Cell Count After Removing Corner Diagonals in Python

Suppose we have a number n representing the length of an n × n board. We need to remove all cells that are diagonal to one of the four corners and count the remaining cells.

In this problem, we identify cells that lie on the main diagonal (top-left to bottom-right) or anti-diagonal (top-right to bottom-left) and mark them as removed. The remaining cells represent our answer.

Problem Visualization

For n = 4, the board looks like this after removing diagonal cells ?

Here, X represents removed diagonal cells, and O represents remaining cells. We can see there are 8 remaining cells.

Mathematical Formula

Instead of simulating the board, we can use a mathematical approach ?

• Total cells = n × n
• Main diagonal cells = n
• Anti-diagonal cells = n
• If n is odd, center cell is counted twice, so subtract 1

The formula becomes: n² - 2n + (n mod 2)

Implementation

def count_remaining_cells(n):
    """
    Count cells remaining after removing corner diagonals.
    
    Args:
        n: Size of the n x n board
    
    Returns:
        Number of remaining cells
    """
    return n * n - 2 * n + (n % 2)

# Test with different board sizes
test_cases = [3, 4, 5, 6]

for size in test_cases:
    result = count_remaining_cells(size)
    print(f"Board size {size}x{size}: {result} remaining cells")

Board size 3x3: 4 remaining cells
Board size 4x4: 8 remaining cells
Board size 5x5: 14 remaining cells
Board size 6x6: 20 remaining cells

Step-by-Step Explanation

Let's trace through the formula for n = 4 ?

n = 4

# Total cells in the board
total_cells = n * n
print(f"Total cells: {total_cells}")

# Cells on main diagonal
main_diagonal = n
print(f"Main diagonal cells: {main_diagonal}")

# Cells on anti-diagonal  
anti_diagonal = n
print(f"Anti-diagonal cells: {anti_diagonal}")

# Check if center cell is counted twice (for odd n)
center_overlap = n % 2
print(f"Center cell overlap: {center_overlap}")

# Apply formula
remaining = total_cells - main_diagonal - anti_diagonal + center_overlap
print(f"Remaining cells: {remaining}")

Total cells: 16
Main diagonal cells: 4
Anti-diagonal cells: 4
Center cell overlap: 0
Remaining cells: 8

Class-Based Solution

class BoardSolver:
    def count_cells_after_diagonal_removal(self, n):
        """
        Calculate remaining cells using the mathematical formula.
        
        Args:
            n: Board dimension (n x n)
            
        Returns:
            Number of cells remaining after diagonal removal
        """
        return n * n - 2 * n + (n % 2)
    
    def visualize_board(self, n):
        """
        Create a visual representation of the board.
        """
        board = []
        for i in range(n):
            row = []
            for j in range(n):
                # Check if cell is on main diagonal or anti-diagonal
                if i == j or i + j == n - 1:
                    row.append('X')  # Removed cell
                else:
                    row.append('O')  # Remaining cell
            board.append(row)
        return board

# Test the solution
solver = BoardSolver()
n = 4

result = solver.count_cells_after_diagonal_removal(n)
board_visual = solver.visualize_board(n)

print(f"Board size: {n}x{n}")
print(f"Remaining cells: {result}")
print("\nBoard visualization:")
for row in board_visual:
    print(' '.join(row))

Board size: 4x4
Remaining cells: 8

Board visualization:
X O O X
O X X O
O X X O
X O O X

Conclusion

The mathematical formula n² - 2n + (n mod 2) efficiently calculates remaining cells without simulating the entire board. This approach has O(1) time complexity and works for any board size.