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.
翻译:我们考虑将汉堡前的算术(PA)扩大为以变法倍增的升格不合理数字($alpha美元),并将这一扩展称为美元(alpha$-Presburger)的计算($alpha$-PA美元)。我们表明,用美元(alpha$-PA美元)决定判决的复杂性比在PA中要大得多。事实上,当美元(alpha)是四面方美元和美元(美元)时,决定用美元交替的量化方块和最多为美元(r美元)的变数和不平等需要至少2美元(cdot ⁇ cdot ⁇ cdot ⁇ ⁇ 2 ⁇ cd ⁇ C\ ⁇ ell(S)美元(美元)的扩展幅度。当常数(美元)美元(K、C>0美元)仅取决于美元(alpha美元)和美元(S&PA美元)的变价时, 美元(美元)仅取决于最高变价(SA-PA-PA)的经常值。