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Optimal Solution Finding Algorithm for Coin Change Problem

Understand the dynamic programming approach for finding the optimal solution to coin change problems, analyzing recursive solutions, speeding up algorithms, and implementing dynamic programming.

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Optimal Solution Finding Algorithm for Coin Change Problem

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  1. 18 = 6 + 6 + 6 = 13 + 2 + 2 + 1 24 = 6 + 6 + 6 + 6 = 13 + 6 + 2 +2 +1 31 = 13 + 18 37 = 13 + 24 44 = 13 + 31 50 = 13 + 37

  2. the greedy algorithm doesn’t work How are we going to find an optimal solution ?

  3. Recursive solution try all possibilities Number(n) if n<0 then RETURN  if n=0 then RETURN 0 a1  1 + Number(n-1) a2  1 + Number(n-2) a6  1 + Number(n-6) a13 1 + Number(n-13) RETURN min(a1,a2,a6,a13)

  4. Recursive solution try all possibilities Number(n) if n<0 then RETURN  if n=0 then RETURN 0 a1  1 + Number(n-1) a2  1 + Number(n-2) a6  1 + Number(n-6) a13 1 + Number(n-13) RETURN min(a1,a2,a6,a13) OK, so now we know the cost of the optimal solution. How do we output the solution?

  5. Outputting solution Solution(n) if n=0 then RETURN ; else if Number(n)=1+Number(n-1) then print(1); Solution(n-1); elif Number(n)=1+Number(n-2) then print(2); Solution(n-2); elif Number(n)=1+Number(n-6) then print(6); Solution(n-6); else print(13); Solution(n-6);

  6. How to make it faster? Time = n Time = Fn 1.618n

  7. How to make it faster? Number[0] 0 for i  1 to n do A  1 + Number[i-1] if i  2 then A  min(A,1+Number[i-2]) if i  6 then A  min(A,1+Number[i-6]) if i  13 then A  min(A,1+Number[i-13]) Number[i]  A Dynamic programming

  8. Example Number[0] 0 for i  1 to n do A  1 + Number[i-1] if i  2 then A  min(A,1+Number[i-2]) if i  6 then A  min(A,1+Number[i-6]) if i  13 then A  min(A,1+Number[i-13]) Number[i]  A $0 - 0 $1 - 1 $2 - 1 $3 - 2 $4 - 2 $5 - 3 $6 - 1 $7 - 2 $8 - 2 $9 - 3 $10 $11 $12 $13 $14 $15

  9. Example Number[0] 0 for i  1 to n do A  1 + Number[i-1] if i  2 then A  min(A,1+Number[i-2]) if i  6 then A  min(A,1+Number[i-6]) if i  13 then A  min(A,1+Number[i-13]) Number[i]  A $0 - 0 $1 - 1 $2 - 1 $3 - 2 $4 - 2 $5 - 3 $6 - 1 $7 - 2 $8 - 2 $9 - 3 $10 - 3 $11 $12 $13 $14 $15

  10. Example Number[0] 0 for i  1 to n do A  1 + Number[i-1] if i  2 then A  min(A,1+Number[i-2]) if i  6 then A  min(A,1+Number[i-6]) if i  13 then A  min(A,1+Number[i-13]) Number[i]  A $0 - 0 $1 - 1 $2 - 1 $3 - 2 $4 - 2 $5 - 3 $6 - 1 $7 - 2 $8 - 2 $9 - 3 $10 - 3 $11 $12 $13 $14 $15

  11. Example Number[0] 0 for i  1 to n do A  1 + Number[i-1] if i  2 then A  min(A,1+Number[i-2]) if i  6 then A  min(A,1+Number[i-6]) if i  13 then A  min(A,1+Number[i-13]) Number[i]  A $0 - 0 $1 - 1 $2 - 1 $3 - 2 $4 - 2 $5 - 3 $6 - 1 $7 - 2 $8 - 2 $9 - 3 $10 - 3 $11 - 4 $12 $13 $14 $15

  12. Heart of the algorithm Let Number[n] be the optimal number of coins needed to pay $n. Let Number[n]= for n<0, and Number[0]=0 For n1 Number[n]=1+min { Number[n-1] Number[n-2] Number[n-6] Number[n-13]

  13. Dynamic programming vs. greedy What is the input size for the coin change problem? Is the dynamic programming algorithm for the coin change polynomial-time? Is the greedy algorithm polynomial-time?

  14. Knapsack INPUT: n items wi = weight of item i vi = value of item i M = capacity of knapsack SOLUTION: a subset S of items whose total weight does not exceed M OBJECTIVE: maximize the total value of S

  15. Knapsack INPUT: n items wi = weight of item i vi = value of item i M = capacity of knapsack SOLUTION: a subset S of items whose total weight does not exceed M OBJECTIVE: maximize the total value of S M = 5lb book 2lb $10 MP3 player 1lb $100 potatoes 3lb $2 oranges 2lb $3

  16. Knapsack INPUT: n items wi = weight of item i vi = value of item i M = capacity of knapsack SOLUTION: a subset S of items whose total weight does not exceed M OBJECTIVE: maximize the total value of S M = 5lb book 2lb $10 MP3 player 1lb $100 potatoes 3lb $2 oranges 2lb $3 Value = $113 Weight = 5lb

  17. Knapsack Algorithm 1 1. sort the items by “price per pound” vi/wi 2. go through the items in this order, if the item fits, put it in the knapsack. Is solution output by Algorithm 1 always optimal ?

