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

To compute the square root of a number in Python without using the `sqrt()` function, you can use the Newton's method (also known as the Newton-Raphson method). This is an iterative algorithm for finding successively better approximations to the roots (or zeroes) of a real-valued function. Here's a basic idea of how Newton's Method works for finding the square root of a number `S`: 1. Start with an initial guess `x0`. A good initial guess can simply be `S / 2`. 2. Improve the guess using the formula: `x1 = 0.5 * (x0 + S / x0)`. This formula is derived from Newton's method when applied to the function `f(x) = x^2 - S`. 3. Repeat step 2 using `x1` to generate the next guess `x2`, and so on, until the guess doesn't change significantly (you can define what "significantly" means by choosing an appropriate tolerance level). Here's a Python function `my_sqrt()` that implements this method: 

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def my_sqrt(S, tolerance=1e-10):
    """
    Calculate the square root of S using Newton's method with a specified tolerance.

    :param S: The number to find the square root of.
    :param tolerance: The tolerance level for the approximation.
    :return: The approximated square root of S.
    """
    if S < 0:
        raise ValueError("Cannot compute the square root of a negative number.")
    elif S == 0:
        return 0
    else:
        x = S / 2  # Initial guess
        while True:
            next_x = 0.5 * (x + S / x)
            if abs(x - next_x) < tolerance:  # Check if the improvement is within the tolerance
                break
            x = next_x
        return x

# Example usage
print(my_sqrt(25))  # Should print a number very close to 5
print(my_sqrt(2))   # Should print a number very close to 1.41421356237
This function will give you a very precise approximation of the square root of any non-negative number `S`. The `tolerance` parameter lets you control how precise the result should be. The lower the tolerance, the more precise the result, but it may require more iterations to converge.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(S, tolerance=1e-10):
    """
    Calculate the square root of S using Newton's method with a specified tolerance.

    :param S: The number to find the square root of.
    :param tolerance: The tolerance level for the approximation.
    :return: The approximated square root of S.
    """
    if S < 0:
        raise ValueError("Cannot compute the square root of a negative number.")
    elif S == 0:
        return 0
    else:
        x = S / 2  # Initial guess
        while True:
            next_x = 0.5 * (x + S / x)
            if abs(x - next_x) < tolerance:  # Check if the improvement is within the tolerance
                break
            x = next_x
        return x

# Example usage
print(my_sqrt(25))  # Should print a number very close to 5
print(my_sqrt(2))   # Should print a number very close to 1.41421356237

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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def my_sqrt(S, tolerance=1e-10):
    """
    Calculate the square root of S using Newton's method with a specified tolerance.

    :param S: The number to find the square root of.
    :param tolerance: The tolerance level for the approximation.
    :return: The approximated square root of S.
    """
    if S < 0:
        raise ValueError("Cannot compute the square root of a negative number.")
    elif S == 0:
        return 0
    else:
        x = S / 2  # Initial guess
        while True:
            next_x = 0.5 * (x + S / x)
            if abs(x - next_x) < tolerance:  # Check if the improvement is within the tolerance
                break
            x = next_x
        return x

# Example usage
print(my_sqrt(25))  # Should print a number very close to 5
print(my_sqrt(2))   # Should print a number very close to 1.41421356237





answer = round(my_sqrt(12345), 1)
expected = 111.1
assert answer == expected, f'Wrong answer; got {answer} instead of {expected}'
print('All tests passed')
And got the output: 5.000000000016778 1.4142135623746899 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(S, tolerance=1e-10): """ Calculate the square root of S using Newton's method with a specified tolerance. :param S: The number to find the square root of. :param tolerance: The tolerance level for the approximation. :return: The approximated square root of S. """ if S < 0: raise ValueError("Cannot compute the square root of a negative number.") elif S == 0: return 0 else: x = S / 2 # Initial guess while True: next_x = 0.5 * (x + S / x) if abs(x - next_x) < tolerance: # Check if the improvement is within the tolerance break x = next_x return x # Example usage print(my_sqrt(25)) # Should print a number very close to 5 print(my_sqrt(2)) # Should print a number very close to 1.41421356237

