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Nov 23, 2015 at 6:47 comment added Brendan McKay The automorphism group acts as a permutation group on the set of perfect matchings, so instead of counting individual matchings you can find the orbits of the action. However this can be hard to do in practice. If you post it as a question you should define your graphs and say what their symmetries are.
Nov 23, 2015 at 3:05 vote accept Sandeep Silwal
Nov 23, 2015 at 1:51 comment added Sandeep Silwal @BrendanMcKay: There are considerable symmetries. What can we say if there are a lot of symmetries? Is it worth posting a new topic about?
Nov 23, 2015 at 1:33 comment added Brendan McKay I'm not aware of anything faster than counting for graphs like yours (I'm assuming that snark $\implies$ cubic and not bipartite). It might be possible to exploit symmetries if there are any.
Nov 23, 2015 at 0:45 history edited Myshkin
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Nov 23, 2015 at 0:44 answer added Myshkin timeline score: 8
Nov 23, 2015 at 0:11 review First posts
Nov 23, 2015 at 0:34
Nov 23, 2015 at 0:09 history asked Sandeep Silwal CC BY-SA 3.0