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Discrete Mathematics with Graph Theory (3rd Edition) 322

# Discrete Mathematics with Graph Theory (3rd Edition) 322 -...

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320 (d) The only abnormal fragment is AG and this does not split, so look for Eulerian trails or circuits whose last vertex is AG. There is a unique Eulerian trail and the chain is AUC- GAAGCGGUAAUAG. AG UAAUAG U Solutions to Review Exercises AU c AUCG GGU G 10. The score of vertex Wi is n - i. Since all scores are different, the result follows from Theorem 11.4.4. 11. (a) No. The sum of all the scores is 23 =1= 21 = m. (b) No. Consider the five players with the lowest scores. Among themselves, they playa total of m = 10 games and yet their total scores add only to 9. 12. Reversing the orientation on all the arcs of T gives a tournament in which v has maximum score. Exercise 6 of Section 11.4 says that the length of the shortest path from v to any other vertex in the new tournament is 1 or 2, so our result follows. 13. Let n be the number of players competing and assume first that the n - 2 players tied for second all
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Unformatted text preview: had winning records. The sum of the outdegrees of these vertices would then exceed the sum of the indegrees by at least n -2. If we add Alice's score to the sums, the sum of the outdegrees would exceed the sum of the indegrees by at least n (since Alice has a better record than the players tied for second). In any digraph, however, the outdegree and indegree sums are the same, so it follows that George's indegree must be at least n greater than his outdegree. This is impossible. A similar argument shows that the players tied for second cannot have losing records, so these players must have an equal number of wins and losses. Since each player has n -1 matches, it follows that n - 1 is even, so n is odd. 14. This is a type I scheduling problem. The shortest time is 20 hours. There are two critical paths: St, V, Q, R, S, T, U and St, V, Q, R, S, X, U. St( -,0) N(Y, 12) T(S, 15) or U(T, 20)...
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