The range of non-linear natural polynomials cannot be context-free

TL;DR

This paper proves that the set of natural values of a non-linear polynomial cannot have a context-free base-q representation, resolving a question about the limitations of context-free languages in describing polynomial-generated sets.

Contribution

It establishes that only linear polynomials produce base-q representations that are context-free, introducing a new criterion combining the Interchange lemma and Pumping lemma for proving non-context-freeness.

Findings

Non-linear polynomials do not produce context-free base-q representations.

A new criterion for context-freeness is developed, combining existing lemmas.

The result answers a previously open question by Shallit.

Abstract

Suppose that some polynomial $f$ with rational coefficients takes only natural values at natural numbers, i.e., $L=\{f(n)\mid n\in \mathbb N\}\subset\mathbb N$. We show that the base-$q$ representation of $L$ is a context-free language if and only if $f$ is linear, answering a question of Shallit. The proof is based on a new criterion for context-freeness, which is a combination of the Interchange lemma and a generalization of the Pumping lemma.

Full text

Call a polynomial $f$ over $\mathbb{Q}$ natural if f(n)\in\mbox{\mathbb{N}} for every n\in\mbox{\mathbb{N}}. For example, $\frac{x^{2}+x}{2}$ is natural. Shallit [5, Reseach problem 3 in Section 4.11, page 138] proposed to study whether the base-$q$ representation of the range, L=\{f(n)\mid n\in\mbox{\mathbb{N}}\}, of a natural polynomial is context-free or not. It is easy to see that if $f$ is linear, i.e., its degree is at most one, then $L$ is context-free for any $q$. It was conjectured that $L$ is not context-free for any other $f$. This conjecture was known to hold only in special cases, though Sándor Horváth had an unpublished manuscript that claimed a solution.111According to Shallit, see https://cstheory.stackexchange.com/a/41864/419.. The goal of this note is to present a simple proof that uses a new lemma, which is a simple combination of two well-known necessary criteria for the context-freeness of a language.

A context-free grammar $G$ is defined as a finite $4$-tuple $G=(V,\Sigma,P,S)$, where $V$ is the set of non-terminal symbols, $\Sigma$ is the set of the terminal symbols, which we also call the letters of the alphabet (where $V\cap\Sigma=\emptyset$), $P$ is the set of production rules and $S\in V$ is the start symbol. Each production rule is of the form $A\to\alpha$ where $A\in V$ and $\alpha\in(V\cup\Sigma)^{*}$ is a string. When such a rule is applied to an occurrence of $A$ in some string $\beta$, that occurrence of the symbol $A$ is replaced with $\alpha$ in $\beta$ to obtain a new string. We say that a string $\gamma\in(V\cup\Sigma)^{*}$ can be derived from another string $\beta\in(V\cup\Sigma)^{*}$ if after applying some rules to certain occurrences of the appropriate non-terminals starting from $\beta$ we can obtain $\gamma$. The language $L(G)$ of the grammar $G$ is the set of words from $\Sigma^{*}$ that can be derived from $S$. A derivation of a word $z\in L(G)$ from $S$ can be described by a derivation tree; this is a rooted ordered tree whose non-leaf nodes are labeled with non-terminal symbols such that the root is labeled with $S$, the labels of the children of any node labeled $A$ are the right side of some rule $A\to\alpha$ in the given order, and leaves are labeled with terminal symbols that give $z$ in the given order. A grammar is in Chomsky normal form if the right side of each production rule is either two non-terminal symbols, or one terminal symbol, or the empty string; every context-free grammar has a Chomsky normal form. A language $L$ is context-free if $L=L(G)$ for some context-free grammar $G$. For other basic definitions and statements about context-free grammars and languages, we direct the reader to [5].

Now we state two lemmas that we later combine.222Note that we here we state them in a slightly weaker form as their original versions, as we do not use some parts of the original statements. The first is known as the Interchange Lemma.

The second is the following generalization of the Pumping Lemma [2].

A straight-forward combination of the proofs of Lemmas 1 and 2 gives the following.

• •

$|u_{i}|$, $|v_{i}|$, $|w_{i}|$, $|x_{i}|$, and $|y_{i}|$ are all independent of $i$,

• •

$v_{i}x_{i}$* has a distinguished position, but no excluded positions,*

• •

$u_{i_{0}}v_{i_{1}}\ldots v_{i_{m}}w_{i_{m+1}}x_{i_{m}}\ldots x_{i_{1}}y_{i_{0}}\in L$* for every sequence of indices $1\leq i_{0},i_{1},\ldots,i_{m+1}\leq k$.*

The proof of Lemma 3 can be found at the end of this note. Now we state an interesting corollary of Lemma 3 that we can apply to Shallit’s problem.

Now we are ready to prove our main result.

