Minimum Cuts can be made in the Chessboard such that it is not divided into 2 parts in Python

Suppose we have an A x B chessboard (matrix), we need to calculate the maximum number of cuts that can be made on this board such that it remains connected as one piece and is not divided into 2 separate parts.

The key insight is that we can cut along the grid lines between squares. For an A x B chessboard, there are (A-1) horizontal lines and (B-1) vertical lines where cuts can be made. To keep the board connected, we can cut at most (A-1) × (B-1) grid intersections.

Understanding the Problem

Consider a 2 x 4 chessboard as shown below ?

For this 2 x 4 board, we can make 3 cuts at the internal grid intersections while keeping the board connected.

Solution Algorithm

The formula to calculate maximum cuts is ?

def max_cuts_count(M, N):
    """
    Calculate maximum cuts in M x N chessboard 
    that keeps the board connected
    """
    return (M - 1) * (N - 1)

# Test with example
M, N = 2, 4
cuts = max_cuts_count(M, N)
print(f"Maximum cuts for {M} x {N} chessboard: {cuts}")

Maximum cuts for 2 x 4 chessboard: 3

Why This Formula Works

The reasoning behind the formula (M-1) × (N-1) ?

def explain_cuts(M, N):
    horizontal_lines = M - 1
    vertical_lines = N - 1
    intersections = horizontal_lines * vertical_lines
    
    print(f"Board size: {M} x {N}")
    print(f"Internal horizontal lines: {horizontal_lines}")
    print(f"Internal vertical lines: {vertical_lines}")
    print(f"Total intersection points: {intersections}")
    print(f"Maximum cuts possible: {intersections}")
    print()

# Test with different board sizes
explain_cuts(2, 4)
explain_cuts(3, 3)
explain_cuts(4, 5)

Board size: 2 x 4
Internal horizontal lines: 1
Internal vertical lines: 3
Total intersection points: 3
Maximum cuts possible: 3

Board size: 3 x 3
Internal horizontal lines: 2
Internal vertical lines: 2
Total intersection points: 4
Maximum cuts possible: 4

Board size: 4 x 5
Internal horizontal lines: 3
Internal vertical lines: 4
Total intersection points: 12
Maximum cuts possible: 12

Complete Implementation

def max_cuts_chessboard(rows, cols):
    """
    Calculate maximum cuts that can be made in a chessboard
    without dividing it into separate parts.
    
    Args:
        rows: Number of rows in chessboard
        cols: Number of columns in chessboard
    
    Returns:
        Maximum number of cuts possible
    """
    if rows <= 0 or cols <= 0:
        return 0
    
    return (rows - 1) * (cols - 1)

# Test cases
test_cases = [(2, 4), (3, 3), (1, 5), (4, 4)]

for rows, cols in test_cases:
    cuts = max_cuts_chessboard(rows, cols)
    print(f"{rows} x {cols} chessboard: {cuts} maximum cuts")

2 x 4 chessboard: 3 maximum cuts
3 x 3 chessboard: 4 maximum cuts
1 x 5 chessboard: 0 maximum cuts
4 x 4 chessboard: 9 maximum cuts

Conclusion

The maximum cuts in an M x N chessboard that keep it connected is (M-1) × (N-1). This represents the number of internal grid intersections where cuts can be made without separating the board into disconnected parts.