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MCS 101: Algorithms

MCS 101: Algorithms. Instructor Neelima Gupta ngupta@cs.du.ac.in. Dynamic Programming. Instructor: Ms. Neelima Gupta. Interval/Job Scheduling –a greedy approach. Weighted Interval Scheduling –. Each job i has (s i , f i , p i ) starting time s i , finishing time f i , and

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MCS 101: Algorithms

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  1. MCS 101: Algorithms Instructor Neelima Gupta ngupta@cs.du.ac.in

  2. Dynamic Programming Instructor: Ms. Neelima Gupta

  3. Interval/Job Scheduling –a greedy approach

  4. Weighted Interval Scheduling – • Each jobi has (si , fi , pi) starting time si , finishing time fi, and profit pi • Aim: Find an optimal schedule of job that makes the maximum profit. • Greedy approach : choose the job which finishes first…..does not work. So, DP Thanks to Neha (16)

  5. i Si Fi Pi • 2 2 4 3 • 1 1 5 10 • 3 4 6 4 • 4 5 8 20 • 5 6 9 2 Thanks to Neha (16)

  6. Weighted Interval Scheduling P(1)=10 P(2)=3 P(3)=4 P(4)=20 P(5)=2 Time 0 1 7 2 3 4 5 6 8 9 Thanks to Neha (16)

  7. Greedy Approach P(1)=10 P(2)=3 P(3)=4 P(4)=20 P(5)=2 Time 0 1 7 2 3 4 5 6 8 9 . Thanks to Neha (16)

  8. Greedy does not work P(1)=10 P(2)=3 Optimal schedule P(3)=4 P(4)=20 Schedule chosen by greedy app P(5)=2 Time 0 1 7 2 3 4 5 6 8 9 Greedy approach takes job 2, 3 and 5 as best schedule and makes profit of 7. While optimal schedule is job 1 and job4 making profit of 30 (10+20). Hence greedy will not work Thanks to Neha (16)

  9. DP Solution for WIS • Let m[j]= optimal schedule solution from the first jth jobs, (jobs are sorted in the increasing order of their finishing times) pj=profit of jth job. p[j] =largest index i<j , such that interval i and j are disjoint i.e. i is the rightmost interval that ends before j begins or the last interval compatible with j and is before j. Thanks to : Neha Mishra (roll no:18)

  10. DP Solution for WIS • Either j is in the optimal solution or it is not. • If it is, then m[j] = pj + profit when considering the jobs before and including the last job compatible with j i.e. m[j] = pj + m[p(j)] • If it is not, then m[j] = m[j-1] • Thus, m[j]= max(pj + m[p(j)], m[j-1])

  11. Recursive Paradigm • Write a recursive function to compute m[n]…afterall we are only interested in that. • some of the problems are solved several times leading to exponential time.

  12. INDEX 1 2 3 4 5 6 P[1]=0 V1=2 V2=4 V3=4 V4=7 V5=2 V6=1 Thanks to : Nikita Khanna(19)

  13. INDEX 1 2 3 4 5 6 P[1]=0 V1=2 P[2]=0 V2=4 V3=4 V4=7 V5=2 V6=1 Thanks to : Nikita Khanna(19)

  14. INDEX 1 2 3 4 5 6 P[1]=0 V1=2 P[2]=0 V2=4 P[3]=1 V3=4 V4=7 V5=2 V6=1 Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19)

  15. INDEX 1 2 3 4 5 6 P[1]=0 V1=2 P[2]=0 V2=4 P[3]=1 V3=4 P[4]=0 V4=7 V5=2 V6=1 Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19)

  16. INDEX 1 2 3 4 5 6 P[1]=0 V1=2 P[2]=0 V2=4 P[3]=1 V3=4 P[4]=0 V4=7 V5=2 P[5]=3 V6=1 Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19)

  17. INDEX 1 2 3 4 5 6 P[1]=0 V1=2 P[2]=0 V2=4 P[3]=1 V3=4 P[4]=0 V4=7 V5=2 P[5]=3 V6=1 P[6]=3 Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19)

  18. Recursive Paradigm : tree of recursion for WIS

  19. Recursive Paradigm • Sometimes some of the problems are solved several times leading to exponential time. For example: Fibonacci numbers. • DP Solution: Solve a problem only once and use it as and when required • Top-down DP …..Memoization….generally storage cost is large • Bottom-Up DP …….Iterative

  20. Principles of DP

  21. m[j] = max{m[j-1], m[p[j] ] + pj} m 0 1 2 3 4 5 6 Thanks to : Nikita Khanna(19)

  22. m[j] = max{m[j-1], m[p[j] ] + pj} 0 m 0 1 2 3 4 5 6 Thanks to : Nikita Khanna(19)

  23. m[j] = max{m[j-1], m[p[j] ] + pj} Max{0,0+2} 0 2 m 0 1 2 3 4 5 6 Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19)

  24. m[j] = max{m[j-1], m[p[j] ] + pj} Max{0,0+2} Max{2,0+4} 0 2 4 m 0 1 2 3 4 5 6 Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19)

