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Simple and fast linear space computation of L ongest c ommon s ubsequences

Simple and fast linear space computation of L ongest c ommon s ubsequences. Claus Rick, 1999. A. What is the LCS problem?. A A B A C. A B C. …Finding a sequence of greatest possible length that can be obtained

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Simple and fast linear space computation of L ongest c ommon s ubsequences

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  1. Simple and fast linear space computation of Longest common subsequences Claus Rick, 1999

  2. A What is the LCS problem? A A B A C A B C …Finding a sequence of greatest possible length that can be obtained From both A and B by deleting zero or more (not necessarily adjacent) symbols.

  3. A Some boring history…

  4. A Pre-Info • Divide and conquer • Midpoint

  5. A Some basic terms Ordered Pair (i,j) A A B A C A B C (2,3)= (A,C)

  6. A Some basic terms Match A A B A C A B C

  7. A Some basic terms Chain A A B A C A B C

  8. A Rank k A A B A C A B C

  9. A Some basic terms c b a b b a c a c Matching Matrix a b a c b c b a

  10. A Some basic terms Dominant matches All Upper-left matches in each rank

  11. A Dominant matches c b a b b a c a c a b a c b c b a 1 2 3 4 5

  12. A A A B A C A B C

  13. A c b a b b a c a c a b a c b c b a

  14. A Backward contours (BC) a b a c b c b a 5 4 3 2 1 c b a b b a c a c

  15. A Some last basic terms FCk BCk

  16. A Forward contours (FC) c b a b b a c a c a b a c b c b a 1 2 3 4 5

  17. A Backward contours (BC) a b a c b c b a 5 4 3 2 1 c b a b b a c a c

  18. A Lemma 1 Let p be the length of an LCS between strings A and B. Then for every match (i,j) the following holds: • There is an LCS containing (i,j) if and only if (i,j) is on the kth forward contour and on the (p-k+1)st backward contour.

  19. A Lemma 1- proof P |BC|- (p-k+1) |FC|= (k) K <(p-k+1) <(p-k+1) P

  20. A Start calculating FC1 BC1 FC2 BC2 Sooner or later…

  21. A Really really last terms Define sets Mi as: M0= M M1= M0\FC1 M2= M1\BC1 M2i-1=M2(i-1) \FCi M2i=M2i-1\BCi

  22. A c b a b b a c a c a b a c b c b a a b a c b c b a M c b a b b a c a c

  23. A c b a b b a c a c a b a c b c b a a b a c b c b a M1 M2 M3 M4 M5 c b a b b a c a c

  24. A Let call the first empty Mi…. M p’

  25. A Lemma 2 • The Length of an LCS is p’ and each match in M(p’-1) is a possible midpoint

  26. A Lemma 2- proof K K-1 K-2 1 0 K=p M k-1 M 0 M 2 M 1 M k

  27. A Little problem… • We can`t keep tracks of each set- very expensive

  28. A c b a b b a c a c a b a c b c b a a b a c b c b a c b a b b a c a c

  29. A What do we do? Keep only dominant matches… When we see a dominant match below- done.

  30. A c b a b b a c a c a b a c b c b a a b a c b c b a c b a b b a c a c

  31. A Lets define: • FCf’ , BCb’ the minimal indices as stated above

  32. A Lemma 3 • The Length of an LCS is b’ + f’ -1.

  33. A Complexity Finding the dominant matches each contour: O(min(m, (n-p)) Number of contours: P O(Min(pm, p(n-p)

  34. A The End

  35. Simple and fast linear space computation of longest common subsequence Written by: Claus Rick,1999 Based on algorithm by: D.Hirschberg, 1975 Cast: Matrices Lines Arrows Squares Blue Red Brown Grey Black String A String B Presentation: Uri Scheiner No Dominant Matches were harmed during the making of this presentation

  36. Appendix What is the LCS Lemma 1 Divided And Conquer Define M… Match Lemma 2 Chain Keep just Dominant… Dominant Matches FC Lemma 3 BC Complexity

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