# A2_Solu - Assignement 2 COSC 3P03 Algorithms Winter 2011...

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Assignement 2 COSC 3P03, Algorithms Winter, 2011 Due: Feb. 18, Friday, noon. 1. (10) The running time of an algorithm A is described by t ( n ) = 7 t ( n/ 2) + n 2 . A competing algorithm B has a running time T ( n ) = aT ( n/ 4) + n 2 . What is the largest integer value for a such that B is asymptotically faster than A? For algorithm A , by the masters method, t ( n ) = O ( n log 2 7 ) . For A p , we can assume that a > 16 since if a = 16, by masters method, T ( n ) = Θ( n 2 log n ) , which is already better than t ( n ). For a > 16, since n 2 = O ( n log 4 a - e ) for some e > 0, the solution for T ( n ) is (Case 1) T ( n ) = O ( n log 4 a ) . To ±nd the largest a such that log 4 a < log 2 7, namely, log 2 a/ log 2 4 < log 2 7, we have log 2 a < log 2 7 2 , i.e., a < 49, therefore, a = 48. 2. (20) Explain why we would still have a linear-time selection algorithm if elements are divided into groups of k elements, for k = 7, 9, 11, . ..? Show that if elements are divided into groups of 3, then we have an Ω( n log n ) algorithm. If the group size is 7, the recurrence becomes t ( n ) = t ( n/ 7) + t (3 n/ 4) + c p n which has a solution t ( n ) = O ( n ) (see A1.Q8, since 1 / 7 + 3 / 4 < 1, or by a proof similar to the one we did when the size is 5 in class). So yes, we will still have a linear-time selction algorithm if the group size is 7 (by the same reasoning, OK if it is 9, 11, . ..... ). When the group size is 3, we would have

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A2_Solu - Assignement 2 COSC 3P03 Algorithms Winter 2011...

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