# Is it possible to tile a $13 \times 13$ board with $4 \times 1$ dominoes such that the center square is left untiled?

Problem Is it possible to tile a $$13 \times 13$$ board with $$4 \times 1$$ dominoes such that the center square is left untiled?

I was not able to find a tiling so I am trying to prove that it is no possible.

I tried the usual way of coloring the board with $$4$$ colors using a chessboard style alternating coloring. Lets say the colors are $$1, 2, 3, 4$$ then I find that we have $$43$$ $$1$$’s, $$42$$ $$2$$‘s, $$42$$ $$3$$‘s, $$42$$ $$4$$’s and the center ($$7^{\text{th}}$$ row and $$7^{\text{th}}$$ column) cell has color $$1$$. But then this meets the demands of the $$4 \times 1$$ dominoes, so there is no contradiction.

Any hint will be helpful. Do I need to do a different kind fo coloring?

• I would like to understand your coloring convention, but I don't (it's even worse when I read the answer given!) Could you make it more explicit ? – Jean Marie Sep 9 '20 at 7:30
• My first row is 1, 2, 3, 4, 1, 2, 4..... – Subham Jaiswal Sep 9 '20 at 9:21
• @Jean my second row is a cyclic left shift of row 1, tht is 2, 3, 4, 1, 2, 3.. n so on for 3rd row n so on – Subham Jaiswal Sep 9 '20 at 9:23
• @JeanMarie I have drawn out the one in the answer. – tkf Sep 9 '20 at 12:51

Any $$4\times 4$$ Latin square gives a colouring by repeating across the board, starting from the top left corner. You just need a Latin square where the $$(1,1)$$ entry is different to the $$(3,3)$$ entry.
For example $$\begin{array}{c} 1234\\2143\\3421\\4312 \end{array}$$
which gives: $$\small\begin{array}{|c|c|c|c|} \hline 1234&1234&1234&1 \\ 2143&2143&2143&2 \\ 3421&3421&3421&3 \\ 4312&4312&4312&4 \\ \hline 1234&1234&1234&1 \\ 2143&2143&2143&2 \\ 3421&34{\tiny\fbox{2}}1&3421&3 \\ 4312&4312&4312&4 \\ \hline 1234&1234&1234&1 \\ 2143&2143&2143&2 \\ 3421&3421&3421&3 \\ 4312&4312&4312&4 \\ \hline 1234&1234&1234&1\\ \hline \end{array}$$
Any $$4\times 1$$ (or $$1\times 4)$$ tetromino will cover one square of each colour, so a collection of tetrominos will cover the same number of squares of each colour. Thus they cannot cover all but the center square (boxed), as you have an extra $$1$$ (e.g. bottom right square of board) and one less $$2$$ (center square).
• A $1 \times 4$ tile is a tetromino, not a domino. Otherwise, a good solution. – Ross Millikan Sep 9 '20 at 15:12