Balanced words in higher dimensions
Siddhartha Bhattacharya

TL;DR
This paper investigates higher-dimensional balanced words, demonstrating the existence of two-dimensional balanced words with irrational densities, thus answering a previously open question in the field.
Contribution
It proves that balanced words in two dimensions can have irrational densities, extending the understanding of their properties beyond rational values.
Findings
Existence of two-dimensional balanced words with irrational densities
Answer to Berthé and Tijdeman's open question
Extension of balanced word theory to higher dimensions
Abstract
For , a word is called balanced if there exists such that for any two rectangles that are translates of each other, the number of occurrences of the symbol in and differ by at most . It is known that for every balanced word , the asymptotic frequency of the symbol ( called the density of ) exists. In this paper we show that there exist two dimensional balanced words with irrational densities, answering a question raised by Berth\'e and Tijdeman.
Peer Reviews
No public reviews on file for this paper yet. If you reviewed it on a platform where reviews are public (OpenReview, ICLR, NeurIPS, ICML), you can paste yours below so the community can read it here.
Videos
No videos yet. Explain this paper in a talk, walkthrough, or lecture? Add one.
Taxonomy
Topicssemigroups and automata theory · Algorithms and Data Compression · Mathematical Dynamics and Fractals
Balanced words in higher dimensions
Siddhartha Bhattacharya
School of Mathhematics, Tata Institute of Fundamental Research, Mumbai 400005, India
Abstract.
For , a word is called balanced if there exists such that for any two rectangles that are translates of each other, the number of occurrences of the symbol in and differ by at most . It is known that for every balanced word , the asymptotic frequency of the symbol ( called the density of ) exists. In this paper we show that there exist two dimensional balanced words with irrational densities, answering a question raised by Berthé and Tijdeman.
Key words and phrases:
Balanced words, irrational density
2010 Mathematics Subject Classification:
5B99
1. Introduction
A word is called -balanced if for any two blocks of equal length, the number of occurrences of the symbol in and differ by at most . We will call a word balanced if it is -balanced for some . In the literature -balanced words are often called balanced. However, following [1], we will reserve this term for the weaker property.
Balanced words occur naturally in many different areas, including ergodic theory, number theory, and theoretical computer science ([2], [3],[6]). The study of balanced words was initiated by Morse and Hedlund ([4], [5]). They obtained a classification theorem for -balanced words involving the density (asymptotic frequency of the symbol 1) of the underlying word. In particular, they proved that any -balanced word admits a density, and -balanced words in with irrational densities are Sturmian words corresponding to certain codings of irrational rotations of the circle.
In [1], Berthé and Tijdeman studied balanced words in for . They showed that higher dimensional balanced words also admit densities, but for , the density of a -balanced word is always rational ([1, the main corollary]). The following question has been raised by several authors ([1],[7, Conjecture 2]) :
Question : *For , does there exist a balanced word in with irrational density ? *
In this paper we give an affirmative answer to this question.
Theorem 1.1**.**
For any there exists a balanced word in with density .
If is an irrational number sufficiently close to then our proof shows that there exists a -balanced word in with that has density .
2. Density of balanced words
For a set , will denote the cardinality of . For with we will denote the set by . A rectangle in is a set of the form . For , will denote the set , the collection of all -dimensional words with as the alphabet set. If , we define by
[TABLE]
A word is said to have a *density * if there exists such that as for every sequence of rectangles with .
Suppose , and is a function from to the two element set . We define by .
Lemma 2.1**.**
Let , and be as above. Suppose there exists such that for every rectangle ,
[TABLE]
Then is a balanced word with density .
Proof. Since for all , it follows that for any rectangle ,
[TABLE]
Hence . This implies that as , i.e., has density . If is a translate of then the above equation also shows that
[TABLE]
Hence is a -balanced word.
For , let denote the ring , the polynomial ring in commuting variables with integer coefficients. For , we will denote the monomial by . Every element of can be expressed as , with and for all but finitely many . For , will denote the space of all bounded functions from to . For and we define by
[TABLE]
Theorem 2.2**.**
Suppose and . If there exists such that for all , then there exists a balanced word with density .
