15b_01f_fs

15b_01f_fs - Math 15B Final Exam Solutions 6 December 2001...

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Math 15B Final Exam Solutions 6 December 2001 1. In each case, give an example or explain why none exists . (a) A permutation f of { 1 , 2 , 3 , 4 , 5 } such that, for some x ∈{ 1 , 2 , 3 , 4 , 5 } , f 20 ( x ) 6 = x . A. In cycle form, choose f to have a 3-cycle and x to belong to the cycle. (There are 120 correct answers.) (b) A permutation f of { 1 , 2 , 3 , 4 , 5 } such that, for every x 1 , 2 , 3 , 4 , 5 } , f 20 ( x ) 6 = x . A. f 20 ( x )= x if and only if x belongs to a cycle of length k for some divisor k of 20. Since the domain has 5 elements, k =1 , 2 , 4, or 5. Thus, f 20 ( x ) 6 = x implies that x belongs to a 3-cycle. Hence every x 1 , 2 , 3 , 4 , 5 } must belong to a 3-cycle, which is impossible. (c) A tree with exactly 10 vertices and exactly 10 edges. A. A tree with n vertices has n - 1 edges, so this is impossible. 2. In each case, give an example or explain why none exists . (a) A function f ( n ) such that f ( n )isin O ( n 2 ) but f ( n ) is not Θ( n 2 ). A. Any function which grows slower than n 2 will work; for example, f ( n n . (b) A function f ( n ) such that f ( n )is O ( n log n ) but f ( n ) is not O ( n 2 ). A. Impossible, since anything that doesn’t grow faster than n log n does not grow faster than n 2 . (Alternatively, you could note that ( n log n ) /n 2 is bounded for large n .) (c) A probability space ( U, P ) and two subsets S and T of U such that P ( S P ( T )=2 / 3 and S 6 = T . A. There are many possible answers. For example, U = { a, b, c } , P is uniform, S = { a, b } and T = { b, c } . Another: S = U = { a, b } , T = { a } and P ( a ) is chosen greater than 1 / 2.
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15b_01f_fs - Math 15B Final Exam Solutions 6 December 2001...

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