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Question 1. Assume that an infinite word $u\in\{0,1\}^{\mathbb Z}$ is not balanced. Is it true that there exists a finite 0-1 word $w$ such that $0w01w1$ or $1w10w0$ is a factor of $u$? Is it true that both are? (Perhaps, with different $w$.) - answered in the negative

Question 1 (modified). Assume that an infinite word $u\in\{0,1\}^{\mathbb Z}$ is not balanced. Is it true that there exists a finite (possibly, empty) 0-1 word $w=w_1\dots w_k$ such that $0w_1\dots w_k01w_k\dots w_11$ or $1w_1\dots w_k10w_k\dots w_10$ is a factor of $u$? Is it true that both are? (Perhaps, with different $w$.) - ANSWERED by Wolfgang

Question 2. Assume now that $u\in\{0,1\}^{\mathbb Z}$ is balanced and aperiodic (i.e., Sturmian). Is it true that for any $n\ge1$ there exists a finite 0-1 word $w=w_1\dots w_k$ with $k\ge n$ such that $w_1\dots w_k01w_k\dots w_1$ or $w_1\dots w_k10w_k\dots w_1$ is a factor of $u$? Again, is it true that both are (with different $w$)? - ANSWERED by domotorp

I am familiar with Lothaire's famous monograph "Algebraic Combinatorics on Words" but couldn't find these results there.

EDIT. Here's what is meant by `balanced'. For a finite 0-1 word $w$ let $\delta(w)$ denote the number of 1s in $w$. A 0-1 word $w$ (finite or infinite) is called balanced if for any $n\ge1$ and any two factors $u$ and $v$ of $w$ of length $n$ we have $|\delta(u)-\delta(v)|\le 1$.

For instance, $0100101$ is balanced whilst $101000$ is not, since $\delta(101)=2$ and $\delta(000)=0$.

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  • $\begingroup$ Would you mint posting the definition of balanced (or a link if it is very complicated)? $\endgroup$ – Daniel Soltész Nov 30 '13 at 20:26
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    $\begingroup$ Daniel, it's done. Hope that helps. $\endgroup$ – Nikita Sidorov Nov 30 '13 at 21:01
  • $\begingroup$ Sorry, I don't have enough reputation to write a comment so I'm writing it into an answer. About the second question, you can replace "or" with "and", since Sturmian words contain all the reversals of their factors. $\endgroup$ – Gabriele Fici Dec 5 '14 at 21:45
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I think Question 2 is true and the proof is as follows. Every Sturmian word is equivalent to the cutting sequence of an irrational number (except, I think, those that have only one extra digit compared to a rational number, for which your statement is true). If you have a line $ax$ that goes through the origin, then its cutting sequence in both directions is the same, i.e., they are each other's reverse. Similarly, whenever the line goes "very close" to a grid point, the part before it will be the reverse of the part after it for a long time. Close to the grid point, you have the 01 or the 10. From Diophantine approximation, it follows that we can indeed achieve "very close" in both situations.

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To produce a counterexample for question 1, you can consider sequences with few $01$ and $10$ subsequences, such as

$...00000.0110000111111110000000000000000111...$

Each $01$ is between a patch of $0$s of length $2^n$ and a patch of $1$s of length $2^{n+1}$. For this to be in the middle of $0w01w1$, either $w$ is all $1$s or it is not. If so, then there would be an isolated $0$ which doesn't happen. If not, then there would be at least $2^{n+1}$ $1$s to the right, but there are fewer than $2^{n}$ $1$s to the left. Similarly, each $10$ is between a subsequence of $1$s of length $2^n$ on the left and $2^{n+1}$ $0$s on the right. For this to be in the middle of $1w10w0$, $w$ can't be all $0$s, but then it must have length at least $2^{n+1}$, and then the first $1$ of $1w10w0$ would be too far to the left.

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  • $\begingroup$ Douglas, thanks. I have modified this question replacing $w$ with its reverse $\widetilde w$ - hope there are no counterexamples for this version. $\endgroup$ – Nikita Sidorov Dec 1 '13 at 0:35
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In case of Question 1 (modified), it might happen that both are not factors, e.g., consider the sequence ...1111010000....

