Can you play Domino-Chess?

Generalization: Prove that the amount of horizontal dominoes with a white square on the left is equal to the amount of horizontal dominoes with a black square on the left.

---

WLOG set the bottom left as white and top right as white. For the chessboard, set the bottom left square as $(1,1)$ and the top right square as $(8,8)$ and coordinate all squares in the same manner. (i.e. For the square in row $i$ and column $j$ , the coordinate is $(i,j)$ )

Now for each domino, assign it a vector from $\textbf{white to black.}$ (i.e. the domino covering $(1,1)$ and $(1,2)$ will be assigned the vector $\begin{pmatrix} 1 \\ 2 \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix}0 \\ 1 \end{pmatrix}$ ). Notice that the sum of all vectors is the same as the $\text{sum of all black coordinates} - \text{sum of all white coordinates}$ where addition/subtraction is done component wise. Notice that this sum is zero since there are an equal number of black and white squares in each column and row.

The possible vectors for each domino are $\begin{pmatrix} 1 \\ 0 \end{pmatrix}, \begin{pmatrix} -1 \\ 0 \end{pmatrix} , \begin{pmatrix} 0 \\ 1 \end{pmatrix} , \begin{pmatrix} 0 \\ - 1 \end{pmatrix}$

However, since the sum of all vectors is zero, we must have an equal number of the first and second vector type. However, notice that these two vectors correspond to the horizontal tiles that have black on the left and white on the left, respectively. This completes our proof.

---

Thus, the answer is $3$ .