Introduction to Greedy Algorithms The greedy technique Problems

Introduction to Greedy Algorithms • The greedy technique • Problems explored – The coin changing problem – Activity selection cs 333/cutler Greedy 1

Optimization problems • An optimization problem: – Given a problem instance, a set of constraints and an objective function. – Find a feasible solution for the given instance for which the objective function has an optimal value – either maximum or minimum depending on the problem being solved. • A feasible solution satisfies the problem’s constraints • The constraints specify the limitations on the required solutions. • For example in the knapsack problem we require that the items in the knapsack will not exceed a given weight cs 333/cutler Greedy 2

The Greedy Technique(Method) • Greedy algorithms make good local choices in the hope that they result in an optimal solution. – They result in feasible solutions. – Not necessarily an optimal solution. • A proof is needed to show that the algorithm finds an optimal solution. • A counter example shows that the greedy algorithm does not provide an optimal solution. cs 333/cutler Greedy 3

Pseudo-code for Greedy Algorithm set Greedy (Set Candidate){ solution= new Set( ); while (Candidate. is. Not. Empty()) { next = Candidate. select(); //use selection criteria, //remove from Candidate and return value if (solution. is. Feasible( next)) //constraints satisfied solution. union( next); if (solution. solves()) return solution} //No more candidates and no solution return null } cs 333/cutler Greedy 4

Pseudo code for greedy cont. • select() chooses a candidate based on a local selection criteria, removes it from Candidate, and returns its value. • is. Feasible() checks whether adding the selected value to the current solution can result in a feasible solution (no constraints are violated). • solves() checks whether the problem is solved. cs 333/cutler Greedy 5

Coin changing problem • Problem: Return correct change using a minimum number of coins. • Greedy choice: coin with highest coin value • A greedy solution (next slide): • American money The amount owed = 37 cents. The change is: 1 quarter, 1 dime, 2 cents. Solution is optimal. • Is it optimal for all sets of coin sizes? • Is there a solution for all sets of coin sizes? (12, D, N, P/15) cs 333/cutler Greedy 6

A greedy solution: Input: Set of coins of different denominations, amountowed change = {} while (more coin-sizes && valueof(change)<amountowed) Choose the largest remaining coin-size // Selection // feasibility check while (adding the coin does not make the valueof(change) exceed the amount-owed ) then add coin to change //check if solved if ( valueof(change) equals amount-owed) return change cs 333/cutler Greedy else delete coin-size 7

Elements of the Greedy Strategy Cast problem as one in which we make a greedy choice and are left with one subproblem to solve. To show optimality: 1. Prove there is always an optimal solution to original problem that makes the greedy choice. cs 333/cutler Greedy 8

Elements of the Greedy Strategy 2. Demonstrate that what remains is a subproblem with property: If we combine the optimal solution of the subproblem with the greedy choice we have an optimal solution to original problem. cs 333/cutler Greedy 9

Activity Selection • Given a set S of n activities with start time si and finish time fi of activity i • Find a maximum size subset A of compatible activities (maximum number of activities). • Activities are compatible if they do not overlap • Can you suggest a greedy choice? cs 333/cutler Greedy 10

Example Activities 1 2 3 4 5 6 7 11 13 2 12 3 10 3 1 0 7 11 15 4 2 0 1 2 3 4 5 cs 333/cutler 6 7 8 9 Greedy 10 11 12 13 14 15 Time 11

Counter Example 1 • Select by start time Activities 1 2 3 11 1 15 4 0 15 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 cs 333/cutler Greedy Time 12

Counter Example 2 • Select by minimum duration Activities 1 2 3 8 1 15 8 7 9 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 cs 333/cutler Greedy Time 13

Select by finishing time Activities 1 2 2 3 3 4 5 3 6 1 4 7 0 2 11 13 12 10 7 11 15 0 1 2 3 4 5 cs 333/cutler 6 7 8 9 Greedy 10 11 12 13 14 15 Time 14

Activity Selection • Assume without loss of generality that we number the intervals in order of finish time. So f 1£. . . £fn. • Greedy choice: choose activity with minimum finish time • The following greedy algorithm starts with A={1} and then adds all compatible jobs. (Theta(n)) • Theta(nlogn) when including sort cs 333/cutler Greedy 15

Greedy-Activity-Selector(s, f) n <- length[s] // number of activities A <- {1} j <- 1 //last activity added for i <- 2 to n //select if si >= fj then //compatible (feasible) add {i} to A j <- i //save new last activity return A cs 333/cutler Greedy 16

Proof that greedy solution is optimal 1. It is easy to see that there is an optimal solution to the problem that makes the greedy choice. Proof of 1. Let A be an optimal solution. Let activity 1 be the greedy choice. If 1 A the proof is done. If 1 A, we will show that A’=A-{a}+{1} is another optimal solution that includes 1. Let a be the activity with minimum finish time in A. Since activities are sorted by finishing time in the algorithm, f(1) f(a). If f(1) s(a) we could add 1 to A and it could not be optimal. So s(1) < f(a), and 1 and a overlap. Since f(1) f(a), if we remove a and add 1 we get another compatible solution A’=A-{a}+{1} and |A’|=|A| cs 333/cutler Greedy 17

Proof that greedy solution is optimal 2. If we combine the optimal solution of the remaining subproblem with the greedy choice we have an optimal solution to the original problem. Proof of 2. Let activity 1 be the greedy choice. Let S’ be the subset of activities that do not overlap with 1. S’={i|i =1, …, n and si f(1)}. Let B be an optimal solution for S’. From the definition of S’, A’={1}+B is compatible, and a solution to the original problem. cs 333/cutler Greedy 18

Proof that greedy solution is optimal 2. If we combine the optimal solution of the remaining subproblem with the greedy choice we have an optimal solution to the original problem. Proof of 2 continued. The proof is by contradiction. Assume that A’ is not an optimal solution to the original problem. Let A be an optimal solution that contains 1. So |A’|<|A|, and |A-{1}|>|A’-{1}|=|B|. But A-{1} is also a solution to the problem of S’, contradicting the assumption that B is an optimal solution to S’. cs 333/cutler Greedy 19