Check whether triangle is valid or not if sides are given in Python

To check if three sides can form a valid triangle, we use the triangle inequality theorem. This states that the sum of any two sides must be greater than the third side for all three combinations.

For three sides a, b, and c, a triangle is valid if:

• a + b > c
• a + c > b
• b + c > a

However, we can simplify this by sorting the sides first. If the sum of the two smaller sides is greater than the largest side, then all conditions are automatically satisfied.

Algorithm

To solve this, we will follow these steps ?

• Sort the list of sides
• If sum of first two sides ? third side, then return False
• Otherwise, return True

Example

Let's implement the triangle validation function ?

def solve(sides):
    sides.sort()
    if sides[0] + sides[1] <= sides[2]:
        return False
    return True

sides = [14, 20, 10]
print(f"Sides {sides}: {solve(sides)}")

The output of the above code is ?

Sides [14, 20, 10]: True

Testing Different Cases

Let's test with multiple examples to understand better ?

def is_valid_triangle(sides):
    sides.sort()
    return sides[0] + sides[1] > sides[2]

# Test cases
test_cases = [
    [14, 20, 10],  # Valid triangle
    [1, 2, 3],     # Invalid: 1 + 2 = 3 (not greater than 3)
    [5, 12, 13],   # Valid triangle
    [1, 1, 3],     # Invalid: 1 + 1 < 3
    [3, 4, 5]      # Valid triangle (right triangle)
]

for sides in test_cases:
    result = is_valid_triangle(sides.copy())  # copy to preserve original
    print(f"Sides {sides}: {'Valid' if result else 'Invalid'}")

The output of the above code is ?

Sides [14, 20, 10]: Valid
Sides [1, 2, 3]: Invalid
Sides [5, 12, 13]: Valid
Sides [1, 1, 3]: Invalid
Sides [3, 4, 5]: Valid

Why This Method Works

After sorting, if a ? b ? c, we only need to check a + b > c because:

• a + c > b is automatically true (since c ? b)
• b + c > a is automatically true (since both b and c are ? a)

Conclusion

Use the triangle inequality theorem to validate triangles. Sort the sides first, then check if the sum of the two smaller sides exceeds the largest side. This approach simplifies the validation from three checks to just one.