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 2031

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

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

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

8. L R M r > r

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

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

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

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

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

14. Analysis If we do more than a single comparison in an iteration then max(, 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 You in fact sort suffixes of a string shorter by a factor of 2 ! Every pair of characters is now a character

18. Change the alphabet a$0 aa1 ab2 b$3 ba4 bb5 $ a b a aa b 2 12

19. But we do not gain anything…

20. Divide into triples $ yab ba da b bad o abb ada bba do$

21. Divide into triples $ yab ba da b bad o abb ada bba do$ $ yab ba da b bad o bba dab bad o$$

22. Sort recursively 2/3 of the suffixes $ yab ba da b bad o abb ada bba do$ bba dab bad o$$ 124645 37 016425 37 $ yab ba da b bad o 142653 78 148275 1011 1234 56 789101112 0 01234 56 7

23. Sort the remaining third $ yab ba da b bad o 142653 78 (b, 2)(a, 5) (a, 7) (y, 1) (b, 2) (a, 5) (a, 7) (y, 1) 3 6 9 0 1234 56 789101112 0 148275 1011 

24. Merge $ yab ba da b bad o 142653 78 1234 56 789101112 1 0 3 6 9 0 148275 1011

25. Merge $ yab ba da b bad o 142653 78 1234 56 789101112 1 0 3 6 9 0 48275 1011 6

26. Merge $ yab ba da b bad o 142653 78 1234 56 789101112 1 0 39 0 48275 1011 64

27. Merge $ yab ba da b bad o 142653 78 1234 56 789101112 1 0 39 0 8275 1011 649

28. Merge $ yab ba da b bad o 142653 78 1234 56 789101112 1 0 3 0 8275 1011 6493

29. Merge $ yab ba da b bad o 142653 78 1234 56 789101112 1 0 0 8275 1011 64938

30. Merge $ yab ba da b bad o 142653 78 1234 56 789101112 1 0 0 275 1011 649382

31. Merge $ yab ba da b bad o 142653 78 1234 56 789101112 1 0 0 75 1011 6493827

32. Merge $ yab ba da b bad o 142653 78 1234 56 789101112 1 0 0 5 1011 64938275

33. Merge $ yab ba da b bad o 142653 78 1234 56 789101112 1 0 0 1011 64938275

34. Merge $ yab ba da b bad o 142653 78 1234 56 789101112 1 0 649382751011 0

35. summary $ yab ba da b bad o 142653 78 1234 56 789101112 1 0 6493827510110 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 $ yab ba da b bad o 1217928 1011 1234 56 789101112 1 0 6493827510110 abbado$ abbadabbado$ adabbado$ ado$ badabbado$ bado$ bbadabbado$ bbado$ dabbado$ do$ o$ yabbadabbado$ 536 4 1 4 8 2 7 5 10 11 0 6 3 9

37. Crucial observation $ yab ba da b bad o 1217928 1011 1234 56 789101112 1 0 6493827510110 abbado$ abbadabbado$ adabbado$ ado$ badabbado$ bado$ bbadabbado$ bbado$ dabbado$ do$ o$ yabbadabbado$ LCP(i,j) = min {LCP(i,i+1),LCP(i+1,i+2),….,LCP(j-1,j)} 536 4 1 4 8 2 7 5 10 11 0 6 3 9

38. $ yab ba da b bad o 1217928 1011 1234 56 789101112 1 0 6493827510110 abbado$ abbadabbado$ adabbado$ ado$ badabbado$ bado$ bbadabbado$ bbado$ dabbado$ do$ o$ yabbadabbado$ LCP(11,0) 536 4 1 6 4 9 3 8 2 7 5 10 11 0 0 Find LCP’s of consecutive suffixes

39. $ yab ba da b bad o 1217928 1011 1234 56 789101112 1 0 6493827510110 abbado$ abbadabbado$ adabbado$ ado$ badabbado$ bado$ bbadabbado$ bbado$ dabbado$ do$ o$ yabbadabbado$ LCP(8,2) 536 4 1 6 4 9 3 8 2 7 5 10 11 0 01

