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Unformatted text preview: B, C, D, E, F) and we also observe that there is a solution using four extra edges. So we have an immediate answer. The table shows the calculations required for the weighted graph and, just for interest's sake, for the unweighted graph as well. 2+5=7 1+3=4 4+3=7 A B / . . ) c .,"', F D / . .... E 1 One of three solutions to the unweighted problem, each requiring the duplication of four edges, is shown above. One of several solutions to the weighted problem, each requiring the duplication of edges of total weight 6, is also shown....
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 Summer '10
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 Graph Theory, Addition

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