2n ~ 3 vertices each of degree d ~ n = 2;. 11. (a) [BB] n edges. (b) If n = 2, there are no cycles, so assume n ~ 3. Numbering the vertices 1,2, ... , n and beginning at 1, there are n - 1 choices for the second vertex in a cycle (since all edges Ii exist), then n - 2 choices for the third vertex and so on. Altogether, there are (n - I)! different Hamiltonian cycles in
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Unformatted text preview: K n . (c) [BB] Kn has n(n 2-1) edges, so the maximum is n 2 1 edge disjoint cycles. (d) If n = 2, there are no Hamiltonian cycles (and therefore no edge disjoint ones). 1 If n = 3, the graph is ~. 2 3 no edge disjoint Hamiltonian cycles. 1 2 Ifn ~ 4, thegmphis ~ 4 3 ~ Ifn= 5, the graph is 5~3. 4 123 and 132 are the only Hamiltonian cycles; so there are The Hamiltonian cycles are 1234, 1243, 1324, 1342, 1423 and 1432. No pair are edge disjoint. The Hamiltonian cycles are 12345, 12354, 12435, 12453,12534,12543,13245,13254,13425,13452, 13524,13542,14235,14253,14325,14352,14523, 14532,15234,15243,15324,15342,15423,15432. Since n;-l = 2, we could possibly have two edge disjoint cycles and we do; for instance, 12345 and 13524. 12. [BB] The cube is indeed Hamiltonian; the labels 1, ... ,8 on the vertices exhibit a Hamiltonian cycle. LQ) : 1 8 J----~ .."" ~ 4 5 13. 1 3...
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## This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.

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