Downloaded 3,125 times
Uploaded byJenny Galino
Knapsack Problem
The document discusses the knapsack problem, which involves selecting a subset of items that fit within a knapsack of limited capacity to maximize the total value. There are two versions - the 0-1 knapsack problem where items can only be selected entirely or not at all, and the fractional knapsack problem where items can be partially selected. Solutions include brute force, greedy algorithms, and dynamic programming. Dynamic programming builds up the optimal solution by considering all sub-problems.
More Related Content
![[ACM-ICPC] Sort](https://cdn.slidesharecdn.com/ss_thumbnails/acm-icpcsort-130227080110-phpapp01-thumbnail.jpg?width=640&height=640&fit=bounds)
Knapsack Problem
- 1.
- 2. Knapsack Problem Item # Size Value 1 1 8 2 3 6 3 5 5
- 3. Knapsack Problem There aretwo versions of the problem: 1. 0-1 Knapsack Problem 2. Fractional Knapsack Problem i. Bounded Knapsack Problem ii. Unbounded Knapsack Problem
- 4. Knapsack Problem Sample Problem A B C D E F G value 7 9 5 12 14 6 12 time 3 4 2 6 7 3 5
- 5. Solutions to Knapsack Problems Brute-Force Approach – solve the problem with a straightforward algorithm
- 6. Solutions to Knapsack Problems {} {1, 2, 3, 4, 5} {} {1} {2, 3, 4, 5} {2, 3, 4, 5} {} {2} {1} {1, 2} {3, 4, 5} {3, 4, 5} {3, 4, 5} {3, 4, 5}
- 7. Solutions to Knapsack Problems
- 8.
- 9.
- 10. Greedy Algorithm A B C D cost 200 240 140 150 weight 1 3 2 5 value 200 80 70 30
- 11. Greedy Algorithm Theoptimal solution to the fractional knapsack Not an optimal solution to the 0-1 knapsack
- 12.
- 13.
- 14.
- 15.
- 16. Example is 0 1 2 3 4 5 0 1 2 3 4
- 17. Example is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 2 3 4
- 18. Example is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 2 0 3 0 4 0
- 19. Example is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 0 2 0 3 0 4 0
- 20. Example is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 0 3 2 0 3 0 4 0
- 21. Example is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 0 3 3 2 0 3 0 4 0
- 22. Example is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 0 3 3 3 2 0 3 0 4 0
- 23. Example is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 0 3 3 3 3 2 0 3 0 4 0
- 24. Example is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 0 3 3 3 3 2 0 0 3 0 4 0
- 25. Example is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 0 3 3 3 3 2 0 0 3 3 0 4 0
- 26. Example is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 0 3 3 3 3 2 0 0 3 4 3 0 4 0
- 27. Example is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 0 3 3 3 3 2 0 0 3 4 4 3 0 4 0
- 28. Example is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 0 3 3 3 3 2 0 0 3 4 4 7 3 0 4 0
- 29. Example is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 0 3 3 3 3 2 0 0 3 4 4 7 3 0 0 3 4 4 0
- 30. Example is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 0 3 3 3 3 2 0 0 3 4 4 7 3 0 0 3 4 5 4 0
- 31. Example is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 0 3 3 3 3 2 0 0 3 4 4 7 3 0 0 3 4 5 7 4 0
- 32. Example is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 0 3 3 3 3 2 0 0 3 4 4 7 3 0 0 3 4 5 7 4 0 0 3 4 5
- 33. Example is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 0 3 3 3 3 2 0 0 3 4 4 7 3 0 0 3 4 5 7 4 0 0 3 4 5 7
- 34.
