Greedy Algorithms WITH Activity Selection Problem.ppt
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The document discusses greedy algorithms, which make the locally optimal choice in hopes of finding a globally optimal solution, illustrated through problems like activity selection and the knapsack problem. It highlights the properties of greedy algorithms, including optimal substructure and greedy-choice property, while differentiating between solvable (fractional knapsack) and unsolvable (0-1 knapsack) scenarios. The document emphasizes that despite their simplicity, greedy algorithms are not always guaranteed to yield optimal solutions.
![Optimal substructures
• Define the following subset of activities which are activities that
can start after ai finishes and finish before aj starts
• Sort the activities according to finish time
• We now define the the maximal set of activities from i to j as
• Let c[i,j] be the maximal number of activities
• We can solve this using dynamic programming, but a simpler
approach exists
• Our recurrence relation for finding c[i, j] becomes](https://image.slidesharecdn.com/greedyalgorithmswithactivityselectionproblem-230929094132-73fc35f9/85/Greedy-Algorithms-WITH-Activity-Selection-Problem-ppt-30-320.jpg)
Greedy Algorithms WITH Activity Selection Problem.ppt
- 2. Greedy algorithms • A greedy algorithm always makes the choice that looks best at the moment – My everyday examples: • Driving in Los Angeles, NY, or Boston for that matter • Playing cards • Invest on stocks • Choose a university – The hope: a locally optimal choice will lead to a globally optimal solution – For some problems, it works • greedy algorithms tend to be easier to code
- 3. An Activity Selection Problem (Conference Scheduling Problem) • Input: A set of activities S = {a1,…, an} • Each activity has start time and a finish time – ai=(si, fi) • Two activities are compatible if and only if their interval does not overlap • Output: a maximum-size subset of mutually compatible activities
- 4. The Activity Selection Problem • Here are a set of start and finish times • What is the maximum number of activities that can be completed? • {a3, a9, a11} can be completed • But so can {a1, a4, a8’ a11} which is a larger set • But it is not unique, consider {a2, a4, a9’ a11}
- 6. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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- 8. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
- 9. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
- 10. Early Finish Greedy • Select the activity with the earliest finish • Eliminate the activities that could not be scheduled • Repeat!
- 11. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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- 13. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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- 17. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
- 18. Assuming activities are sorted by finish time
- 19. Why it is Greedy? • Greedy in the sense that it leaves as much opportunity as possible for the remaining activities to be scheduled • The greedy choice is the one that maximizes the amount of unscheduled time remaining
- 20. Why this Algorithm is Optimal? • We will show that this algorithm uses the following properties • The problem has the optimal substructure property • The algorithm satisfies the greedy-choice property • Thus, it is Optimal
- 21. Greedy-Choice Property • Show there is an optimal solution that begins with a greedy choice (with activity 1, which as the earliest finish time) • Suppose A S in an optimal solution – Order the activities in A by finish time. The first activity in A is k • If k = 1, the schedule A begins with a greedy choice • If k 1, show that there is an optimal solution B to S that begins with the greedy choice, activity 1 – Let B = A – {k} {1} • f1 fk activities in B are disjoint (compatible) • B has the same number of activities as A • Thus, B is optimal
- 22. Optimal Substructures – Once the greedy choice of activity 1 is made, the problem reduces to finding an optimal solution for the activity-selection problem over those activities in S that are compatible with activity 1 • Optimal Substructure • If A is optimal to S, then A’ = A – {1} is optimal to S’={i S: si f1} • Why? – If we could find a solution B’ to S’ with more activities than A’, adding activity 1 to B’ would yield a solution B to S with more activities than A contradicting the optimality of A – After each greedy choice is made, we are left with an optimization problem of the same form as the original problem • By induction on the number of choices made, making the greedy choice at every step produces an optimal solution
- 23. Elements of Greedy Strategy • An greedy algorithm makes a sequence of choices, each of the choices that seems best at the moment is chosen – NOT always produce an optimal solution • Two ingredients that are exhibited by most problems that lend themselves to a greedy strategy – Greedy-choice property – Optimal substructure
- 24. Greedy-Choice Property • A globally optimal solution can be arrived at by making a locally optimal (greedy) choice – Make whatever choice seems best at the moment and then solve the sub-problem arising after the choice is made – The choice made by a greedy algorithm may depend on choices so far, but it cannot depend on any future choices or on the solutions to sub-problems • Of course, we must prove that a greedy choice at each step yields a globally optimal solution
- 25. Optimal Substructures • A problem exhibits optimal substructure if an optimal solution to the problem contains within it optimal solutions to sub-problems – If an optimal solution A to S begins with activity 1, then A’ = A – {1} is optimal to S’={i S: si f1}
- 26. Knapsack Problem • One wants to pack n items in a luggage – The ith item is worth vi dollars and weighs wi pounds – Maximize the value but cannot exceed W pounds – vi , wi, W are integers • 0-1 knapsack each item is taken or not taken • Fractional knapsack fractions of items can be taken • Both exhibit the optimal-substructure property – 0-1: If item j is removed from an optimal packing, the remaining packing is an optimal packing with weight at most W-wj – Fractional: If w pounds of item j is removed from an optimal packing, the remaining packing is an optimal packing with weight at most W-w that can be taken from other n-1 items plus wj – w of item j
- 27. Greedy Algorithm for Fractional Knapsack problem • Fractional knapsack can be solvable by the greedy strategy – Compute the value per pound vi/wi for each item – Obeying a greedy strategy, take as much as possible of the item with the greatest value per pound. – If the supply of that item is exhausted and there is still more room, take as much as possible of the item with the next value per pound, and so forth until there is no more room – O(n lg n) (we need to sort the items by value per pound) – Greedy Algorithm? – Correctness?
- 28. O-1 knapsack is harder! • 0-1 knapsack cannot be solved by the greedy strategy – Unable to fill the knapsack to capacity, and the empty space lowers the effective value per pound of the packing – We must compare the solution to the sub-problem in which the item is included with the solution to the sub- problem in which the item is excluded before we can make the choice – Dynamic Programming
- 30. Optimal substructures • Define the following subset of activities which are activities that can start after ai finishes and finish before aj starts • Sort the activities according to finish time • We now define the the maximal set of activities from i to j as • Let c[i,j] be the maximal number of activities • We can solve this using dynamic programming, but a simpler approach exists • Our recurrence relation for finding c[i, j] becomes