Ed, surely you know this already, but: Spot It (the original one, not Jr) has 57 symbols, 8 per card, and the arrangement is the projective plane over the field of 7 elements (with 57 = 7^2 + 7^1 + 7^0). The Jr version is the projective plane over F_5. So there's much more structure here than cyclic difference sets. The thing that makes me gnash my teeth is that Spot It only contains 55 of the 57 cards!! Because they had a set-up that printed 60 cards per sheet, but they wanted to reserve 5 of them for instructions on five different games you could play with the deck! Arrgh -- it's like selling a deck of cards for War and omitting the 3 of diamonds, because War works just the same without it. --Michael On Wed, May 15, 2019 at 2:39 PM Ed Pegg Jr <ed@mathpuzzle.com> wrote:
The game Spot It Jr https://smile.amazon.com/SpotIt/dp/B0075XNGGK/ has 31 cards each with 6 of 31 animals. Every pair of cards shares 1 animal. This is a clever usage of the {31,6,2} cyclic difference set. Here are the first few difference sets.
{{{7, 3}, {1, 2, 4}}, {{13, 4}, {0, 1, 3, 9}}, {{21, 5}, {3, 6, 7, 12, 14}}, {{31, 6}, {1, 5, 11, 24, 25, 27}}, {{57, 8}, {0, 1, 6, 15, 22, 26, 45, 55}}, {{73, 9}, {0, 1, 12, 20, 26, 30, 33, 35, 57}}, {{91, 10}, {0, 2, 6, 7, 18, 21, 31, 54, 63, 71}}, {{133, 12}, {1, 10,11, 13, 27, 31, 68, 75, 83, 110, 115, 121}}, {{183, 14}, {1, 13, 20, 21, 23, 44, 61, 72, 77, 86, 90, 116, 122, 169}}
What if there are 7 symbols per card? The best possible is 40. And a solution for that was found by Alessandro Jurcovich. C(40,7,2) = 40 https://ljcr.dmgordon.org/cover/show_cover.php?v=40&k=7&t=2
What about 11, 13, 15, 16? I don't know, and the La Jolla Covering repository doesn't help. What are the lower bounds for these? Have solutions been found?
I think this would be a good sequence for OEIS.
--Ed Pegg Jr _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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