- 35. Example is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 0 3 3 3 3 2 0 0 3 4 4 7 3 0 0 3 4 5 7 4 0 0 3 4 5 7
- 36. Example is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 0 3 3 3 3 2 0 0 3 4 4 7 3 0 0 3 4 5 7 4 0 0 3 4 5 7
- 37. Example is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 0 3 3 3 3 2 0 0 3 4 4 7 3 0 0 3 4 5 7 4 0 0 3 4 5 7
- 38. Example is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 0 3 3 3 3 2 0 0 3 4 4 7 3 0 0 3 4 5 7 4 0 0 3 4 5 7
- 39. Example is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 0 3 3 3 3 2 0 0 3 4 4 7 3 0 0 3 4 5 7 4 0 0 3 4 5 7
- 40. Example is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 0 3 3 3 3 2 0 0 3 4 4 7 3 0 0 3 4 5 7 4 0 0 3 4 5 7
- 41. Example The optimal knapsackshould contain {1,2} = 7 is 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 0 3 3 3 3 2 0 0 3 4 4 7 3 0 0 3 4 5 7 4 0 0 3 4 5 7
- 42. Knapsack Problem Common Applications •Resource allocation with financial constraints • Construction and scoring of heterogeneous test • Selection of capital investments
Editor's Notes
- #4 1. Items are indivisible: you either take an item or not. Solved with dynamic programming2.Items are divisible: you can take any fraction of an item. Solved with a greedy algorithm
- #5 Imagine you have a problem set with different parts labelled A through G. Each part has a “value” (in points) and a “size” (time in hours to complete). Say the value and time for the problem set are as follows… And say you have a total of 15 hours – the knapsack – : which parts should you do? With the 0-1 Knapsack, you need to know which parts you should do to get the best total value possible. We want maximizing our chance to get more points.If there was partial credit that was proportional to the amount of work done (e.g., one hour spent on problem C earns you 2.5 points), that is what we call now the Fractional Knapsack the best approach is to work on problems in order of points/hour (a greedy strategy).
- #7 Since there are nitems, there are 2𝑛 possible combinations of itemsWe need to go through all combinations to find the one with maximum value with total size <= SNOT OPTIMALSince there are nitems, there are 2^𝑛 possible combinations of itemsWe need to go through all combinations to find the one with max value with total weight <= W
- #10 Legend:Red – walapudkokasabot
- #11 Solution is 1 pds A3 pds B1 pd C
- #13 This means that the best subset of 𝑆𝑘 that has the total size 𝑆,can either contains item k or not.First case: 𝑠𝑘>𝑠. Item k can’t be part of the solution, since if it was, the total size would be >s, which is unacceptableSecond case: 𝑤𝑘≤𝑤. Then the item k can be in the solution, and we choose the case with greater value.It means, that the best subset of 𝑆_𝑘 that has total weight w is:The best subset of 𝑆_(𝑘−1) that has total weight 𝑠,orThe best subset of 𝑆_(𝑘−1) that has total weight s−𝑠_𝑘 plus the item 𝑘The best subset of 𝑆_𝑘that has the total size 𝑆, either contains item k or not.First case: 𝑠_𝑘>𝑠. Item k can’t be part of the solution, since if it was, the total size would be >s, which is unacceptableSecond case: 𝑤_𝑘≤𝑤. Then the item k can be in the solution, and we choose the case with greater value.
- #14 First for-loop: 𝑠=0 𝑡𝑜 𝑆. We go through all the possible sizes of our knapsack until S and if item i is equal to 0, which is “V[0,s]” its corresponding maximum value is of course, 0. Because when i = 0, this means that we are not taking any item.Second for-loop: i=1 𝑡𝑜 𝑛. We go through all the items from 1 to n and if the knapsack’s size is equal to 0, which is “V[i, 0]” its corresponding values is again 0. Because when s = 0, this means that we can’t put anything in the knapsack.It means, that the best subset of 𝑆_𝑘 that has total weight w is:The best subset of 𝑆_(𝑘−1) that has total weight 𝑠,orThe best subset of 𝑆_(𝑘−1) that has total weight s−𝑠_𝑘 plus the item 𝑘The best subset of 𝑆_𝑘that has the total size 𝑆, either contains item k or not.First case: 𝑠_𝑘>𝑠. Item k can’t be part of the solution, since if it was, the total size would be >s, which is unacceptableSecond case: 𝑤_𝑘≤𝑤. Then the item k can be in the solution, and we choose the case with greater value.
- #15 Again, we go through all the items and:Outer if: 𝑠𝑖≤𝑠. This means that the size of the item can fit in the current size of the knapsack and we should consider its possible maximum value.Outer else: 𝑠𝑖>𝑠. Item i can’t be part of the solution, since its size is bigger than the knapsack’s current limit. Then, we’ll just copy the value above it.Inner if: 𝑣𝑖+𝑉𝑖−1, 𝑠−𝑠𝑖>𝑉[𝑖−1, 𝑠]. This means that if the current item’s value + 𝑉𝑖−1, 𝑠−𝑠𝑖 is greater than the value above it, we must use the current item’s value + 𝑉𝑖−1, 𝑠−𝑠𝑖.Inner else:𝑣𝑖+𝑉𝑖−1, 𝑠−𝑠𝑖≤𝑉[𝑖−1, 𝑠]. This means that if the current item’s value + 𝑉𝑖−1, 𝑠−𝑠𝑖 is less than or equal to the value above it, we must use the value on the previous item (or simple the value above it).The outer if and else conditions check if the knapsack can hold the current item or not.The inner if and else conditions check if the current value is bigger than the previous value so as to maximize the values the knapsack can hold.It means, that the best subset of 𝑆_𝑘 that has total weight w is:The best subset of 𝑆_(𝑘−1) that has total weight 𝑠,orThe best subset of 𝑆_(𝑘−1) that has total weight s−𝑠_𝑘 plus the item 𝑘The best subset of 𝑆_𝑘that has the total size 𝑆, either contains item k or not.First case: 𝑠_𝑘>𝑠. Item k can’t be part of the solution, since if it was, the total size would be >s, which is unacceptableSecond case: 𝑤_𝑘≤𝑤. Then the item k can be in the solution, and we choose the case with greater value.