Output

To determine if the student cheated by directly calling the `sqrt` function from a library such as `math`, let's analyze the provided code for `my_sqrt`. 1. **Methodology**: The code uses Newton's method (also known as the Newton-Raphson method) for finding successively better approximations to the roots (or zeroes) of a real-valued function. In this case, the function is \(f(x) = x^2 - S\), where \(S\) is the number whose square root we want to find. The iterative formula used is \(x_{n+1} = 0.5 \cdot (x_n + \frac{S}{x_n})\), which is a classic application of Newton's method to find square roots. 2. **Implementation Details**: The code starts with an initial guess, `x = S / 2`, and iteratively improves this guess using the formula mentioned above. It stops iterating once the difference between successive approximations is less than the specified `tolerance`. This is a standard approach to implementing Newton's method for this problem. 3. **Direct Call to `sqrt`**: There is no direct call to any `sqrt` function from a standard library such as `math.sqrt` within the provided code. The calculation is done purely through the iterative process described, without relying on an external function to compute the square root. 4. **Conclusion**: Based on the analysis, the student did not cheat by directly calling a `sqrt` function. Instead, they implemented the square root calculation from scratch using Newton's method, adhering to the assignment's requirements. **Final Answer**: No they did not cheat.

Check this condition is not true:

Substring Evaluation

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

Final Answer: True

Final Answer: True

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

To compute the square root of a number in Python without using the `sqrt()` function, you can use several methods, including the Newton-Raphson method (also known as the Newton Method). This is an efficient iterative method to approximate roots of a real-valued function. For finding the square root of a number, `a`, you want to solve for `x` in the equation `x^2 - a = 0`. The formula to iterate in the Newton-Raphson method is: \[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\] For our specific case, where \(f(x) = x^2 - a\), the derivative \(f'(x) = 2x\). Substituting these into the formula gives: \[x_{n+1} = x_n - \frac{x_n^2 - a}{2x_n} = \frac{1}{2} \left( x_n + \frac{a}{x_n} \right)\] This formula will iteratively approach the square root of `a`. The iteration can start with an initial guess `x_0`. A common choice for `x_0` is `a / 2` or simply `1`, especially if `a` is not zero. Here's a Python function implementing this: 

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def my_sqrt(a, tolerance=1e-10):
    if a < 0:
        raise ValueError("Cannot compute the square root of a negative number.")
    if a == 0:
        return 0
    x = a / 2  # Initial guess
    while True:
        next_x = 0.5 * (x + a / x)
        if abs(x - next_x) < tolerance:  # Check if the change is smaller than the tolerance
            break
        x = next_x
    return x

# Example usage
number = 9
print(f"The square root of {number} is approximately {my_sqrt(number)}")
This function, `my_sqrt(a)`, computes the square root of `a` with an internal tolerance level set to `1e-10`, which dictates the precision of the approximation. You can adjust the `tolerance` value to make the result more or less precise according to your needs. Remember, this implementation does not use the built-in `math.sqrt()` function and provides a high degree of precision for most practical purposes.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(a, tolerance=1e-10):
    if a < 0:
        raise ValueError("Cannot compute the square root of a negative number.")
    if a == 0:
        return 0
    x = a / 2  # Initial guess
    while True:
        next_x = 0.5 * (x + a / x)
        if abs(x - next_x) < tolerance:  # Check if the change is smaller than the tolerance
            break
        x = next_x
    return x

# Example usage
number = 9
print(f"The square root of {number} is approximately {my_sqrt(number)}")

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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def my_sqrt(a, tolerance=1e-10):
    if a < 0:
        raise ValueError("Cannot compute the square root of a negative number.")
    if a == 0:
        return 0
    x = a / 2  # Initial guess
    while True:
        next_x = 0.5 * (x + a / x)
        if abs(x - next_x) < tolerance:  # Check if the change is smaller than the tolerance
            break
        x = next_x
    return x

# Example usage
number = 9
print(f"The square root of {number} is approximately {my_sqrt(number)}")





answer = round(my_sqrt(12345), 1)
expected = 111.1
assert answer == expected, f'Wrong answer; got {answer} instead of {expected}'
print('All tests passed')
And got the output: The square root of 9 is approximately 3.0000000000393214 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(a, tolerance=1e-10): if a < 0: raise ValueError("Cannot compute the square root of a negative number.") if a == 0: return 0 x = a / 2 # Initial guess while True: next_x = 0.5 * (x + a / x) if abs(x - next_x) < tolerance: # Check if the change is smaller than the tolerance break x = next_x return x # Example usage number = 9 print(f"The square root of {number} is approximately {my_sqrt(number)}")

