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Oct 2, 2018 at 13:34 history edited kne CC BY-SA 4.0
Undid a previous edit which had incorrect contents.
Oct 2, 2018 at 11:55 history edited kne CC BY-SA 4.0
added 232 characters in body
Oct 2, 2018 at 11:47 comment added kne Yes, my memory of the strong homomorphism case was faulty. Now corrected.
Oct 2, 2018 at 11:41 history edited kne CC BY-SA 4.0
Fixed a mistake with respect to strong homomorphisms.
Oct 2, 2018 at 8:29 comment added Thinh D. Nguyen Anyway, I have to admit that breaking the algebraic nature of these kinds of problems seems to push us into an unknown universe.
Oct 2, 2018 at 2:13 comment added Thinh D. Nguyen If you allow the fixed graph $H$ to have loops (which are not allowed in our problem here), then instead of $3$ independent sets, you may have something like $1$ independent set and $2$ cliques, each two of them form a biclique.
Oct 2, 2018 at 2:11 comment added Thinh D. Nguyen That would be a surprise if testing for being strongly homomorphic to a graph $H$ with only $3$ vertices is $NP$ complete. Note that that is testing if $G$ can be partitioned into $3$ independent sets each two of which form a biclique.
Oct 1, 2018 at 13:38 history answered kne CC BY-SA 4.0