Is this counting language context free?

Let $$\Sigma = \left\{ 0,\,1,\,2\right\}$$. I want to look at the following language: $$L=\left\{ xyz \, | \, |x|_0 + |z|_0 = |x|_2 +|z|_2 \wedge y \in \left\{ 1 \right\} ^{*} \right\}$$.

I would like to prove or disprove $$L$$ being context-free

I'm having a very hard time to construct a context-free grammer for $$L$$, so I attempted to build a non-deterministic PDA.

My attempt goes in the direction of: If i want to sum the zeroes from $$x$$ and $$z$$, and fron this sum to substract the sum of $$2$$s from both $$x$$ and $$z$$, I can do it differently:

I can count the number of $$0$$s from $$x$$ by inserting a #, then remove one # for every $$2$$ I count, if negative I'll push @ . Then I will do the same with $$z$$.

The points are:

1. I'm not sure if this idea will work?
2. If this language is CFL, I have no idea how to begin constructing a CFL grammer for it?
• @dkaeae That's a pretty invasive edit that seems to go beyond clarification into the terrain of stylistic adaptions. Your first line of action should be to ask the OP for clarification, and then to only edit if they don't and/or there are obvious typos or formatting mishaps. The style of the presentation should be in the OP's responsibility.
– Raphael
Jan 31 '19 at 16:52

I can count the number of 0s from $$x$$ by inserting a #, then remove one # for every 2 I count, if negative I'll push @ . Then I will do the same with $$𝑧$$.

This PDA is pretty good. You can certainly continue to convert this PDA to a context-free grammar according to some algorithm such as the one given in the proof of the language of a PDA is context-free, although that procedure is usually rather heavy and less enlightening.

If this language is CFL, I have no idea how to begin constructing a CFL grammer for it?

I would agree that it might not be easy. I would recommend that you should read this answer to how to produce a context-free grammar by Hendrik Jan if you have not yet.

Here is the simplest context-free grammar for $$L$$. $$S \to 0S2S \mid 2S0S \mid 1S \mid \epsilon$$

Note that $$L=\{ xz \mid |x|_0 + |z|_0 = |x|_2 +|z|_2\}=\{ w \mid |w|_0 = |w|_2 \}$$. That is, $$L$$ is the language of words with equal number of 0s and 2s, as you have observed.

Here is a brief reasoning that shows the language of above grammar is $$L$$.

• It is immediate that all words generated by $$S$$ has equal number of $$0$$'s and $$2$$'s.

• Suppose $$1^kw\in L$$ for some $$w$$ that does not start with 1.

• Suppose $$w$$ starts with 0. If we count the number of 0s and the number of 2s in $$w$$ starting from the starting 0, there will be a time that the number of 2s catches up with the number of 0s. So $$w=0w_12w_2$$ for some $$w_1, w_2\in L$$.
• Suppose $$w$$ starts with 2. Similarly, $$w=2w_10w_2$$ for some $$w_1, w_2\in L$$.

Combining the above two cases with the base case $$1^k$$ for $$k\ge0$$, which can be generated, we see by mathematical induction that all words in $$L$$ can be generated.

1. Your idea is correct. After all, $$|x|_0 + |z|_0 = |x|_2 + |z|_2$$ if and only if $$|x|_0 + |z|_0 - (|x|_2 + |z|_2) = |xz|_0 - |xz|_2 = 0$$. Note, however, you must (ab)use nondeterminism correctly so as to guess where $$y$$ is so as to know what belongs to $$x$$ and what to $$z$$.

2. Producing a CFG from a PDA is, unfortunately, not easy in general. There are cases in which, given a CFL, it is (much) easier to construct one than it is to construct the other. Having experience in these kind of exercises might give you a sudden flash of inspiration; failing that, you can always resort to the standard construction.

In this particular case, your "flash of inspiration" could have come from taking a closer look at the equation on the number of zeroes and two's. The idea is the equation holds for the empty word and we can produce symbols while conserving it; that is, for example, if you generate a zero for $$x$$ (i.e., $$|x|_0$$ is increased by one), then you also want to generate a $$2$$ somewhere so the equation still holds. Hence, a CFG with the following set of productions suffices: \begin{align*} S &\to X_0 S X_2 \mid X_2 S X_0 \mid X S X \mid Y \mid \varepsilon, \\ X_0 &\to X 0 \mid 0 X, \\ X_2 &\to X 2 \mid 2 X, \\ X &\to 0 X 2 \mid 2 X 0 \mid 1 X \mid X 1 \mid \varepsilon, \\ Y &\to 1 Y \mid \varepsilon \end{align*} A brief reasoning to make sure every word $$w = x y z \in L$$ is generated: The ones in $$x$$ and $$z$$ are generated by $$X$$, so we can exclude them. Further, any neighboring pair $$02$$ or $$20$$ may be generated by $$X$$ (and we include $$X$$ everywhere it is possible for such a pair to appear). Removing all such pairs, the only possible word left is of the form $$x = 0^n$$ and $$z = 2^n$$ or $$x = 2^n$$ and $$z = 0^n$$, and we can generate $$w$$ by using the appropriate rules for $$S$$ containing $$X_0$$ and $$X_2$$.

• Nice answer. Note that $1012\in L$ is not generated. Feb 1 '19 at 2:56
• @Apass.Jack Whoops. I was under the impression $x, z \in \{ 0, 2 \}^\ast$. (It seems there is much more nondeterminism at work than I'd expected.) Patched by adding production rules for ones everywhere. Still, your answer looks far more elegant :) Feb 1 '19 at 8:20