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Lecture 3: Greedy Method

Lecture 3: Greedy Method. Greedy Matching     Coin Changing     Minimum Spanning Tree     Fractional Knapsack     Dijkstra's Single-Source Shortest-Path. Greedy Method. The Greedy Concept Makes the choice that looks best at the moment.

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Lecture 3: Greedy Method

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  1. Lecture 3: Greedy Method Greedy Matching    Coin Changing    Minimum Spanning Tree    Fractional Knapsack    Dijkstra's Single-Source Shortest-Path

  2. Greedy Method • The Greedy Concept • Makes the choice that looks best at the moment. • Hopes that ''local optimal'' choices lead to ''global optimal'. solution. • Two basic properties of optimal greedy algorithms • Optimal Substructure Property: A problem has optimal substructure if an • optimal solution to the problem contains within it optimal solutions to its • sub problems. • Greedy Choice Property: If a local greedy choice is made, then an optimal • solution including this choice is possible. Advanced Algorithms, Feodor F. Dragan, Kent State University

  3. A B C D F E G H Greedy Matching M = { } V = { A B C D E F G H } E = {(AB),(AD),(AE),(BC),(BE),(CE),(CF), (DE),(DG),(DH),(EF),(EH),(FH),(GH)} The Annals of Applied Probability, Vol. 3, No. 2. (May, 1993), pp. 526-552

  4. A B C D F E G H Greedy Matching M = {(CE)} V = { A B C D E F G H } E = {(AB),(AD),(AE),(BC),(BE),(CE),(CF), (DE),(DG),(DH),(EF),(EH),(FH),(GH)}

  5. A B C D F E G H Greedy Matching M = {(CE)} V = { A B C D E F G H } E = {(AB),(AD),(AE),(BC),(BE),(CE),(CF), (DE),(DG),(DH),(EF),(EH),(FH),(GH)}

  6. A B C D F E G H Greedy Matching M = {(CE),(AB)} V = { A B C D E F G H } E = {(AB),(AD),(AE),(BC),(BE),(CE),(CF), (DE),(DG),(DH),(EF),(EH),(FH),(GH)}

  7. A B C D F E G H Greedy Matching M = {(CE),(AB)} V = { A BC D E F G H } E = {(AB),(AD),(AE),(BC),(BE),(CE),(CF), (DE),(DG),(DH),(EF),(EH),(FH),(GH)}

  8. A B C D F E G H Greedy Matching M = {(CE),(AB),(GH)} V = { A BC D E F G H } E = {(AB),(AD),(AE),(BC),(BE),(CE),(CF), (DE),(DG),(DH),(EF),(EH),(FH),(GH)}

  9. A B C D F E G H Greedy Matching M = {(CE),(AB),(GH)} V = { A BC D E F G H } E = {(AB),(AD),(AE),(BC),(BE),(CE),(CF), (DE),(DG),(DH),(EF),(EH),(FH),(GH)}

  10. 25 25 25 25 10 5 5 1 1 1 1 Coin Changing An optimal solution to the coin changing problem is the minimum number of coins whose total value equals a specified amount. For example what is the minimum number of coins (current U.S. mint) needed to total 83 cents. A Greedy Algorithm for Coin Changing 1. Set remval=initial_value 2. Choose largest coin that is less than remval. 3. Add coin to set of coins and set remval:=revamal-coin_value 4. repeat Steps 2 and 3 until remval = 0; 83 - 25 = 58 58 - 25 = 33 33 - 25 = 8 8 - 5 = 3 3 - 1 = 2 2 - 1 = 1 1 - 1 = 0

  11. 26 26 26 26 26 26 10 10 10 10 5 1 1 1 1 1 Coin Changing - Another Example The greedy method works for the U.S. minted coin set, but will it work for any coin set? Consider the coin set below and a change value of 82 cents.. 82 - 26 = 56 56 - 26 = 30 30 - 26 = 4 4 - 1 = 3 3 - 1 = 2 2 - 1 = 1 1 - 1 = 0 By the coin-changing algorithm we obtain 3 26 cent pieces and 4 pennies, but we can use 2 26 cent pieces and 3 dimes to total 82 cents. Hence the greedy approach does not work for this coin set.

  12. When Does the Greedy Method Work? When using the U.S. minted coin set, we find that the coin-changing algorithm gives us the minimum number of coins. However, this greedy algorithm does not work when we replace the quarter with a 26-cent piece. See if you can determine the necessary relationships between coin values in a coin set for which the greedy coin-changing algorithm will result in a minimum number of coins for any initial_value. For your rules to be correct they must hold up under the following evaluations: 1. See if you can find a coin set that follows the rules but doesn't support the greedy algorithm. 2. Alternately you can attempt to find a coin set that supports the greedy algorithm but is not represented by the rules.

