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Unformatted text preview: Math 55: Discrete Mathematics, Fall 2008 Homework 6 Solutions 4.3: 6(b) The function is welldefined and given by the formula f ( n ) = 2 n/ 3 for n ≡ (mod 3) for n ≡ 1 (mod 3) 2 ( n +1) / 3 for n ≡ 2 (mod 3) To prove it, observe that this gives the correct initial values f (0) = 1, f (1) = 0, f (2) = 2, and that for n ≥ 3 the above formula satisfies the recurrence f ( n ) = 2 f ( n 3). * 12. We prove that f 2 1 + ··· + f 2 n = f n f n +1 by induction on n . Basis step n = 1 is true, as both sides are equal to 1. For n > 1, assume by induction that f 2 1 + ··· + f 2 n 1 = f n 1 f n . Adding f 2 n to both sides gives f 2 1 + ··· + f 2 n = f n 1 f n + f 2 n = f n ( f n 1 + f n ) = f n f n +1 . 30. Let’s denote the number of occurences of 01 in a bit string w by b ( w ) and the number of occurences of 10 by a ( w ). We are to prove that b ( w ) ≤ a ( w ) + 1 for any given bit string w . We may assume by strong induction that the inequality holds for all shorter bit strings. If w is empty or has length 1, then a ( w ) = b ( w ) = 0, so we can assume w has length at least 2. If w ends with 0, say w = v 0, then b ( w ) = b ( v ) ≤ a (...
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This note was uploaded on 08/16/2010 for the course MATH 55 taught by Professor Strain during the Fall '08 term at Berkeley.
 Fall '08
 STRAIN
 Math

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