2010
DOI: 10.1007/978-3-642-12200-2_52
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Counting Hexagonal Patches and Independent Sets in Circle Graphs

Abstract: A hexagonal patch is a plane graph in which inner faces have length 6, inner vertices have degree 3, and boundary vertices have degree 2 or 3. We consider the following counting problem: given a sequence of twos and threes, how many hexagonal patches exist with this degree sequence along the outer face? This problem is motivated by the enumeration of benzenoid hydrocarbons and fullerenes in computational chemistry. We give the first polynomial time algorithm for this problem. We show that it can be reduced to … Show more

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Cited by 3 publications
(4 citation statements)
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References 26 publications
(48 reference statements)
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“…Solving the subproblem The problem of finding variables with negative reduced costs is almost equivalent to finding a maximum weight independent set in a circle graph, which can be solved in polynomial time by dynamic programming [1,12]. A circle graph is an intersection graph of chords of a a circle, two vertices/chords are adjacent if and only if they intersect.…”
Section: Formulation As An Ilpmentioning
confidence: 99%
“…Solving the subproblem The problem of finding variables with negative reduced costs is almost equivalent to finding a maximum weight independent set in a circle graph, which can be solved in polynomial time by dynamic programming [1,12]. A circle graph is an intersection graph of chords of a a circle, two vertices/chords are adjacent if and only if they intersect.…”
Section: Formulation As An Ilpmentioning
confidence: 99%
“…Our algorithm has complexity n O( f 5 ) . An algorithm with complexity f ( f 5 )n O (1) for some computable function f (a fixed parameter tractable (FPT) algorithm) would be preferable, but we do not know whether such an algorithm is possible.…”
Section: Question 4 Is Fullerene Patch -Boundary Code Np-hard?mentioning
confidence: 99%
“…Their construction can be extended to show that exponentially many solutions are possible. Nevertheless, we have shown in [1] that in this case counting can be done in polynomial time:…”
Section: Introductionmentioning
confidence: 99%
“…Without condition (b) this problem would be equivalent to finding a maximum weight independent set in a circle graph, which can be solved in polynomial time by dynamic programming[10,75].A circle graph is an intersection graph of chords of a circle, and two vertices/chords are adjacent if and only if they intersect. Circle graphs can equivalently be defined as the overlap graph of a set of intervals.…”
mentioning
confidence: 99%