| Abstract: |
This paper introduces Lasso, a new family of lookup arguments, which allow an untrusted prover to commit to a vector $a \in \mathbb{F}^m$ and prove that all entries of $a$ reside in some predetermined table $t \in \mathbb{F}^n$. Lasso's performance characteristics unlock the so-called ``lookup singularity''. Lasso works with any multilinear polynomial commitment scheme, and provides the following efficiency properties.
(1) For $m$ lookups into a table of size $n$, Lasso's prover commits to just $m+n$ field elements. Moreover, the committed field elements are \emph{small}, meaning that, no matter
how big the field $\mathbb{F}$ is, they are all in the set $\{0, \dots, m\}$. When using a multiexponentiation-based commitment scheme, this
results in the prover's costs dominated by only $O(m+n)$ group \emph{operations} (e.g., elliptic curve point additions), plus the cost to prove an evaluation of a multilinear polynomial whose evaluations over the Boolean hypercube are the table entries. This represents a significant improvement in prover costs over prior lookup arguments (e.g., plookup, Halo2's lookups, lookup arguments based on logarithmic derivatives).
(2) Unlike all prior lookup arguments, if the table $t$ is structured (in a precise sense that we define), then no party needs to commit to $t$, enabling the use of much larger tables than prior works (e.g., of size $2^{128}$ or larger). Moreover, Lasso's prover only ``pays'' in runtime for table entries that are
accessed by the lookup operations. This applies to tables commonly used to implement range checks, bitwise operations, big-number arithmetic, and even transitions of a full-fledged CPU such as RISC-V. Specifically, for any integer parameter $c>1$, Lasso's prover's dominant cost is committing to $3 \cdot c \cdot m + c \cdot n^{1/c}$ field elements. Furthermore, all these field elements are ``small'', meaning they are in the set $\{0, \dots, \max\{m, n^{1/c}, q\}-1\}$, where $q$ is the maximum value in any of the sub-tables that collectively capture $t$ (in a precise manner that we define). |
