Twins in words M. Axenovich 1 Y. Person 2 S. Puzynina 3 1 Iowa State - - PowerPoint PPT Presentation

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Twins in words M. Axenovich 1 Y. Person 2 S. Puzynina 3 1 Iowa State - - PowerPoint PPT Presentation

May 21, 2023 327 likes •577 views

Twins in words M. Axenovich 1 Y. Person 2 S. Puzynina 3 1 Iowa State University, U.S.A. and Karlsruher Institut f ur Technologie, Germany 2 Freie Universit at Berlin, Institut f ur Mathematik, Germany 3 University of Turku, Finland, and

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Twins in words

  • M. Axenovich1
  • Y. Person2
  • S. Puzynina3

1Iowa State University, U.S.A. and Karlsruher Institut f¨

ur Technologie, Germany

2Freie Universit¨

at Berlin, Institut f¨ ur Mathematik, Germany

3University of Turku, Finland, and Sobolev Institute of Mathematics, Novosibirsk,

Russia

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words

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Twins

S = s1 . . . sn a word of length n A (scattered) subword of S is a word S′ = si1si2 . . . sil, where i1 < i2 < · · · < il. disjoint subwords si1si2 . . . sil and sj1sj2 . . . sjt: {i1, . . . , il} ∩ {j1, . . . , jt} = ∅. Definition 1 Twins: two disjoint identical subwords of S.

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words

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Twins

S = s1 . . . sn a word of length n A (scattered) subword of S is a word S′ = si1si2 . . . sil, where i1 < i2 < · · · < il. disjoint subwords si1si2 . . . sil and sj1sj2 . . . sjt: {i1, . . . , il} ∩ {j1, . . . , jt} = ∅. Definition 1 Twins: two disjoint identical subwords of S. Example 2 S = s1s2s3s4s5s6s7 = 0001010

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words

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Twins

S = s1 . . . sn a word of length n A (scattered) subword of S is a word S′ = si1si2 . . . sil, where i1 < i2 < · · · < il. disjoint subwords si1si2 . . . sil and sj1sj2 . . . sjt: {i1, . . . , il} ∩ {j1, . . . , jt} = ∅. Definition 1 Twins: two disjoint identical subwords of S. Example 2 S = s1s2s3s4s5s6s7 = 0001010 S1 = s1s2s4 = 001 and S2 = s3s5s6 = 001 are twins.

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words

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Twins

S = s1 . . . sn a word of length n A (scattered) subword of S is a word S′ = si1si2 . . . sil, where i1 < i2 < · · · < il. disjoint subwords si1si2 . . . sil and sj1sj2 . . . sjt: {i1, . . . , il} ∩ {j1, . . . , jt} = ∅. Definition 1 Twins: two disjoint identical subwords of S. Example 2 S = s1s2s3s4s5s6s7 = 0001010 S1 = s1s2s4 = 001 and S2 = s3s5s6 = 001 are twins. S′

1 = s1s4s5 = 010 and S′ 2 = s2s6s7 = 010 are also twins.

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words

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Twins

Problem How large could the twins be in any word over a given alphabet?

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words

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Twins

Problem How large could the twins be in any word over a given alphabet? More precisely: f (S): the largest integer m such that there are twins of length m In the example S = 0001010 one has f (S) = 3.

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words

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Twins

Problem How large could the twins be in any word over a given alphabet? More precisely: f (S): the largest integer m such that there are twins of length m In the example S = 0001010 one has f (S) = 3. We are interested in f (n, Σ) = min{f (S) : S ∈ Σn}

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words

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Twins

Problem How large could the twins be in any word over a given alphabet? More precisely: f (S): the largest integer m such that there are twins of length m In the example S = 0001010 one has f (S) = 3. We are interested in f (n, Σ) = min{f (S) : S ∈ Σn} Trivial lower bound f (n, {0, 1}) ≥ ⌊(1/3)n⌋

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words

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Twins

Problem How large could the twins be in any word over a given alphabet? More precisely: f (S): the largest integer m such that there are twins of length m In the example S = 0001010 one has f (S) = 3. We are interested in f (n, Σ) = min{f (S) : S ∈ Σn} Trivial lower bound f (n, {0, 1}) ≥ ⌊(1/3)n⌋ S = 001 101 111 010 twins equal to 0110: S = 001 101 111 010

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words

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Twins

Our main result is Theorem 3 There exists an absolute constant C such that

  • 1 − C
  • log n

log log n −1/4 n ≤ 2f (n, {0, 1}) ≤ n − log n.

