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Babylonian Square Root

Write a Python function `babylonian_squareroot(number: float, tolerance: float = 1e-6) -> float` that computes the square root of a given positive number using the Babylonian method (also known as Heron's method). The function should iteratively approximate the square root until the difference between successive approximations is less than the specified tolerance.

#### Function Signature
```python [main.nopy]
def babylonian_squareroot(number: float, tolerance: float = 1e-6) -> float:
```

#### Parameters
- `number` (float): The number for which the square root is to be calculated. Must be non-negative.
- `tolerance` (float): The precision of the result. Defaults to `1e-6`.

#### Returns
- `float`: The approximated square root of the input number.

#### Example Usage
```python [main.nopy]
print(babylonian_squareroot(10))  # Output: Approximately 3.162
print(babylonian_squareroot(2))   # Output: Approximately 1.414
print(babylonian_squareroot(9))   # Output: 3.0
```

#### Notes
- If the input number is `0`, the function should return `0` immediately.
- Use an initial guess of `number / 2` for the iterative process.
- Ensure the function handles edge cases, such as very small or very large numbers.
def babylonian_squareroot(number: float, tolerance: float = 1e-6) -> float:
    """
    Compute the square root of a number using the Babylonian method.

    Args:
        number (float): The number to compute the square root of.
        tolerance (float): The precision of the result.

    Returns:
        float: The approximated square root.
    """
    # Handle the edge case for zero
    if number == 0:
        return 0.0

    # Initial guess
    guess = number / 2.0

    # Iterative process
    while True:
        # Placeholder for the iterative computation
        pass

    # Return the computed square root
    return guess
Intermediate mathfunctionsloops
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