In fact we can do a bit better; 3^floor(K/2)*2, by using triangles instead of rhombi; Warren's degree-of-freedom argument is evaded here in a similar way. I'm pretty sure that 1, 2, 3, 6 are the best achievable for K=0, 1, 2, 3 respectively. We can do K=4 with 9 vertices but I'm not sure we can't get away with fewer. On Tue, May 10, 2016 at 8:55 PM, Warren D Smith <warren.wds@gmail.com> wrote:
Allan Wechsler is of course correct, K-regular matchstick graphs exist with 2^K vertices and graphically the same as the K-cube. (if edge-crossings allowed.)
These evade my degree of freedom counting argument because his graph drawings "really" have very few, about 2K or fewer, degrees of freedom.
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