Spot It Deck & Projective Planes

Two cards, exactly one symbol in common. Tap it. There is always exactly one — never none and never two — and that is not a design achievement but a theorem, which is what the rest of this page is about. Nothing is uploaded.

One rule fixes the entire deck

Decide only that every card carries n symbols and any two cards share exactly one. Everything else follows: there are n² − n + 1 cards, exactly as many symbols as cards, and every symbol appears on exactly n cards. This is a finite projective plane — symbols are points, cards are lines, and "two lines meet once" is the game's rule.

Every deck above is built and then every pair of cards in it is compared — 34,524 pairs in total, none sampled. The cards column and the symbols column are equal in every row, which is the striking part: a deck is its own dual. Swap the words "card" and "symbol" throughout and you have a perfectly good deck again.

A retail deck is 2 cards short, and the gap has a fingerprint

Eight symbols per card gives 57 cards. Spot It ships 55. Nothing about play breaks — removing cards cannot create a bad pair, so every two cards still share exactly one symbol. What breaks is the balance, and it breaks in precisely one way.

All 1,596 ways of removing two cards from the 57 were tried, and they give 1 distinct outcome. It could not be otherwise: the two discarded cards share exactly one symbol, so that symbol loses two appearances and the other fourteen lose one each — 16 slots gone, split 1 + 14. The same argument at any size predicts one symbol on q − 1 cards, 2q on q, and the rest untouched, which is what the test checks at four different deck sizes.

Some decks cannot be built, and one size nobody can settle

There is no seven-symbol deck. Not "none has been found" — none can exist. Bruck-Ryser says that if the order n is 1 or 2 modulo 4 then a plane needs n to be a sum of two squares; 6 is 2 mod 4 and is not a sum of two squares, so it is out. That arithmetic is checked here rather than quoted.

Note the asymmetry between the two impossible rows. Order 6 falls to a theorem anyone can check in a line of arithmetic. Order 10 does not — Bruck-Ryser permits it, and it took an exhaustive computer search in 1989 to rule it out. That result is cited here, not reproduced: this page verifies the arithmetic it can and says plainly where it is trusting someone else.

And then there is order 12. A deck with 13 symbols per card and 157 cards is not known to exist and is not known to be impossible. It is the smallest such case, and it has been open for decades.

The obvious construction is wrong, and it does not tell you

The standard construction lays symbols on a grid and reads cards off as lines: y = mx + b for every slope and intercept, plus the verticals, plus one line at infinity. It needs a field of size q — and arithmetic modulo q is a field only when q is prime. Use it at a composite size and nothing throws.

At 5 symbols per card, modulo-4 arithmetic hands you 21 cards of 5 symbols — exactly the right shape — of which 32 pairs share two symbols or none. The deck looks correct in every summary statistic and fails the only rule that matters. Building GF(4) properly, as polynomials modulo an irreducible, repairs all of it.

The same trap has a second floor: choosing a reducible polynomial for the field gives a ring with zero divisors that also looks fine. Our first attempt at GF(9) used x² = 1, which factors, and at GF(27) x³ = 2x, which also factors. Both were caught by checking that every non-zero element has an inverse before building anything on top — a check the test suite makes sure can still fail.