If the degree of $f$ is $d$, then at most $d$ numbers can take the same value, thus we can select $x_{1},\ldots,x_{N}$ from the first $dN$ numbers, which means that they have $O(\log n)$ digits (since $d$ is a constant). In this case $f(x_{i})=O((dN)^{d})$, thus each $f(x_{i})$ will also have $O(\log n)$ digits. If we pick $s$ to be some number with $n^{2}$ digits, then $f(s)$ will have $D=dn^{2}+\Theta(1)$ digits, and each $f(x_{i}+s)$ will have $D$ or $D+1$ digits, thus at least half, i.e., $N/2$ of them have the same length; these will be the words we input to Corollary 4. We still need to show that for these $f(x_{i}+s)$ their first $\Omega(n^{2})$ digits are the same and that their last $O(\log n)$ digits differ.

Let M\in\mbox{\mathbb{N}} be such that $f(x)=\sum_{i=0}^{d}\frac{\alpha_{i}}{M}x^{i}$ for \alpha_{i}\in\mbox{\mathbb{Z}}. If $s$ is a multiple of $Mq^{m}$, then the last $m$ digits of $f(x)$ and $f(x+s)$ are the same for any $x$. This way it is easy to ensure that the last $O(\log n)$ digits in base-$q$ stay different. Since $f(x+s)=\sum_{i=0}^{d}\frac{\alpha_{i}}{M}(x+s)^{i}=\frac{\alpha_{d}}{M}s^{d}+O(s^{d-1}(dN)^{d})$, the first $n^{2}-O(\log n)$ digits can take only two possible values (depending on whether there is a carry or not), thus one of these values is the same for $N/2$ of the $f(x_{i}+s)$. Thus we have shown that the condition of Corollary 4 is satisfied

If L=\{f(x)\mid x\in\mbox{\mathbb{N}}\} was context-free, then from the conclusion of Corollary 4 we would obtain infinitely many pairs of numbers, $a_{i},b_{i}\in L$, such that $|a_{i}-b_{i}|\leq 2^{B}$, but this is impossible for non-linear polynomials. ∎

We end with the omitted proof.

Call a node of the derivation tree an e-branch node if both of its children have an excluded descendant. There are exactly $e-1$ e-branch nodes in the derivation tree (if $e\geq 1$).

Call a node of the derivation tree a d-branch node if both of its children have a distinguished descendant. There are exactly $d-1$ d-branch nodes in the derivation tree. Say that a d-branch node is the d-parent of its descendant d-branch node if there are no d-branch nodes between them. With this structure the d-branch nodes form a binary tree. The $i^{\mathrm{th}}$ d-parent of a d-branch node is the $i$-times iteration of the d-parent operator.

Call a d-branch node bad if there is an e-branch node between it and its $(2t+3)^{\mathrm{th}}$ d-parent (excluding the node, but including its $(2t+3)^{\mathrm{th}}$ d-parent), or if it does not have a $(2t+3)^{\mathrm{th}}$ d-parent. Because of the binary structure of the d-branch nodes, each e-branch node can cause at most $2^{2t+3}$ d-branch nodes to be bad, and a further $2^{2t+3}$ d-branch nodes might not have a $(2t+3)^{\mathrm{th}}$ d-parent. Therefore in total there are at most $e2^{2t+3}$ bad d-branch nodes, so there is a d-branch node that is not bad if $p>2^{2t+3}$. Consider the path from the $(2t+3)^{\mathrm{th}}$ d-parent of a non-bad d-branch node to the non-bad d-branch node. Note that the nodes on this path might have an excluded descendant, but since there is no e-branching node along the path, we can conclude that there is a subpath with $t+1$ d-branch nodes on it such that no sibling of any node along the subpath has an excluded descendant. (The worst case is when the excluded descendant(s) belong to the sibling of the $(t+2)^{\mathrm{nd}}$ d-parent of a non-bad d-branch node.)

By the pigeonhole principle some non-terminal $A$ appears twice on the left side of a rule along this subpath. While we reach one node from the other, some string $\alpha A\beta$ is derived from $A$. Apply the corresponding rules from the derivation tree to $\alpha$ and $\beta$ to obtain the string $vAx$ where $v,x\in\Sigma^{*}$. Thus, $z$ can be written as $z=uvwxy$ such that $vAx$ can be derived from $A$, $w$ can be derived from $A$, the subwords $v$ and $x$ have no excluded position (since they are descendants of siblings of nodes along the path), but at least one of them has a distinguished position. For each $z\in R$ we fix such a decomposition $z=uvwxy$.

We partition $R$ into at most $t\binom{n+4}{4}$ groups depending on which non-terminal $A$ appeared on the left side of the rule, and the lengths of $u,v,w,x$ and $y$. Let $c=t\max_{n}\binom{n+4}{4}/n^{4}=5t$ (if we only care about large $n$, then $c$ would be close to $t/24$). By the pigeonhole principle one of the groups will have at least $|R|/(cn^{4})$ words in it; this will be $Z$. Since we can arbitrarily apply the rules for $A$, the conclusion follows with $p\geq\max(c,2^{2t+3}+1)$. ∎

Remark

This note started as a CSTheory.SE answer.333See my two answers for https://cstheory.stackexchange.com/questions/41863/base-k-representations-of-the-co-domain-of-a-polynomial-is-it-context-free; note that at that time I didn’t know about Lemmas 1 and 2.