Team selection tests for GMO 2018

Let us define the distance between any two cells of the board to be the minimum number of steps required to get from one of the cells to the other one provided that one moves between adjacent cells in each step. Consequently, the distance of any cell to itself is $0$, the distance between adjacent cells is $1$ and the largest distance on the $13\times 13$ board (that between two opposite corners) is $24$. It is easy to observe that the numbers written on any two cells cannot differ by more than the distance between the cells.

Let us now assign coordinates to the cells of the board in the usual way: each cell is denoted $(i,j)$ where $i,j\in \{0,1,\dots ,12\}$. Let us also denote the number on the cell $(i,j)$ by $x_{(i,j)}$. Thus the inequality mentioned above is expressed as $$\left|x_{\left(i_1,j_1\right)}-x_{\left(i_2,j_2\right)}\right|\le \left|i_1-i_2\right|+\left|j_1-j_2\right|.$$ Now there must be a distance of at least $22$ between a $2$ and a $24$ and this places a strong restriction on the possible locations of $2$s and $24$s. Namely, if $x_{(i,j)}\in \{2,24\}$ then we have $$\min (i,12-i)+\min (j,12-j)\le 2.$$ This roughly means that $2$s and $24$s can be found only very near the corners. Without loss of generality (by symmetry of the board), let us assume that there exists a cell $(i,j)$ with $x_{(i,j)}=2$ and $i+j\le 2$. Now, $x_{(i,j)}=24$ implies that $i+j\ge 22$, that is, both of the $24$s must be near the opposite corner. Hence $x_{(i,j)}=2$ thus $i+j\le 2$, that is, both of the $2$s must be near the same corner.

If there exists a cell $(i,j)$ with $x_{(i,j)}=2$ and $i+j=2$, then $x_{(i,j)}=24$ then $i+j\ge 24$, so $i=j=12$. Thus there cannot be two distinct $24$s on the board. We therefore conclude that $$x_{(i,j)}=2\Rightarrow i+j\le 1\Rightarrow (i,j)\in \{(0,0),(1,0),(0,1)\}.$$ Similarly, $$x_{(i,j)}=24\Rightarrow i+j\ge 23\Rightarrow (i,j)\in \{(12,12),(11,12),(12,11)\}.$$ If $x_{(0,0)}=2$, then $x_{(1,0)}\ne 2$ and $x_{(0,1)}\ne 2$, thus there cannot be two $2$s on the board. We therefore conclude that $$x_{(0,0)}\ne 2,x_{(1,0)}=x_{(0,1)}=2$$ and similarly $$x_{(12,12)}\ne 24,x_{(11,12)}=x_{(12,11)}=24.$$ Now, if we consider a path (a sequence of adjacent cells) of length $22$ between the cells $(1,0)$ and $(11,12)$ or between the cells $(0,1)$ and $(12,11)$, we see that the numbers on the cells on this path are uniquely determined. Since any cell of the board except $(0,0)$ and $(12,12)$ can be placed on such a path, almost the whole board is uniquely determined. We obtain the following formula: $$1\le i+j\le 23\Rightarrow x_{(i,j)}=i+j+1$$ Furthermore, $x_{(0,0)}\in \{1,3\}$ and $x_{(12,12)}\in \{23,25\}$, therefore $x_{(i,j)}=13\iff i+j=12$. Hence there are exactly thirteen $13$s on the board.