1. Semi-dynamic compact index for short patterns and succinct van Emde Boas tree 1 Yoshiaki Matsuoka 1, Tomohiro I 2, Shunsuke Inenaga 1, Hideo Bannai 1, Masayuki Takeda 1 ( 1 Kyushu University) ( 2 TU Dortmund)

2. Overview  There exist many space-efficient indices (e.g. FM-index [Ferragina&Manzini, 2000]) but most of them are static.  Some (e.g. Dynamic FM-index [Salson et al., 2010]) are dynamic but consume more space than static counterparts. 2

3. Overview  There exist many space-efficient indices (e.g. FM-index [Ferragina&Manzini, 2000]) but most of them are static.  Some (e.g. Dynamic FM-index [Salson et al., 2010]) are dynamic but consume more space than static counterparts.  We propose a self-index for searching patterns of limited length, which: is theoretically and practically efficient in terms of construction, updates (adding characters at the end of the text) and searches, is compact, i.e., requires only O(n log σ) bits of space, where n is the text size and σ is the alphabet size, and can be constructed in online manner. 3

4. Problem  Preprocess : text T of length n over an alphabet of size σ.  Query : pattern P of length at most r.  Answer : all occurrences of P in T. 4

5. Problem  Preprocess : text T of length n over an alphabet of size σ.  Query : pattern P of length at most r.  Answer : all occurrences of P in T.  Example. 5 0123456789101112131415161718192021 T = abbbabaaabaaabbaaaabaa If P = baa, then we output {5, 9, 14, 19} (in any order).

6. A naïve algorithm  Since we would like to search for any pattern of length at most r, a naïve solution would be to store all occurrences of all r-grams in T.  This naïve algorithm requires at least n log n bits.  Example. 6 r-gramsOccurrences aaa 6, 10, 15, 16 aab 7, 11, 17 aba 4, 8, 18 0123456789101112131415161718192021 T = abbbabaaabaaabbaaaabaa r=3

7. Sampling of q-grams  To reduce the space, we only store the beginning positions divisible by some k (> 1).  We also sample longer substrings (of length r + k − 1 = q) so that occurrences of substrings of length at most r are not missed.  Example. 7 r=3 k=4 q=6 0123456789101112131415161718192021 T = abbbabaaabaaabbaaaabaa q-gramsOccurrences at positions divisible by k aaabaa 16 abaaab 4, 8 abbaaa 12 abbbab 0

8. Sampling of q-grams  For any pattern P of length at most r, if w is a sampled q-gram at position x in T and P has an occurrence in w with relative position d (i.e., w[d.. d+|P|−1] = P), then x + d is an occurrence of P in T. 8 P = baa r=3 k=4 q=6 0123456789101112131415161718192021 T = abbbabaaabaaabbaaaabaa occurrence at 4+1 occurrence at 8+1 occurrence at 12+2 occurrence at 16+3

9. Set of q-grams Q P,d  Let Q P,d be the set of (not only sampled but) all q-grams w in T where P has an occurrence in w with relative position d, i.e., w[d.. d+|P|−1] = P.  For example, consider the following string T: In this example, if k = 4, q = 6 and P = baa, then Q P,0 = { baaaab, baaaba, baaabb }, Q P,1 = { abaaab, bbbaab }, Q P,2 = { aabaaa, abbaaa, babaaa }, and Q P,3 = { aaabaa, aabbaa, bbabaa }. 9 0123456789101112131415161718192021 T = abbbabaaabaaabbaaaabaa

10. Set of q-grams Q P,d 10  For example, consider the following string T: In this example, if k = 4, q = 6 and P = baa, then Q P,0 = { baaaab, baaaba, baaabb }, Q P,1 = { abaaab, bbbaab }, Q P,2 = { aabaaa, abbaaa, babaaa }, and Q P,3 = { aaabaa, aabbaa, bbabaa }. 0123456789101112131415161718192021 T = abbbabaaabaaabbaaaabaa Q P,0 ∪ Q P,1 ∪ … ∪ Q P,k−1 contains all sampled q-grams which contain P (with its offset). |Q P,d | ≤ #occ for any 0 ≤ d < k. Observation

