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Aug 23, 2017 at 17:07 history edited MTyson CC BY-SA 3.0
added 48 characters in body
Aug 23, 2017 at 17:03 comment added MTyson @ToddTrimble You're right, I guess I was. I'll make an edit. Thanks.
Aug 23, 2017 at 16:55 comment added Todd Trimble Are you assuming graphs are finite? It seems you are, since for $V = \mathbb{N}$ and $E= \{(n, n+1): n \in V\}$, the successor function $s: \mathbb{N} \to \mathbb{N}$ has no induced cycles. (Of course the fix is obvious: an edge $(x, y)$ belongs to the maximal $F$-compatible graph iff for all $f \in F$ and all $n \in \mathbb{N}$ we have $f^n(x) \neq f^n(y)$.)
Aug 22, 2017 at 10:19 vote accept Dominic van der Zypen
Aug 21, 2017 at 17:20 history answered MTyson CC BY-SA 3.0