Placing numbers $1,2,...,7$ on an infinite square grid with constraint on each $3 \times 3$ block

Suppose there is an infinite tile space (like infinitely large chess board). On each tile, I'm going to write a number among $1,2,...,7$. The question is

Is there a possible numbering on each tile on infinite tile space so that every time I look $3\times 3$ block, all numbers $1$ to $7$ appear on the boundary $8$ tiles ?

My conjecture is No since I tried some examples but all failed. But I don't know how to prove. Could you help?

Solution 1:

It is possible.

We'll number the tiles with $(i, j)$ with $i, j \in \mathbb{Z}$. Suppose we fill the tile $(i, j)$ with number $i + j + 2\lfloor j/2 \rfloor$ for every integer $i$ and $j$.

Claim. For any $3 \times 3$ block of tiles centred in tile $(i, j)$, the eight numbers in the boundary consist of seven consecutive integers (and a duplicate).

Proof. We consider two cases.

• If $j = 2k+1$ for some $k \in \mathbb{Z}$, then the boundary tiles are

  • $(i-1, 2k)$, with number $(i-1) + 2k + 2k = i + 4k - 1$,
  • $(i-1, 2k+1)$, with number $(i-1) + (2k+1) + 2k = i + 4k$,
  • $(i-1, 2k+2)$, with number $(i-1) + (2k+2) + (2k+2) = i + 4k + 3$,
  • $(i, 2k)$, with number $i + 2k + 2k = i + 4k$,
  • $(i, 2k+2)$, with number $i + (2k+2) + (2k+2) = i + 4k + 4$,
  • $(i+1, 2k)$, with number $(i+1) + 2k + 2k = i + 4k + 1$,
  • $(i+1, 2k+1)$, with number $(i+1) + (2k+1) + 2k = i + 4k + 2$,
  • $(i+1, 2k+2)$, with number $(i+1) + (2k+2) + (2k+2) = i + 4k + 5$.

  So, the numbers in the boundaries are the seven integers from $i + 4k - 1$ to $i + 4k + 5$ (with one duplicate $i + 4k$).

• If $j = 2k$ for some $k \in \mathbb{Z}$, then the boundary tiles are

  • $(i-1, 2k-1)$, with number $(i-1) + (2k-1) + (2k-2) = i + 4k - 4$,
  • $(i-1, 2k)$, with number $(i-1) + 2k + 2k = i + 4k - 1$,
  • $(i-1, 2k+1)$, with number $(i-1) + (2k+1) + 2k = i + 4k$,
  • $(i, 2k-1)$, with number $i + (2k-1) + (2k-2) = i + 4k - 3$,
  • $(i, 2k+1)$, with number $i + (2k+1) + 2k = i + 4k + 1$,
  • $(i+1, 2k-1)$, with number $(i+1) + (2k-1) + (2k-2) = i + 4k - 2$,
  • $(i+1, 2k)$, with number $(i+1) + 2k + 2k = i + 4k + 1$,
  • $(i+1, 2k+1)$, with number $(i+1) + (2k+1) + 2k = i + 4k + 2$.

  Here, the numbers in the boundaries are the seven integers from $i + 4k - 4$ to $i + 4k + 2$ (with one duplicate $i + 4k + 1$).

In either case, we know that the boundary tiles contains $7$ consecutive integers. $\square$

To complete the proof, we simply consider modulo $7$. It should be easy to finish the proof from here.

Here is a portion of a desired construction.

Edit: Added border lines to illustrate how the pattern repeats. \begin{array}{|cccccccccccccc|cccccccccccccc|} \hline 1&2&5&6&2&3&6&7&3&4&7&1&4&5&1&2&5&6&2&3&6&7&3&4&7&1&4&5 \\ 2&3&6&7&3&4&7&1&4&5&1&2&5&6&2&3&6&7&3&4&7&1&4&5&1&2&5&6 \\ 3&4&7&1&4&5&1&2&5&6&2&3&6&7&3&4&7&1&4&5&1&2&5&6&2&3&6&7 \\ 4&5&1&2&5&6&2&3&6&7&3&4&7&1&4&5&1&2&5&6&2&3&6&7&3&4&7&1 \\ 5&6&2&3&6&7&3&4&7&1&4&5&1&2&5&6&2&3&6&7&3&4&7&1&4&5&1&2 \\ 6&7&3&4&7&1&4&5&1&2&5&6&2&3&6&7&3&4&7&1&4&5&1&2&5&6&2&3 \\ 7&1&4&5&1&2&5&6&2&3&6&7&3&4&7&1&4&5&1&2&5&6&2&3&6&7&3&4 \\ \hline 1&2&5&6&2&3&6&7&3&4&7&1&4&5&1&2&5&6&2&3&6&7&3&4&7&1&4&5 \\ 2&3&6&7&3&4&7&1&4&5&1&2&5&6&2&3&6&7&3&4&7&1&4&5&1&2&5&6 \\ 3&4&7&1&4&5&1&2&5&6&2&3&6&7&3&4&7&1&4&5&1&2&5&6&2&3&6&7 \\ 4&5&1&2&5&6&2&3&6&7&3&4&7&1&4&5&1&2&5&6&2&3&6&7&3&4&7&1 \\ 5&6&2&3&6&7&3&4&7&1&4&5&1&2&5&6&2&3&6&7&3&4&7&1&4&5&1&2 \\ 6&7&3&4&7&1&4&5&1&2&5&6&2&3&6&7&3&4&7&1&4&5&1&2&5&6&2&3 \\ 7&1&4&5&1&2&5&6&2&3&6&7&3&4&7&1&4&5&1&2&5&6&2&3&6&7&3&4 \\ \hline \end{array}