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Discrete Mathematics with Graph Theory (3rd Edition) 296

Discrete Mathematics with Graph Theory (3rd Edition) 296 -...

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294 Solutions to Exercises 2. [BB; K5] K5 is Eulerian, so no additional edges are needed. The solution for K6 is on the left and for K 3,5 on the right. ~ ~ 3. [BB] There are four odd vertices and the table shows the requisite calculations. Partition into pairs Sum of lengths of shortest paths {A,B}, {C,D} {A,C}, {B,D} {A, D}, {B, C} The unique solution is shown. 4. [BB] There are six odd vertices, labeled A, ... , F in the figures to the right. In the unweighted case, we know that at least four additional edges are required (since A is not adjacent to any of
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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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