2023 SSMO Team Round Problems/Problem 9

Problem

Let $B$ , $K$ , and $R$ be the total number of possible moves for a bishop, knight, or rook from any position of a $9$ by $9$ grid. Find $B + K + R$ .

(A bishop moves along diagonals, a rook moves along rows, and a knight moves in the form of a $2 \times 1$ "L" shape)

Solution

We first compute $R$ . Note that from any of the 81 squares, the rook can move to another 16 squares. Thus, $R = 16 \cdot 81 = 1296.$

We now present an alternative method that generalizes well to bishops and knights. Define an $(i,j)$ -move as a move that shifts $i$ squares to the right and $j$ squares up.

There are exactly $(9 - |i|)(9 - |j|)$ possible ways to perform such a move, since the piece must start in a position where it can legally make that move within the bounds of the board.

The total number of possible moves is the sum over all legal $(i,j)$ -moves.

The rook can perform moves of the type $(0,i)$ and $(i,0)$ for integers $i \ne 0$ , with $-8 \le i \le 8$ .

So, $R = 4 \cdot \sum_{i=1}^{8} 9(9 - i) = 4 \cdot 9 \cdot \frac{8 \cdot 9}{2} = 1296,$ matching our earlier computation.

The bishop can perform moves of the type $(i,i)$ and $(-i,i)$ for $1 \le i \le 8$ , so $B = 4 \cdot \sum_{i=1}^{8} (9 - i)^2 = 4 \cdot \frac{8 \cdot 9 \cdot 17}{6} = 4 \cdot 12 \cdot 17 = 816.$

The knight can perform moves of the type $(\pm1, \pm2)$ and $(\pm2, \pm1)$ . For each such pair, the number of legal positions is: $K = 8 \cdot (9 - 2)(9 - 1) = 8 \cdot 7 \cdot 8 = 448.$

The final answer is: $B + K + R = 816 + 448 + 1296 = \boxed{2560}.$

~SMO_Team

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2023 SSMO Team Round Problems/Problem 9

Problem

Solution