- #20 i = 1vi = 3si = 2s = 1s - si = -1
- #21 i = 1vi = 3si = 2s = 2s - si = 0
- #22 i = 1vi = 3si = 2s = 3s - si = 1
- #23 i = 1vi = 3si = 2s = 4s - si = 2
- #24 i = 1vi = 3si = 2s = 5s - si = 3
- #25 i = 2vi = 4si = 3s = 1s - si = -2
- #26 i = 2vi = 4si = 3s = 2s - si = -1
- #27 i = 2vi = 4si = 3s = 3s - si = 0
- #28 i = 2vi = 4si = 3s = 4s - si = 1
- #29 i = 2vi = 4si = 3s = 5s - si = 2
- #30 i = 3vi = 5si = 4s = 1…3
- #31 i = 3vi = 5si = 4s = 4s - si = 0
- #32 i = 3vi = 5si = 4s = 5s - si = 1
- #33 i = 4vi = 6si= 5s = 1…4
- #34 i = 4vi = 6si = 5s = 5s - si = 0
- #35 To identify which items to include in the knapsack…All of the information we need is in the tableV[n,S] is the maximal value of items that can be placed in the Knapsackif...mark the ith items as in the knapsackelse...assume the ith item is not in the knapsack
- #36 𝑖 = 4𝑘 = 5𝑣𝑖=6𝑠𝑖 = 5𝑉𝑖,𝑘=7𝑉𝑖−1,𝑘=7𝑖 = 4𝑘 = 5𝑣_𝑖=6𝑠_𝑖 = 5𝑉[𝑖,𝑘]=7𝑉[𝑖−1,𝑘]=7
- #37 𝑖 = 4𝑘 = 5𝑣𝑖=6𝑠𝑖 = 5𝑉𝑖,𝑘=7𝑉𝑖−1,𝑘=7𝑖 = 4𝑘 = 5𝑣_𝑖=6𝑠_𝑖 = 5𝑉[𝑖,𝑘]=7𝑉[𝑖−1,𝑘]=7
- #38 𝑖 = 3𝑘 = 5𝑣𝑖=6𝑠𝑖 = 4𝑉𝑖,𝑘=7𝑉𝑖−1,𝑘=7𝑖 = 3𝑘 = 5𝑣_𝑖=6𝑠_𝑖 = 4𝑉[𝑖,𝑘]=7𝑉[𝑖−1,𝑘]=7
- #39 𝑖 = 1𝑘 = 5𝑣𝑖=4𝑠𝑖 = 3𝑉𝑖,𝑘=7𝑉𝑖−1,𝑘=3𝑘−𝑠𝑖=2𝑖 = 1𝑘 = 5𝑣_𝑖=4𝑠_𝑖 = 3𝑉[𝑖,𝑘]=7𝑉[𝑖−1,𝑘]=3𝑘−𝑠_𝑖=2
- #40 𝑖 = 1𝑘 = 2𝑣𝑖=3𝑠𝑖 = 2𝑉𝑖,𝑘=3𝑉𝑖−1,𝑘=0𝑘−𝑠𝑖=0𝑖 = 1𝑘 = 2𝑣_𝑖=3𝑠_𝑖 = 2𝑉[𝑖,𝑘]=3𝑉[𝑖−1,𝑘]=0𝑘−𝑠_𝑖=0
- #41 𝑖 = 0𝑘 = 0The optimal knapsack should contain {1, 2}𝑖 = 0𝑘 = 0The optimal knapsack should contain {1, 2}
- #42 𝑖 = 0𝑘 = 0The optimal knapsack should contain {1, 2}𝑖 = 0𝑘 = 0The optimal knapsack should contain {1, 2}