Output

To determine whether the student cheated by directly calling a built-in `sqrt` function or not, let's analyze the provided code for calculating the square root of a number. The code defines a function `my_sqrt(a, tolerance=1e-10)` that calculates the square root of a number `a` with a specified tolerance level. The method used in this function does not directly call any built-in `sqrt` function. Instead, it implements an algorithm to find the square root. Let's break down the steps of the algorithm: 1. **Input Validation**: The function first checks if the input `a` is less than 0, throwing a `ValueError` if true, since the square root of a negative number is not defined in the realm of real numbers. It also checks if `a` is 0, returning 0 immediately, which is correct since the square root of 0 is 0. 2. **Initial Guess**: It initializes the guess for the square root as `a / 2`. This is a common starting point for iterative methods to find square roots, although the initial guess can vary depending on the method. 3. **Iterative Calculation**: The function then enters a loop where it applies an iterative formula to refine the guess for the square root. The formula used is `next_x = 0.5 * (x + a / x)`, which is a part of the Newton-Raphson method for finding roots of a function. This method is well-known for its efficiency in finding square roots among other roots. 4. **Tolerance Check**: The loop continues until the change between the new guess and the old guess is smaller than the specified tolerance. This ensures that the function returns a value that is close enough to the actual square root, within the bounds of the specified tolerance. 5. **Return Value**: Once the change is less than the tolerance, the loop breaks, and the function returns the current guess for the square root. The method implemented here is a legitimate numerical method for finding square roots and does not involve cheating by calling a built-in `sqrt` function. The student has implemented an iterative algorithm (specifically, a form of the Newton-Raphson method) to approximate the square root of a given number to a specified degree of accuracy. **Conclusion**: No they did not cheat.

Check this condition is not true:

Substring Evaluation

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

Final Answer: True

Final Answer: True

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

To compute the square root of a number in Python without directly using the `sqrt()` function, you can use the Newton's method (also known as the Newton-Raphson method). It's an efficient iterative method to approximate the square root of a number, \(x\), by finding the zero of the function \(f(y) = y^2 - x\). The method starts with an initial guess, \(y_0\), and iteratively improves the guess according to the formula: \[y_{n+1} = \frac{1}{2} \left( y_n + \frac{x}{y_n} \right)\] This formula comes from the Newton-Raphson iteration for solving \(y^2 - x = 0\). Here's how you can define the `my_sqrt` function: 

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def my_sqrt(x, tolerance=1e-10):
    if x < 0:
        raise ValueError("Cannot compute square root of a negative number.")
    if x == 0:
        return 0
    # Initial guess for the square root
    y = x
    # Loop until the absolute difference between y*y and x is within the desired tolerance
    while abs(y * y - x) > tolerance:
        # Newton's method formula
        y = (y + x / y) / 2
    return y

# Example usage
number = 25
print(f"The square root of {number} is approximately {my_sqrt(number)}")
This function defines `my_sqrt(x, tolerance=1e-10)` that computes the square root of `x` with a precision defined by `tolerance`. The default tolerance is `1e-10`, but you can adjust it to achieve the desired precision. The function uses a loop with Newton's method to iteratively improve the guess until the difference between the square of the current guess and `x` is within the specified tolerance.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(x, tolerance=1e-10):
    if x < 0:
        raise ValueError("Cannot compute square root of a negative number.")
    if x == 0:
        return 0
    # Initial guess for the square root
    y = x
    # Loop until the absolute difference between y*y and x is within the desired tolerance
    while abs(y * y - x) > tolerance:
        # Newton's method formula
        y = (y + x / y) / 2
    return y

# Example usage
number = 25
print(f"The square root of {number} is approximately {my_sqrt(number)}")

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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def my_sqrt(x, tolerance=1e-10):
    if x < 0:
        raise ValueError("Cannot compute square root of a negative number.")
    if x == 0:
        return 0
    # Initial guess for the square root
    y = x
    # Loop until the absolute difference between y*y and x is within the desired tolerance
    while abs(y * y - x) > tolerance:
        # Newton's method formula
        y = (y + x / y) / 2
    return y