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words

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Twins

Our main result is Theorem 3 There exists an absolute constant C such that

  • 1 − C
  • log n

log log n −1/4 n ≤ 2f (n, {0, 1}) ≤ n − log n. i.e., a binary word of length n has twins of length n/2 − o(n)

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words

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k-twins

Definition 4 k-twins in S ∈ Σ∗: k disjoint identical subwords of S f (S, k): the largest m so that S contains k-twins of length m each f (n, k, Σ) = min{f (S, k) : S ∈ Σn} Theorem 5 For any integer k ≥ 2, and alphabet Σ, |Σ| ≤ k,

  • 1 − C|Σ|
  • log n

log log n −1/4 n ≤ kf (n, k, Σ) ≤ n − log n.

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words

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ε-regular words

The density of the letter q in S: dq(S) = |S|q/|S|. S[i, i + m] = sisi+1 · · · si+m Definition 6 (ε-regular word) For a positive ε, ε < 1/3, call a word S of length n over an alphabet Σ ε-regular if for every i, εn + 1 ≤ i ≤ n − 2εn + 1 and every q ∈ Σ it holds that |dq(S) − dq(S[i, i + εn − 1])| < ε.

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words

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ε-regular words

Example 7 Word S of length n = 60, density 1/2:

011101010101000101001100110101010100110101010101111000010100

is ε-regular for ε = 1/5 Verification by definition: consider factors of length εn = (1/5) · 60 = 12 starting at positions 13, 14 . . . , 37 compare their densities with the density 1/2 of S

S′ = S[13, 24] = 000101001100, d(S′) = 8/12, |8/12 − 1/2| < ǫ = 1/5 S′′ = S[14, 25] = 001010011001, d(S′′) = 7/12, |5/12 − 1/2| < 1/5 etc.

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words

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ε-regular partition

S := (S1, . . . , St): a partition of S if S = S1S2 . . . St Definition 8 (ε-regular partition) A partition S is an ε-regular partition of a word S ∈ Σn if

  • i∈[t]

Si is not ε−regular

|Si| ≤ εn, i.e., the total length of ε-irregular factors is at most εn.

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words

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Regularity lemma

Key lemma: Lemma 9 (Regularity Lemma for Words) For every ε, t0 and n such that 0 < ε < 1/3, t0 > 0 and n > n0 = t0ε−ε−4, any word S ∈ Σn admits an ε-regular partition into t parts with t0 ≤ t ≤ T0 = t031/ε4.

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words

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twins in ε-regular word

S = 01110 11001 10110 11010 11100 10110 10101 S′ = 11 1 11 1 00 11 1 S′′ = 00 1 11 0 0 111 0 0 1 1 1

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words

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Large alphabets and small k-twins

Theorem 10 For any integer k ≥ 2, and alphabet Σ of size l, |Σ| > k,

  • k

|Σ| − C|Σ|

  • log n

log log n

  • − 1

4

  • n ≤ kf (n, k, Σ) ≤ n−max{αn, log n},

where α ∈ [0, 1/k] is the solution of the equation l−(k−1)αα−kα(1 − kα)kα−1 = 1, whenever such solution exists and 0 otherwise.

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words

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Large alphabets and small k-twins

Theorem 10 For any integer k ≥ 2, and alphabet Σ of size l, |Σ| > k,

  • k

|Σ| − C|Σ|

  • log n

log log n

  • − 1

4

  • n ≤ kf (n, k, Σ) ≤ n−max{αn, log n},

where α ∈ [0, 1/k] is the solution of the equation l−(k−1)αα−kα(1 − kα)kα−1 = 1, whenever such solution exists and 0 otherwise. k = 2, l = 5: α < 0.49 ⇒ no twins of length n/2 − o(n)

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words

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Summary

We studied the following Question: is it true that any given word of length n over alphabet Σ has k-twins of size n(1 − o(1))/k each? Informally: the k-twins cover almost all letters of the word

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words

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Summary

We studied the following Question: is it true that any given word of length n over alphabet Σ has k-twins of size n(1 − o(1))/k each? Informally: the k-twins cover almost all letters of the word We have shown that the answer is: YES for k ≥ |Σ| NO for some pairs (k, |Σ|) with k < |Σ|, the smallest such pair we know is (k, |Σ|) = (2, 5)

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words

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Summary

We studied the following Question: is it true that any given word of length n over alphabet Σ has k-twins of size n(1 − o(1))/k each? Informally: the k-twins cover almost all letters of the word We have shown that the answer is: YES for k ≥ |Σ| NO for some pairs (k, |Σ|) with k < |Σ|, the smallest such pair we know is (k, |Σ|) = (2, 5) Open question Is it true for (k, |Σ|) = (k, k + 1)? We do not know even for (2, 3).

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words

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Reference

Maria Axenovich, Yury Person, Svetlana Puzynina: A regularity lemma and twins in words. J. Comb. Theory, Ser. A 120(4): 733-743 (2013)

  • M. Axenovich, Y. Person, S. Puzynina

Twins in words