Lecture 11-14

# Introduction to Algorithms, Second Edition

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The University of Texas at Austin Lectures 11-14 Department of Computer Sciences Professor Vijaya Ramachandran Greedy, MST CS357: ALGORITHMS, Spring 2006 1 The Greedy Framework Rcall that an optimization problem is one for which an input has a collection of feasible solutions, each with an associated cost, and we need to find a feasible solution that optimizes the cost; here ‘optimizes’ would mean either minimizes or maximizes , depending on the nature of the problem. Such a solution is called an optimal solution. A greedy strategy for an optimization problem constructs an optimal solution (which is typ- ically a set of elements) incrementally by making a locally optimal choice at each step to decide the next element to be added to the solution. If this strategy generates an optimal solution, then this represents a greedy algorithm . 2 A Toy Problem Activity Selection Problem Given a collection of activities S = { ( s i , f i ), 1 i n } , where s i and f i are the start and finish times of the i th activity, and f 1 f 2 ≤ · · · ≤ f n , a feasible solution is a subset of activities whose durations do not overlap. A feasible solution with the maximum number of activities is an optimal solution . Algorithm Greedy Activity Selection Start with an initially empty solution A . Let T := S repeat Select an activity α in T with smallest finish time and set A := A ∪ { α } . Delete from T those activities whose durations overlap with the selected activity. until T = φ Correctness Lemma 1 (Greedy Choice) : Let X be a subset of an optimal solution for the activity selection problem on S . Let S be the set of activities in S that do not overlap with any activity in X , and let α = ( s, f ) be the activity with smallest finish time in S . Then, X = X ∪ { α } is a subset of an optimal solution for S .

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Proof. Clearly X is a feasible solution since none of the activities in X overlap with one another. Consider an optimal solution A for S that contains X . (We are given that such an optimal solution exists.) If A contains α then we are done. Otherwise let β = ( s , f ) be the activity with smallest finish time in A X . Then, f > f since ( s, f ) is the activity with smallest finish time that does not overlap with any activity in X . Since α is in S , it does not overlap with any activity in X . Also, α does not overlap with any activity in A X − { β } since any activity in A X − { β } must have a start time greater than f and the finish time f of α is smaller than f . Hence α does not overlap with any activity in A − { β } .
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