A note on the number of squares in a partial word with one hole

Francine Blanchet-Sadri; Robert Mercaş

RAIRO - Theoretical Informatics and Applications (2009)

  • Volume: 43, Issue: 4, page 767-774
  • ISSN: 0988-3754

Abstract

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A well known result of Fraenkel and Simpson states that the number of distinct squares in a word of length n is bounded by 2n since at each position there are at most two distinct squares whose last occurrence starts. In this paper, we investigate squares in partial words with one hole, or sequences over a finite alphabet that have a “do not know” symbol or “hole”. A square in a partial word over a given alphabet has the form uv where u is compatible with v, and consequently, such square is compatible with a number of words over the alphabet that are squares. Recently, it was shown that for partial words with one hole, there may be more than two squares that have their last occurrence starting at the same position. Here, we prove that if such is the case, then the length of the shortest square is at most half the length of the third shortest square. As a result, we show that the number of distinct squares compatible with factors of a partial word with one hole of length n is bounded by 7 n 2 .

How to cite

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Blanchet-Sadri, Francine, and Mercaş, Robert. "A note on the number of squares in a partial word with one hole." RAIRO - Theoretical Informatics and Applications 43.4 (2009): 767-774. <http://eudml.org/doc/250618>.

@article{Blanchet2009,
abstract = { A well known result of Fraenkel and Simpson states that the number of distinct squares in a word of length n is bounded by 2n since at each position there are at most two distinct squares whose last occurrence starts. In this paper, we investigate squares in partial words with one hole, or sequences over a finite alphabet that have a “do not know” symbol or “hole”. A square in a partial word over a given alphabet has the form uv where u is compatible with v, and consequently, such square is compatible with a number of words over the alphabet that are squares. Recently, it was shown that for partial words with one hole, there may be more than two squares that have their last occurrence starting at the same position. Here, we prove that if such is the case, then the length of the shortest square is at most half the length of the third shortest square. As a result, we show that the number of distinct squares compatible with factors of a partial word with one hole of length n is bounded by $\frac\{7n\}\{2\}$. },
author = {Blanchet-Sadri, Francine, Mercaş, Robert},
journal = {RAIRO - Theoretical Informatics and Applications},
keywords = {Combinatorics on words; partial words; squares.; combinatorics on words; squares},
language = {eng},
month = {10},
number = {4},
pages = {767-774},
publisher = {EDP Sciences},
title = {A note on the number of squares in a partial word with one hole},
url = {http://eudml.org/doc/250618},
volume = {43},
year = {2009},
}

TY - JOUR
AU - Blanchet-Sadri, Francine
AU - Mercaş, Robert
TI - A note on the number of squares in a partial word with one hole
JO - RAIRO - Theoretical Informatics and Applications
DA - 2009/10//
PB - EDP Sciences
VL - 43
IS - 4
SP - 767
EP - 774
AB - A well known result of Fraenkel and Simpson states that the number of distinct squares in a word of length n is bounded by 2n since at each position there are at most two distinct squares whose last occurrence starts. In this paper, we investigate squares in partial words with one hole, or sequences over a finite alphabet that have a “do not know” symbol or “hole”. A square in a partial word over a given alphabet has the form uv where u is compatible with v, and consequently, such square is compatible with a number of words over the alphabet that are squares. Recently, it was shown that for partial words with one hole, there may be more than two squares that have their last occurrence starting at the same position. Here, we prove that if such is the case, then the length of the shortest square is at most half the length of the third shortest square. As a result, we show that the number of distinct squares compatible with factors of a partial word with one hole of length n is bounded by $\frac{7n}{2}$.
LA - eng
KW - Combinatorics on words; partial words; squares.; combinatorics on words; squares
UR - http://eudml.org/doc/250618
ER -

References

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  1. J. Berstel and L. Boasson, Partial words and a theorem of Fine and Wilf. Theoret. Comput. Sci.218 (1999) 135–141.  
  2. F. Blanchet-Sadri, Algorithmic Combinatorics on Partial Words. Chapman & Hall/CRC Press, Boca Raton, FL (2008).  
  3. F. Blanchet-Sadri and D.K. Luhmann, Conjugacy on partial words. Theoret. Comput. Sci.289 (2002) 297–312.  
  4. F. Blanchet-Sadri, R. Mercaş and G. Scott, Counting distinct squares in partial words, edited by E. Csuhaj-Varju, Z. Esik, AFL 2008, 12th International Conference on Automata and Formal Languages, Balatonfüred, Hungary (2008) 122–133, www.uncg.edu/cmp/research/freeness  
  5. F. Blanchet-Sadri, D.D. Blair and R.V. Lewis, Equations on partial words. RAIRO-Theor. Inf. Appl.43 (2009) 23–39, www.uncg.edu/cmp/research/equations  
  6. A.S. Fraenkel and J. Simpson, How many squares can a string contain? J. Combin. Theory Ser. A82 (1998) 112–120.  
  7. L. Ilie, A simple proof that a word of length n has at most 2n distinct squares. J. Combin. Theory Ser. A112 (2005) 163–164.  
  8. L. Ilie, A note on the number of squares in a word. Theoret. Comput. Sci.380 (2007) 373–376.  

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