hw06sol - Math 55 Discrete Mathematics Fall 2008 Homework 6...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 55: Discrete Mathematics, Fall 2008 Homework 6 Solutions 4.3: 6(b) The function is well-defined 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 (...
View Full Document

This note was uploaded on 08/16/2010 for the course MATH 55 taught by Professor Strain during the Fall '08 term at Berkeley.

Page1 / 2

hw06sol - Math 55 Discrete Mathematics Fall 2008 Homework 6...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online