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  • $\begingroup$ But $111010=1w10\widetilde w0$ is a factor, with $w=1$. $\endgroup$ – Nikita Sidorov Dec 2 '13 at 22:59
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    $\begingroup$ Yes, I mean that it might happen that we do not have both forms as factors but only one of them. $\endgroup$ – domotorp Dec 2 '13 at 23:12
  • $\begingroup$ Sorry, I misunderstood your answer. Indeed, you are right. As a matter of fact, I can see how your sequence can be used to construct uncountably many counterexamples. $\endgroup$ – Nikita Sidorov Dec 2 '13 at 23:40
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I'll prove that the modified question 1 has an affirmative answer by showing conversely that an infinite word $u\in\{0,1\}^{\mathbb Z}$ which avoids all $0w01\widetilde w1 $ and $1w10\widetilde w0 $ (call these factors forbidden) is necessarily balanced.

If $u\ne...0101...$, there is wlog a factor 00. (Otherwise, swap all 0's and 1's). For $k\ge1$, define a $k$-patch (or $p_k$ for short) as a 'maximal' run of $k$ consecutive zeros in $u$, i.e. it is preceded and followed by a 1. As 0011 and 1100 are forbidden, these 1's are isolated if $k\ge 2$, so each $10_k1$ (where indices mean repetition) is preceded and followed by patches $p_a$ and $p_b$, and the forbidden factors imply $k-1\le a,b\le k+1$. The same holds trivially for $k=1$. So two 'neighboring' patches have a difference of length of at most 1.
Now suppose that there is a $k\ge 1$ such that $u$ contains a $p_{k-1}$ and a $p_{k+1}$. (It makes sense to define a $0$-patch $p_0$ as the empty set in the middle of a factor 11 forcing those two 1's.) Suppose they are closest possible to each other, i.e. between them there is only a number, say $n$, of $k$-patches (each separated by an isolated 1).

So $u$ has a factor $$ 10_{k-1}\overbrace{10_k\cdots10_k}^n10_{k+1}1$$ (or the other way round, in which case inverse the whole sequence wlog). Which entries must follow to the right? In fact we must have

$$ 10_{k-1}\overbrace{10_k\cdots10_k}^n\underbrace{10}_{(x)}0_{k-1} \underbrace{01}_{(y)}\overbrace{\color{red}{0_k1\cdots0_k1}}^{n-1}\color{red}{0_k}$$ because each red 0 is forced by the fact that the string 01 marked by $(y)$ must not be the center of a forbidden factor, and each red 1 is forced by the fact that the string 10 marked by $(x)$ must not be the center of a forbidden factor. But due to the initial 1, the whole string above is a forbidden one with center $(x)$. Contradiction.

So there is a $k$ such that $u$ consists only of patches of lengths $k$ and $k+1$, separated by isolated 1's. It is easy to see that such a $u$ is balanced. qed.

EDIT: this was wrong, but the proof can be rescued by the following completion.

Suppose $u$ is not balanced. Then take the smallest $r$ such that there are two factors $s$ and $t$ of length $r$ where $t$ has 2 more 1's than $s$. So $s$ starts and ends with $0_{k+1}$ and $t$ starts and ends with $10_k1$.

Let $n$ be the number of interior 1's of each $s$ and $t$. Suppose there are $i$ instances of $p_k$ and $j$ instances of $p_{k+1}$ with both $i,j>0$, then the interior of $t$ must have the same numbers of each (note that $s$ and $t$ have the same number of 1's in their interior).

EDIT after the last remark of Harry Altman (the one starting with "Yay") : I have found a way to repare the proof again, making it at the same time more elegant.

We'll now introduce the process of reduction: put $a^0=1,b^0=0$ and $u^0=u$. For $\nu=0,1,...$ proceed as follows: As the $a^\nu$-$b^\nu$-word $u^\nu$ contains no forbidden factors (in terms of $a^\nu$ and $b^\nu$), the boldface statement above about 'neighboring' patches applies also to $u^\nu$, so there must be a $k^\nu$ such that $u^\nu$ is composed either
(1) of runs $a^\nu_{k^\nu}$ and $a^\nu_{k^\nu+1}$, interspearsed by isolated $b^\nu$'s, or
(2) of runs $b^\nu_{k^\nu}$ and $b^\nu_{k^\nu+1}$, interspearsed by isolated $a^\nu$'s.

