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Let $V = (v_1,...,v_n)$ be a set of vertices for a directed graph. We denote an edge between the two vertices $v_i$ and $v_j$ by $(i,j)$. The edge set $E$ for a graph $(V,E)$ is then a set of tuples $E \in P(\{(i,j) \mid i,j \in \{1,...,n\}\})$.

My Question is if for fixed $n \in \mathbb{N}$ we can efficiently define a map $$ F_n : P(\{(i,j) \mid i,j \in \{1,...,n\}\}) \rightarrow \mathbb{R} $$ such that $F_n(E)=F_n(E')$ if and only if the two Graphs are the same modulo permutation of vertices, i.e. if there exists a permutation $\sigma \in S_n$ such that $E = \{(\sigma(i),\sigma(j)) \mid (i,j) \in E'\}$.

Such a map should be relatively simple to construct with sums over all permutations. I am interested in simpler known invariants, which would ideally require less than $\mathcal{O}(n!)$ amount of work to calculate.

I'm new to graph-theory and thought someone must have already thought about this, since it is fundamental to classifying graphs. Any references or pointers in the right direction are much appreciated.

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  • $\begingroup$ Aren't these exactly (real valued) graph invariants? That is, anything from graph diameter and girth, over chromatic number and independence number, to properties like planarity, connectedness etc? $\endgroup$
    – M. Winter
    Commented Feb 19, 2022 at 12:01

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You are essentially asking about the graph isomorphism problem. Or, more precisely, you are asking about canonical representation of graphs. There is a huge literature about this. See https://en.wikipedia.org/wiki/Graph_canonization.

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  • $\begingroup$ I would argue he is just asking for graph invariants. After all, $F$ could map the edge set into, say, the chromatic number or so. $\endgroup$
    – M. Winter
    Commented Feb 19, 2022 at 12:03
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    $\begingroup$ Thanks! I must have googled precisely the wrong words to not have found this. It's exactly what I was asking about. $\endgroup$ Commented Feb 19, 2022 at 12:05
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    $\begingroup$ @M.Winter OP used the words "if and only if", so $F$ should be a complete graph invariant, which is basically the same as canonization. $\endgroup$ Commented Feb 19, 2022 at 12:20
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    $\begingroup$ @Sean You are right. I missed this. $\endgroup$
    – M. Winter
    Commented Feb 19, 2022 at 12:20

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