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Unformatted text preview: 1 CMPS 101 Summer 2009 Homework Assignment 4 Solutions 1. (3 Points) Consider the function ) ( n T defined by the recurrence formula ≥ + < ≤ = 3 ) 3 / ( 2 3 1 6 ) ( n n n T n n T a. (1 Points) Use the iteration method to write a summation formula for ) ( n T . Solution: ) 3 / ( 2 ) ( n T n n T + = ) ) 3 / 3 / ( 2 3 / ( 2 n T n n + + = ) 3 / ( 2 3 / 2 2 2 n T n n + + = ) 3 / ( 2 3 / 2 3 / 2 3 3 2 2 n T n n n + + + = etc.. After substituting the recurrence into itself k times, we get ) 3 / ( 2 3 2 ) ( 1 k k k i i i n T n n T + = ∑ = . This process terminates when the recursion depth k is chosen so that 3 3 / 1 < ≤ k n , which is equivalent to 3 3 / 1 < ≤ k n , whence 1 3 3 + < ≤ k k n , so 1 ) ( log 3 + < ≤ k n k , and hence ) ( log 3 n k = . With this value of k we have 6 ) 2 or 1 ( ) 3 / ( = = T n T k . Therefore ) ( log 1 ) ( log 3 3 2 6 3 2 ) ( n n i i i n n T ⋅ + = ∑ = . b. (1 Points) Use the summation in (a) to show that ) ( ) ( n O n T = Solution: Using the above summation, we have ) ( log 1 ) ( log 3 3 2 6 ) 3 / 2 ( ) ( n n i i n n T ⋅ + ≤ ∑ = since x x ≤ for any x ) 2 ( log 3 6 ) 3 / 2 ( n n i i + ≤ ∑ ∞ = adding ∞many positive terms ) 2 ( log 3 6 ) 3 / 2 ( 1 1 n n +  = by a well known formula ) ( 6 3 ) 2 ( log 3 n O n n = + = ) ( 1 ) 2 ( log 3 2 ) 2 ( log 3 3 n o n = ⇒ < ⇒ < Therefore ) ( ) ( n O n T = . 2 c. (1 Points) Use the Master Theorem to show that ) ( ) ( n n T Θ = Solution: Let ) 2 ( log 1 3 > = ε . Then 1 ) 2 ( log 3 = + ε , and ) ( ) 2 ( log ) 2 ( log 3 3 ε ε + + Ω = = n n n . Also for any c in the range 1 3 / 2 < ≤ c , and any positive n , we have cn n n ≤ = ) 3 / 2 ( ) 3 / ( 2 , so the regularity...
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This note was uploaded on 12/03/2009 for the course CS CS101 taught by Professor Agoreback during the Spring '09 term at American College of Gastroenterology.
 Spring '09
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