40. $ yab ba da b bad o 1217928 1011 1234 56 789101112 1 0 6493827510110 abbado$ abbadabbado$ adabbado$ ado$ badabbado$ bado$ bbadabbado$ bbado$ dabbado$ do$ o$ yabbadabbado$ LCP(9,3) 536 4 1 6 4 9 3 8 2 7 5 10 11 0 01 0

41. $ yab ba da b bad o 1217928 1011 1234 56 789101112 1 0 6493827510110 abbado$ abbadabbado$ adabbado$ ado$ badabbado$ bado$ bbadabbado$ bbado$ dabbado$ do$ o$ yabbadabbado$ LCP(6,4) 536 4 1 6 4 9 3 8 2 7 5 10 11 0 101 0

42. $ yab ba da b bad o 1217928 1011 1234 56 789101112 1 0 6493827510110 abbado$ abbadabbado$ adabbado$ ado$ badabbado$ bado$ bbadabbado$ bbado$ dabbado$ do$ o$ yabbadabbado$ LCP(7,5) 536 4 1 6 4 9 3 8 2 7 5 10 11 0 0101 0

43. $ yab ba da b bad o 1217928 1011 1234 56 789101112 1 0 6493827510110 abbado$ abbadabbado$ adabbado$ ado$ badabbado$ bado$ bbadabbado$ bbado$ dabbado$ do$ o$ yabbadabbado$ LCP(1,6) 536 4 1 6 4 9 3 8 2 7 5 10 11 0 5 0101 0

44. $ yab ba da b bad o 1217928 1011 1234 56 789101112 1 0 6493827510110 abbado$ abbadabbado$ adabbado$ ado$ badabbado$ bado$ bbadabbado$ bbado$ dabbado$ do$ o$ yabbadabbado$ LCP(2,7) 536 4 1 6 4 9 3 8 2 7 5 10 11 0 45 0101 0

45. $ yab ba da b bad o 1217928 1011 1234 56 789101112 1 0 6493827510110 abbado$ abbadabbado$ adabbado$ ado$ badabbado$ bado$ bbadabbado$ bbado$ dabbado$ do$ o$ yabbadabbado$ 536 4 1 6 4 9 3 8 2 7 5 10 11 0 45 0101 0 LCP(3,8) 3

46. $ yab ba da b bad o 1217928 1011 1234 56 789101112 1 0 6493827510110 abbado$ abbadabbado$ adabbado$ ado$ badabbado$ bado$ bbadabbado$ bbado$ dabbado$ do$ o$ yabbadabbado$ 536 4 1 6 4 9 3 8 2 7 5 10 11 0 45 0101 0 LCP(4,9) 3 2

47. $ yab ba da b bad o 1217928 1011 1234 56 789101112 1 0 6493827510110 abbado$ abbadabbado$ adabbado$ ado$ badabbado$ bado$ bbadabbado$ bbado$ dabbado$ do$ o$ yabbadabbado$ 536 4 1 6 4 9 3 8 2 7 5 10 11 0 45 0101 0 LCP(5,10) 3 21

48. $ yab ba da b bad o 1217928 1011 1234 56 789101112 1 0 6493827510110 abbado$ abbadabbado$ adabbado$ ado$ badabbado$ bado$ bbadabbado$ bbado$ dabbado$ do$ o$ yabbadabbado$ 536 4 1 6 4 9 3 8 2 7 5 10 11 0 45 0101 0 LCP(10,11) 3 210

49. $ yab ba da b bad o 1217928 1011 1234 56 789101112 1 0 6493827510110 abbado$ abbadabbado$ adabbado$ ado$ badabbado$ bado$ bbadabbado$ bbado$ dabbado$ do$ o$ yabbadabbado$ 536 4 1 6 4 9 3 8 2 7 5 10 11 0 45 0101 0 The starting position deceases by 1 in every iteration. So it cannot increase more than O(n) times 3 210 Analysis

50. $ yab ba da b bad o 1217928 1011 1234 56 789101112 1 0 6493827510110 536 4 45 0101 0 3 210 We need more LCPs for search Linearly many, calculate the all bottom up