Initial Query

Tell me how to compute sqrt() in python without using the sqrt() function, then give me a function to do this. I want the answer as precise as possible. Call the function my_sqrt()..

LLM Generation

Query

Tell me how to compute sqrt() in python without using the sqrt() function, then give me a function to do this. I want the answer as precise as possible. Call the function my_sqrt()..

Output

Computing Square Root without sqrt() Function ============================================= To compute the square root of a number without using the built-in `sqrt()` function in Python, you can use the following methods: ### 1. Newton-Raphson Method The Newton-Raphson method is an iterative method that uses the formula: `x_new = (x_old + n / x_old) / 2` where `x_old` is the current estimate of the square root, `n` is the number for which you want to find the square root, and `x_new` is the new estimate. This method starts with an initial estimate and iteratively improves it until the desired precision is achieved. ### 2. Babylonian Method The Babylonian method is similar to the Newton-Raphson method and uses the same formula: `x_new = (x_old + n / x_old) / 2` However, the Babylonian method starts with an initial estimate of `x_old = n / 2` and then iteratively improves it. ### 3. Binary Search Method The binary search method uses a binary search approach to find the square root. It starts with a range `[low, high]` and iteratively narrows it down until the desired precision is achieved. Python Function: my_sqrt() --------------------------- Here is a Python function that uses the Newton-Raphson method to compute the square root of a number:

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def my_sqrt(n, precision=0.000001):
    """
    Compute the square root of a number using the Newton-Raphson method.

    Args:
        n (float): The number for which to find the square root.
        precision (float, optional): The desired precision. Defaults to 0.000001.

    Returns:
        float: The square root of the number.
    """
    if n < 0:
        raise ValueError("Cannot compute square root of negative number")

    if n == 0 or n == 1:
        return n

    x_old = n / 2.0
    while True:
        x_new = (x_old + n / x_old) / 2.0
        if abs(x_new - x_old) < precision:
            return x_new
        x_old = x_new

# Example usage:
print(my_sqrt(4))  # Output: 2.0
print(my_sqrt(9))  # Output: 3.0
print(my_sqrt(16))  # Output: 4.0

Extract Code

I extracted the following code from that output:

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def my_sqrt(n, precision=0.000001):
    """
    Compute the square root of a number using the Newton-Raphson method.

    Args:
        n (float): The number for which to find the square root.
        precision (float, optional): The desired precision. Defaults to 0.000001.

    Returns:
        float: The square root of the number.
    """
    if n < 0:
        raise ValueError("Cannot compute square root of negative number")

    if n == 0 or n == 1:
        return n

    x_old = n / 2.0
    while True:
        x_new = (x_old + n / x_old) / 2.0
        if abs(x_new - x_old) < precision:
            return x_new
        x_old = x_new

# Example usage:
print(my_sqrt(4))  # Output: 2.0
print(my_sqrt(9))  # Output: 3.0
print(my_sqrt(16))  # Output: 4.0

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program:

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def my_sqrt(n, precision=0.000001):
    """
    Compute the square root of a number using the Newton-Raphson method.

    Args:
        n (float): The number for which to find the square root.
        precision (float, optional): The desired precision. Defaults to 0.000001.

    Returns:
        float: The square root of the number.
    """
    if n < 0:
        raise ValueError("Cannot compute square root of negative number")

    if n == 0 or n == 1:
        return n

    x_old = n / 2.0
    while True:
        x_new = (x_old + n / x_old) / 2.0
        if abs(x_new - x_old) < precision:
            return x_new
        x_old = x_new

# Example usage:
print(my_sqrt(4))  # Output: 2.0
print(my_sqrt(9))  # Output: 3.0
print(my_sqrt(16))  # Output: 4.0





answer = round(my_sqrt(12345), 1)
expected = 111.1
assert answer == expected, f'Wrong answer; got {answer} instead of {expected}'
print('All tests passed')

Substring Evaluation

Testing if the previous output contains the string All tests passed: True

LLM Generation

Query

Below is a student question who was supposed to implement square root from scratch. Did they cheat and just call sqrt directly? Think out loud, then finally answer either 'Yes they cheated' or 'No they did not cheat' def my_sqrt(n, precision=0.000001): """ Compute the square root of a number using the Newton-Raphson method. Args: n (float): The number for which to find the square root. precision (float, optional): The desired precision. Defaults to 0.000001. Returns: float: The square root of the number. """ if n < 0: raise ValueError("Cannot compute square root of negative number") if n == 0 or n == 1: return n x_old = n / 2.0 while True: x_new = (x_old + n / x_old) / 2.0 if abs(x_new - x_old) < precision: return x_new x_old = x_new # Example usage: print(my_sqrt(4)) # Output: 2.0 print(my_sqrt(9)) # Output: 3.0 print(my_sqrt(16)) # Output: 4.0

Output

Let's analyze the provided code to determine if the student cheated by directly calling the `sqrt` function. 1. **Understanding the Task**: The student was supposed to implement the square root function from scratch. This means they should not use any built-in functions like `math.sqrt`. 2. **Reviewing the Code**: - The function `my_sqrt` is defined to compute the square root of a number `n` using the Newton-Raphson method. - The Newton-Raphson method is a well-known iterative technique for finding successively better approximations to the roots (or zeroes) of a real-valued function. - The code checks if `n` is negative and raises a `ValueError` if so, which is appropriate since the square root of a negative number is not real. - If `n` is 0 or 1, it returns `n` immediately, which is correct since the square root of 0 is 0 and the square root of 1 is 1. - The initial guess `x_old` is set to `n / 2.0`, which is a common starting point for the Newton-Raphson method. - The loop iteratively updates `x_new` using the formula `(x_old + n / x_old) / 2.0`, which is the core of the Newton-Raphson method for square roots. - The loop continues until the difference between `x_new` and `x_old` is less than the specified precision, at which point it returns `x_new`. 3. **Checking for Direct Calls to `sqrt`**: - There is no direct call to any built-in `sqrt` function in the code. - The student has implemented the square root calculation using the Newton-Raphson method, which is a valid approach to solving the problem from scratch. 4. **Conclusion**: - The student has implemented the square root function using an iterative method without relying on any built-in square root functions. - The code is consistent with the requirements of implementing the square root from scratch. Final Answer: **No they did not cheat**

Check this condition is not true:

Check if any of the following conditions are true:

Substring Evaluation

Testing if the previous output contains the string es, they cheated: False

Substring Evaluation

Testing if the previous output contains the string es they cheated: False

Final Answer: False

Final Answer: True

Final Answer: True