# Example usage
number = 25
print(f"The square root of {number} is approximately {my_sqrt(number)}")





answer = round(my_sqrt(12345), 1)
expected = 111.1
assert answer == expected, f'Wrong answer; got {answer} instead of {expected}'
print('All tests passed')
And got the output: The square root of 25 is approximately 5.0 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(x, tolerance=1e-10): if x < 0: raise ValueError("Cannot compute square root of a negative number.") if x == 0: return 0 # Initial guess for the square root y = x # Loop until the absolute difference between y*y and x is within the desired tolerance while abs(y * y - x) > tolerance: # Newton's method formula y = (y + x / y) / 2 return y # Example usage number = 25 print(f"The square root of {number} is approximately {my_sqrt(number)}")

Output

To determine whether the student cheated by directly calling a built-in `sqrt` function or not, let's analyze the provided code. The function `my_sqrt` is designed to compute the square root of a given number `x` with a specified tolerance level. The approach used in this function does not directly call any built-in `sqrt` function but instead implements an algorithm to find the square root. The key points in the code are: 1. **Error Handling for Negative Input**: The function starts by checking if the input `x` is negative, which is a good practice since the square root of a negative number is not defined in the realm of real numbers. 2. **Special Case for Zero**: It correctly handles the special case where `x` is 0, immediately returning 0, which is accurate. 3. **Initial Guess**: The initial guess for the square root is taken as the number itself (`y = x`). This is a reasonable starting point, although other strategies might choose a different starting guess. 4. **Iterative Approach Using Newton's Method**: The core of the function uses a while loop to iteratively refine the guess for the square root. It employs Newton's method (also known as the Newton-Raphson method) for finding successively better approximations to the roots (or zeroes) of a real-valued function. The specific formula used here, `y = (y + x / y) / 2`, is a classic application of Newton's method to find the square root of `x`. 5. **Tolerance Check**: The loop continues until the absolute difference between `y*y` (the square of the current guess) and `x` is within a specified tolerance, which ensures that the result is as accurate as desired within the limits of floating-point arithmetic. 6. **Return Value**: Once the loop condition is satisfied, the function returns the current guess `y`, which is the approximation of the square root of `x`. Given this analysis, it's clear that the student implemented an algorithm to compute the square root of a number from scratch using Newton's method. They did not cheat by directly calling a built-in `sqrt` function. **Final Answer**: No they did not cheat.

Check this condition is not true:

Substring Evaluation

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

Final Answer: True

Final Answer: True

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

To compute the square root of a number in Python without using the `sqrt()` function, you can use the Newton's Method (also known as the Newton-Raphson method). This method is an efficient way to find successively better approximations to the roots (or zeroes) of a real-valued function. The formula for Newton's Method, when applied to finding the square root of a number `x`, is: 

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new_guess = 0.5 * (old_guess + x / old_guess)
You start with an initial guess for the square root, and then iteratively apply this formula until the guess does not change significantly anymore (indicating that you've found a sufficiently precise approximation of the square root). Below is a Python function `my_sqrt()` that implements this method. This function takes two parameters: `x`, the number you want to find the square root of, and `tolerance`, which defines how close the square root approximation should be before the function returns the result. The smaller the `tolerance`, the more precise the result but potentially at the cost of more iterations. 
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def my_sqrt(x, tolerance=1e-10):
    if x < 0:
        raise ValueError("Cannot compute the square root of a negative number")
    
    if x == 0:
        return 0
    
    guess = x
    while True:
        new_guess = 0.5 * (guess + x / guess)
        if abs(new_guess - guess) < tolerance:
            break
        guess = new_guess
    return new_guess

# Example usage
number = 25
print(f"The square root of {number} is approximately {my_sqrt(number)}")
This function works for positive numbers. It starts with an initial guess equal to the number itself (for `x > 1`, a smaller starting guess like `x / 2` might converge faster, but starting with `x` itself is a simple and generally effective choice). It then iteratively refines the guess using Newton's Method until the difference between successive guesses is less than the specified `tolerance`. This ensures that the result is as precise as the user desires. Remember, choosing a very small tolerance can lead to more computation time, but it will give you a more precise result. The example above sets a default tolerance of `1e-10`, which should be sufficient for most applications.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(x, tolerance=1e-10):
    if x < 0:
        raise ValueError("Cannot compute the square root of a negative number")
    
    if x == 0:
        return 0
    
    guess = x
    while True:
        new_guess = 0.5 * (guess + x / guess)
        if abs(new_guess - guess) < tolerance:
            break
        guess = new_guess
    return new_guess

