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Suffix arrays

Suffix arrays. Suffix array. We loose some of the functionality but we save space. Let s = abab. Sort the suffixes lexicographically: ab, abab, b, bab. The suffix array gives the indices of the suffixes in sorted order. 2. 0. 3. 1. How do we build it ?. Build a suffix tree

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Suffix arrays

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  1. Suffix arrays

  2. Suffix array • We loose some of the functionality but we save space. Let s = abab Sort the suffixes lexicographically: ab, abab, b, bab The suffix array gives the indices of the suffixes in sorted order 2 0 3 1

  3. How do we build it ? • Build a suffix tree • Traverse the tree in DFS, lexicographically picking edges outgoing from each node and fill the suffix array. • O(n) time

  4. How do we search for a pattern ? • If P occurs in T then all its occurrences are consecutive in the suffix array. • Do a binary search on the suffix array • Takes O(mlogn) time

  5. 10 7 4 1 0 9 8 6 3 5 2 Example Let S = mississippi i L ippi issippi Let P = issa ississippi mississippi pi M ppi sippi sisippi ssippi ssissippi R

  6. How do we accelerate the search ? Maintain l = LCP(P,L) Maintain r = LCP(P,R) Assume l ≥ r r l L M R

  7. If l = r then start comparing M to P at l + 1 r l L M R

  8. l > r r l L M R

  9. Someone whispers LCP(L,M) LCP(L,M)> l r l L M R

  10. Continue in the right half LCP(L,M)> l r l L M R

  11. LCP(L,M)< l r l L M R

  12. Continue in the left half LCP(L,M)< l r l L M R

  13. LCP(L,M)= l start comparing M to P at l + 1 r l L M R

  14. Analysis If we do more than a single comparison in an iteration then max(l, r ) grows by 1 for each comparison  O(m + logn) time

  15. Construct the suffix array without the suffix tree

  16. Linear time construction Recursively ? Say we want to sort only suffixes that start at even positions ?

  17. Change the alphabet Every pair of characters is now a character You in fact sort suffixes of a string shorter by a factor of 2 !

  18. Change the alphabet a a b a a b $ 2 1 2

  19. But we do not gain anything…

  20. Divide into triples y a b b a b o d a b a d $ abb ada bba do$

  21. Divide into triples y a b b a b o d a b a d $ abb ada bba do$ y a b b a b o d a b a d $ bba dab bad o$$

  22. 3 7 0 1 6 4 2 5 10 11 1 4 8 2 7 5 y a b b a b o d a b a d $ 7 8 1 4 2 6 5 3 Sort recursively 2/3 of the suffixes 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 0 1 2 3 4 7 5 6 abb ada bba do$ bba dab bad o$$ 3 7 1 2 4 6 4 5

  23. 10 11 1 4 8 2 7 5 Sort the remaining third 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 7 8 1 4 2 6 5 3 (a, 7) (y, 1) (b, 2) (a, 5)  (y, 1) (a, 7) (b, 2) (a, 5) 0 9 3 6

  24. Merge 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 7 8 1 4 2 6 5 3 0 9 3 6 10 11 1 4 8 2 7 5 1

  25. Merge 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 7 8 1 4 2 6 5 3 0 9 3 6 10 11 4 8 2 7 5 1 6

  26. Merge 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 7 8 1 4 2 6 5 3 0 9 3 10 11 4 8 2 7 5 1 6 4

  27. Merge 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 7 8 1 4 2 6 5 3 0 9 3 10 11 8 2 7 5 1 6 4 9

  28. Merge 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 7 8 1 4 2 6 5 3 0 3 10 11 8 2 7 5 1 6 4 9 3

  29. Merge 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 7 8 1 4 2 6 5 3 0 10 11 8 2 7 5 1 6 4 9 3 8

  30. Merge 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 7 8 1 4 2 6 5 3 0 10 11 2 7 5 1 6 4 9 3 8 2

  31. Merge 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 7 8 1 4 2 6 5 3 0 10 11 7 5 1 6 4 9 3 8 2 7

  32. Merge 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 7 8 1 4 2 6 5 3 0 10 11 5 1 6 4 9 3 8 2 7 5

  33. Merge 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 7 8 1 4 2 6 5 3 0 10 11 1 6 4 9 3 8 2 7 5

  34. Merge 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 7 8 1 4 2 6 5 3 0 1 6 4 9 3 8 2 7 5 10 11

  35. summary 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 7 8 1 4 2 6 5 3 1 6 4 9 3 8 2 7 5 10 11 0 When comparing to a suffix with index 1 (mod 3) we compare the char and break ties by the ranks of the following suffixes When comparing to a suffix with index 2 (mod 3) we compare the char, the next char if there is a tie, and finally the ranks of the following suffixes

  36. Compute LCP’s 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 4 10 11 12 1 7 5 3 9 2 8 6 1 6 4 9 3 8 2 7 5 10 11 0 yabbadabbado$ 0 o$ 11 do$ 10 dabbado$ 5 bbado$ 7 bbadabbado$ 2 bado$ 8 badabbado$ 3 ado$ 9 adabbado$ 4 abbado$ 6 abbadabbado$ 1

  37. Crucial observation 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 4 10 11 12 1 7 5 3 9 2 8 6 1 6 4 9 3 8 2 7 5 10 11 0 yabbadabbado$ 0 o$ 11 do$ 10 dabbado$ 5 bbado$ 7 bbadabbado$ 2 LCP(i,j) = min {LCP(i,i+1),LCP(i+1,i+2),….,LCP(j-1,j)} bado$ 8 badabbado$ 3 ado$ 9 adabbado$ 4 abbado$ 6 abbadabbado$ 1

