Well, you have to do some refractory period math, but it's doable. The bigger problem is trying to partition the graph so everyone is willing to sleep with everyone in their partition, and the partitions are evenly sized. I'm willing to bet this is an NP-hard problem.
Reminds me of Stable Matching[1] but the groups aren't the same size. I guess if we can relax the "opposite sex" requirement then we reduce the problem to a Stable Roommates Problem[2] ;)
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Well, you have to do some refractory period math, but it's doable. The bigger problem is trying to partition the graph so everyone is willing to sleep with everyone in their partition, and the partitions are evenly sized. I'm willing to bet this is an NP-hard problem.
Reminds me of Stable Matching[1] but the groups aren't the same size. I guess if we can relax the "opposite sex" requirement then we reduce the problem to a Stable Roommates Problem[2] ;)
[1] http://en.wikipedia.org/wiki/Stable_matching [2] http://en.wikipedia.org/wiki/Stable_roommates_problem