Sachs graphs

Fig- 18 illustrates another interesting case where the natural logarithms of the total number of acyclic Sachs graphs [63] are plotted against connectivity indices up to sixth order for a series of benzenoid hydrocarbons-... 

In our case we can distinguish in the given graph, for example, 5 Sachs graphs shown in Fig. 14. 

This graph is a Sachs graph and it cannot be splitted into other Sachs graphs. 

Clearly, Si is a void set since there are no Sachs graphs with one vertex, and therefore ai = 0. Note that Si is always void in accordance with the definition of the Sachs graph and = 0 all the time. v... 

This graph corresponds to the following Sachs graphs with indicated indices c (s) and r (s) ... 

Corollary five the number of Sachs graphs of the type is equal to the number v of the edges in the... 

Indeed, since a bipartite graph has no subgraphs with odd-membered cycles, the set of the Sachs graphs iS , where n is odd, is void (iS = 0) and a = 0. 

Sachs graph (= basic graph, mutation graph, characteristic graph)... 

Here Sn is the set of all Sachs graphs of the graph G with n vertices and, hence, the summation goes over all Sachs graphs with n vertices. In the case when the set Sn is an empty set (0), there is no Sachs graph with n vertices, and hence an = 0. 

A cycle with N vertices has three Sachs graphs with N vertices, i. e. ... 

Proof op 4.5.2.1 and 4.4.1,9. Suppose G is an acyclic graph and ply Sachs theorem 4.4.3.2. Since G contains no odd cycles we may immediately conclude that G contains no Sachs graphs with an odd number of vertices. Therefore... 

If is even G may contain Sachs gxhpha having k vertices. All these Sachs gr >hs must be composed kf2 components (since G is assumed to be acyclic). In other words p(s) = kj2 and c(s) = 0 for all the Sachs grs hs of G. Every such Sachs graph corresponds in an obvious manner to a selection of kj2 independent edges in G. Therefore... 

The idea partial ne ect of cyclic Sachs graphs in Sachs formula (4.4.S.2) led to the discovery of a polynomial whidi generalizes both Ch(x) and Ma(x). Since the idea was bom in the Institut fur Strahienchemie in Mdlheim (F.R.G.) during a discussion between I. Gutman and O.E. Polansky, the name Molheim pdynomial seems appropriate. 

J. R. Dias, An example molecular orbital calculation using Sachs graph method, J. Chem. Educ. 69 (1992) 695-700. 

Consider applying eq 17 to the case of benzene. Benzene contains no Sachs graphs with an odd number of vertexes so ai = 0 for all odd 1. To get az, construct all Sachs graphs with 2 atoms. These are simply the bonds 1—6. Each has 1 component and 0 rings. There are six of these therefore... 

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