Efficient Data Structures for the Factor Periodicity Problem

Transcription

1 Efficient Data Structures for the Factor Periodicity Problem Tomasz Kociumaka Wojciech Rytter Jakub Radoszewski Tomasz Waleń University of Warsaw, Poland SPIRE 2012 Cartagena, October 23, 2012 Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 1/15

2 Factor Periodicity Problem w: a b a a b a b a a b a a b a b a a b a b a a b a a b a b a a b a a b Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 2/15

3 Factor Periodicity Problem w: a b a a b a b a a b a a b a b a a b a b a a b a a b a b a a b a a b Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 2/15

4 Factor Periodicity Problem w: a b a a b a b a a b a a b a b a a b a b a a b a a b a b a a b a a b a a b a b a a b a b a a Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 2/15

5 Factor Periodicity Problem w: a b a a b a b a a b a a b a b a a b a b a a b a a b a b a a b a a b a a b a b a a b a b a a Periods of w[11..22] are 5 Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 2/15

6 Factor Periodicity Problem w: a b a a b a b a a b a a b a b a a b a b a a b a a b a b a a b a a b a a b a b a a b a b a a Periods of w[11..22] are 5, 10 Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 2/15

7 Factor Periodicity Problem w: a b a a b a b a a b a a b a b a a b a b a a b a a b a b a a b a a b a a b a b a a b a b a a Periods of w[11..22] are 5, 10, 11 Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 2/15

8 Factor Periodicity Problem w: a b a a b a b a a b a a b a b a a b a b a a b a a b a b a a b a a b a a b a b a a b a b a a Periods of w[11..22] are 5, 10, 11 and 12. Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 2/15

9 Factor Periodicity Problem w: a b a a b a b a a b a a b a b a a b a b a a b a a b a b a a b a a b a a b a b a a b a b a a Periods of w[11..22] are 5, 10, 11 and 12. Notation P er(w[11..22]) ={5, 10, 11, 12}, per(w[11..22]) = 5. Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 2/15

10 Arithmetic sets A word of length m might have Θ(m) periods, e.g. a m. Definition A set A = {a, a + d, a + 2d,..., a + kd} Z is called arithmetic. An integer d is called the difference of A. Observe that an arithmetic set can be represented by three integers: a, d and k. Fact Let v be a word of length m. Then P er(v) is a union of at most log m disjoint arithmetic sets. For example P er(w[11..22]) = {5} {10, 11, 12} = {5, 10} {11, 12}. Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 3/15

11 Formal problem statement Problem (Periods Queries) Design a data structure that for a fixed word w of length n answers the following queries. Given integers i, j (1 i j n) compute P er(w[i..j]) respresented as a union of O(log w ) arithmetic sets. Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 4/15

12 Formal problem statement Problem (Periods Queries) Design a data structure that for a fixed word w of length n answers the following queries. Given integers i, j (1 i j n) compute P er(w[i..j]) respresented as a union of O(log w ) arithmetic sets. Definition We say that p is an (1 + δ)-period of v if v (1 + δ)p. Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 4/15

13 Formal problem statement Problem (Periods Queries) Design a data structure that for a fixed word w of length n answers the following queries. Given integers i, j (1 i j n) compute P er(w[i..j]) respresented as a union of O(log w ) arithmetic sets. Definition We say that p is an (1 + δ)-period of v if v (1 + δ)p. Problem ((1 + δ)-period Queries) Let us fix a real number δ > 0. Design a data structure that for a fixed word w of length n answers the following queries. Given integers i, j (1 i j n) compute all (1 + δ)-periods of w[i..j] respresented as a union of O(1) arithmetic sets. Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 4/15

14 Related work To the best of our knowledge no previous research on the general case of Period Queries. Even for computing the maximal period, only straightforward solutions: memorize all answers O(n 2 ) space, O(1) query time compute the answer from scratch for each query no extra space, O(n) query time Efficient data structures for primitivity testing (generalized by (1 + δ)-period Queries with δ = 1) Karhumäki, Lifshits & Rytter; CPM 2007 O(n log n) space, O(1) query time, Crochemore et. al; SPIRE 2010 O(n log ε n) space, O(log n) query time. Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 5/15