Then denote in case (1) $a^{\nu+1}:=b^\nu a^\nu_{k^\nu}, b^{\nu+1}:=b^\nu a^\nu_{k^\nu+1}$, and
in case (2) $a^{\nu+1}:=a^\nu b^\nu_{k^\nu}, b^{\nu+1}:=a^\nu b^\nu_{k^\nu+1}$,
further define $u^{\nu+1}$ as the sequence $u$ written as a $a^{\nu+1}$-$b^{\nu+1}$-word. This reduction can be traced back, and it is easy to see that that $u^{\nu+1}$ cannot contain $a^{\nu+1}$-$b^{\nu+1}$-words as forbidden factors. (Looking at the $a^{\nu+1}b^{\nu+1}$ or $b^{\nu+1}a^{\nu+1}$ in the center of a forbidden factor, it is clear that this gives rise to a forbidden factor in $u^\nu$, which is excluded.)

$s^\nu$ and $t^\nu$ can be written as factors of $u^\nu$, one of them starts and ends with $a^\nu$’s, the other one of them starts and ends with $b^\nu$’s. By the minimality of $s$ and $t$ and the fact that they have the same number of letters, it is easy to see that, up to an initial or final isolated letter (like in the $\nu=0$ case : a missing 1 before $s$ and the final 1of $t$), they must factor also in terms of the ‘new letters’ $a^{\nu+1}$’s and $b^{\nu+1}$, and their numbers of occurrences in the interior must again coincide. (To illustrate this, think of $s=bbbabbbabbabbabbb$ and $t=abbabbabbabbbabba$ omitting the $\nu$’s. Here $k=2$, as there are runs in $b$ of lengths 2 and 3.. Then the reduction would be $abbb\to b$ and $abb\to a$ with $s$ (or rather $as$) $\to bbaab$ ant $t$ (or rather $t$ with the final $a$ omitted) $\to aaaba$.)

It is clear that $s^{\nu+1}$ has less letters $a^{\nu+1}$ and $b^{\nu+1}$ than $s^\nu$ has letters $a^\nu$ and $b^\nu$, and that by minimality, one of them starts and ends with $a^{\nu+1}$’s, the other one of them starts and ends with $b^{\nu+1}$’s. Iterating this reduction, we must thus come to a point where (omitting again the $\nu$’s) either $s$ or $t$, say $s$, consists only of letters of the same kind, say $a$. But at that moment, $t$ must have the form $baa\cdots ab$, thus it has a run of two less $a$’s. Contradiction.

We conclude that the initial $s$ and $t$ cannot exist, thus $u$ is balanced. Qed.

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  • $\begingroup$ Wolfgang, I like your idea about patches, but I don't see how the fact that 0011 and 1100 are forbidden can lead to this. Take, for instance, the periodic sequence with period 00101101 - it contains both 00 and 11 as factors but neither 0011 nor 1100. $\endgroup$ – Nikita Sidorov Dec 4 '13 at 14:25
  • $\begingroup$ Also, the last sentence of your comment is somewhat perplexing. Surely, there are plenty of unbalanced words which are concatenations of the blocks $10_k$ and $10_{k+1}$. For instance, the set of all such words has an exponential growth whilst the number of balanced words is known to grow polynomially. $\endgroup$ – Nikita Sidorov Dec 4 '13 at 14:40
  • $\begingroup$ for your first comment: 00101101 is also forbidden. I have said that the argument 0011 and 1100 only applies for $k\ge 2$, of course. But I realize that you are right with your second comment. Sure enough I missed that e.g. factors 10101 and 00100 might co-occur. What a pity! $\endgroup$ – Wolfgang Dec 5 '13 at 7:35
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    $\begingroup$ I think I have been able to rescue my proof! Please check. $\endgroup$ – Wolfgang Dec 5 '13 at 13:42
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    $\begingroup$ @Harry's 2nd comment: Thank you for checking that. You are absolutely right, this is (was) still a flaw. Again it's rescuable, see my last edit. I am quite sure the proof is waterproof now. $\endgroup$ – Wolfgang Dec 6 '13 at 9:11
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Too late, but here's my proof that the answer to Question 1 (modified) is yes.

