sample-exam1-solns

# sample-exam1-solns - Illinois Institute of Technology...

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Illinois Institute of Technology Department of Computer Science Solutions to First Examination CS 430 Introduction to Algorithms Fall, 2007 10am–11:15am, Wednesday, September 26, 2007 121 Life Sciences 1. (a) n 2 (2 n ) 2 = 4 n 2 so the algorithm takes 4 times as long (b) n 3 (2 n ) 3 = 8 n 3 so the algorithm takes 8 times as long (c) 100 n 100 · (2 n ) = 200 n so the algorithm takes 2 times as long (d) n lg n (2 n ) lg(2 n ) = 2 n lg n + 2 n so the algorithm takes a bit more than 2 times as long AS n - > (e) 2 n 2 2 n = 4 n = 2 n · 2 n so the algorithm takes 2 n times as long (or, the time is SQUARED). For an increase by 1: (a) n 2 ( n + 1) 2 = n 2 + 2 n + 1 so the algorithm takes an extra (2 n + 1) units of time (b) n 3 ( n + 1) 3 = n 3 + 3 n 2 + 3 n + 1 so the algorithm takes an extra (3 n 2 + 3 n + 1) units of time (c) 100 n 100( n + 1) = 100 n + 100 so the algorithm takes 100 units of time (d) n lg n ( n + 1) lg( n + 1) n lg n + lg n so the algorithm takes about an extra lg n units of time (e) 2 n 2 n +1 = 2 · 2 n so the algorithm takes 2 times as long 2. We modify merge sort to count the number of inversions(pair ( i, j ) is called inversion if and only if i < j and A [ i ] > A [ j ], the number of inversions is the same as the number of comparisons in insertion sort) in Θ( n lg n ) time. To start, let us de±ne a merge-inversion as a situation within the execution of merge sort in which the MERGE procedure, after copying A [ p..q ] to L and A [ q + 1 ..r ] to R , has values x in L and y in R such that x > y . Consider an inversion (

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## This note was uploaded on 04/07/2009 for the course CS 430 taught by Professor Kapoor during the Spring '08 term at Illinois Tech.

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sample-exam1-solns - Illinois Institute of Technology...

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