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Oct 21, 2021 at 23:29 comment added Tony Huynh Good point. I assumed the OP meant that $u_i$ is a copy of $v_i$.
Oct 21, 2021 at 23:11 comment added Gordon Royle Not quite, surely. If mapping $u_i$ to $v_i$ is an isomorphism, then it is exactly the Cartesian product with $K_2$, but otherwise it has no particular name.
Oct 21, 2021 at 11:17 vote accept W. Paul Liu
Oct 21, 2021 at 11:08 history answered Tony Huynh CC BY-SA 4.0