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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def my_sqrt(x, tolerance=1e-10):
    if x < 0:
        raise ValueError("Cannot compute the square root of a negative number")
    
    if x == 0:
        return 0
    
    guess = x
    while True:
        new_guess = 0.5 * (guess + x / guess)
        if abs(new_guess - guess) < tolerance:
            break
        guess = new_guess
    return new_guess





answer = round(my_sqrt(12345), 1)
expected = 111.1
assert answer == expected, f'Wrong answer; got {answer} instead of {expected}'
print('All tests passed')
And got the output: 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(x, tolerance=1e-10): if x < 0: raise ValueError("Cannot compute the square root of a negative number") if x == 0: return 0 guess = x while True: new_guess = 0.5 * (guess + x / guess) if abs(new_guess - guess) < tolerance: break guess = new_guess return new_guess

Output

To determine if the student cheated by directly calling a built-in `sqrt` function or not, let's analyze the provided implementation of `my_sqrt`. 1. **Initial Checks**: The function starts by handling edge cases, such as negative inputs and zero. This is a good practice and doesn't involve calling any square root function directly. It's a logical step to ensure the function behaves correctly for all inputs. 2. **Guess and Check Method**: The core of the function uses a guess and check method, specifically a form of the Newton-Raphson method for finding successively better approximations to the roots (or zeroes) of a real-valued function. The formula used here, `new_guess = 0.5 * (guess + x / guess)`, is a specific application of the Newton-Raphson method to find the square root of `x`. This method starts with an initial guess (in this case, the value of `x` itself) and iteratively refines that guess. The refinement is done by averaging the guess with the quotient of the input `x` and the guess, which is a well-known technique for square root approximation. 3. **Tolerance Check**: The loop continues until the difference between the new guess and the previous guess is less than a specified tolerance. This ensures that the function doesn't run indefinitely and stops when the approximation is sufficiently close to the actual square root. The use of a tolerance value for stopping the iteration is a common technique in numerical methods to decide when an approximation is "good enough." 4. **No Direct Call to `sqrt`**: Throughout the provided code, there is no direct call to any built-in `sqrt` function or any other external library function that computes the square root. The student has implemented the square root calculation using a numerical method that is well-documented and widely taught in courses on numerical analysis or even basic programming courses that cover algorithmic problem solving. **Conclusion**: Based on the analysis, the student did **not** cheat by calling the `sqrt` function directly. Instead, they implemented a legitimate and well-known numerical method (the Newton-Raphson method) to approximate the square root of a number. So, the final answer is: **No they did not cheat**

Check this condition is not true:

Substring Evaluation

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

Final Answer: True

Final Answer: True

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

To compute the square root of a number in Python without using the built-in `sqrt()` function, you can use the Newton's method (also known as the Newton-Raphson method). This method is an iterative algorithm that approximates the roots of a real-valued function. When applied to find the square root of a number `x`, the method iteratively improves the guess `a` for the square root according to the formula: \[a_{n+1} = \frac{1}{2} \left(a_n + \frac{x}{a_n}\right)\] This formula will iteratively converge to the square root of `x`. For most practical purposes, this method is sufficiently fast and accurate. Here's how you could implement it: 

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def my_sqrt(x, tolerance=1e-10):
    if x < 0:
        raise ValueError("Cannot compute the square root of a negative number.")
    if x == 0:
        return 0  # The square root of 0 is 0
    # Starting with an initial guess for the square root
    # A good initial guess can be half of the number, but 1 works for any number
    guess = x / 2.0
    while True:
        # Newton's method formula
        next_guess = (guess + x / guess) / 2
        # Check if the difference between the current guess and the next is within the desired tolerance
        if abs(guess - next_guess) < tolerance:
            return next_guess
        guess = next_guess

