In a Merlin–Arthur proof system, the proof verifier (Arthur) accepts valid proofs (from Merlin) with probability 1, and rejects invalid proofs with probability arbitrarily close to 1. The running time of such a system is defined to be the length of Merlin’s proof plus the running time of Arthur. We provide new Merlin–Arthur proof systems for some key problems in fine-grained complexity. In several cases our proof systems have optimal running time. Our main results include:Certifying that a list of n integers has no 3-SUM solution can be done in Merlin–Arthur time O~(n)\documentclass[12pt]{minimal}
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\begin{document}$$\tilde{O}(n)$$\end{document}. Previously, Carmosino et al. [ITCS 2016] showed that the problem has a nondeterministic algorithm running in O~(n1.5)\documentclass[12pt]{minimal}
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\begin{document}$$\tilde{O}(n^{1.5})$$\end{document} time (that is, there is a proof system with proofs of length O~(n1.5)\documentclass[12pt]{minimal}
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\begin{document}$$\tilde{O}(n^{1.5})$$\end{document} and a deterministic verifier running in O~(n1.5)\documentclass[12pt]{minimal}
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\begin{document}$$\tilde{O}(n^{1.5})$$\end{document} time).Counting the number of k-cliques with total edge weight equal to zero in an n-node graph can be done in Merlin–Arthur time O~(n⌈k/2⌉)\documentclass[12pt]{minimal}
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\begin{document}$${\tilde{O}}(n^{\lceil k/2\rceil })$$\end{document} (where k≥3\documentclass[12pt]{minimal}
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\begin{document}$$k\ge 3$$\end{document}). For odd k, this bound can be further improved for sparse graphs: for example, counting the number of zero-weight triangles in an m-edge graph can be done in Merlin–Arthur time O~(m)\documentclass[12pt]{minimal}
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\begin{document}$${\tilde{O}}(m)$$\end{document}. Previous Merlin–Arthur protocols by Williams [CCC’16] and Björklund and Kaski [PODC’16] could only count k-cliques in unweighted graphs, and had worse running times for small k.Computing the All-Pairs Shortest Distances matrix for an n-node graph can be done in Merlin–Arthur time O~(n2)\documentclass[12pt]{minimal}
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\begin{document}$$\tilde{O}(n^2)$$\end{document}. Note this is optimal, as the matrix can have Ω(n2)\documentclass[12pt]{minimal}
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\begin{document}$$\Omega (n^2)$$\end{document} nonzero entries in general. Previously, Carmosino et al. [ITCS 2016] showed that this problem has an O~(n2.94)\documentclass[12pt]{minimal}
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\begin{document}$$\tilde{O}(n^{2.94})$$\end{document} nondeterministic time algorithm.Certifying that an n-variable k-CNF is unsatisfiable can be done in Merlin–Arthur time 2n/2-n/O(k)\documentclass[12pt]{minimal}
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\begin{document}$$2^{n/2 - n/O(k)}$$\end{document}. We also observe an algebrization barrier for the previous 2n/2·poly(n)\documentclass[12pt]{minimal}
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\begin{document}$$2^{n/2}\cdot \textrm{poly}(n)$$\end{document}-time Merlin–Arthur protocol of R. Williams [CCC’16] for #\documentclass[12pt]{minimal}
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\begin{document}$$\#$$\end{document}SAT: in particular, his protocol algebrizes, and we observe there is no algebrizing protocol for k-UNSAT running in 2n/2/nω(1)\documentclass[12pt]{minimal}
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\begin{document}$$2^{n/2}/n^{\omega (1)}$$\end{document} time. Therefore we have to exploit non-algebrizing properties to obtain our new protocol.Certifying a Quantified Boolean Formula is true can be done in Merlin–Arthur time 24n/5·poly(n)\documentclass[12pt]{minimal}
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\begin{document}$$2^{4n/5}\cdot \textrm{poly}(n)$$\end{document}. Previously, the only nontrivial result known along these lines was an Arthur–Merlin–Arthur protocol (where Merlin’s proof depends on some of Arthur’s coins) running in 22n/3·poly(n)\documentclass[12pt]{minimal}
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\begin{document}$$2^{2n/3}\cdot \textrm{poly}(n)$$\end{document} time. Due to the centrality of these problems in fine-grained complexity, our results have consequences for many other problems of interest. For example, our work implies that certifying there is no Subset Sum solution to n integers can be done in Merlin–Arthur time 2n/3·poly(n)\documentclass[12pt]{minimal}
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\begin{document}$$2^{n/3}\cdot \textrm{poly}(n)$$\end{document}, improving on the previous best protocol by Nederlof [IPL 2017] which took 20.49991n·poly(n)\documentclass[12pt]{minimal}
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\begin{document}$$2^{0.49991n}\cdot \textrm{poly}(n)$$\end{document} time.