15 Our results Several results based on the common idea but different tools. Space All periods (1 + δ)-periods O(n) O(log 1+ε n) O(log ε n) O(n log log n) O(log n(log log n) 2 ) O((log log n) 2 ) O(n log ε n) O(log n log log n) O(log log n) O(n log n) O(log n) O(1) Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 6/15

16 Our results Several results based on the common idea but different tools. Space All periods (1 + δ)-periods O(n) O(log 1+ε n) O(log ε n) O(n log log n) O(log n(log log n) 2 ) O((log log n) 2 ) O(n log ε n) O(log n log log n) O(log log n) O(n log n) O(log n) O(1) Standard assumption on the model of computation integer alphabet, i.e. Σ { 0, 1,..., n O(1)}, word RAM model with w = Ω(log n), randomization. Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 6/15

17 Our approach Let Borders(v) = { u : u is a border of v}. Fact P er(v) = v Borders(v) = { v b : b Borders(v)}. Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 7/15

18 Our approach Let Borders(v) = { u : u is a border of v}. Fact P er(v) = v Borders(v) = { v b : b Borders(v)}. We compute Borders(v) {2 k,..., 2 k+1 1} separately for each k {0,..., log v }. 2 k 2 k 1 2 k 1 2 k v Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 7/15

19 Our approach Let Borders(v) = { u : u is a border of v}. Fact P er(v) = v Borders(v) = { v b : b Borders(v)}. We compute Borders(v) {2 k,..., 2 k+1 1} separately for each k {0,..., log v }. 2 k 2 k 1 2 k 1 2 k v prefix suffix Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 7/15

20 Our approach Let Borders(v) = { u : u is a border of v}. Fact P er(v) = v Borders(v) = { v b : b Borders(v)}. We compute Borders(v) {2 k,..., 2 k+1 1} separately for each k {0,..., log v }. 2 k 2 k 1 2 k 1 2 k v prefix border suffix Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 7/15

21 Our approach Let Borders(v) = { u : u is a border of v}. Fact P er(v) = v Borders(v) = { v b : b Borders(v)}. We compute Borders(v) {2 k,..., 2 k+1 1} separately for each k {0,..., log v }. 2 k 2 k 1 2 k 1 2 k v prefix border suffix Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 7/15

22 Close occurrences Let Occ(u, v) be the set of positions of v where an occurrence of u starts. Arithmetic sets naturally appear as the Occ sets. Fact Let v 2 u. Then Occ(u, v) is arithmetic. Moreover, if Occ(u, v) 3 then its difference is equal to per(u). Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 8/15

23 Close occurrences Let Occ(u, v) be the set of positions of v where an occurrence of u starts. Arithmetic sets naturally appear as the Occ sets. Fact Let v 2 u. Then Occ(u, v) is arithmetic. Moreover, if Occ(u, v) 3 then its difference is equal to per(u). Case with Occ(u, v) 2 is trivial. Assume Occ(u, v) 2 Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 8/15

24 Close occurrences Let Occ(u, v) be the set of positions of v where an occurrence of u starts. Arithmetic sets naturally appear as the Occ sets. Fact Let v 2 u. Then Occ(u, v) is arithmetic. Moreover, if Occ(u, v) 3 then its difference is equal to per(u). Case with Occ(u, v) 2 is trivial. Assume Occ(u, v) 2 v u periods of u Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 8/15

25 Close occurrences Let Occ(u, v) be the set of positions of v where an occurrence of u starts. Arithmetic sets naturally appear as the Occ sets. Fact Let v 2 u. Then Occ(u, v) is arithmetic. Moreover, if Occ(u, v) 3 then its difference is equal to per(u). Case with Occ(u, v) 2 is trivial. Assume Occ(u, v) 2 v u period of u Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 8/15

26 A formula for border lengths 2 k 2 k 1 2 k 1 2 k v p p s s Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 9/15

27 A formula for border lengths 2 k 2 k 1 2 k 1 2 k v p p s s P = Occ(p, s s) Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 9/15

28 A formula for border lengths 2 k 2 k 1 2 k 1 2 k v p p s s S = Occ(s, pp ) P = Occ(p, s s) Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 9/15

29 A formula for border lengths 2 k 2 k 1 2 k 1 2 k l l v p p s s l S = Occ(s, pp ) P = Occ(p, s s) l Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 9/15