*Proof. * Let denote the homomorphism defined by
[TABLE]
and let denote the ring homomorphism from to induced by , i.e., . We note that is an element of , and for any ,
[TABLE]
Hence for all . We define and by
[TABLE]
It is easy to see that . Since the image of is contained in , we deduce that the image of is also contained in . Let be an arbitrary rectangle in . Let denote the polynomial . Then
[TABLE]
We note that , where . As , this implies that
[TABLE]
We note that for any and . Since the right hand side of the above equality has terms of the form , we conclude that . In particular, . As is arbitrary, the given assertion follows from the previous lemma.
It is easy to see that for any the value of at depends only on . Hence if denotes the space of all sequences taking values in then is a well defined map from to . We note the following consequence of the previous result :
Corollary 2.3**.**
Suppose and . If there exists a bounded sequence such that and lies in for all , then there exists a balanced word with density .
Proof. In view of the previous theorem it is enough to construct such that for all . We note that , where . Suppose is even. We define by if , and if . Clearly is an element of . Since lies in for all , and for all , it follows that lies in for all . If is odd, we define if , and if . Since for all , we deduce that lies in for all .
3. Two dimensional balanced words
Now we consider the case when . Let be an arbitrary function from to . We define a function by
[TABLE]
We define , and . Since , it follows that . Hence for all ,
[TABLE]
As , Theorem 1.1 follows from Corollary 2.3 and the following result :
Theorem 3.1**.**
For any , there exists such that the set is bounded.
*Proof. * We define as follows : if , or if both and are positive. Otherwise, . We define and claim that for all . Suppose this is not the case. If there exists such that , we set . Then . From the minimality of we deduce that . As the range of is , this shows that . But this can happen only if , i.e., . Since this contradicts , we conclude that for all . Now suppose for some . We define . Clearly , i.e., is negative. Hence and . But this implies that
[TABLE]
which is not possible since by construction whenever . This proves the claim.
Now we will show that for all . Suppose this is not the case. We pick any such that and define
[TABLE]
Clearly for . We look at the finite sequence . If for some , then is also positive. Since , this implies that . Hence . If then there are two possibilities. In the first case, when , from the definition of it follows that . In particular, . In the second case, when , it is easy to see that irrespective of whether or . Combining these two cases, we observe that whenever . Therefore the sequence decreases by at each step as long as is positive, and once it becomes non-positive it remains non-positive. So there exists a unique such that for and for . Since , it follows that
[TABLE]
We note that , , and for . This shows that . From the previous claim we see that for all . As , we obtain that
[TABLE]
. This contradiction shows that for all .
To complete the proof of the given assertion we need to show that the sequence is also bounded from below. We set and claim that for all . Suppose this is not the case. We pick such that and define
[TABLE]
Clearly for . As before, we look at the finite sequence . If for some , then . Since , this implies that . Hence . If then there are two possibilities. In the first case, when , from the definition of it follows that . In particular, . In the second case, when , it is easy to see that irrespective of whether or . Combining these two cases, we observe that whenever . Therefore the sequence increases by at each step as long as is negative, and once it becomes non-negative it remains non-negative. So there exists a unique such that for and for . Since , it follows that
[TABLE]
We note that , , and for . This shows that . From the previous claim we see that for all . As , we obtain that
[TABLE]
This contradiction shows that for all .
The reference list from the paper itself. Each links out to its DOI / PubMed record.
- 1[1] Valérie Berthé and Robert Tijdeman, Balance properties of multi-dimensional words , WORDS (Rouen, 1999), Theoret. Comput. Sci. 273 (2002), no. 1-2, 197-224.
- 2[2] Amy Glen, Jacques Justin, Steve Widmer and Luca Q. Zamboni, Palindromic richness , European J. Combin. 30 (2009), no. 2, 510-531.
- 3[3] Gwénaël Richomme, Kalle Saari and Luca Q. Zamboni, Balance and abelian complexity of the Tribonacci word , Adv. in Appl. Math. 45 (2010), no. 2, 212-231.
- 4[4] M. Morse and G.A. Hedlund, Symbolic dynamics , Amer. J. Math. 60 (1938) 815-866.
- 5[5] M. Morse and G.A. Hedlund, Symbolic dynamics II: Sturmian trajectories , Amer. J. Math. 62 (1940) 1-42.
- 6[6] Robert Tijdeman, Periodicity and almost-periodicity , More sets, graphs and numbers, 381-405, Bolyai Soc. Math. Stud., 15, Springer, Berlin, 2006.
- 7[7] Laurent Vuillon, Balanced words , Bull. Belg. Math. Soc. Simon Stevin 10 (2003), suppl., 787-805.