11. Basic strategy of our search algorithm  To compute all occurrences of P in T, we incrementally compute Q P,0, Q P,1, …, Q P,k−1 and output occurrences of P when we encounter sampled q-grams in each Q P,d. 11 Q P,0 ∪ Q P,1 ∪ … ∪ Q P,k−1 contains all sampled q-grams which contain P (with its offset). |Q P,d | ≤ #occ for any 0 ≤ d < k. Observation

12. q-gram transition graph  To compute Q P,1,…, Q P,k−1, we consider a directed graph G = (Σ q, E), which we call a q-gram transition graph. A q-gram transition graph is a subgraph of the de Bruijn graph of T s.t. the indegree of each vertex is at most 1. 12

13. q-gram transition graph 13 r=3 k=4 q=6 abbbabbbbababbabaababaaaabaaabbaaaba baaabb aaabbaaabbaaabbaaabbaaaabaaaabaaaaba aaabaaaabaaa 0123456789101112131415161718192021 T = abbbabaaabaaabbaaaabaa We limit the indegree at most 1, so this edge is not constructed.

14. q-gram transition graph 14 r=3 k=4 q=6 abbbabbbbababbabaababaaaabaaabbaaaba baaabb aaabbaaabbaaabbaaabbaaaabaaaabaaaaba aaabaaaabaaa 0123456789101112131415161718192021 T = abbbabaaabaaabbaaaabaa 0 4, 8 12 16 Positions of sampled q-grams.

15. Computing Q P,0, …, Q P,k−1 15 baaabb baaaba Q P,0 P = baa This edge does not exist, therefore abaaba is enumerated only once. r=3 k=4 q=6 0123456789101112131415161718192021 T = abbbabaaabaaabbaaaabaa baaaabbbaaaa Q P,1 abaaab abbaaa aabaaa Q P,2 babaaa aabbaa bbabaa Q P,3 aaabaa 16 12 4, 8

16. Computing Q P,0, …, Q P,k−1 16 baaabb baaaba Q P,0 P = baa This edge does not exist, therefore abaaba is enumerated only once. r=3 k=4 q=6 0123456789101112131415161718192021 T = abbbabaaabaaabbaaaabaa baaaabbbaaaa Q P,1 abaaab abbaaa aabaaa Q P,2 babaaa aabbaa bbabaa Q P,3 aaabaa 16 12 4, 8

17. Computing Q P,0, …, Q P,k−1 17 baaabb baaaba Q P,0 P = baa This edge does not exist, therefore abaaba is enumerated only once. r=3 k=4 q=6 baaaabbbaaaa Q P,1 abaaab abbaaa aabaaa Q P,2 babaaa aabbaa bbabaa Q P,3 aaabaa 16 12 4, 8 0123456789101112131415161718192021 T = abbbabaaabaaabbaaaabaa

18. Computing Q P,0, …, Q P,k−1 18 baaabb baaaba Q P,0 P = baa This edge does not exist, therefore abaaba is enumerated only once. r=3 k=4 q=6 baaaabbbaaaa Q P,1 abaaab abbaaa aabaaa Q P,2 babaaa aabbaa bbabaa Q P,3 aaabaa 16 12 4, 8 0123456789101112131415161718192021 T = abbbabaaabaaabbaaaabaa

19. Computing Q P,0, …, Q P,k−1 19 baaabb baaaba Q P,0 P = baa This edge does not exist, therefore abaaba is enumerated only once. r=3 k=4 q=6 baaaabbbaaaa Q P,1 abaaab abbaaa aabaaa Q P,2 babaaa aabbaa bbabaa Q P,3 aaabaa 16 12 4, 8 0123456789101112131415161718192021 T = abbbabaaabaaabbaaaabaa

20. Computing Q P,0, …, Q P,k−1 20 baaabb baaaba Q P,0 P = baa This edge does not exist, therefore abaaba is enumerated only once. r=3 k=4 q=6 baaaabbbaaaa Q P,1 abaaab abbaaa aabaaa Q P,2 babaaa aabbaa bbabaa Q P,3 aaabaa 16 12 4, 8 0123456789101112131415161718192021 T = abbbabaaabaaabbaaaabaa

