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Tad, do you imply that a solution is known for k = 1? Isn't this NP? --Moses

2011/2/23  <[log in to unmask]>:
> *****  To join INSNA, visit http://www.insna.org  *****
>
> Dear Socnet Members
>
> Let me address a  question to those of you who have been watching
> developments  in GRAPH THEORY.
>
> Is the following statement true?
>
> 'No nontrivial complete set collection of invariants has yet been found
> for directed graphs
>
> Definitions
>
> A (finite) DIRECTED GRAPH is a mathematical object of the form G=(X,R),
> where X is a finite set (of points) and R is  a binary relation in X.
>
> Two digraphs G=(X,R) and G =(X ,R ) are ISOMORPHIC if there exists a 1-1
> mapping F of X onto X  such that xRy if and only if  F(x)R F(y), for any
> x,y in X.
>
> A STRUCTURAL PARAMETER of a digraph is any mapping P which assigns a
> numerical value to any digraph in such a way that  P(G)=P(G ) if G and G
> are isomorphic.
>
> A COMPLETE SET OF INVARIANTS is any finite collection P1, ,Pk of
> structural parameters such that, for any two digraphs G and G , the
> equations  P1(G)=P1(G ), ,Pk(G)=Pk(G ) imply that G and G  are isomorphic.
>
> For any n, let  gn denote the number of equivalence classes of  the
> isomorphism relation in  the set of digraphs with X={1, ,n}. The numbers
> 1, ,gn can be used as  code symbols of equivalence classes. A 'trivial'
> set of invariants is obtained by assigning to any G=(X,R) the number n of
> points (n=|X| ) and an appropriate symbol from the range from 1 to gn.
> Hence, by saying that a set of invariants is  NONTRIVIAL I mean that  the
> values of P1, ,Pk for two n-point digraphs G and G   can be computed
> without prior knowledge of the list of isomorphism classes.
>
> Tad Sozanski
> www.cyf-kr.edu.pl/~ussozans/
>
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