Section 10.1 265 12. (a) [BB] Kn is Eulerian +-+ n is odd. (b) [BB] Kn has an Eulerian trail if and only if n = 2. For n = 2, certainly 0-0 has an Eulerian trail. For n > 2, if two vertices have odd degree, then there are other vertices of odd degree so no Eulerian trail can exist. 13. (a) Since Km,n is connected, we seek only necessary and sufficient conditions for every vertex to be even. In Km,n, every vertex in the bipartition set with m vertices has degree n and every vertex in the other bipartition set has degree m. Hence, Km,n is Eulerian if and only if m and n are both even. (b) Km,n, m :::; n has an Eulerian trail if and only if m = 1, n = 2 or m = 2 and n is odd. By Theorem to.1.5, complete bipartite graphs with the specified conditions on m and n have Eulerian trails. On the other hand, since Km, n has n vertices of degree m and m of degree n, if m > 2, then each bipartition set contains more than two vertices (since n ~ m) so it is impossible for there to exist precisely two vertices of odd degree.
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