  38. Find LCP’s of consecutive suffixes 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 4 10 11 12 1 7 5 3 9 2 8 6 1 6 4 9 3 8 2 7 5 10 11 0 yabbadabbado$ 0 0 o$ 11 do$ 10 dabbado$ 5 bbado$ 7 bbadabbado$ 2 LCP(11,0) bado$ 8 badabbado$ 3 ado$ 9 adabbado$ 4 abbado$ 6 abbadabbado$ 1

  39. 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 4 10 11 12 1 7 5 3 9 2 8 6 1 6 4 9 3 8 2 7 5 10 11 0 yabbadabbado$ 0 1 0 o$ 11 do$ 10 dabbado$ 5 bbado$ 7 bbadabbado$ 2 LCP(8,2) bado$ 8 badabbado$ 3 ado$ 9 adabbado$ 4 abbado$ 6 abbadabbado$ 1

  40. 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 4 10 11 12 1 7 5 3 9 2 8 6 1 6 4 9 3 8 2 7 5 10 11 0 yabbadabbado$ 0 0 1 0 o$ 11 do$ 10 dabbado$ 5 bbado$ 7 bbadabbado$ 2 LCP(9,3) bado$ 8 badabbado$ 3 ado$ 9 adabbado$ 4 abbado$ 6 abbadabbado$ 1

  41. 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 4 10 11 12 1 7 5 3 9 2 8 6 1 6 4 9 3 8 2 7 5 10 11 0 yabbadabbado$ 0 0 1 1 0 o$ 11 do$ 10 dabbado$ 5 bbado$ 7 bbadabbado$ 2 LCP(6,4) bado$ 8 badabbado$ 3 ado$ 9 adabbado$ 4 abbado$ 6 abbadabbado$ 1

  42. 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 4 10 11 12 1 7 5 3 9 2 8 6 1 6 4 9 3 8 2 7 5 10 11 0 yabbadabbado$ 0 0 1 1 0 0 o$ 11 do$ 10 dabbado$ 5 bbado$ 7 bbadabbado$ 2 LCP(7,5) bado$ 8 badabbado$ 3 ado$ 9 adabbado$ 4 abbado$ 6 abbadabbado$ 1

  43. 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 4 10 11 12 1 7 5 3 9 2 8 6 1 6 4 9 3 8 2 7 5 10 11 0 yabbadabbado$ 0 0 5 1 1 0 0 o$ 11 do$ 10 dabbado$ 5 bbado$ 7 bbadabbado$ 2 LCP(1,6) bado$ 8 badabbado$ 3 ado$ 9 adabbado$ 4 abbado$ 6 abbadabbado$ 1

  44. 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 4 10 11 12 1 7 5 3 9 2 8 6 1 6 4 9 3 8 2 7 5 10 11 0 yabbadabbado$ 0 0 5 4 1 1 0 0 o$ 11 do$ 10 dabbado$ 5 bbado$ 7 bbadabbado$ 2 LCP(2,7) bado$ 8 badabbado$ 3 ado$ 9 adabbado$ 4 abbado$ 6 abbadabbado$ 1

  45. 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 4 10 11 12 1 7 5 3 9 2 8 6 1 6 4 9 3 8 2 7 5 10 11 0 yabbadabbado$ 0 0 5 4 1 3 1 0 0 o$ 11 do$ 10 dabbado$ 5 bbado$ 7 bbadabbado$ 2 LCP(3,8) bado$ 8 badabbado$ 3 ado$ 9 adabbado$ 4 abbado$ 6 abbadabbado$ 1

  46. 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 4 10 11 12 1 7 5 3 9 2 8 6 1 6 4 9 3 8 2 7 5 10 11 0 yabbadabbado$ 0 0 5 4 1 3 1 0 0 2 o$ 11 do$ 10 dabbado$ 5 bbado$ 7 bbadabbado$ 2 LCP(4,9) bado$ 8 badabbado$ 3 ado$ 9 adabbado$ 4 abbado$ 6 abbadabbado$ 1

  47. 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 4 10 11 12 1 7 5 3 9 2 8 6 1 6 4 9 3 8 2 7 5 10 11 0 yabbadabbado$ 0 0 5 4 1 3 1 0 0 2 1 o$ 11 do$ 10 dabbado$ 5 bbado$ 7 bbadabbado$ 2 LCP(5,10) bado$ 8 badabbado$ 3 ado$ 9 adabbado$ 4 abbado$ 6 abbadabbado$ 1

  48. 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 4 10 11 12 1 7 5 3 9 2 8 6 1 6 4 9 3 8 2 7 5 10 11 0 yabbadabbado$ 0 0 5 4 1 3 1 0 0 2 1 0 o$ 11 do$ 10 dabbado$ 5 bbado$ 7 bbadabbado$ 2 LCP(10,11) bado$ 8 badabbado$ 3 ado$ 9 adabbado$ 4 abbado$ 6 abbadabbado$ 1

  49. Analysis 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 4 10 11 12 1 7 5 3 9 2 8 6 1 6 4 9 3 8 2 7 5 10 11 0 yabbadabbado$ 0 0 5 4 1 3 1 0 0 2 1 0 o$ 11 do$ 10 dabbado$ 5 bbado$ 7 bbadabbado$ 2 The starting position deceases by 1 in every iteration. So it cannot increase more than O(n) times bado$ 8 badabbado$ 3 ado$ 9 adabbado$ 4 abbado$ 6 abbadabbado$ 1

  50. We need more LCPs for search 1 2 3 4 7 8 9 10 11 12 5 6 0 y a b b a b o d a b a d $ 4 10 11 12 1 7 5 3 9 2 8 6 1 6 4 9 3 8 2 7 5 10 11 0 0 5 4 1 3 1 0 0 2 1 0 Linearly many, calculate the all bottom up

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