30 A formula for border lengths 2 k 2 k 1 2 k 1 2 k l l v p p s s l S = Occ(s, pp ) P = Occ(p, s s) l Fact Let 0 l < 2 k. Then the word v has a border of length 2 k + l if and only if l + 1 S and s s l P. Consequently Borders(v) {2 k,..., 2 k+1 1} is arithmetic. Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 9/15

31 Intersecting arithmetic sets Intersecting to arithmetics set in general is performed using the extended Euclid s algorithm, so takes O(log n) time. But if this sets share a common difference, constant time is sufficient. Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 10/15

32 Intersecting arithmetic sets Intersecting to arithmetics set in general is performed using the extended Euclid s algorithm, so takes O(log n) time. But if this sets share a common difference, constant time is sufficient. Lemma If P 3 and S 3, then per(p) = per(s). Consequently P and S are arithmetic of common difference. Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 10/15

33 Intersecting arithmetic sets Intersecting to arithmetics set in general is performed using the extended Euclid s algorithm, so takes O(log n) time. But if this sets share a common difference, constant time is sufficient. Lemma If P 3 and S 3, then per(p) = per(s). Consequently P and S are arithmetic of common difference. v p p s s S P Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 10/15

34 Intersecting arithmetic sets Intersecting to arithmetics set in general is performed using the extended Euclid s algorithm, so takes O(log n) time. But if this sets share a common difference, constant time is sufficient. Lemma If P 3 and S 3, then per(p) = per(s). Consequently P and S are arithmetic of common difference. v p p s s S P 2per(s) 2per(p) Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 10/15

35 Intersecting arithmetic sets Intersecting to arithmetics set in general is performed using the extended Euclid s algorithm, so takes O(log n) time. But if this sets share a common difference, constant time is sufficient. Lemma If P 3 and S 3, then per(p) = per(s). Consequently P and S are arithmetic of common difference. v p p s s S P 2per(s) of period both per(p) and per(s) 2per(p) Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 10/15

36 Summary of the combinatorial part Problem (Occurrence Queries) Design a data structure that for a word w can answer the following queries. Given a basic factor u and a factor v of w such that v 2 u (both represented by one of their occurrences) compute the arithmetic set Occ(u, v). Theorem Assume there is a data structure answering the Occurrence Queries in O(f(n)) time. Then this data structure can answer Period Queries in O(f(n) log n) time and (1 + δ)-period Queries is O(f(n)) time. Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 11/15

37 Range Predecessor/Successor Queries Problem (Range Predecessor/Successor Queries) Design a data structure that for a word w can answer the following queries. Given a factor u of w (represented by an occurrence in w) and i {1... n} find P RED(u, i) the last occurrence of u ending at a position i, SUCC(u, i) the first occurrence of u starting at a position i. The Occurrence Queries can be reduced to three Range Predecessor/Successor Queries, where u is a basic factor of w. w i i v j Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 12/15

38 Range Predecessor/Successor Queries Problem (Range Predecessor/Successor Queries) Design a data structure that for a word w can answer the following queries. Given a factor u of w (represented by an occurrence in w) and i {1... n} find P RED(u, i) the last occurrence of u ending at a position i, SUCC(u, i) the first occurrence of u starting at a position i. The Occurrence Queries can be reduced to three Range Predecessor/Successor Queries, where u is a basic factor of w. w i i v j Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 12/15

39 Range Predecessor/Successor Queries Problem (Range Predecessor/Successor Queries) Design a data structure that for a word w can answer the following queries. Given a factor u of w (represented by an occurrence in w) and i {1... n} find P RED(u, i) the last occurrence of u ending at a position i, SUCC(u, i) the first occurrence of u starting at a position i. The Occurrence Queries can be reduced to three Range Predecessor/Successor Queries, where u is a basic factor of w. w SUCC(u, i) i i v j Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 12/15

40 Range Predecessor/Successor Queries Problem (Range Predecessor/Successor Queries) Design a data structure that for a word w can answer the following queries. Given a factor u of w (represented by an occurrence in w) and i {1... n} find P RED(u, i) the last occurrence of u ending at a position i, SUCC(u, i) the first occurrence of u starting at a position i. The Occurrence Queries can be reduced to three Range Predecessor/Successor Queries, where u is a basic factor of w. w SUCC(u, i) i i SUCC(u, i + 1) v j Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 12/15

