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**Unformatted text preview: **Assignment #1 CS4/531 Due Date: Monday, Sep. 19, 2011 UNSUPPORTED SOLUTIONS RECEIVE NO CREDIT. Total points: 55 1. (4pts) We want to prove the function f ( n ) = 3 n 2 + 4 n − 20 = Ω( n 2 ) by using the definition of Ω. Namely we need to find c > 0 and n ≥ 0 such that: 3 n 2 + 4 n − 20 ≥ cn 2 for all n ≥ n There are many combinations of c and n that will satisfy the definition. Try the following values for c . (a) Pick c = 2, determine the smallest n that satisfies the definition. (b) Pick c = 3, determine the smallest n that satisfies the definition. (c) Pick c = 4, what happens? Answer: (a) When we pick c = 2 then equation becomes: 3 n 2 + 4 n − 20 ≥ 2 n 2 which can be written as n 2 + 4 n − 20 ≥ Roots of this equation are n ≈ − 6 . 89 and n 1 ≈ 2 . 89. Since we look for n ≥ 0 then smallest n is 3 (b) When we pick c = 3 then equation becomes: 3 n 2 + 4 n − 20 ≥ 3 n 2 which can be written as n ≥ 5 Smallest n is 5 (c) When we pick c = 2 then equation becomes: 3 n 2 + 4 n − 20 ≥ 4 n 2 which can be written as − n 2 + 4 n − 20 ≥ Since this equation has no roots then we cannot find any positive rational n that satisfies equation. a50 2. (16 pts) Rank the following functions by the order of growth rate (that is, list them in a list f 1 ( n ) ,f 2 ( n ) ,f 3 ( n ) ,... such that f 1 ( n ) = O ( f 2 ( n )), f 2 ( n ) = O ( f 3 ( n )) ,... ). Partition your list into equivalent classes such that f ( n ) and g ( n ) are in the same class if and only if f ( n ) = Θ( g ( n )). You must prove your answer (by limit test or other means). ( √ 3) lg 3 n , n 2 , ln 2 n 2 , ln ln n, (ln n ) ln n , n ln ln n , 4 lg n , n lg 2 n, n ! , ( n − 1)! , (4 / 3) n , n 1 / 2 Note: lg is the log function with base 2. ln is the log function with base e . Answer: We rank the following functions by the order of growth rate as follows: ln ln n , ln 2 n 2 , [ n 1 / 2 , ( √ 3) lg 3 n ], n lg 2 n , [4 lg n , n 2 ], [ n ln ln n , (ln n ) ln n ],(4 / 3) n , ( n − 1)!, n !. The functions in the same class are in [ ... ]. 1 (a) lim n →∞ ln ln n ln 2 n 2 = lim n →∞ 1 ln n 1 n 4 n ln n (by L’Hopital rule) = lim n →∞ 1 4 ln 2 n = 0 . Thus ln ln n = o (ln 2 n 2 ). (b) lim n →∞ ln 2 n 2 n 1 / 2 = lim n →∞ (4 ln n ) /n 1 / 2 n- 1 / 2 (by L’Hopital rule) = lim n →∞ 8 ln n n 1 / 2 = lim n →∞ 8 /n 1 / 2 n- 1 / 2 (by L’Hopital rule) = lim n →∞ 16 n 1 / 2 = 0 . Thus ln 2 n 2 = o ( n 1 / 2 ). (c) ( √ 3) lg 3 n = 3 (lg 3 n ) / 2 = n 1 / 2 . (d) lim n →∞ n 1 / 2 n lg 2 n = lim n →∞ 1 n 1 / 2 lg 2 n = 0. Thus n 1 / 2 = o ( n lg 2 n ). (e) lim n →∞ n n lg 2 n = lim n →∞ 1 lg 2 n = 0 . Thus n = o ( n lg 2 n )....

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