21. Computing Q P,0  Given pattern P, first we need to compute the source Q P,0 of the q-gram transition graph, i.e., all q-grams in T which begin with P. 21

22. w aaaaaa 0 aaaaab 1 aaaaba 2 ・・・ abbbbb 31 baaaaa 32 baaaab 33 baaaba 34 baaabb 35 ・・・ baabbb 39 ・・・ bbbbbb 63 Computing Q P,0  Given pattern P, first we need to compute the source Q P,0 of the q-gram transition graph, i.e., all q-grams in T which begin with P.  Consider all q-grams in lexicographical order. For any w ∈ Σ q (not necessary appearing in T), we denote by the lexicographical rank of w.  For any pattern P, there exists a single range [sp(P), ep(P)] s.t. a q-gram w begins with P iff. This range can be computed easily. 22 q-grams that begin with baa. sp( baa ) = 32 ep( baa ) = 39

23. Computing Q P,0  Consider a bit array B of size σ q s.t. iff w appears in T. Then, w ∈ Q P,0 iff and.  Hence we need to output all w s.t. and. 23 w aaaaaa 00 aaaaab 11 aaaaba 20 ・・・ abbbbb 311 baaaaa 320 baaaab 331 baaaba 340 baaabb 351 ・・・ baabbb 390 ・・・ bbbbbb 630 q-grams that begin with baa. sp( baa ) = 32 ep( baa ) = 39

24. Summary of our index  We need to store: a) q-gram transition graph, b) bit array B[0.. σ q − 1] for computing Q P,0, and c) positions of sampled q-grams. 24 n : length of T. σ : alphabet size. q : length of sampled substrings. k : sampling distance.

25. Summary of our index  We need to store: a) q-gram transition graph, b) bit array B[0.. σ q − 1] for computing Q P,0, and c) positions of sampled q-grams.  We can represent a)in O(σ q log σ) bits, b)in σ q + O(σ q / ω) bits, and c)in (n / k + σ q ) log(n / k) bits.  We can search any pattern in O(k × #occ + log σ n) time. 25 n : length of T. σ : alphabet size. q : length of sampled substrings. k : sampling distance. ω : machine word size.

26. Summary of our index  We need to store: a) q-gram transition graph, b) bit array B[0.. σ q − 1] for computing Q P,0, and c) positions of sampled q-grams.  We can represent a)in O(σ q log σ) bits, b)in σ q + O(σ q / ω) bits, and c)in (n / k + σ q ) log(n / k) bits.  We can search any pattern in O(k × #occ + log σ n) time. 26 n : length of T. σ : alphabet size. q : length of sampled substrings. k : sampling distance. ω : machine word size. I will explain these next.

27. Representation of (a)  Since q-gram transition graph is a subgraph of de Bruijn graph, from each node u, it is enough to store the character c s.t. v = c u[0..q−2] if an edge (u,v) exists. 27 abaaabbaaaba baaaabaaaaba aaabaaaabaaa b a … … b a a a a

28. Representation of (a)  Since q-gram transition graph is a subgraph of de Bruijn graph, from each node u, it is enough to store the character c s.t. v = c u[0..q−2] if an edge (u,v) exists.  Since the number of vertices is σ q and the indegree of each vertex is at most 1, the number of edges is at most σ q. We can represent this graph in O(σ q log σ) bits by using some tables. 28 abaaabbaaaba baaaabaaaaba aaabaaaabaaa b a … … b a a a a

29. Representation of (b)  By data structure (b), we output all w s.t. and.  So, using a fast successor data structure, we can compute all such q-grams w. 29 w aaaaaa 00 aaaaab 11 aaaaba 20 ・・・ abbbbb 311 baaaaa 320 baaaab 331 baaaba 340 baaabb 351 ・・・ baabbb 390 ・・・ bbbbbb 630 q-grams that begin with baa. sp( baa ) = 32 ep( baa ) = 39