41 Range Predecessor/Successor Queries Problem (Range Predecessor/Successor Queries) Design a data structure that for a word w can answer the following queries. Given a factor u of w (represented by an occurrence in w) and i {1... n} find P RED(u, i) the last occurrence of u ending at a position i, SUCC(u, i) the first occurrence of u starting at a position i. The Occurrence Queries can be reduced to three Range Predecessor/Successor Queries, where u is a basic factor of w. w SUCC(u, i) i i SUCC(u, i + 1) v j P RED(u, j) Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 12/15

42 Range Predecessor/Successor Queries Problem (Range Predecessor/Successor Queries) Design a data structure that for a word w can answer the following queries. Given a factor u of w (represented by an occurrence in w) and i {1... n} find P RED(u, i) the last occurrence of u ending at a position i, SUCC(u, i) the first occurrence of u starting at a position i. The Occurrence Queries can be reduced to three Range Predecessor/Successor Queries, where u is a basic factor of w. w SUCC(u, i) i i SUCC(u, i + 1) v j P RED(u, j) Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 12/15

43 Simple solution Recall that the Dictionary of Basic Factors is a data structure, which assigns integer identifiers to all basic factors of w. DBF k [i] = DBF k [j] w[i..i + 2 k 1] = w[j..j + 2 k 1]. Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 13/15

44 Simple solution Recall that the Dictionary of Basic Factors is a data structure, which assigns integer identifiers to all basic factors of w. DBF k [i] = DBF k [j] w[i..i + 2 k 1] = w[j..j + 2 k 1]. For each k and identifier id we store a set A k,id = {i : DBF k [i] = id}. Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 13/15

45 Simple solution Recall that the Dictionary of Basic Factors is a data structure, which assigns integer identifiers to all basic factors of w. DBF k [i] = DBF k [j] w[i..i + 2 k 1] = w[j..j + 2 k 1]. For each k and identifier id we store a set A k,id = {i : DBF k [i] = id}. If u = w[i..i + 2 k 1] and DBF k [i] = id, then A k,id = Occ(u, w). Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 13/15

46 Simple solution Recall that the Dictionary of Basic Factors is a data structure, which assigns integer identifiers to all basic factors of w. DBF k [i] = DBF k [j] w[i..i + 2 k 1] = w[j..j + 2 k 1]. For each k and identifier id we store a set A k,id = {i : DBF k [i] = id}. If u = w[i..i + 2 k 1] and DBF k [i] = id, then A k,id = Occ(u, w). A single binary search in A k,id allows to answer P RED(u, i) and SUCC(u, i). Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 13/15

47 Simple solution Recall that the Dictionary of Basic Factors is a data structure, which assigns integer identifiers to all basic factors of w. DBF k [i] = DBF k [j] w[i..i + 2 k 1] = w[j..j + 2 k 1]. For each k and identifier id we store a set A k,id = {i : DBF k [i] = id}. If u = w[i..i + 2 k 1] and DBF k [i] = id, then A k,id = Occ(u, w). A single binary search in A k,id allows to answer P RED(u, i) and SUCC(u, i). Corollary There exists a (simple) data structure of O(n log n) size that answer the Occurrence Queries in O(log n) time, and consequently the Period Queries in O(log 2 n) time. Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 13/15

48 O(1) query time solution Fix 2 k n and consider a basic factor u, u = 2 k. Split w into fragments of length 2 k+1 with overlaps of size 2 k. 2 k+1 w 2 k+1 Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 14/15

49 O(1) query time solution Fix 2 k n and consider a basic factor u, u = 2 k. Split w into fragments of length 2 k+1 with overlaps of size 2 k. 2 k+1 w 2 k+1 Each occurrence of u occurs within a fragment. Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 14/15

50 O(1) query time solution Fix 2 k n and consider a basic factor u, u = 2 k. Split w into fragments of length 2 k+1 with overlaps of size 2 k. 2 k+1 w 2 k+1 arithmetic sets Each occurrence of u occurs within a fragment. Occurrences in single fragments form arithmetic sets. Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 14/15

51 O(1) query time solution Imagine a (large) array indexed by identifiers and fragments. u. u u [ ] [ 0, 2 k+1 2 k, 3 2 k] [ 2 k+1, 2 k+2]... [ m 2 k, n ]. This array has Θ.. ( n 2 2 k ) cells Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 15/15

