Inapproximability of the Tutte polynomial of a planar graph

TL;DR

This paper investigates the computational difficulty of approximating the Tutte polynomial for planar graphs, establishing that no efficient approximation scheme exists in many cases unless NP equals RP.

Contribution

It extends the understanding of Tutte polynomial approximation complexity, showing non-existence of FPRAS in broad regions of the (x,y) plane under standard complexity assumptions.

Findings

No FPRAS for x>1, y<-1 or y>1, x<-1

No FPRAS for x<0, y<0 with q>5

No FPRAS for x<1, y<1 with q=3

Abstract

The Tutte polynomial of a graph G is a two-variable polynomial T(G;x,y) that encodes many interesting properties of the graph. We study the complexity of the following problem, for rationals x and y: given as input a planar graph G, determine T(G;x,y). Vertigan completely mapped the complexity of exactly computing the Tutte polynomial of a planar graph. He showed that the problem can be solved in polynomial time if (x,y) is on the hyperbola H_q given by (x-1)(y-1)=q for q=1 or q=2 or if (x,y) is one of the two special points (x,y)=(-1,-1) or (x,y)=(1,1). Otherwise, the problem is #P-hard. In this paper, we consider the problem of approximating T(G;x,y), in the usual sense of "fully polynomial randomised approximation scheme" or FPRAS. Roughly speaking, an FPRAS is required to produce, in polynomial time and with high probability, an answer that has small relative error. Assuming that NP is different from RP, we show that there is no FPRAS for the Tutte polynomial in a large portion of the (x,y) plane. In particular, there is no FPRAS if x>1, y<-1 or if y>1, x<-1 or if x<0, y<0 and q>5. Also, there is no FPRAS if x<1, y<1 and q=3. For q>5, our result is intriguing because it shows that there is no FPRAS at (x,y)=(1-q/(1+epsilon),-epsilon) for any positive epsilon but it leaves open the limit point epsilon=0, which corresponds to approximately counting q-colourings of a planar graph.