30. Representation of (b)  By data structure (b), we output all w s.t. and.  So, using a fast successor data structure, we can compute all such q-grams w.  We need a dynamic successor data structure to support online updates to T. 30 w aaaaaa 00 aaaaab 11 aaaaba 20 ・・・ abbbbb 311 baaaaa 320 baaaab 331 baaaba 340 baaabb 351 ・・・ baabbb 390 ・・・ bbbbbb 630 q-grams that begin with baa. sp( baa ) = 32 ep( baa ) = 39

31. Representation of (b)  By data structure (b), we output all w s.t. and.  So, using a fast successor data structure, we can compute all such q-grams w.  We need a dynamic successor data structure to support online updates to T.  We can use van Emde Boas tree but it requires Θ(σ q ) words = Θ(σ q ω) bits. We want to reduce the space. 31 w aaaaaa 00 aaaaab 11 aaaaba 20 ・・・ abbbbb 311 baaaaa 320 baaaab 331 baaaba 340 baaabb 351 ・・・ baabbb 390 ・・・ bbbbbb 630 q-grams that begin with baa. sp( baa ) = 32 ep( baa ) = 39

32. Representation of (b)  We present a succinct variant of van Emde Boas tree.  We divide B into blocks of size ω h where ω is the machine word size and h (> 1) is some constant integer.  We maintain an ω-ary tree of height h (bottom tree) for each block, and a van Emde Boas tree (top tree) over the bottom trees. 32 10101100……100000000……000100000……0 ω-ary trees of height h …… van Emde Boas tree 101 ωhωh Corresponds to B. ……

33. Representation of (b) : bottom tree  Each bottom tree is a complete ω-ary tree.  Each node has a bit array A of length ω s.t. A[ j] = 1 iff the j-th child of the node contains 1. 33 10101100010000000000111110110001… 01234567... 1110aaaaaaaaaaaaaaaa0111…… 1101 Block of size ω h. … … … A

34. Representation of (b)  Data structure (b) can be represented in σ q + o(σ q ) bits. The bottom trees require σ q + O(σ q / ω) = σ q + o(σ q ) bits and the top tree requires O(σ q / ω h−1 ) = o(σ q ) bits, assuming the machine word size ω = Θ(log n).  Updates of a single bit in B and successor queries can be done in O(h + log log σ q ) = O(log log σ q ) time. If σ q ≤ n then O(log log n) time. 34

35. Complexities  We represent each q-gram by an integer, and we do not store the original text T.  We assume that σ = polylog(n), k ≥ 1, q = k + r − 1 and q ≤ log σ n − log σ log σ n.  If we choose k = Θ(log σ n), then the space complexity is O(n log σ) bits, and hence our index is compact. 35 Complexities Construction timeO(n)O(n) Searching timeO(k × #occ + log σ n) Space (in bits)(n / k + σ q ) log(n / k) + o(n)

36. Experimental results of construction Time for construction (in seconds). Text size n (in megabytes). 36

37. Experimental results of construction Time for construction (in seconds). Text size n (in megabytes). Our index is the fastest to construct. 37

38. Experimental results of searching Average time for searching, using100 patterns of length 6 (in seconds). Text size n (in megabytes). 38

39. Experimental results of searching Average time for searching, using100 patterns of length 6 (in seconds). Text size n (in megabytes). Ours is the fastest compact/compressed index to search. 39

40. Experimental results of memory usage Memory usage (in megabytes). Text size n (in megabytes). 40

41. Experimental results of memory usage Memory usage (in megabytes). Text size n (in megabytes). Ours is much more space-efficient than Dynamic FM-index 41

42. Conclusion  We proposed a q-gram based self-index for searching patterns of limited length. Our self-index: is theoretically and practically efficient in terms of construction, updates (adding characters at the end of the text) and searches, is compact, i.e., requires only O(n log σ) bits of space, where n is the text size and σ is the alphabet size, and can be constructed in online manner.  When the text is DNA sequence of human (i.e., σ = 4 and n ~ 10 9 ), the practical limit of pattern length is about 10 for our index.  Can we further reduce the space complexity? 42