52 O(1) query time solution Imagine a (large) array indexed by identifiers and fragments. u. u u [ ] [ 0, 2 k+1 2 k, 3 2 k] [ 2 k+1, 2 k+2]... [ m 2 k, n ]. This array has Θ. ( ) n 2 2 k cells.. All factors of a fixed length have n occurrences in total, so n non-empty fields perfect hashing can be used Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 15/15

53 O(1) query time solution Imagine a (large) array indexed by identifiers and fragments. u. u u [ ] [ 0, 2 k+1 2 k, 3 2 k] [ 2 k+1, 2 k+2]... [ m 2 k, n ]. This array has Θ. ( ) n 2 2 k cells.. All factors of a fixed length have n occurrences in total, so n non-empty fields perfect hashing can be used. This gives O(n log n) in total for all values of k Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 15/15

54 O(1) query time solution Answering queries: v w 2 k+1 2 k+1 v lies within at most two consecutive fragments, Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 16/15

55 O(1) query time solution Answering queries: v w 2 k+1 u 2 k+1 v lies within at most two consecutive fragments, get the occurrences of u from the hash table, Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 16/15

56 O(1) query time solution Answering queries: v w 2 k+1 u 2 k+1 v lies within at most two consecutive fragments, get the occurrences of u from the hash table, Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 16/15

57 O(1) query time solution Answering queries: v w 2 k+1 u 2 k+1 v lies within at most two consecutive fragments, get the occurrences of u from the hash table, crop and merge these arithmetic sets to obtain the result. Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 16/15

58 O(1) query time solution Answering queries: v w 2 k+1 u 2 k+1 v lies within at most two consecutive fragments, get the occurrences of u from the hash table, crop and merge these arithmetic sets to obtain the result. Corollary There exists a data structure of O(n log n) size that answers the Occurrence Queries in O(1) time. Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 16/15

59 Space-efficient solution Theorem (Nekrich, Navarro; 2012) There exists data structures that given the locus of u in the suffix tree of w answer the Range Predecessor/Successor queries in and satisfy the following space and time bounds: Space Query time O(n) O(log ε n) O(n log log n) O((log log n) 2 ) O(n log ε n) O(log log n) Theorem (Weighted LA Kopelovitz, Lewenstein; 2007) There exists a data structure of size O(n), which given an interval [i..j] finds the locus of w[i..j] in the suffix tree of w in O(log log n) time. Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 17/15

60 Summary Theorem (this paper) There exists data structures that satisfy following time and space bounds for size, Period Queries query time and (1 + δ)-period Queries query time: Space Period Queries (1 + δ)-period Q. O(n) O(log 1+ε n) O(log ε n) O(n log log n) O(log n(log log n) 2 ) O((log log n) 2 ) O(n log ε n) O(log n log log n) O(log log n) O(n log n) O(log n) O(1) Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 18/15

61 Further research Currently in progress: Space Period Queries (1 + δ)-period Q. O(n) O(log 1+ε n) O(log ε n) O(n log log n) O(log n(log log n) 2 ) O((log log n) 2 ) O(n log ε n) O(log n log log n) O(log log n) O(n log n) O(log n) O(1) Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 19/15

62 Further research Currently in progress: Space Period Queries 2-Period Queries O(n) O(log 1+ε n) O(log ε n) O(n log log n) O(log n(log log n) 2 ) O((log log n) 2 ) O(n log ε n) O(log n log log n) O(log log n) O(n log n) O(log n) O(1) O(n) O(1) Open problems: Can the O(n log n) time preprocessing improved with o(n) queries? Can the maximum period be found faster than O(log n) with o(n 2 ) space? Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 19/15

63 Further research Currently in progress: Space Period Queries 2-Period Queries O(n) O(log 1+ε n) O(log ε n) O(n log log n) O(log n log log n) O(log log n) O(n log n) O(log n) O(1) O(n) O(1) Open problems: Can the O(n log n) time preprocessing improved with o(n) queries? Can the maximum period be found faster than O(log n) with o(n 2 ) space? Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 19/15

64 Thank you Thank you for your attention! Tomasz Kociumaka Efficient Data Structures for the Factor Periodicity Problem 20/15