  13. Given a weighted graph G consisting of a set of vertices V and a set of edges E with weights, where Prepare a vertex set and an edge set to hold elements selected by Prim's Algorithm. B 1 A 1 C 2 5 2 2 4 D 1 3 E G 1 2 F Prim's Algorithm 1. Choose an arbitrary starting vertex vj 2. Find the smallest edge e incident with with a vertex in the vertex set whose inclusion in the edge set does not create a cycle. 3. Include this edge in the edge list and its vertices in the vertex list. 4. Repeat Steps 2 and 3 until all vertices are in the vertex list. 2 Exercise Prim's Algorithm on this example weighted graph starting with vertex B and again starting at vertex F. Did you get the same spanning tree? If the trees were distinct, did they have the same value?

  14. Kruskal's Algorithm The minimum spanning tree problem can also be solved using Kruskal's Algorithm. In this approach, we simply choose minimum-weight edges from the graph so long as an edge does not create a cycle in the edge set. We stop choosing edges when every vertex is a node for at least one of the edges in the set and the tree is connected. B B B 1 1 1 A A A C 1 C 1 C 1 2 2 2 5 5 5 2 2 2 2 2 2 D D D 4 4 4 1 2 1 2 1 2 3 3 3 E E E G G G 1 2 1 2 1 2 F F F B B 1 B 1 1 A A C 1 A C 1 C 1 2 2 2 5 2 5 2 2 5 2 2 D 2 D 4 D 4 1 4 2 1 2 1 2 3 E 3 G E 3 G E G 1 2 1 2 1 2 F F F

  15. Knapsack Problem A thief can carry a maximum weight in his knapsack and he wants to maximize the amount of value he carries. There is a given amount of each type of item and each item has a specified total value. How much of each item should he carry to get the maximum value? Maximum weight = 75. Item # Total Total Weight Value 1 20 10 2 40 25 3 10 15 4 50 30 5 30 10 V/W 1/2 5/8 3/2 3/5 1/3 Order Fract Taken Taken 4 0.0 2 1.0 1 1.0 3 0.5 5 0.0 Total Total Weight Value 75 55 If we compute the ratio of the total value to the total weight, we can arrange the items in the order of preference. Then we can take the most valuable items first. Eventually we will fill our knapsack by taking as much of the last item as will fit.

  16. Total Total Weight Value 70 50 0/1 Knapsack Problem We now consider the case in which fractional portions of an item are not allowed. That is the thief must take all or none of each item. Item # Total Total Weight Value 1 20 10 2 40 25 3 10 15 4 50 30 5 30 10 V/W 1/2 5/8 3/2 3/5 1/3 Rank Order 4 2 1 3 5 Item Value 10 25 15 X This is called the 0/1 knapsack problem. As shown in this example, the greedy method does not apply to this problem. The third most valuable item is too heavy to be added to the knapsack. Develop an example 0/1 knapsack problem in which the most valuable item is not an item in the optimal solution. (Weight of most valuable item cannot exceed weight limit by itself.)

  17. 2 v3 v2 v1 v0 v5 v4 5 1 3 1 1 4 1 6 3 The single source shortest path problem is as follows. We are given a directed graph with nonnegative edge weights G = (V,E) and a distinguished source vertex, . The problem is to determine the distance from the source vertex to every vertex in the graph. Single Source Shortest Path Given a weighted graph G find the minimum weight path from a specified vertex v0 to every other vertex.

  18. 2 v3 v4 v5 v0 v1 v2 v1 v2 v3 v4 v5 node minimum list path v1 v2 v3 v4 v5 5 1 4 - 6 {2} 3 42 6 {24} 3 3 5 {241} 35 {2413} 4 5 1 3 1 1 4 1 6 3 Dijkstra's Algorithm for SSSP v1 v2 v3 v4 v5 5 1 4 - 6 {2} 3 4 2 6 {24} 3 3 5 {241} 3 5 {2413} 4 v1 v2 v3 v4 v5 5 1 4 - 6 {2} 3 4 2 6 {24} 3 3 5 {241} 3 5 v1 v2 v3 v4 v5 5 1 4 - 6 {2} 3 4 2 6 {24} 3 3 5 v1 v2 v3 v4 v5 5 1 4 - 6 {2} 3 4 2 6 v1 v2 v3 v4 v5 5 1 4 - 6

  19. Summary Greedy Matching Coin Changing Greedy Optimal Greedy Non-Optimal Minimum Spanning Tree Prim's Algorithm Kruskal's Algorithm Knapsack Fractional Knapsack 0/1 Knapsack Diskstra's Single-Source Shortest Path Algorithm

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