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ps1sol - Introduction to Algorithms Massachusetts Institute...

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Unformatted text preview: Introduction to Algorithms September 18, 2006 Massachusetts Institute of Technology 6.046J/18.410J Professors Erik Demaine and Madhu Sudan Problem Set 1 Solutions Problem Set 1 Solutions Problem 1-1. Asymptotic Notation For each of the following relationships, find two nonnegative functions f ( n ) and g ( n ) that satisfy it. If no such functions exist, write “NONE” and briefly justify your answer. (a) f ( n ) = O ( g ( n )) and g ( n ) = O ( f ( n )) . Solution: f ( n ) = g ( n ) = n. (b) f ( n ) = o ( g ( n )) and g ( n ) = o ( f ( n )) . Solution: NONE : If f ( n ) = o ( g ( n )) , then there exists an n such that, for all n > n , f ( n ) < 1 2 g ( n ) . Similarly, g ( n ) = o ( f ( n )) implies that there exists an n 1 such that, for all n > n 1 , g ( n ) < 1 2 f ( n ) . Thus, for all n > max { n , n 1 } , we have f ( n ) < 1 4 f ( n ) , which cannot hold if f ( n ) is nonnegative. (c) f ( n ) = ω ( f ( n )) Solution: NONE : f ( n ) = ω ( f ( n )) implies that there exists an n such that, for all n > n , f ( n ) > f ( n ) . But this is clearly impossible. (d) f ( n ) = Ω( g ( n )) and g ( n ) = ω ( f ( n ) + g ( n )) . Solution: NONE : If we add g ( n ) to both sides of f ( n ) = Ω( g ( n )) , we obtain that f ( n ) + g ( n ) = Ω( g ( n )) . Substituting this relation into the right-hand side of g ( n ) = ω ( f ( n ) + g ( n )) , we obtain that g ( n ) = ω ( g ( n )) , which is impossible by part (c). (e) f ( n ) = Θ(1) and g ( n ) = o ( f ( n )) Solution: f ( n ) = 1 , g ( n ) = 1 /n . (f) f ( n ) = O (1) and g ( n ) = o ( f ( n ) + g ( n )) . Solution: f ( n ) = 1 , g ( n ) = 1 /n . 2 Problem Set 1 Solutions Problem 1-2. Recurrences Give asymptotic upper and lower bounds for T ( n ) in each of the following recurrences. Assume that T ( n ) is a nonnegative constant for n ≤ 10 . Make your bounds as tight as possible, and justify your answers. (a) T ( n ) = 3 T ( n/ 3) + n Solution: Using the notation of the Master Theorem, we have a = 3 , b = 3 , so n log b a = n , and f ( n ) = n . Thus, by Case 2 of the Master Theorem, T ( n ) = Θ( n lg n ) . (b) T ( n ) = 11 T ( n/ 7) + n 3 Solution: Using the notation of the Master Theorem, we have a = 11 , b = 7 , so n log b a ≈ n 1 . 23 , and f ( n ) = n 3 . The regularity condition is easy to verify: af ( n/b ) = 11( n/ 7) 3 = (11 / 343) n 3 ≤ 1 2 n 3 . Thus, by Case 3 of the Master Theorem, T ( n ) = Θ( n 3 ) ....
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