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Dominos of size $2 × 1$ can be placed on a $m × n$ board so as to cover two squares exactly. Two players alternate placing dominos. The first one who is unable to place a domino is the loser.

I can show that if $n$ and $m$ are even, then player 2 can secure winning by replicating any movement from player 1 with respect to the center of the board. Moreover, I can show that if one is even and one is odd, then the roles reverse and player 1 can secure a win by blocking the center in the first move (if the board is $4×5$, he places his first domino covering (2,3),(3,3)) and then he replicates any move by player 2.

The question I face is: who wins when $m=1$? I have tried and showed that:

  1. 1 is L, 2 is W, 3 is W, 4 is W, 5 is L, 6 is W, 7 is W, 8 is W, 9 is L. Here W means a winning position: if when you are going to play there are 2 contiguous spaces on the board, you win (or 3,4,6,7,8,...).

  2. A partition of the board into two non-contiguous sets of size ($x,1$) is W if and only if $x$ is W.

  3. A partition of the board into two non-contiguous sets of size ($x,x$) is L for the second player can copy the moves of the first one.

  4. The partition ($x,2$) is W for any $x$ that is L.

I have the feeling that the losing positions are 1,5,9, and so on; in general you will lose if $n-1$ can be divided by 4. Can anybody prove it?

Asaf Karagila
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fox
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    I think the questions has been answered before here http://math.stackexchange.com/questions/615242/game-involving-tiling-a-1-by-n-board-with-1-x-2-tiles?rq=1, my apologies. I will read the answer and check if it is correct. – fox May 21 '16 at 22:57
  • I think you should change your question to odd $m,n$, and not restrict to $m=1$. Then your question will not be a duplicate and will be far more interesting! (And probably unsolvable too...) – user21820 May 22 '16 at 04:32
  • This is Nim in disguise and the Sprague-Grundy theorem applies. You "just" need to make a correspondence between contiguous strings of squares and nim-heaps, then add the heaps. Any L position (the usual is P for previous player winning) is $0$. $2$ and $3$ are $*1$ as you can move to $0$. – Ross Millikan May 22 '16 at 04:44
  • My answer (below the accepted one) to the question Josué Ortega linked two days ago analyzes this game (1-dimensional Cram) completely. I just added to my answer there a link to the community wiki post with the prerequisite background knowledge. Unfortunately the pattern of "lose if $n-1$ can be divided by $4$" doesn't continue forever, but after the first several, a pattern that repeats every $34$ does work. @RossMillikan, the Octal Periodicity theorem (or indirectly, OEIS) turns out to be key to proving the full pattern. – Mark S. May 24 '16 at 11:49

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