If $u\in\{0,1\}^\mathbb{Z}$ is not balanced, there exists a palindrome $w$ such that $0w0$ and $1w1$ are factors of $u$ (see Prop. 2.1.3 in Lothaire 2); if $w$ is of minimal length $n$, then $u$ has at most $k+1$ factors of each length $k\leq n+1$ (Prop. 2.1.2). Moreover $\{0w0,1w1\}$ is the only imbalance among factors of length $\leq n+2$ in $u$. It follows that the set of factors of length $n+2$ is contained in $\{0w0,1w1\}\cup [w01]$, where [x] denotes the conjugacy class of $x$ (all of its cyclic shifts). Also, every factor of length $n+1$ except $0w$ and $1w$ are not right special, i.e., they always occur followed by the same letter.

From this we can derive that there exists $m\geq 0$ such that every occurrence of $0w0$ is followed by $(1w0)^m$ and every occurrence of $1w1$ is followed by $(0w1)^m$. If such $m$ is chosen to be maximal, then either $$0w0(1w0)^m1w1(0w1)^m=0(w01)^mw01w(10w)^m1$$ or $$1w1(0w1)^m0w0(1w0)^m=1(w10)^mw10w(01w)^m0$$ is a factor of $u$.

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    $\begingroup$ Even if Wolfgang beat me to it, maybe someone might be interested in my proof. I know it is much denser; I suggest drawing the Rauzy graph for length $n+1$ to see it more clearly. $\endgroup$ – Ale De Luca Dec 5 '13 at 23:38
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For Question 1, it seems that the word $\dots 001001001001\,011011011011\dots$ has neither such factor.

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  • $\begingroup$ Ilya, thanks - that works. I have modified this question in the spirit of Question 2, this time hoping for the affirmative answer. $\endgroup$ – Nikita Sidorov Nov 30 '13 at 23:17
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For Question 1, I believe the result you are looking for is the following by Dulucq and Gouyou-Beauchamps (1990).

A finite word $w$ over $\{0,1\}$ is not balanced (and so not a factor of a Sturmian word) if and only if it can be written $w = x0u0y1\tilde{u}1z$ or $w = x1u1y0\tilde{u}0z$ for some words $u, x, y, z$ (where $\tilde{u}$ is the reverse of $u$).

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  • $\begingroup$ Amy, I am not sure how this helps unless $y$ is the empty word - in which case $0u01\tilde{u}1$ or $1u10\tilde{u}0$ is indeed a factor of $w$. Could you elaborate, please? $\endgroup$ – Nikita Sidorov Dec 4 '13 at 3:51
  • $\begingroup$ Also, I couldn't find the result in question in the paper you mention. Could you give the exact reference, please? $\endgroup$ – Nikita Sidorov Dec 4 '13 at 14:36
  • $\begingroup$ The result is implicitly proved in Section 3 of the paper that I mentioned. Alternatively, see Problem 2.1.4 on page 90 of the 2nd Lothaire book "Algebraic Combinatorics on Words". By this result, a (finite or infinite) word over $\{0,1\}$ is not balanced if and only if it contains a factor of the form $0u0y1\tilde{u}1$ or $1u1y0\tilde{u}0$ for some (possibly empty) finite words $u, y$. So it seems to me that the answer to your modified Question 1 is NO, unless you modify it to include the possibility of a non-empty word between the 1 & 0 (or 0 & 1) in the middle ... $\endgroup$ – Amy Glen Dec 4 '13 at 16:22
  • $\begingroup$ And here's a possible counterexample to your modified Question 1: $\cdots 010101(00100010101)101010\cdots$ which is not balanced because of the "bracketed" factor that takes the form $0u0y1\tilde{u}1$ with $u=010$ and $y=0$, for instance. It also contains the non-balanced factor $00010101$ with $u=0$ and $y=10$. I think you'll find, however, that it does not contain a factor of the form $0u01\tilde{u}1$ or $1u10\tilde{u}0$. Am I right? $\endgroup$ – Amy Glen Dec 4 '13 at 16:46
  • $\begingroup$ Amy, thanks. I remember I've seen it somewhere, and now I know where. A non-empty word in the middle is not an option, I'm afraid. However, I don't see immediately why their characterization implies the NO answer. I still think the answer is YES, it's just that this particular result doesn't help, that's all. ;) $\endgroup$ – Nikita Sidorov Dec 4 '13 at 16:48

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