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[object Object],[object Object],[object Object]
Dynamic programming ,[object Object],[object Object],[object Object]
Longest Common Subsequence (LCS) ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
LCS Algorithm ,[object Object],[object Object],[object Object],[object Object]
LCS Algorithm ,[object Object],[object Object],[object Object],[object Object]
LCS recursive solution ,[object Object],[object Object],[object Object]
LCS recursive solution ,[object Object],[object Object]
LCS recursive solution ,[object Object],[object Object],Why not just take the length of LCS(X i-1 , Y j-1 ) ?
LCS Length Algorithm ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
LCS Example ,[object Object],[object Object],[object Object],LCS(X, Y) = BCB X = A B C B Y = B D C A B What is the Longest Common Subsequence of X and Y?
LCS Example (0) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D X = ABCB; m = |X| = 4 Y = BDCAB; n = |Y| = 5 Allocate array c[5,4] ABCB BDCAB
LCS Example (1) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 for i = 1 to m c[i,0] = 0 for j = 1 to n c[0,j] = 0 ABCB BDCAB
LCS Example (2) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 0 A BCB B DCAB
LCS Example (3) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 0 0 0 A BCB B DC AB
LCS Example (4) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 0 0 0 1 A BCB BDC A B
LCS Example (5) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 0 0 0 1 1 A BCB BDCA B
LCS Example (6) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 0 0 1 0 1 1 A B CB B DCAB
LCS Example (7) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 1 1 1 1 A B CB B DCA B
LCS Example (8) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 1 1 1 1 2 A B CB BDCA B
LCS Example (10) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 2 1 1 1 1 1 1 AB C B BD CAB
LCS Example (11) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 1 2 1 1 1 1 1 2 AB C B BD C AB
LCS Example (12) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 1 2 1 1 1 1 2 1 2 2 AB C B BDC AB
LCS Example (13) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 1 2 1 1 1 1 2 1 2 2 1 ABC B B DCAB
LCS Example (14) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 1 2 1 1 1 1 2 1 2 2 1 1 2 2 ABC B B DCA B
LCS Example (15) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 1 2 1 1 1 1 2 1 2 2 1 1 2 2 3 ABC B BDCA B
LCS Algorithm Running Time ,[object Object],[object Object],O(m*n) since each c[i,j] is calculated in constant time, and there are m*n elements in the array
How to find actual LCS ,[object Object],[object Object],[object Object],[object Object],[object Object],2 2 3 2 For example, here c[i,j] = c[i-1,j-1] +1 = 2+1=3
How to find actual LCS - continued ,[object Object],[object Object],[object Object],[object Object]
Finding LCS j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C Yj B B A C D 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 2 1 1 1 1 2 1 2 2 1 1 2 2 3 B
Finding LCS (2) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C Yj B B A C D 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 2 1 1 1 1 2 1 2 2 1 1 2 2 3 B B C B LCS (reversed order): LCS (straight order): B C B (this string turned out to be a palindrome)
Knapsack problem ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Knapsack problem There are two versions of the problem: (1) “0-1 knapsack problem” and (2) “Fractional knapsack problem” (1) Items are indivisible; you either take an item or not. Solved with dynamic programming (2) Items are divisible: you can take any fraction of an item. Solved with a greedy algorithm .

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lecture 24

  • 1.
  • 2.
  • 3.
  • 4.
  • 5.
  • 6.
  • 7.
  • 8.
  • 9.
  • 10.
  • 11. LCS Example (0) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D X = ABCB; m = |X| = 4 Y = BDCAB; n = |Y| = 5 Allocate array c[5,4] ABCB BDCAB
  • 12. LCS Example (1) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 for i = 1 to m c[i,0] = 0 for j = 1 to n c[0,j] = 0 ABCB BDCAB
  • 13. LCS Example (2) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 0 A BCB B DCAB
  • 14. LCS Example (3) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 0 0 0 A BCB B DC AB
  • 15. LCS Example (4) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 0 0 0 1 A BCB BDC A B
  • 16. LCS Example (5) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 0 0 0 1 1 A BCB BDCA B
  • 17. LCS Example (6) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 0 0 1 0 1 1 A B CB B DCAB
  • 18. LCS Example (7) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 1 1 1 1 A B CB B DCA B
  • 19. LCS Example (8) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 1 1 1 1 2 A B CB BDCA B
  • 20. LCS Example (10) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 2 1 1 1 1 1 1 AB C B BD CAB
  • 21. LCS Example (11) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 1 2 1 1 1 1 1 2 AB C B BD C AB
  • 22. LCS Example (12) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 1 2 1 1 1 1 2 1 2 2 AB C B BDC AB
  • 23. LCS Example (13) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 1 2 1 1 1 1 2 1 2 2 1 ABC B B DCAB
  • 24. LCS Example (14) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 1 2 1 1 1 1 2 1 2 2 1 1 2 2 ABC B B DCA B
  • 25. LCS Example (15) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 1 2 1 1 1 1 2 1 2 2 1 1 2 2 3 ABC B BDCA B
  • 26.
  • 27.
  • 28.
  • 29. Finding LCS j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C Yj B B A C D 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 2 1 1 1 1 2 1 2 2 1 1 2 2 3 B
  • 30. Finding LCS (2) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C Yj B B A C D 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 2 1 1 1 1 2 1 2 2 1 1 2 2 3 B B C B LCS (reversed order): LCS (straight order): B C B (this string turned out to be a palindrome)
  • 31.
  • 32. Knapsack problem There are two versions of the problem: (1) “0-1 knapsack problem” and (2) “Fractional knapsack problem” (1) Items are indivisible; you either take an item or not. Solved with dynamic programming (2) Items are divisible: you can take any fraction of an item. Solved with a greedy algorithm .