The computational power of parsing expression grammars

TL;DR

This paper explores the computational capabilities of parsing expression grammars (PEGs), introducing a new model and demonstrating PEGs' surprising power, including universality and limitations like the absence of a pumping lemma.

Contribution

It introduces the scaffolding automaton model and characterizes PEGs' computational power, revealing their unexpected universality and semantic complexity.

Findings

PEGs can define complex languages like palindromes of power-of-two length

PEGs are computationally universal for certain functions

No pumping lemma exists for PEGs

Abstract

We study the computational power of parsing expression grammars (PEGs). We begin by constructing PEGs with unexpected behaviour, and surprising new examples of languages with PEGs, including the language of palindromes whose length is a power of two, and a binary-counting language. We then propose a new computational model, the scaffolding automaton, and prove that it exactly characterises the computational power of parsing expression grammars (PEGs). Using this characterisation we show that: (*) PEGs have unexpected power and semantics. We present several PEGs with surprising behaviour, and languages which, unexpectedly, have PEGs, including a PEG for the language of palindromes whose length is a power of two. (*) PEGs are computationally `universal', in the following sense: take any computable function $f:\{0,1\}^\ast\to \{0,1\}^\ast$; then there exists a computable function $g: \{0,1\}^\ast \to \mathbb{N}$ such that $\{ f(x) \#^{g(x)} x \mid x \in \{0,1\}^\ast \}$ has a PEG. (*) There can be no pumping lemma for PEGs. There is no total computable function $A$ with the following property: for every well-formed PEG $G$, there exists $n_0$ such that for every string $x \in \mathcal{L}(G)$ of size $|x| \ge n_0$, the output $y = A(G, x)$ is in $\mathcal{L}(G)$ and has $|y| > |x|$. (*) PEGs are strongly non real-time for Turing machines. There exists a language with a PEG, such that neither it nor its reverse can be recognised by any multi-tape online Turing machine which is allowed to do only $o(n/\log n)$ steps after reading each input symbol.