sol5 - AMS 301 Fall, 2009 Homework Set # 5 Solution Notes...

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AMS 301 Fall, 2009 Homework Set # 5 — Solution Notes Problem A: (a). How many internal nodes does the tree have? n = mi + 1, n = 81, m = 5, so i = 16. n = i + l , l = 81 16 = 65. (b). How many edges does the tree have? 81 1 = 80 (c). What is the height of T ? log 5 65 = 3 # 16, Section 3.1: Let e i (resp., v i ) be the number of edges (resp., vertices) in tree i ( i = 1 , 2 , . . . , t ). In tree i , we know that e i = v i 1. Sum this equation over all i to get: i e i = i ( v i 1). But, i e i = e , and i ( v i 1) = ( i v i ) ( i 1) = v t (since there are t terms in the summation). Thus, e = v t . # 26(a), Section 3.1: Number the coins 1 , 2 , . . . , 20. There are 21 possible outcomes: “All Fair”, (since problem states that AT MOST 1 coin is too light) 1L (“coin 1 is too light”), 2L, . .., 20L. In any decision tree that solves the problem, there must be (at least) 21 leaves corresponding to the possible outcomes. Now, each internal vertex of the tree will correspond to placing some number,
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sol5 - AMS 301 Fall, 2009 Homework Set # 5 Solution Notes...

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