Corner to Corner Minimum Moves

The problem requires finding the minimum number of specific moves to get from one corner of an 8x8 board to the diagonally opposite corner.

A standard 8x8 board has corners like (1,1) and (8,8). The required displacement is $(\Delta X, \Delta Y) = (8-1, 8-1) = (7, 7)$.

A move involves changing position by $(\pm 1, \pm 2)$ or $(\pm 2, \pm 1)$, similar to a knight's move in chess.

Board Move Analysis

We analyze the properties of the moves and the required displacement.

• Parity Check: The starting corner (1,1) has a coordinate sum $1+1=2$ (even). The opposite corner (8,8) has a coordinate sum $8+8=16$ (even).
• Each move changes the coordinate sum by $\pm 1 \pm 2 = \pm 3$ or $\pm 2 \pm 1 = \pm 3$. All these changes are odd numbers.
• To transition from an even coordinate sum to another even sum, an even number of odd changes (moves) is necessary. Thus, the total number of moves, $k$, must be even.

Displacement Calculation

Consider the total displacement needed: $(\Delta X, \Delta Y) = (7, 7)$.

Let's analyze the effect of moves on the sum of coordinates $u = x+y$. Each move changes $u$ by $\Delta u = \Delta x + \Delta y$, which can be $\pm 1$ or $\pm 3$. The total change needed is $\Delta u_{total} = 7+7=14$.

Let $k$ be the number of moves. The maximum possible value for $\Delta u$ in one move is 3. Therefore, the maximum possible total change in $u$ over $k$ moves is $3k$.

To achieve $\Delta u_{total} = 14$, we must have $3k \ge 14$. This means $k \ge 14/3 \approx 4.67$. Since $k$ must be an integer, $k \ge 5$.

Combining this with the parity requirement ($k$ must be even), the minimum possible number of moves is the smallest even integer greater than or equal to 5, which is 6.

6-Move Path Example

We need to confirm that 6 moves are sufficient. A path exists using 6 moves:

• 4 moves of type (+1, +2)
• 1 move of type (+2, +1)
• 1 move of type (+1, -2)

Total Moves: $4 + 1 + 1 = 6$. This is an even number.

Total Displacement:

• X-displacement: $(4 \times 1) + (1 \times 2) + (1 \times 1) = 4 + 2 + 1 = 7$.
• Y-displacement: $(4 \times 2) + (1 \times 1) + (1 \times -2) = 8 + 1 - 2 = 7$.

The net displacement is $(7, 7)$, which reaches the opposite corner.

A possible sequence of moves starting from (1,1) is:

(1,1) $\rightarrow$ (+2,+1) $\rightarrow$ (3,2) $\rightarrow$ (+1,+2) $\rightarrow$ (4,4) $\rightarrow$ (+1,+2) $\rightarrow$ (5,6) $\rightarrow$ (+1,+2) $\rightarrow$ (6,8) $\rightarrow$ (+1,-2) $\rightarrow$ (7,6) $\rightarrow$ (+1,+2) $\rightarrow$ (8,8).

Final Minimum Moves

Based on the parity and displacement analysis, fewer than 6 moves are impossible. The constructed 6-move path proves that 6 moves are sufficient. Therefore, the minimum number of moves is 6.