254 Solutions to Exercises puts 1 and 5 in the same set, a contradiction. R I§} 3 w~: R W (e) This is bipartite, with bipartition sets indicated R and W on the graph at the right above. (f) This is not bipartite because it contains a triangle. 22. (a) [BB] At least two of the three vertices must lie in one of the bipartition sets. Since these two are joined by an edge, the graph cannot be bipartite. (b) A graph which contains an n-cycle with n odd cannot be partite. To see why, let VI V2V3 ... V2k+IVI be an odd cycle. Then VI and V2 must lie in different bipartition sets, so VI and V3 lie in the same bipartition set. Continuing this way, VI, V3, V5, ... ,V2k+1 all lie in the same bipartition set. Since VI and V2k+1 are adjacent, we have a contradiction. 23. (a) Yes. Assign to each vertex of the subgraph the bipartition set it had in the larger graph. Since each edge in the subgraph is an edge in the graph, the sub graph can only contain edges joining vertices in different bipartition sets. (b) Again we say "yes".
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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.