Greedy Algorithms. Algorithms

Transcription

1 Greedy Algorithms Algorithms

2 Greedy Algorithms Many algorithms run from stage to stage At each stage, they make a decision based on the information available A Greedy algorithm makes decisions At each stage, using locally available information, the greedy algorithm makes an optimal choice Sometimes, greedy algorithms give an overall optimal solution Sometimes, greedy algorithms will not result in an optimal solution but often in one good enough

3 Divisible Items Knapsack Given a set of items Each item has a weight Each item has a value Select a subset Constraint: Problem S M S x M w(x) v(x) w(x) < W Objective Function: x M v(x) max

4 Divisible Items Knapsack Order all items by impact Problem impact(x) = v(x) w(x) In order of impact (highest first), ask whether you want to include the item And you include it if the sum of the weights of the items already selected is smaller than W

5 Set of activities Each activity has a start time and a finish time 0 s i < f i < S = {a 1, a 2,, a n } Each activity needs to use your facility Only one activity at a time Make the rental agreements that maximize the number of rentals

6 Two activities and are compatible iff a i [s i, f i ) [s j, f j ) = a j This means that activity i < j finishes before activity j

7 Example: i si fi A compatible set is Another compatible set is {A 1, A 5, A 8, A 10 } {A 3, A 9 }

8 Optimal rental with a dynamic programming algorithm Subproblems: Define to be the set of activities that start after finishes and finish before starts a i S ik a k i si fi S 1,8 = {a 5 }

9 We want to find an optimal rental plan for Assume that there is an optimal solution that contains activity a j S i,k a j a j By selecting, we need to decide what to do with the time before starts and after finishes a j S ik These sets are S ij and S jk

10 Assume that is part of an optimal solution for a j A i,k S i,k Then A i,k is divided into the ones that end before a j and the ones that start after a j A i,j = A i,k S i,k A j,k = A i,k S j,k A i,k = A i,j {a j } A j,k

11 S ij S jk a i A ij a j A jk a k S i,k S i,j S j,k

12 Clearly, A i,j is an optimal solution for is an optimal solution for S i,j A j,k S j,k S i,k For if not, we could construct a better solution for S ij S jk a i A ij a j A jk a k S i,k S i,j S j,k

13 We can therefore solve recursively the problem for looking at all possible activities for a j S i,k by Define in S i,k Then: C[i, k] = Max number of compatible activities C[i, k] = max(0, max (C[i, j] + C[j, k] + 1 a j S i,k )) The 0 is necessary because there might be no activity in S i,k

14 The recursion leads to a nice dynamic programming problem C[i, k] = max(0, max (C[i, j] + C[j, k] + 1 a j S i,k ))

15 But can we do better?

16 Start out with the initial problem Select the activity that finishes first this would be This leaves most space for all other activities Call S the set of activities compatible with a 1 1 These are those starting after Similarly, call a k a 1 S k a 1 the set of activities starting after

17 Theorem: For any non-empty problem let be the activity with the smallest end time. Then a m is contained in an optimal solution S k a m Proof: Let be a solution i.e. the maximum sized compatible subset in Let If A k a 1 A k a m = a 1 S k be the activity with earliest finish time then we are done

18 Theorem: For any non-empty problem let be the activity with the smallest end time. Then a m is contained in an optimal solution S k a m Proof: Otherwise replace with in a 1 a m A k A k = A k {a 1 } {a m } a m Since is the first to finish, this is a set of compatible activities Therefore, there exists an optimal solution with a m

19 Result of the Theorem: We can find an optimal solution (but not necessarily all optimal solutions) by always picking the first one to finish.

20 Example i si fi Select a 1 a 2 a 3 a 4 Exclude,, and as incompatible a 5 a 8 a 10 Choose,, and for the complete solution

21 Greedy Algorithms Greedy algorithms Determine the optimal substructure Develop a recursive solution Show that making the greedy choice is best Show that making the greedy choice leads to a similar subproblem Obtain a recursive algorithm Convert the recursive algorithm to an iterative algorithm

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