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Feb 1, 2014 at 22:51 comment added dspyz I found one with nine vertices. See the updated answer
Feb 1, 2014 at 20:02 comment added G. Bach @dspyz Thank you very much for putting this much effort into this! I was quite a bit confused since just like you, I couldn't see an error in the construction. At least that bit is settled, then.
Feb 1, 2014 at 19:10 comment added dspyz I'm sorry. I constructed the graph wrong. I fixed it and clingo now reports no TFAS.
Jan 30, 2014 at 14:02 comment added G. Bach @dspyz I thought the construction was clearer than just a list of the edges, since if my reasoning isn't wrong, all that would be required to verify is whether the tournament above the construction actually has those implication properties. Thanks for pointing out the mistake about edge (3,6)! I also got 8 TFAS for that gadget, the one you listed being one of them.
Jan 30, 2014 at 14:00 history edited G. Bach CC BY-SA 3.0
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Jan 30, 2014 at 6:32 comment added dspyz I figured it out, see my updated answer: cstheory.stackexchange.com/a/20778/13643
Jan 30, 2014 at 5:20 comment added dspyz I just noticed you mentioned the edge (6,3) in the gadget graph, but the image you provided has the edge (3,6)
Jan 30, 2014 at 4:13 comment added dspyz I'm sorry, I don't follow. Is there any reason you can't just post a list of the edges so I can run it through an ASP solver and try to verify it? According to clingo, your gadget graph has 8 different TFAS's. Here's the smallest one: tfas(edge(5,0)) tfas(edge(6,0)) tfas(edge(7,0)) tfas(edge(6,2)) tfas(edge(7,3)) tfas(edge(1,2)) tfas(edge(1,3)) tfas(edge(7,5))
Jan 30, 2014 at 2:30 history edited G. Bach CC BY-SA 3.0
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Jan 30, 2014 at 2:10 history answered G. Bach CC BY-SA 3.0