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15b_00f_fs

15b_00f_fs - Math 15B Final Exam Solutions 8 December 2000...

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Math 15B Final Exam Solutions 8 December 2000 1. (60 pts) In each case, give an example or explain why none exists. (a) A tree with exactly nine vertices and exactly nine edges. Ans : Impossible since a tree always has one less edge than it has vertices. (b) A permutation f on { 1 , 2 , 3 , 4 } such that f 100 6 = f . Notes : Remember that f 6 = g for functions with the same domain means there is at least one x such that f ( x ) 6 = g ( x ). Also remember that f 100 ( x ) means f ( f ( · · · f ( x ) · · · )), NOT ( f ( x )) 100 . Ans : The question is equivalent to asking that f 99 not be the identity function. Such a permutation must have at least one cycle whose length does not divide 9. Since we are permuting a 4-set, the possible cycle lengths are 1, 2, 3 and 4. Thus f must be either a 4-cycle (there are six possibilities) or a product of two 2-cycles (there are three possibilities). Any of the nine possible answers is acceptable. (c) A simple graph with exactly five vertices that has a cycle containing six edges. Ans : Impossible, if we had six edges, the cycle would pass through at least one vertex twice. (d) A sample space U with a probability function P and two different elements s and t of U such that P ( s ) > 1 / 2 and also P ( t ) > 1 / 2. Note : P ( s ) means the same thing as P ( { s } ). Ans : Impossible since P ( x ) 0 and 1 = x U P ( x ) P ( s ) + P ( t ). (e) A sample space U with a probability function P and two different subsets S and T of U such that P ( S ) > 1 / 2 and also P ( T ) > 1 / 2. Ans : There are many possibilities. For example, the uniform distribution on U = { 1 , 2 , 3 } with S = { 1 , 2 } and T = { 1 , 2 , 3 } . (f) Two functions f ( n ) and g ( n ) such that “ f ( n ) is O ( g ( n ))” is TRUE and, at the same time , “ g ( n ) is O ( f ( n ))” is FALSE.

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15b_00f_fs - Math 15B Final Exam Solutions 8 December 2000...

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