1993
DOI: 10.1017/cbo9780511752506
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Complexity: Knots, Colourings and Countings

Abstract: These notes are based on a series of lectures given at the Advanced Research Institute of Discrete Applied Mathematics held at Rutgers University. Their aim is to link together algorithmic problems arising in knot theory, statistical physics and classical combinatorics. Apart from the theory of computational complexity concerned with enumeration problems, introductions are given to several of the topics treated, such as combinatorial knot theory, randomised approximation algorithms, percolation and random clus… Show more

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Cited by 295 publications
(319 citation statements)
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“…The Tutte polynomial is known as a basic invariant of a graph structure. For a given graph G = (V, E), the Tutte polynomial is defined as [41]:…”
Section: Frontier-based Methods For Various Graph Problemsmentioning
confidence: 99%
“…The Tutte polynomial is known as a basic invariant of a graph structure. For a given graph G = (V, E), the Tutte polynomial is defined as [41]:…”
Section: Frontier-based Methods For Various Graph Problemsmentioning
confidence: 99%
“…[27] it was further shown that there is an equivalence between classical partition functions with real couplings and quantum amplitudes for a certain certain class of quantum circuits known as Clifford circuits. When this mapping exists the graph underling the classical theory is planar with no magnetic fields and can be estimated with a polynomial time classical algorithm [28]. Also, deciding if a certain quantum circuit belongs to this equivalence class is classically easy.…”
Section: Computational Power Of Classical Modelsmentioning
confidence: 99%
“…yFðG À eÞ if e is a loop; aFðG=eÞ þ bFðG À eÞ if e is an ordinary edge: The Tutte polynomial is an example of Tutte-Gröthendieck invariant for a ¼ b ¼ c ¼ 1 [26] and so, TðG; x; yÞ…”
Section: The Tutte Polynomialmentioning
confidence: 99%
“…Also, evaluating TðG; x; yÞ for specific points ðx; yÞ is NP-hard except for the points on the hyperbola ðx À 1Þðy À 1Þ¼ 1 or when ðx; yÞ equals ð1; 1Þ; ðÀ1; À1Þ; ð0; À1Þ; ðÀ1; 0Þ; ði; ÀiÞ; ðÀi; iÞ; ðj; j 2 Þ; ðj 2 ; jÞ where j ¼ e ð2pi=3Þ for which it can be done in polynomial time [16,26]. Vertigan [29] has extended this by showing that a similar result holds for planar graphs except that, in this case, for the additional points lying on the hyperbola ðx À 1Þðy À 1Þ¼ 2, the problem can be solved again in polynomial time.…”
Section: Introductionmentioning
confidence: 99%