Presburger Arithmetic with algebraic scalar multiplications

TL;DR

This paper explores the increased computational complexity of Presburger arithmetic extended with algebraic irrational scalar multiplication, revealing significant hardness and undecidability results depending on the nature of the algebraic number.

Contribution

It introduces $ ext{α}$-Presburger arithmetic, analyzing its decision complexity and proving hardness and undecidability results based on the algebraic properties of $ ext{α}$.

Findings

Deciding $ ext{α}$-PA with quadratic $ ext{α}$ and multiple quantifiers requires extremely high space complexity.

Existence of PSPACE-hardness for certain $ ext{α}$-PA sentences with fixed quantifier structure.

Undecidability arises with four quantifier blocks when $ ext{α}$ is non-quadratic.

Abstract

We consider Presburger arithmetic (PA) extended by scalar multiplication by an algebraic irrational number $\alpha$, and call this extension $\alpha$-Presburger arithmetic ($\alpha$-PA). We show that the complexity of deciding sentences in $\alpha$-PA is substantially harder than in PA. Indeed, when $\alpha$ is quadratic and $r\geq 4$, deciding $\alpha$-PA sentences with $r$ alternating quantifier blocks and at most $c\ r$ variables and inequalities requires space at least $K 2^{\cdot^{\cdot^{\cdot^{2^{C\ell(S)}}}}}$ (tower of height $r-3$), where the constants $c, K, C>0$ only depend on $\alpha$, and $\ell(S)$ is the length of the given $\alpha$-PA sentence $S$. Furthermore deciding $\exists^{6}\forall^{4}\exists^{11}$ $\alpha$-PA sentences with at most $k$ inequalities is PSPACE-hard, where $k$ is another constant depending only on~$\alpha$. When $\alpha$ is non-quadratic, already four alternating quantifier blocks suffice for undecidability of $\alpha$-PA sentences.