The Fine-Grained Complexity of Computing the Tutte Polynomial of a Linear Matroid

preprint2020arXivOpen access

Abstract

We show that computing the Tutte polynomial of a linear matroid of dimension kk on kO(1)k^{O(1)} points over a field of kO(1)k^{O(1)} elements requires kΩ(k)k^{Ω(k)} time unless the \#ETH---a counting extension of the Exponential Time Hypothesis of Impagliazzo and Paturi [CCC 1999] due to Dell {\em et al.} [ACM TALG 2014]---is false. This holds also for linear matroids that admit a representation where every point is associated to a vector with at most two nonzero coordinates. We also show that the same is true for computing the Tutte polynomial of a binary matroid of dimension kk on kO(1)k^{O(1)} points with at most three nonzero coordinates in each point's vector. This is in sharp contrast to computing the Tutte polynomial of a kk-vertex graph (that is, the Tutte polynomial of a {\em graphic} matroid of dimension kk---which is representable in dimension kk over the binary field so that every vector has two nonzero coordinates), which is known to be computable in 2kkO(1)2^k k^{O(1)} time [Björklund {\em et al.}, FOCS 2008]. Our lower-bound proofs proceed via (i) a connection due to Crapo and Rota [1970] between the number of tuples of codewords of full support and the Tutte polynomial of the matroid associated with the code; (ii) an earlier-established \#ETH-hardness of counting the solutions to a bipartite (d,2)(d,2)-CSP on nn vertices in do(n)d^{o(n)} time; and (iii) new embeddings of such CSP instances as questions about codewords of full support in a linear code. We complement these lower bounds with two algorithm designs. The first design computes the Tutte polynomial of a linear matroid of dimension~kk on kO(1)k^{O(1)} points in kO(k)k^{O(k)} operations. The second design generalizes the Björklund~{\em et al.} algorithm and runs in qk+1kO(1)q^{k+1}k^{O(1)} time for linear matroids of dimension kk defined over the qq-element field by kO(1)k^{O(1)} points with at most two nonzero coordinates each.

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