AI Agents & Reasoning• Published on July 7, 2026

When Does Tool Use Increase the Expressive Power of Finite-Precision Recurrent Models?

Nikola ZubićQian LiYuyi WangDavide Scaramuzza

Abstract

Modern sequence models are increasingly deployed as agents that interleave token generation with calls to external tools. We give an exact, architecture-level account of when such tool access increases computational expressivity. We model any fixed finite-precision recurrent sequence model, including finite-precision state-space models (SSMs) with $B$ bits of internal state, as a deterministic finite-state controller interacting with an oracle through a finite command/observation interface. Our results form a sharp dichotomy. First, tools that are themselves finite-state add essentially nothing: a product-state simulation internalizes any finite-state bounded-interface oracle with finite memory set $M$ at a cost of only $\log_2 |M| + O(1)$ additional bits, so the augmented system remains finite-state. Second, a single minimal infinite-state tool, namely a tape supporting only local $\mathtt{read}$, $\mathtt{write}$, and $\mathtt{move}$ commands, makes the system Turing complete: for every single-tape Turing machine with state set $Q$ and tape alphabet $Γ$, a controller with $O(\log |Q| + \log |Γ|)$ bits of internal memory simulates it, and we exhibit a concrete exponential separation: $\mathrm{EQ}_n$ requires $2^n$ states without tools but a single constant-size controller with the tape tool. Third, we show that this construction is realized exactly by a natural one-layer finite-precision selective affine SSM controller with binary one-hot hidden states, $\{0,1\}$ transition matrices, and zero biases. Selectivity is essential to the construction. In the supplementary material, we make all constants explicit, prove a logarithmic oracle-assisted universal simulation, where $O(\log B)$ recurrent bits suffice to simulate any $B$-state Turing machine, and prove a matching impossibility result.