Yhyjyk

I am working through a hard exercise in a textbook, and I just can't figure out how to proceed. Here is the problem. Suppose we have the language $L = a^ib^j: i leq j gamma, igeq 0, jgeq 1$ where $gamma$ is some irrational number. How would I prove that $L$ is not a context-free language?

In the case when $gamma$ is rational, it's pretty easy to construct a grammar that accepts the language. But because $gamma$ is irrational, I don't really know what to do. It doesn't look like any of the pumping lemmas would work here. Maybe Parikh's theorem would work here, since it would intuitively seem like this language doesn't have an accompanying semilinear Parikh image.

This exercise is from "A Second Course in Formal Languages and Automata Theory" by Jeffrey Shallit, Exercise 25 of Chapter 4.

I would really appreciate any help, or nudges in the right direction. Thank you!

According to Parikh's theorem, if $L$ were context-free then the set $M = (a,b) : a leq gamma b $ would be semilinear, that is, it would be the union of finitely many sets of the form $S = u_0 + mathbbN u_1 + cdots + mathbbN u_ell$, for some $u_i = (a_i,b_i)$.

Obviously $u_0 in M$, and moreover $u_i in M$ for each $i > 0$, since otherwise $u_0 + N u_i notin M$ for large enough $N$. Therefore $g(S) := max(a_0/b_0,ldots,a_ell/b_ell) < gamma$ (since $g(S)$ is rational). This means that every $(a,b) in S$ satisfies $a/b leq g(S)$.

Now suppose that $M$ is the union of $S^(1),ldots,S^(m)$, and define $g = max(g(S^(1)),ldots,g(S^(m))) < gamma$. The foregoing shows that every $(a,b)$ in the union satisfies $a/b leq g < gamma$, and we obtain a contradiction, since $sup a/b : (a,b) in M = gamma$.

When $gamma$ is rational, the proof fails, and indeed $M$ is semilinear:
$$
(a,b) : a leq tfracst b = bigcup_a=0^s-1 (a,lceil tfracts a rceil) + mathbbN (s,t) + mathbbN (0,1).
$$
Indeed, by construction, any pair $(a,b)$ in the right-hand side satisfies $a leq tfracst b$ (since $s = tfracst t$). Conversely, suppose that $a leq fracst b$. While $a geq s$ and $b geq t$, subtract $(s,t)$ from $(a,b)$. Eventually $a < s$ (since $b < t$ implies $a leq fracstb < s$). Since $a leq fracst b$, necessarily $b geq lceil tfracts a rceil$. Hence we can subtract $(0,1)$ from $(a,b)$ until we reach $(a,lceil tfracts a rceil)$.

Every variable except $gamma$ in this answer stands for a positive integer. It is well-known that given an irrational $gamma>0$, there is a sequence of rational numbers $dfraca_1b_1ltdfraca_2b_2ltdfraca_3b_3ltcdots ltgamma$ such that $dfraca_ib_i$ is nearer to $gamma$ than any other rational number smaller than $gamma$ whose denominator is less than $b_i$.

It turns out that the pumping lemma does work!

For the sake of contradiction, let $p$ be the pumping length of $L$ as a context-free language. Let $s=a^a_pb^b_p$, a word that is $L$ but "barely". Note that $|s|>b_pge p$. Consider
$s=uvwxy$, where $|vx|> 1$ and $s_n=uv^nwx^nyin L$ for all $nge0$.

Let $t_a$ and $t_b$ be the number of $a$s and $b$s in $vx$ respectively.

• If $t_b=0$ or $dfract_at_bgtgamma$, for $n$ large enough, the ratio of the number of $a$s to that of $b$s in $s_n$ will be larger than $gamma$, i.e., $s_nnotin L$.


• Otherwise, $dfract_at_bltgamma$. Since $t_b<b_p$, $dfract_at_blt dfraca_pb_p$. Hence,
  $dfraca_p-t_ab_p-t_b>dfraca_pb_p$
  Since $b_p-t_b<b_p$, $dfraca_p-t_ab_p-t_b>gamma,$
  which says that $s_0notin L$.

The above contradiction shows that $L$ cannot be context-free.

Here are two related easier exercises.

Exercise 1. Show that $L_gamma=a^lfloor i gammarfloor: iinBbb N$ is not context-free where $gamma$ is an irrational number.

Exercise 2. Show that $L_gamma=a^ib^j: i leq j gamma, i ge0, jge 0$ is context-free where $gamma$ is a rational number.