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Sorting suffixes of two-pattern strings

Sorting suffixes of two-pattern strings. F. Franek & W.F. Smyth Algorithms Research Group Computing and Software McMaster University Hamilton, Ontario Canada. PSC04, Praha, Czech Republic, August-September 2004. Slide 1.

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Sorting suffixes of two-pattern strings

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  1. Sorting suffixes of two-pattern strings F. Franek & W.F. SmythAlgorithms Research GroupComputing and SoftwareMcMaster University Hamilton, Ontario Canada PSC04, Praha, Czech Republic, August-September 2004 Slide 1

  2. In 2003 several very different linear-time (recursive) algorithms to sort suffixes of strings appeared. All work in four basic steps: • . Split all suffixes into two sets • . Sort the first set of suffixes by recursion (recursive reduction of the problem) • . Sort the second set of suffixes based on the order of the first set • . Merge both sorted sets together Slide 2

  3. Our question --- will two-pattern strings exhibit a “natural” tendency to reduce the problem in a recursive fashion? Two-pattern strings were introduced by us as a generalization of Sturmian (and hence Fibonacci) strings. Let p, q be binary strings. σ = [p,q,i,j]λ is an expansion of scope λ if |p|, |q| ≤ λ and i ≠ j non-negative integers. We require p and q to be dissimilar enough to be efficiently recognizable (see the paper for the details). Slide 3

  4. σ(a)=piq, σ(b)=pjq, σ(x[1..n])=σ(x[1])σ(x[2..n])σ1○σ2(x)=σ1(σ2(x))x is two-pattern string of scope λ iff there is a sequence σ1, σ2,...,σn of expansions of scope λ so that x=σ1○σ2○ … ○σn(a) • The "nice" properties of two-pattern strings (see a series of papers by Franek, Smyth and others): • can be recognized in linear time Slide 4

  5. when recognized, the canonical expansion sequence is computed • repetitions and near repetitions can be effectively computed in linear time using recursive approach • generalize finite fragments of the Fibonacci string and Sturmian strings • can easily be generated and represented in recursive fashion • exhibit rich yet comprehensible recursive structure Slide 5

  6. they occur relatively frequently among binary strings An illustration of a very simple two-pattern string; will be used later to illustrate the workings of the algorithm: [a,b,2,3] apply to a:a → aab [ba,ab,1,2] apply to aab: aab → baab baab babaab baabbaabbabaab is a two-pattern string of scope 2 Slide 6

  7. Now we can rephrase our question: Given an expansion σ and knowing the order of suffixes of a two-pattern string x, can we efficiently determine the order of suffixes of σ(x)? The answer is yes and in the following we describe the algorithm. So let x be a two-pattern string of scope λ and let σ = [p,q,i,j]λ be an expansion and let y = σ(x). Let ρ1 < ρ2 < … < ρ|x| the sorted suffixes of x.We are assuming that q —p (since then x1 < x2 iff σ(x1) < σ(x2), otherwise we work with Slide 7

  8. complements and reverse the resulting order of suffixes while taking complements of the suffixes). First we assign all suffixes of y into various buckets: Aδ,k = {δpkqσ(ρ) : ρ is a proper suffix of x or ρ=ε}δ is a suffix of p and 0 < k < i ………. δ pk q σ(ρ) Slide 8

  9. ………. ………. Aδ,i= {δpiqσ(ρ) : ρ is a proper suffix of x or ρ=ε}δ is a suffix of p and also a suffix of qorAδ,i= {δpiqσ(ρ) : bρ proper suffix of x, ρ can be ε}δ is a suffix of p and not a suffix of q Slide 9

  10. Aδ,k = {δpkqσ(ρ) : bρ proper suffix of x, ρ can be ε} δ is a suffix of p and i < k < j ………. δ pk q σ(ρ) Bδ = {δqσ(ρ) : ρ proper nontrivial suffix of x} δ is a suffix of p and i < k < j ………. δ q σ(ρ) Slide 10

  11. Cδ = {δpiqσ(ρ) : aρ proper suffix of x, ρ can be ε} δ is a suffix of q but not of p ………. δ piq σ(ρ) Dδ= {δpjqσ(ρ) : bρ proper suffix of x, ρ can be ε} δ is a suffix of q ………. δ pjq σ(ρ) Slide 11

  12. E = {δ: δ is a nontrivial suffix of p or q} ………. δ ………. δ • All suffixes are covered by A-E ! • Order of suffixes in buckets A-D determined by ρ ! • A-D buckets are order invariant ! Slide 12

  13. So, if we can determine the order of buckets, we can determine the order of all suffixes in buckets A-D. To merge in the suffixes from E is easy (brute force only requires ≤ 4λ2|y| steps). The main results is based on the fact that the order of buckets A-D can be efficiently determined using 5 cases: (C1) δ1 —δ2 (C2) δ2 —δ1(C3) δ1 is a proper prefix ofδ2(C4) δ2 is a proper prefix ofδ1(C5) δ1=δ2=δ Slide 13

  14. (C1) Aδ1,k1n Aδ2,k2(C2) Aδ2,k2n Aδ1,k1(C3) δ2=δ1μ (a) if μ — p, then Aδ2,k2n Aδ1,k1 (b) otherwise Aδ1,k1n Aδ2,k2(C4) δ1=δ2μ (a) if μ — p, then Aδ1,k1n Aδ2,k2 (b) otherwise Aδ2,k2n Aδ1,k1(C5) (a) if k1 < k2, then Aδ,k1n Aδ,k2(b) if k1 = k2, then Aδ,k1=Aδ,k2(c) if k1 > k2, then Aδ,k2n Aδ,k1 Slide 14

