Graphical Enumeration 1973
DOI: 10.1016/b978-0-12-324245-7.50005-8
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Cited by 358 publications
(484 citation statements)
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“…There are as yet no formulas, direct or recursive, by means of which the number for a given E, or a given V or F, or a given combination of these elements, can be calculated [18], [25]. Enumeration of any of these classes must proceed by first constructing the members and then counting them one by one.…”
mentioning
confidence: 99%
“…There are as yet no formulas, direct or recursive, by means of which the number for a given E, or a given V or F, or a given combination of these elements, can be calculated [18], [25]. Enumeration of any of these classes must proceed by first constructing the members and then counting them one by one.…”
mentioning
confidence: 99%
“…To estimate the number of ways to construct a given structure S, denoted as N (S), we need to consider the automorphisms of a graph. An automorphism of a graph G is an adjacency preserving permutation of vertices of G. The collection Aut(G) of all automorphisms of G is called the automorphism group of G. In the sequel, Aut(S) of a structure S denotes Aut(G) for some labeled graph G such that G ∼ = S. In group theory, it is well known that N (S) = n!/|Aut(S)| [11]. With these definitions, we proved in [4] that…”
Section: Structural Entropymentioning
confidence: 99%
“…The total number of inequivalent orbits for a given weight X n 1 Y n 2 can be determined using the Pólya-Redfield theorem, by reading out the coefficient of X n 1 Y n 2 from a so-called Cycle Index computed using appropriate figure inventories and the cyclestructure of the permutation representation of C L on ∆. 18 However, such approach will not help us here to determine the length of each orbit, which is the key piece of information to determine the Hückel annulene lengths, thus their energies. The problem can instead be solved by counting the orbits of weight X n 1 Y n 2 for each allowed symmetry describing the orbits of necklace configurations in F M S .…”
Section: Solution Of the Necklace Enumeration Problemmentioning
confidence: 99%