288 Solutions to Review Exercises Case 3a. AD is in 1i while AE is not. Since 1i contains precisely two edges incident with E, both BE and DE are in 1i and we have a proper cycle in 1i. Case 3b. AE is in 1i while AD is not. This time BE is not in 1i (else, 1i would contain a proper cycle). We need two edges incident with E and with B, so both DE and BD are in 1i, so 1i does not contain a path from D to any ofthe vertices F, G, H, I, J. This concludes the proof. 11. (a) 2n edges. (b) If a graph is Hamiltonian, any vertex is the starting point of a Hamiltonian cycle. So let v be a fixed vertex. There are n choices for the second vertex in the cycle, then n -1 for the third, n -1 for the fourth (which is in the bipartition set not containing v), and so on. Altogether, there are n[(n -1)!]2 different Hamiltonian cycles in Kn,n. 2 (c) Kn n has n2 edges and each cycle requires 2n edges, so the maximum is ~ = !!: edge disjoint , 2n 2
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Glossary of graph theory, hamiltonian cycle, Graph theory objects, hamiltonian cycles, Eulerian