La lingua che coinvolge il numero irrazionale non è un CFL

Sto lavorando a un duro esercizio in un libro di testo e non riesco proprio a capire come procedere. Ecco il problema Supponiamo di avere la lingua $L = \{a^ib^j: i \leq j \gamma, i\geq 0, j\geq 1\}$ dove $\gamma$ è un numero irrazionale. Come potrei dimostrare che $L$ non è un linguaggio privo di contesto?

Nel caso in cui $\gamma$ sia razionale, è abbastanza facile costruire una grammatica che accetta la lingua. Ma poiché $\gamma$ è irrazionale, non so davvero cosa fare. Non sembra che nessuno dei lemmi di pompaggio funzionerebbe qui. Forse il teorema di Parikh avrebbe funzionato qui, poiché sembrerebbe intuitivamente che questa lingua non abbia un'immagine semilineare di Parikh di accompagnamento.

Questo esercizio è tratto da "Un secondo corso in lingue formali e teoria degli automi" di Jeffrey Shallit, esercizio 25 del capitolo 4.

Gradirei davvero qualsiasi aiuto, o spingendo nella giusta direzione. Grazie!

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 $S = u_0 + \mathbb{N} u_1 + \cdots + \mathbb{N} u_\ell$, 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 $\sup \{ a/b : (a,b) \in M \} = \gamma$.

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When $\gamma$ is rational, the proof fails, and indeed $M$ is semilinear:

$$\{ (a,b) : a \leq \tfrac{s}{t} b \} = \bigcup_{a=0}^{s-1} (a,\lceil \tfrac{t}{s} a \rceil) + \mathbb{N} (s,t) + \mathbb{N} (0,1).$$ Indeed, by construction, any pair $(a,b)$ in the right-hand side satisfies $a \leq \tfrac{s}{t} b$ (since $s = \tfrac{s}{t} t$). Conversely, suppose that $a \leq \frac{s}{t} b$. While $a \geq s$ and $b \geq t$, subtract $(s,t)$ from $(a,b)$. Eventually $a < s$ (since $b < t$ implies $a \leq \frac{s}{t}b < s$). Since $a \leq \frac{s}{t} b$, necessarily $b \geq \lceil \tfrac{t}{s} a \rceil$. Hence we can subtract $(0,1)$ from $(a,b)$ until we reach $(a,\lceil \tfrac{t}{s} a \rceil)$.

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 $\dfrac{a_1}{b_1}\lt\dfrac{a_2}{b_2}\lt\dfrac{a_3}{b_3}\lt\cdots \lt\gamma$ such that $\dfrac{a_i}{b_i}$ is nearer to $\gamma$ than any other rational number smaller than $\gamma$ whose denominator is less than $b_i$.

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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 $s=a^{a_p}b^{b_p}$, a word that is $L$ but "barely". Note that $|s|>b_p\ge p$. Consider $s=uvwxy$, where $|vx|> 1$ and $s_n=uv^nwx^ny\in L$ for all $n\ge0$.

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

• If $t_b=0$ or $\dfrac{t_a}{t_b}\gt\gamma$, 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., $s_n\not\in L$.
• Otherwise, $\dfrac{t_a}{t_b}\lt\gamma$. Since $t_b<b_p$, $\dfrac{t_a}{t_b}\lt \dfrac{a_p}{b_p}$. Hence, $\dfrac{a_p-t_a}{b_p-t_b}>\dfrac{a_p}{b_p}$ Since $b_p-t_b<b_p$, $\dfrac{a_p-t_a}{b_p-t_b}>\gamma,$ which says that $s_0\not\in L$.

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

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Here are two related easier exercises.

Exercise 1. Show that $L_\gamma=\{a^{\lfloor i \gamma\rfloor}: i\in\Bbb N\}$ is not context-free where $\gamma$ is an irrational number.

Exercise 2. Show that $L_\gamma=\{a^ib^j: i \leq j \gamma, i \ge0, j\ge 0\}$ is context-free where $\gamma$ is a rational number.