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Think of a one-dimensional infinite walk as a map $$w\colon \mathbb{N}\to \{-1,1\}.$$ (If it is more convenient, you can think of a walk as a subset of $\mathbb{N}$, or as a binary word, or as any other convenient model.) Similarly, a finite walk, say of $n$-steps (for some integer $n\geq 1$), is a map $$v\colon [0,n-1]\cap \mathbb{N}\to\{-1,1\}.$$

Given a finite walk $v$, with exactly $n$ steps, let us say that is is nondecreasing if $\sum_{i=0}^{n-1}v(i)\geq 0$. Define nonincreasing finite walks dually, using $\leq$.

Given an infinite walk $w$, and a finite (nonempty) set $S$ of finite walks, let us say that $w$ avoids using $S$ too often, if there is some positive integer $N$ (depending on $w$ and $S$) so that if we concatenate any $N$ elements of $S$ (allowing repetitions), then such a concatenated finite walk does not appear as a (consecutive) subwalk in $w$.

For example, if $S=\{v_0\}$ where $v_0$ is the $1$ step walk where $v_0(0)=1$, and if $w$ is the walk where $w(i)=(-1)^{i}$, then we can take $N=2$, and we see that $w$ avoids using $S$ too often. Or in in terms of binary words, the infinite binary word $01010101\ldots$ does not have arbitrarily long strings of consecutive $1$s.

Of course, there exists an infinite walk that avoids using too often any finite set of nondecreasing finite walks. Just take $w$ to be the strictly decreasing walk where $w(i)=-1$.

My question is whether the following balancing act is possible.

Does there exist an infinite walk $w$, such that for every finite set $S$ of nondecreasing finite walks, as well as for every finite set $S$ of nonincreasing finite walks, $w$ avoids using $S$ too often?

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  • $\begingroup$ Doesn't $010011000111\dots$ work? $\endgroup$ Apr 24 at 18:01
  • $\begingroup$ @CommandMaster No, because if we take $S=\{1\}$, then $N$-fold concatenations of $S$ occur for every positive integer $N$. $\endgroup$ Apr 24 at 18:06
  • $\begingroup$ Right. The Thue-Morse sequence is cube free, so it should work with $N=3$. $\endgroup$ Apr 24 at 18:58
  • $\begingroup$ Oh, nevermind, it only works for $|S|=1$, it fails for $10,01$. $\endgroup$ Apr 24 at 19:14
  • $\begingroup$ @CommandMaster Right, this is a sort of generalization of cube-free-ness, which gets tricky for sets of words. $\endgroup$ Apr 24 at 20:21

1 Answer 1

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The limit of the following sequences probably works:

  • $A_0 = \varepsilon$
  • $A_i = (A_{i-1}0)^{2^{(i^2)}}A_{i-1}(1A_{i-1})^{2^{(i^2)}}$

Intuition: For every $S$, there exists some $A_i$ which is not a factor of a word in $S^*$, because it has parts that are increasing/decreasing too much. However note that $A_n \in (A_i(0|1))^*A_i$ for all $n \geq i$. Thus we can pick $N = 2|A_i|$.

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    $\begingroup$ This is a beautiful answer! (I think you can remove the word "probably", and change "Intuition" to "Proof".) Also, am I right in thinking that $2^{(i^2)}$ can be replaced by any unbounded positive sequence? $\endgroup$ Apr 24 at 21:48
  • $\begingroup$ Maybe it is not necessary. However, the words in $S$ can have different sizes, so I would like the fluctuation within $A_i$ to be more than any constant factor greater than the fluctuation in $A_{i-1}$ just to be safe. Something like $2^{\omega(i)}$ should be sufficient. $\endgroup$
    – 1001
    Apr 24 at 21:54
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    $\begingroup$ On second thought, yes you are right. Any unbounded positive sequence will do. $\endgroup$
    – 1001
    Apr 24 at 22:03

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