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Jul 21, 2015 at 14:02 vote accept Dominic van der Zypen
Jul 21, 2015 at 13:41 answer added Will Brian timeline score: 3
Jul 21, 2015 at 13:02 comment added Ben Barber More generally, given examples on $n_1$ vertices and $n_2$ vertices you can find an example on $n_1 + n_2$ vertices by taking a vertex-disjoint union. So it remains only to check whether such graphs exist for small odd $n$.
Jul 21, 2015 at 11:21 answer added Ben Barber timeline score: 4
Jul 21, 2015 at 10:59 comment added Gordon Royle I think we can do all even numbers $2k \geq 4$ by taking $k$ disjoint edges. The partition that takes one vertex from each edge is nasty as each vertex has 0 neighbours in its own part, and 1 in the other. But the partition that takes some edges in one part and the remaining edges in the other part is nice, because each vertex has 1 neighbour in its own part and 0 in the other.
Jul 21, 2015 at 9:20 history asked Dominic van der Zypen CC BY-SA 3.0