# MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 36, Monday, December 1.

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MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 36, Monday, December 1

7.2. Divide-and-Conquer Relations Homework (MATH 310#11M): Read 7.2,7.3 Do 7.1,7.2: all odd numbered problems Turn in 7.2: 2,4,6,8

Divide-and-Conquer Divide the problem in two smaller parts, solve each subproblem nad combine solutions. a n = ca n/2 + f(n).

Divide-and-Conquer Divide the problem in two smaller parts, solve each subproblem nad combine solutions. a n = ca n/2 + f(n). cf(n)anan 1dd d log 2 n e + A 2dAn – d >2dnan log2c + (2d/(2-c)) n 2dndn( d log 2 n e + A)

Example 1: Rounds in a Tournament Number of rounds in a tennis tournament. Number of players: n = 2 k, for some k. a n = a n/2 + 1. a n = log 2 n + A. Observe that a 1 = 0, and so A = 0. On the left: n = 8, a 8 = 3.

Example 2: Finding the Largest and the Smallest Element in a Set a n = 2a n/2 + 2 a 2 = 1 a n = (3/2)n – 2. m,M m 1,M 1 m 2,M 2 m = min{m 1,m 2 } M = max{M 1,M 2 }

Example 3: Efficient Multidigit Multiplication g = [g 1,g 2 ] = g 1 £ 10 n/2 + g 2. h = [h 1,h 2 ] = h 1 £ 10 n/2 + h 2. g £ h = (g 1 £ h 1 ) £ 10 n + (g 1 £ h 2 + g 2 £ h 1 ) £ 10 n/2 +(g 2 £ h 2 ). We need only (g 1 £ h 1 ), (g 2 £ h 2 ) and (g 1 + g 2 ) £ (h 1 + h 2 ). a n = 3a n/2. a n = n log2 3 = n 1.5849....

Theorem Let a n = ca n/k + f(n) be recurrence relation with positive constant c and the positive function f(n). (a) If for large n, f(n) grows proportional to n logk c, then a n grows proportional to n logk c log 2 c. (b) If for large n f(n) · pn q, where p is a positive constant and q < logk c, then a n grows at most at rate proportional to n logk c.

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