Ultimately, my question is whether it's possible to exactly characterize the regular languages in terms of fooling sets. To help explain my motivation for asking this, here's a quick exposition.

Let $L$ be a language. A set of strings $S$ is called pairwise distinguishable relative to $L$ if for any two distinct strings $x, y \in S$, there is a string $w \in \Sigma^*$ such that exactly one of $xw$ and $yw$ belongs to $L$.

Assuming the axiom of choice, the Myhill-Nerode theorem can be framed in terms of pairwise distinguishable sets:

Theorem: let $L$ be a language. Then $L$ is nonregular if and only if there is an infinite set $S$ that is pairwise distinguishable relative to $L$.

Now, let's consider a related definition. If $L$ is a language, a set $S \subseteq \Sigma^* \times \Sigma^*$ is called a fooling set for $L$ if for any distinct pairs $(x_i, y_i), (x_j, y_j) \in S$, then $x_i y_i \in S, x_j y_j \in S$, and at least one of $x_i y_j$ and $x_j y_i$ isn't in $L$.

I'm curious whether the following statement is true:

Conjecture: let $L$ be a language. Then $L$ is nonregular if and only if there is an infinite set $S$ that is a fooling set for $L$.

I can prove one direction of this using the Myhill-Nerode theorem: if there is an infinite fooling set $S$ for a language $L$, then we can take the first halves of each pair in $S$, gather them into a set $S'$, and we'll have an infinite set $S$ that is pairwise distinguishable relative to $L$, so $L$ is nonregular.

However, I keep getting stuck trying to go the other way. I'm not sure how to show that if $L$ is nonregular, then there must be an infinite fooling set for $L$. The equivalent direction of the Myhill-Nerode theorem follows by the contrapositive and the fact that the binary relation "is indistinguishable relative to $L$" is an equivalence relation. We don't - I believe - have a similar equivalence relation when it comes to fooling sets.

Does the other direction of implication hold here? Or is this a strictly weaker criterion than the Myhill-Nerode theorem?

Thanks!

  • 1
    Try it out for the following language over $\Sigma = \{0\}$: $L = \{0^n : n \text{ is not a power of $2$}\}$. – Yuval Filmus Jun 4 '15 at 1:45
  • @YuvalFilmus I know it's been a while, but do you want to convert that into an answer? – templatetypedef Nov 3 '15 at 19:58
up vote 2 down vote accepted

Your conjecture is refuted by the language $\{ 0^n : \text{$n$ is not a power of $2$}\}$. Let $\{(x_i,y_i) : i \in \mathbb{N}\}$ be an infinite fooling set for this language. We can identify $x_i,y_i$ with integers. These pairs have to satisfy the following conditions:

  1. $x_i + y_i$ is not a power of $2$.
  2. For each $i \neq j$, either $x_i + y_j$ is a power of $2$ or $x_j + y_i$ is a power of $2$.

Color $\mathbb{N}^2$ with two colors according to whether $x_i + y_j$ is a power of $2$ or $x_j + y_i$ is a power of $2$, where $i < j$. Ramsey's theorem shows that without loss of generality we can assume that for all $i < j$, $x_i + y_j$ is a power of $2$.

If $x_i = x_j$ for $i < j$ then $x_j + y_j$ is a power of $2$, contradicting (1) above. Similarly, if $y_i = y_j$ for $i < j$ then $x_i + y_i$ is a power of $2$, contradicting (1) above. Thus all $x_i$ are distinct and all $y_j$ are distinct.

Assume without loss of generality that $x_1 < x_2$. Let $x_1 + y_3 = 2^a$ and $x_2 + y_3 = 2^b$. Choose $j$ so that $y_j > 2^b$. Let $x_1 + y_j = 2^c$. Then $2^c < x_2 + y_j < 2^c + 2^b < 2^{c+1}$, and we obtain a contradiction.

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