336 (c) The edge ofleast weight incident with E is CEo The edge of least weight among those adjacent to CE is SC. The edge of least weight among those which, together with CE and SC, fonn a tree is BC. Then we must add, in order, DE, AD, and DT. Finally, so as not to duplicate the tree we obtained in Exercise 1, we select C F. (d) Starting at vertex E, we select first edge G E of weight 1 and then GF (to change the tree we had before). Of the edges incident with G, E and F, those of least weight are FE and GC. We cannot select FE because it is incident with two of the vertices already chosen, so we choose GC, then BC, CD and AF. We obtain another spanning tree of weight 10. Solutions to Exercises A 2 D ~T C 6 F B C D F E 3. [BB] We need five edges because there are six vertices in the graph. We use Kruskal's algorithm. First select BC of weight 5, then DF of weight 6. Next we select BF and AB, both of weight 7. We would like to select AF next, but cannot since it completes the circuit
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