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Sorting suffixes of two-pattern strings F. Franek & W.F. Smyth Algorithms Research Group Computing and Software McMaster University Hamilton, Ontario Canada.

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Presentation on theme: "Sorting suffixes of two-pattern strings F. Franek & W.F. Smyth Algorithms Research Group Computing and Software McMaster University Hamilton, Ontario Canada."— Presentation transcript:

1 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

2 In 2003 several very different linear-time (recursive) algorithms to sort suffixes of strings appeared. All work in four basic steps: 1. Split all suffixes into two sets 2. Sort the first set of suffixes by recursion (recursive reduction of the problem) 3. Sort the second set of suffixes based on the order of the first set 4. 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 Slide 4 σ(a)=p i q, σ(b)=p j q, σ(x[1..n])=σ(x[1])σ(x[2..n]) σ 1 σ 2 (x)=σ 1 ( σ 2 (x) ) x is two-pattern string of scope λ iff there is a sequence σ 1, σ 2,..., σ n of expansions of scope λ so that x= σ 1 σ 2 … σ n (a) The "nice" properties of two-pattern strings (see a series of papers by Franek, Smyth and others): can be recognized in linear time

5 Slide 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

6 Slide 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

7 Slide 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 ρ 1 < ρ 2 < … < ρ |x| the sorted suffixes of x. We are assuming that q p (since then x 1 < x 2 iff σ(x 1 ) < σ(x 2 ), otherwise we work with

8 Slide 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: ………. A δ,k = {δp k qσ(ρ) : ρ is a proper suffix of x or ρ=ε} δ is a suffix of p and 0 < k < i δ p k q σ(ρ)

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

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

11 Slide 11 ………. δ p i q σ(ρ) C δ = {δp i qσ(ρ) : aρ proper suffix of x, ρ can be ε} δ is a suffix of q but not of p ………. δ p j q σ(ρ) D δ = {δp j qσ(ρ) : bρ proper suffix of x, ρ can be ε} δ is a suffix of q

12 Slide 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 !

13 Slide 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 4λ 2 |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) δ 1 δ 2 (C2) δ 2 δ 1 (C3) δ 1 is a proper prefix of δ 2 (C4) δ 2 is a proper prefix of δ 1 (C5) δ 1 =δ 2 =δ

14 Slide 14 (C1) A δ 1,k 1 n A δ 2,k 2 (C2) A δ 2,k 2 n A δ 1,k 1 (C3) δ 2 =δ 1 μ (a) if μ p, then A δ 2,k 2 n A δ 1,k 1 (b) otherwise A δ 1,k 1 n A δ 2,k 2 (C4) δ 1 =δ 2 μ (a) if μ p, then A δ 1,k 1 n A δ 2,k 2 (b) otherwise A δ 2,k 2 n A δ 1,k 1 (C5) (a) if k 1 k 2, then A δ,k 2 n A δ,k 1

15 Slide 15 (C1) A δ 1,k n B δ 2 (C2) B δ 2 n A δ 1,k (C3) δ 2 =δ 1 μ (a) if μ p, then B δ 2 n A δ 1,k (b) otherwise A δ 1,k n B δ 2 (C4) δ 1 =δ 2 μ (a) if μp k q pq, then A δ 1,k n B δ 2 (b) otherwise B δ 2 n A δ 1,k (C5) B δ n A δ,k No bucket comparison requires more than 3λ steps.

16 Slide 16 Similarly A~C, A~D, B~B, B~C, B~D, C~C, and C~D. One more example: (C1) B δ 1 n B δ 2 (C2) B δ 2 n B δ 1 (C3) δ 2 =δ 1 μ (a) if μqp qp, then B δ 2 n B δ 1 (b) otherwise B δ 1 n B δ 2 (C4) δ 1 =δ 2 μ (a) if μqp qp, then B δ 1 n B δ 2 (b) otherwise B δ 2 n B δ 1 (C5) B δ 1 = B δ 2

17 The High-level logic of the algorithm: 1. 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.) 2. Sort the names according to the comparisons of the four A buckets (according to (C1)-(C4)). (This requires at most 2λ 3 steps as we are sorting λ names and each comparison requires at most 2λ steps.) 3. Replace every name (A,δ) by a sequence of names (A,δ,k), 0< k < j. Let us call the resulting Slide 17

18 Slide 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|.) 4. Create names (B,δ) for every suffix δ of p. (This requires at most λ steps.) 5. Merge into BUCKETS all names (B,δ) according to comparisons. (This requires at most |BUCKETS|3λ 2 steps, as we are merging in λ names and each comparison requires 3λ steps)

19 Slide Create names (C,δ) for every suffix δ of q that is not a suffix of p. (This requires at most λ 2 steps.) 7. Merge into BUCKETS all names (C,δ) according to comparisons. (This requires at most |BUCKETS|3λ 2 steps.) 8. Create names (D,δ) for every suffix δ of q. (This requires at most λ steps.) 9. Merge into BUCKETS all names(D,δ) according to comparisons. (Now we have all required bucket names, except E, in proper order.

20 Slide 20 This requires at most |BUCKETS|3λ 2 steps.) 10. 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.) 11. Merge into SUFFIXES the suffixes from the bucket E. (This requires at most |y|4λ 2 steps.) Done in less than (2λ 3 +14λ 2 +3λ+2)|y| steps!

21 Slide 21 The algorithm works in 2(2λ 3 +14λ 2 +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: ordered suffixes of y:

22 Slide 22 A ba,1 = {babaabσ(ρ) : bρ proper suffix of x, ρ can be ε}= {babaab}={9} A a,1 = {abaabσ(ρ) : bρ proper suffix of x, ρ can be ε}= {abaab}={10} B ba = {baabσ(ρ) : ρ proper suffix of x}={baabσ(ab), baabσ(b)}={baabbaabbabaab, baabbabaab}={1,5} B a = {aabσ(ρ) : ρ proper suffix of x}={aabσ(ab), aabσ(b)}={aabbaabbabaab, aabbabaab}={2,6} C ab = {abbaabσ(ρ) : aρ proper suffix of x}= {abbaabσ(b)}={abbaabbabaab}={3} C b = {bbaabσ(ρ) : bρ proper suffix of x}= {bbaabσ(b)}={bbaabbabaab}={4}

23 Slide 23 D ab = {abbabaabσ(ρ) : bρ proper suffix of x, ρ can be ε}= {abbabaab}={7} D b = {bbabaabσ(ρ) : bρ proper suffix of x, ρ can be ε}= {bbabaab}={8} E = {baab, aab, ab, b}={11, 12, 13, 14} A ba,1 o A a,1 (by C2) A ba,1 o B ba (by C5) A ba,1 o B a (by C2) A ba,1 o C ab (by C2) A ba,1 n C b (by C4a)

24 Slide 24 A ba,1 o D ab (by C2) A ba,1 n D b (by C4a) A a,1 n B ba (by C1) A a,1 o B a (by C5) A a,1 n C ab (by C3b) A a,1 n C b (by C1) A a,1 n D ab (by C3b) A a,1 n D b (by C1)

25 Slide 25 B ba o B a (by C2) B ba o C ab (by C2) B ba n C b (by C4a) B ba o D ab (by C2) B ba n D b (by C4a) B a n C ab (by C3b) B a n C b (by C1) B a n D ab (by C3b) B a n D b (by C1)

26 Slide 26 C ab n C b (by C1) C ab n D ab (by C5) C ab n D b (by C1) C b o D ab (by C1) C b n D b (by C5) D ab n D b (by C1) B a n A a,1 n C ab n D ab n B ba n A ba,1 n C b n D b

27 Slide 27


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