  18. Knapsack K [0..M] where K[ j ] = optimal value of items in knapsack of size j K[ j ] = ?

  19. Knapsack K [0..M,0..n] where K[ j,i ] = optimal value of items in knapsack of size j, using only items 1,…,i K[ j,0 ] = ?

  20. Knapsack K [0..M,0..n] where K[ j,i ] = optimal value of items in knapsack of size j, using only items 1,…,i K[ j,0 ] = 0

  21. Knapsack K [0..M,0..n] where K[ j,i ] = optimal value of items in knapsack of size j, using only items 1,…,i K[ j,i ] = ?

  22. Knapsack K [0..M,0..n] where K[ j,i ] = optimal value of items in knapsack of size j, using only items 1,…,i K[ j,i ] = ? use item i do not use item i

  23. Knapsack K [0..M,0..n] where K[ j,i ] = optimal value of items in knapsack of size j, using only items 1,…,i K[ j,i ] = ? use item i K[ j-wi,i-1]+vi do not use item i K[ j,i-1]

  24. Knapsack K [0..M,0..n] where K[ j,i ] = optimal value of items in knapsack of size j, using only items 1,…,i K[ j,0] = 0 K[ j,i ] = max( K[ j-wi,i-1] +vi , K[ j,i-1] ) if j  wi K[ j,i ] = K[ j, i-1 ] if j < wi Heart of the algorithm

  25. Knapsack K [0..M,0..n] where K[ j,i ] = optimal value of items in knapsack of size j, using only items 1,…,i K[ j,0] = 0 K[ j,i ] = max( K[ j-wi,i-1] +vi , K[ j,i-1] ) if j  wi K[ j,i ] = K[ j, i-1 ] if j < wi What order shall we compute the K [j,i] in?

  26. K[ j,0] = 0 K[ j,i ] = max( K[ j-wi,i-1] +vi , K[ j,i-1] ) if j  wi K[ j,i ] = K[ j, i-1 ] if j < wi for j  0 to M do K[ j,0 ]  0 for i  1 to n do for j  0 to K[ j,i ] K[ j,i-1 ] if j  wi and vi+K[ j-wi,i-1 ] > K[ j,i] then K[ j,i ] vi+K[ j-wi,i-1 ]

  27. INPUT: n items wi = weight of item i vi = value of item i M = capacity of knapsack SOLUTION: a subset S of items whose total weight does not exceed M OBJECTIVE: maximize the total value of S for j  0 to M do K[ j,0 ]  0 for i  1 to n do for j  0 to K[ j,i ] K[ j,i-1 ] if j  wi and vi+K[ j-wi,i-1 ] > K[ j,i] then K[ j,i ] vi+K[ j-wi,i-1 ] Ok, know value of the optimal solution, how to output it?

  28. for j  0 to M do K[ j,0 ]  0 for i  1 to n do for j  0 to K[ j,i ]  K[ j,i-1 ] Get [ j,i]  false if j  wi and vi+K[ j-wi,i-1 ] > K[ j,i] then K[ j,i ] vi+K[ j-wi,i-1 ] Get [ j,I ]  true j  M for i  n downto 1 do if Get[ j,I ] then print(i); j j-wi

  29. What if the thief had 2 knapsacks?

  30. Back to 1 knapsack What if the weights are real numbers and values are integers. Can we still use dynamic programming? K[0..M,0..n] where K[ j,i ] = optimal value of items in knapsack of size j, using only items 1,…,i

  31. Back to 1 knapsack What if the weights are real numbers and values are integers. Can we still use dynamic programming? L[0..S,0..n] where L[ j,i ] = minimal weight of items in knapsack of total value  j, using only items 1,…,i S = total value of all items

  32. Knapsack S = total value of all items L[0..S,0..n] where L[ j,i ] = minimal weight of items in knapsack of total value  j, using only items 1,…,i

  33. Knapsack S = total value of all items L[0..S,0..n] where L[ j,i ] = minimal weight of items in knapsack of total value  j, using only items 1,…,i L[ j,0 ] = ? L[ 0,i ] = ?

  34. Knapsack S = total value of all items L[0..S,0..n] where L[ j,i ] = minimal weight of items in knapsack of total value  j, using only items 1,…,i L[ j,0 ] = 0 L[ 0,i ] = ?

  35. Knapsack S = total value of all items L[0..S,0..n] where L[ j,i ] = minimal weight of items in knapsack of total value  j, using only items 1,…,i L[ j,i ] = get item i don’t get item i

  36. Knapsack S = total value of all items L[0..S,0..n] where L[ j,i ] = minimal weight of items in knapsack of total value  j, using only items 1,…,i L[ j,i ] = get item i don’t get item i L[ j-vi , i-1] + wi L[ j,i-1]

  37. Knapsack S = total value of all items L[0..S,0..n] where L[ j,i ] = minimal weight of items in knapsack of total value  j, using only items 1,…,i L[ j,i ] = min ( L[j,i-1] , L[j-vi,i-1] + wi ) get item i don’t get item i if j  vi L[ j-vi , i-1] + wi L[ j,i-1]

  38. Knapsack for j 0 to S do L[ j,0 ] 0 for i 0 to n do for j 0 to S do L[ j,i ] = L[ j,i-1] if j vi and L[ j-vi,i-1 ] + wi < L[ j,i ] then L[ j,i ]  L [ j-vi, i-1 ] + wi L[0..S,0..n] where L[ j,i ] = minimal weight of items in knapsack of total value  j, using only items 1,…,i What is the value of the optimal solution?

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