  25. m[j] = max{m[j-1], m[p[j] ] + pj} Max{0,0+2} Max{2,0+4} 0 2 4 6 m 0 1 2 3 4 5 6 Max{4,2+4} Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19) )

  26. m[j] = max{m[j-1], m[p[j] ] + pj} Max{0,0+2} Max{6,0+7} Max{2,0+4} 0 2 4 6 7 m 0 1 2 3 4 5 6 Max{4,2+4} Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19)

  27. m[j] = max{m[j-1], m[p[j] ] + pj} Max{0,0+2} Max{6,0+7} Max{2,0+4} 0 2 4 6 7 8 m 0 1 2 3 4 5 6 Max{7,6+2} Max{4,2+4} Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19)

  28. m[j] = max{m[j-1], m[p[j] ] + pj} Max{0,0+2} Max{8,6+1} Max{6,0+7} Max{2,0+4} 0 2 4 6 7 8 8 m 0 1 2 3 4 5 6 Max{7,6+2} Max{4,2+4} Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19) Thanks to : Nikita Khanna(19)

  29. FRACTIONAL KNAPSACK PROBLEM Given a set S of n items, with value vi and weight wi and a knapsack with capacity W. Aim: Pick items with maximum total value but with weight at mostW. You may choose fractions of items.

  30. GREEDY APPROACH • Pick the items in the decreasing order of value per unit weight i.e. highest first.

  31. Example knapsack capacity 50 Item 2 item 3 Item 1 • vi = 60vi = 100 vi = 120 vi/wi = 6 vi/wi = 5 vi/wi = 4 30 20 10 Thanks to: Neha Katyal

  32. Example knapsack capacity 50 Item 2 item 3 • 60 vi = 100 vi = 120 vi/wi = 5 vi/wi = 4 30 20 10 Thanks to: Neha Katyal

  33. Example knapsack capacity 50 item 3 100 • + • 60 vi = 120 vi/wi = 4 20 30 10 Thanks to: Neha Katyal

  34. Example knapsack capacity 50 $80 + 100 + 60 = 240 20/30 20 10 Thanks to: Neha Katyal

  35. 0-1 Kanpsack • example to show that the above greedy approach will not work, So, DP

  36. GREEDY APPROACH DOESN’T WORK FOR 0-1 KNAPSACK Counter Example: knapsack Item 2 item 3 Item 1 • vi = $60vi = $100 vi = $120 vi/wi = 6 vi/wi = 5 vi/wi = 4 30 20 10 Thanks to: Neha Katyal

  37. Example knapsack Item 2 item 3 • $60 vi = $100 vi = $120 vi/wi = 5 vi/wi = 4 30 20 10 Thanks to: Neha Katyal

  38. Example knapsack item 3 $100 • + • $60 vi = $120 = $160 vi/wi = 4 suboptimal 20 30 10 Thanks to: Neha Katyal

  39. Fractional Knapsack –a greedy approach

  40. DP Solution for 0-1KS Let m[I,w] be the optimal value obtained when considering objects upto I and filling a knapsack of capacity w • m[0,w] = 0 • m[i,0] = 0 • m[i,w] = m[i-1,w] if wi > w • m[i,w] = max{m[i-1, w-wi] + vi , m[i-1, w]} if wi <= w Running Time : Pseudo-polynomial

  41. Example • n = 4 • W = 5 • Elements (weight, value): (2,3), (3,4), (4,5), (5,6) Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

  42. (2,3), (3,4), (4,5), (5,6) W 0 1 2 3 4 5 i 0 1 2 3 4 Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

  43. (2,3), (3,4), (4,5), (5,6) As w<w1 ; m[1,1] = m[1-1,1] = m[0,1] W 0 1 2 3 4 5 i 0 1 2 3 4 Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

  44. (2,3), (3,4), (4,5), (5,6) As w<w2 ; m[2,1] = m[2-1,1] = m[1,1] W 0 1 2 3 4 5 i 0 1 2 3 4 Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

  45. (2,3), (3,4), (4,5), (5,6) As w<w3 ; m[3,1] = m[3-1,1] = m[2,1] W 0 1 2 3 4 5 i 0 1 2 3 4 Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

  46. (2,3), (3,4), (4,5), (5,6) As w<w4 ; m[4,1] = m[4-1,1] = m[3,1] W 0 1 2 3 4 5 i 0 1 2 3 4 Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

  47. (2,3), (3,4), (4,5), (5,6) As w>=w1 ; m[1,2] = max{m[1-1,2] , m[1-1,2-2]+3} =max{ 0,0+3} W 0 1 2 3 4 5 i 0 1 2 3 4 Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

  48. (2,3), (3,4), (4,5), (5,6) As w<w2 ; m[2,2] = m[2-1,2] = m[1,2] W 0 1 2 3 4 5 i 0 1 2 3 4 Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

  49. (2,3), (3,4), (4,5), (5,6) As w<w3 ; m[3,2] = m[3-1,2] = m[2,2] W 0 1 2 3 4 5 i 0 1 2 3 4 Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

  50. (2,3), (3,4), (4,5), (5,6) As w<w4 ; m[4,2] = m[4-1,2] = m[3,2] W 0 1 2 3 4 5 i 0 1 2 3 4 Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

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