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Feb 22, 2016 at 7:14 comment added Aaron Meyerowitz Maryam. I found a set with 204. If I can't improve it I'll put it up. If one could do 224 that would be the exact number. Otherwise I don't know. There are 60 points which use neither 1 nor 8 and another 120 (as I mentioned) that use both, but in differant parts. Together, 180.
Feb 19, 2016 at 13:34 comment added Maryam Aaron, Is it possible to get the exact number not a bound?
Feb 19, 2016 at 13:32 comment added Maryam Flo, how you get the number 180? I can not understand?
Feb 18, 2016 at 22:03 history edited Aaron Meyerowitz CC BY-SA 3.0
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Feb 18, 2016 at 18:15 comment added Flo Pfender starting with your set of size 120, you can add all sets made from $\{2,3,...,7\}$, bringing your count up to $120+{6\choose 3}{3\choose 2}=180$.
Feb 18, 2016 at 18:11 comment added Flo Pfender Since it is 4-regular and not $K_5$, it is 4-colorable, so it has an independent set of size at least 560/4=140.
Feb 17, 2016 at 8:47 comment added Christian Stump And only that interpretation is computationally hard. The indepentence number for the simple interpretation is 6.
Feb 17, 2016 at 1:03 comment added Brendan McKay Although a literal reading matches Gerhard and Tony's interpretation, my money is on Aaron's interpretation being what the OP intended.
Feb 16, 2016 at 19:27 comment added Tony Huynh I thought the same thing as Gerhard. The description seems to be a complicated way of saying that the vertices are the 5-sets of an 8-set, and two 5-sets are adjacent if they intersect in exactly 2 or 3 elements.
Feb 16, 2016 at 19:23 comment added Gerhard Paseman I think the vertex set is 5-sets of an 8-set. If not, union is not a good choice of notation for describing A, B, and C. Gerhard "Perhaps Original Poster Will Clarify" Paseman, 2016.02.16.
Feb 16, 2016 at 19:09 history answered Aaron Meyerowitz CC BY-SA 3.0