# Example usage
number = 25
print(f"The square root of {number} is approximately {my_sqrt(number)}")
This function takes a number `x` and an optional parameter `tolerance` which defines how close the estimated square root should be to the actual square root before the function returns the estimate. The lower the tolerance, the more precise the result but potentially at the cost of more iterations. The example usage demonstrates finding the square root of 25. You can adjust the `number` variable to compute the square root of any non-negative number you choose.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(x, tolerance=1e-10):
    if x < 0:
        raise ValueError("Cannot compute the square root of a negative number.")
    if x == 0:
        return 0  # The square root of 0 is 0
    # Starting with an initial guess for the square root
    # A good initial guess can be half of the number, but 1 works for any number
    guess = x / 2.0
    while True:
        # Newton's method formula
        next_guess = (guess + x / guess) / 2
        # Check if the difference between the current guess and the next is within the desired tolerance
        if abs(guess - next_guess) < tolerance:
            return next_guess
        guess = next_guess

# Example usage
number = 25
print(f"The square root of {number} is approximately {my_sqrt(number)}")

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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def my_sqrt(x, tolerance=1e-10):
    if x < 0:
        raise ValueError("Cannot compute the square root of a negative number.")
    if x == 0:
        return 0  # The square root of 0 is 0
    # Starting with an initial guess for the square root
    # A good initial guess can be half of the number, but 1 works for any number
    guess = x / 2.0
    while True:
        # Newton's method formula
        next_guess = (guess + x / guess) / 2
        # Check if the difference between the current guess and the next is within the desired tolerance
        if abs(guess - next_guess) < tolerance:
            return next_guess
        guess = next_guess

# Example usage
number = 25
print(f"The square root of {number} is approximately {my_sqrt(number)}")





answer = round(my_sqrt(12345), 1)
expected = 111.1
assert answer == expected, f'Wrong answer; got {answer} instead of {expected}'
print('All tests passed')
And got the output: The square root of 25 is approximately 5.0 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(x, tolerance=1e-10): if x < 0: raise ValueError("Cannot compute the square root of a negative number.") if x == 0: return 0 # The square root of 0 is 0 # Starting with an initial guess for the square root # A good initial guess can be half of the number, but 1 works for any number guess = x / 2.0 while True: # Newton's method formula next_guess = (guess + x / guess) / 2 # Check if the difference between the current guess and the next is within the desired tolerance if abs(guess - next_guess) < tolerance: return next_guess guess = next_guess # Example usage number = 25 print(f"The square root of {number} is approximately {my_sqrt(number)}")

Output

To determine if the student cheated by directly calling a built-in `sqrt` function or not, let's analyze the provided code step by step: 1. **Error Handling for Negative Inputs**: The function starts by checking if the input `x` is negative, which is a valid step in implementing a square root function since the square root of a negative number is not a real number. This is a thoughtful implementation detail and not indicative of cheating. 2. **Special Case for Zero**: The function correctly returns 0 when the input is 0, as the square root of 0 is 0. This is another correct implementation detail. 3. **Initial Guess**: The student chooses to start with an initial guess for the square root, which is `x / 2.0`. This is a common approach in numerical methods for finding square roots and shows an understanding of the problem. 4. **Newton's Method**: The core of the function uses Newton's method (also known as the Newton-Raphson method) for finding successively better approximations to the roots (or zeroes) of a real-valued function. The formula `next_guess = (guess + x / guess) / 2` is a specific application of Newton's method for finding square roots. This method is iterative and refines the guess for the square root by averaging the guess and the quotient of the number and the guess. 5. **Tolerance Check**: The loop continues until the difference between successive guesses is less than a specified tolerance, which is a common approach in numerical methods to decide when an approximation is sufficiently accurate. This shows an understanding of the need for a stopping criterion in iterative methods. 6. **No Direct Call to `sqrt`**: Throughout the function, there is no direct call to any built-in `sqrt` function or any other external library function that computes the square root directly. The student implements the square root calculation from scratch using Newton's method, which is a legitimate and well-known numerical method for this purpose. Based on this analysis, the final answer is: **No they did not cheat** The student implemented the square root function from scratch using Newton's method, demonstrating an understanding of numerical methods and proper implementation techniques without resorting to calling a built-in `sqrt` function.

Check this condition is not true:

Substring Evaluation

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

Final Answer: True

Final Answer: True