English (United Kingdom)

https://curated-unify.zendy.io/wp-json/zendy-region/v1/featured_content/oa?rat=en

https://curated-unify.zendy.io/wp-json/zendy-region/v1/highlighted_journal/

Global: Zendy Plus annual

Presents the access of premium content as premium feature

Premium Content

Presents the keyphrase highlighting as premium feature

Keyphrase Highlighting

Presents the summarisation as premium feature

Summarisation

Insights

Presents the pdf analysis as premium feature

PDF Analysis

Presents the zaia usage as premium feature

ZAIA

Global: Zendy Tools annual

Global: Zendy Free Plan

Global: Zendy Tools monthly

Global: Zendy Plus monthly

We consider Presburger arithmetic (PA) extended by scalar multiplication byan algebraic irrational number $\alpha$, and call this extension$\alpha$-Presburger arithmetic ($\alpha$-PA). We show that the complexity ofdeciding 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 inequalitiesrequires space at least $K 2^{\cdot^{\cdot^{\cdot^{2^{C\ell(S)$ (tower ofheight $r-3$), where the constants $c, K, C>0$ only depend on $\alpha$, and$\ell(S)$ is the length of the given $\alpha$-PA sentence $S$. Furthermoredeciding $\exists^{6}\forall^{4}\exists^{11}$ $\alpha$-PA sentences with atmost $k$ inequalities is PSPACE-hard, where $k$ is another constant dependingonly on~$\alpha$. When $\alpha$ is non-quadratic, already four alternatingquantifier blocks suffice for undecidability of $\alpha$-PA sentences.

Presburger Arithmetic with algebraic scalar multiplications