Find the number of rectangles of size 2x1 which can be placed inside a rectangle of size n x m in Python

Finding the number of 2x1 rectangles that can be placed inside an n x m rectangle is a classic combinatorial problem. We need to consider different cases based on whether the dimensions are even or odd.

Problem Understanding

Given a rectangle of size n x m, we want to find the maximum number of non-overlapping 2x1 rectangles that can be placed inside it. The constraints are ?

• No two small rectangles can overlap
• Every small rectangle must lie completely inside the larger one
• Small rectangles can touch the edges of the larger rectangle

Algorithm

The solution depends on the parity (even/odd nature) of the dimensions ?

• If n is even: We can place (n/2) × m rectangles
• If m is even: We can place (m/2) × n rectangles
• If both are odd: We can place (n×m-1)/2 rectangles

Implementation

def count_rectangles(n, m):
    if n % 2 == 0:
        return (n // 2) * m
    elif m % 2 == 0:
        return (m // 2) * n
    return (n * m - 1) // 2

# Test with the example
n = 3
m = 3
result = count_rectangles(n, m)
print(f"For a {n}x{m} rectangle, we can place {result} rectangles of size 2x1")

For a 3x3 rectangle, we can place 4 rectangles of size 2x1

Testing Different Cases

def count_rectangles(n, m):
    if n % 2 == 0:
        return (n // 2) * m
    elif m % 2 == 0:
        return (m // 2) * n
    return (n * m - 1) // 2

# Test various cases
test_cases = [(2, 3), (4, 4), (3, 5), (6, 2)]

for n, m in test_cases:
    result = count_rectangles(n, m)
    print(f"{n}x{m} rectangle can fit {result} rectangles of size 2x1")

2x3 rectangle can fit 3 rectangles of size 2x1
4x4 rectangle can fit 8 rectangles of size 2x1
3x5 rectangle can fit 7 rectangles of size 2x1
6x2 rectangle can fit 6 rectangles of size 2x1

How It Works

The algorithm works by analyzing the area coverage ?

• Even dimension case: When one dimension is even, we can perfectly tile that direction
• Both odd case: We lose exactly one unit square that cannot be covered by 2x1 rectangles

Conclusion

This problem demonstrates how parity affects tiling solutions. The key insight is that 2x1 rectangles cover exactly 2 unit squares, so odd total areas will always have one uncovered square.