# The question

I recently bought a board game called Novem in which the set up rules indicates that 9 numbered tiles have to be organized in a square with this indication:

The sum of the values of the tiles in a row or column must always equal 15 at the beginning of the game.

Which basically means that you must create a magic square that may have diagonals not adding up to 15.

Two examples are given in the rulebook:

$$\begin{matrix} 7 & 6 & 2 \\ 5 & 1 & 9 \\ 3 & 8 & 4 \\ \end{matrix}$$

and

$$\begin{matrix} 1 & 6 & 8 \\ 9 & 2 & 4 \\ 5 & 7 & 3 \\ \end{matrix}$$

I would like to be able to create multiple different set ups without having to use the examples (i.e. opening the rule book at the beginning of each game) or over-thinking it. I know I can already "modify" the examples by adding rotations or replacing the tile $$x$$ by the tile $$10-x$$, but I'd like to forge a solution from scratch.

Is there an easy way that I could come up with a working starting position on the spot, just in my head?

# The complete rules

For the game enthusiastic out there, the other rules are:

• Under each tile is a second one, which value is its complement to 10.
• One player plays the rows, the other one plays the columns, each having 3 tokens with the names of their lines (1, 2, 3 or A, B, C).
• Each player will pick a tile at their turn. The tile is chosen by both players, each secretly selecting the row or column in which the active player must pick the tile (using one of their tokens).
• If the given spot is empty, the active player doesn't win a tile.
• When a row or column is empty, the game ends and the player with the biggest total on the collected tiles wins.

Two simple rules can help achieving this:

• 9 just always have the pairs $$(1, 5)$$ and $$(2, 4)$$ on the lines where it is located.
• 4, 5 and 6 cannot be on the same line.

Start placing a 9 wherever you want and choose freely how you place 1, 5 and 2, 4.

Example:

$$\begin{matrix} 1 & 9 & 5 \\ . & 4 & . \\ . & 2 & . \\ \end{matrix}$$

Then, place the number 6 where it is not aligned with 4 or 5.

Example:

$$\begin{matrix} 1 & 9 & 5 \\ . & 4 & . \\ 6 & 2 & . \\ \end{matrix}$$

Finally, just fill the rest of the board like this: the 7 and the 8 go on the same line as the 6; the 7 aligned with the 2 and the 8 with the 1. The 3 goes in the last spot.

Example:

$$\begin{matrix} 1 & 9 & 5 \\ 8 & 4 & 3 \\ 6 & 2 & 7 \\ \end{matrix}$$

I unfortunately have no mathematical proof that this can lead to any working solution (except by brute force).