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Unformatted text preview: vertex has degree 2 must be a cycle. But if the vertices of the cycle, written in order, are V!, V2, • .. ,V2n, then edges Vl V2, V3V4, ... , V2nl Vn give a perfect matching. 11. (a) The sum of the degrees of the vertices ofg is 31VI. This is also 21&1, so 211VI. (b) No, and we present a counterexample at the right. If this graph had a perfect matching, the middle vertex V would have to be joined to an edge which is part of the matching, say vw as shown. Then the five vertices in the leftmost group (or in the bottom group) could not all be saturated, since they could only be joined to each other....
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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.
 Summer '10
 any
 Graph Theory

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