  15. (C1) Aδ1,k n Bδ2(C2) Bδ2 n Aδ1,k(C3) δ2=δ1μ (a) if μ — p, then Bδ2n Aδ1,k (b) otherwise Aδ1,kn Bδ2(C4) δ1=δ2μ (a) if μpkq— pq, then Aδ1,kn Bδ2 (b) otherwise Bδ2n Aδ1,k(C5) Bδ n Aδ,k No bucket comparison requires more than 3λ steps. Slide 15

  16. Similarly A~C, A~D, B~B, B~C, B~D, C~C, and C~D. One more example:(C1) Bδ1n Bδ2(C2) Bδ2 n Bδ1(C3) δ2=δ1μ (a) if μqp— qp, then Bδ2n Bδ1 (b) otherwise Bδ1 n Bδ2(C4) δ1=δ2μ (a) if μqp— qp, then Bδ1 n Bδ2 (b) otherwise Bδ2n Bδ1(C5) Bδ1=Bδ2 Slide 16

  17. The High-level logic of the algorithm: • . Create names (A,δ) for every suffix δ of p. (This requires at most λ steps. Each name will be eventually replaced by a sequence of buckets.) • . Sort the names according to the comparisons of the four A buckets (according to (C1)-(C4)). (This requires at most 2λ3 steps as we are sorting λ names and each comparison requires at most 2λ steps.) • . Replace every name (A,δ) by a sequence of names (A,δ,k), 0< k < j. Let us call the resulting Slide 17

  18. BUCKETS. (Now we have the names of A buckets • in the proper order. This requires at most |y| steps as the size of BUCKETS is ≤ |y|.) • . Create names (B,δ) for every suffix δ of p. (This requires at most λ steps.) • . Merge into BUCKETS all names (B,δ) according to comparisons. (This requires at most |BUCKETS|3λ2steps, as we are merging in λ names and each comparison requires ≤ 3λ steps) Slide 18

  19. . Create names (C,δ) for every suffix δ of q that is not a suffix of p. (This requires at most λ2 steps.) • . Merge into BUCKETS all names (C,δ) according to comparisons. (This requires at most |BUCKETS|3λ2 steps.) • . Create names (D,δ) for every suffix δ of q. (This requires at most λ steps.) • . Merge into BUCKETS all names(D,δ) according to comparisons. (Now we have all required bucket names, except E, in properorder. Slide 19

  20. This requires at most |BUCKETS|3λ2steps.) • 0. Traverse BUCKETS and replace each name by a sequence of suffixes according to the sequence of suffixes of x. Let us call this sequence SUFFIXES. (Now we have all suffixes from buckets A-D in proper order. This requires at most |y| steps.) • 1. Merge into SUFFIXES the suffixes from the bucket E. (This requires at most |y|4λ2steps.)Done in less than (2λ3+14λ2+3λ+2)|y| steps! Slide 20

  21. The algorithm works in 2(2λ3+14λ2+3λ+2)n steps, where n is the size of the input string. An example: x = aab$ y = baabbaabb a b a a b $σ=[ba,ab,1,2] ordered suffixes of x:1 2 3ordered suffixes of y:12 2 6 13 10 3 7 14 11 1 5 9 4 8 1 2 3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Slide 21

  22. Aba,1 = {babaabσ(ρ) : bρ proper suffix of x, ρ can be ε}= {babaab}={9}Aa,1 = {abaabσ(ρ) : bρ proper suffix of x, ρ can be ε}= {abaab}={10}Bba = {baabσ(ρ) : ρ proper suffix of x}={baabσ(ab), baabσ(b)}={baabbaabbabaab, baabbabaab}={1,5}Ba = {aabσ(ρ) : ρ proper suffix of x}={aabσ(ab), aabσ(b)}={aabbaabbabaab, aabbabaab}={2,6}Cab = {abbaabσ(ρ) : aρ proper suffix of x}= {abbaabσ(b)}={abbaabbabaab}={3}Cb = {bbaabσ(ρ) : bρ proper suffix of x}= {bbaabσ(b)}={bbaabbabaab}={4} Slide 22

  23. Dab = {abbabaabσ(ρ) : bρ proper suffix of x, ρ can be ε}= {abbabaab}={7}Db = {bbabaabσ(ρ) : bρ proper suffix of x, ρ can be ε}= {bbabaab}={8}E = {baab, aab, ab, b}={11, 12, 13, 14} Aba,1 o Aa,1 (by C2)Aba,1 oBba (by C5)Aba,1 oBa (by C2)Aba,1 oCab (by C2)Aba,1 n Cb (by C4a) Slide 23

  24. Aba,1 oDab (by C2)Aba,1 n Db (by C4a)Aa,1 n Bba (by C1)Aa,1 oBa (by C5)Aa,1 n Cab (by C3b)Aa,1 n Cb (by C1)Aa,1 n Dab (by C3b)Aa,1 n Db (by C1) Slide 24

  25. BbaoBa (by C2)BbaoCab (by C2)Bban Cb (by C4a)BbaoDab (by C2)Bban Db (by C4a)Ban Cab (by C3b)Ban Cb (by C1)Ban Dab (by C3b)Ban Db (by C1) Slide 25

  26. Cabn Cb (by C1)Cabn Dab (by C5)Cabn Db (by C1)CboDab (by C1)Cbn Db (by C5)Dabn Db (by C1) BanAa,1 nCab nDab nBba nAba,1nCb nDb2 6 10 3 7 1 5 9 4 8 12 13 14 11 Slide 26

  27. www.cas.mcmaster.ca/~franek Slide 27

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