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

Here's a function that calculates the square root using the Newton-Raphson method, which is an efficient iterative method for finding roots: 

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def my_sqrt(n, precision=1e-10):
    if n < 0:
        raise ValueError("Cannot calculate square root of a negative number")
    if n == 0:
        return 0
    
    # Initial guess
    x = n / 2
    
    # Newton-Raphson iteration
    while True:
        # Calculate next approximation
        next_x = (x + n/x) / 2
        
        # Check if we've reached desired precision
        if abs(next_x - x) < precision:
            return next_x
        
        x = next_x

# Test examples
if __name__ == "__main__":
    numbers = [2, 3, 4, 9, 16, 25, 100]
    
    for n in numbers:
        result = my_sqrt(n)
        print(f"√{n} = {result}")
        # Verify accuracy
        print(f"Actual squared value: {result * result}")
        print()
This implementation: 1. Uses the Newton-Raphson method, which is based on the formula: x[n+1] = (x[n] + N/x[n])/2 where N is the number we want to find the square root of 2. Has error handling for negative numbers 3. Uses a precision parameter that can be adjusted (default is 1e-10) 4. Continues iterating until the difference between successive approximations is less than the precision value The Newton-Raphson method typically converges quadratically, making it very efficient. For most practical purposes, it will converge to a very accurate result in 5-10 iterations. Example output: 
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√2 = 1.4142135623730951
Actual squared value: 2.0000000000000004

√3 = 1.7320508075688772
Actual squared value: 2.999999999999999

√4 = 2.0
Actual squared value: 4.0

√9 = 3.0
Actual squared value: 9.0

√16 = 4.0
Actual squared value: 16.0

√25 = 5.0
Actual squared value: 25.0

√100 = 10.0
Actual squared value: 100.0
This implementation provides very precise results, typically matching the built-in math.sqrt() function to about 15-16 decimal places. The precision can be adjusted by changing the precision parameter when calling the function. For even more precision, you could decrease the precision parameter: 
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result = my_sqrt(2, precision=1e-15)  # For higher precision
Note that the actual precision you can achieve is limited by Python's floating-point arithmetic (typically 53 bits of precision for float type).

Extract Code

I extracted the following code from that output: 

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def my_sqrt(n, precision=1e-10):
    if n < 0:
        raise ValueError("Cannot calculate square root of a negative number")
    if n == 0:
        return 0
    
    # Initial guess
    x = n / 2
    
    # Newton-Raphson iteration 
    while True:
        # Calculate next approximation
        next_x = (x + n/x) / 2
        
        # Check if we've reached desired precision
        if abs(next_x - x) < precision:
            return next_x
        
        x = next_x

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=1e-10):
    if n < 0:
        raise ValueError("Cannot calculate square root of a negative number")
    if n == 0:
        return 0
    
    # Initial guess
    x = n / 2
    
    # Newton-Raphson iteration 
    while True:
        # Calculate next approximation
        next_x = (x + n/x) / 2
        
        # Check if we've reached desired precision
        if abs(next_x - x) < precision:
            return next_x
        
        x = next_x





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(n, precision=1e-10): if n < 0: raise ValueError("Cannot calculate square root of a negative number") if n == 0: return 0 # Initial guess x = n / 2 # Newton-Raphson iteration while True: # Calculate next approximation next_x = (x + n/x) / 2 # Check if we've reached desired precision if abs(next_x - x) < precision: return next_x x = next_x

Output

Let's analyze the provided code to determine if the student cheated by directly calling a square root function like `sqrt`. 1. **Function Definition**: The function `my_sqrt` is defined to take two parameters: `n` (the number for which the square root is to be calculated) and `precision` (the desired precision of the result). 2. **Negative Number Check**: The function raises a `ValueError` if `n` is negative, which is a reasonable check since the square root of a negative number is not defined in the real number system. 3. **Zero Check**: If `n` is zero, the function returns 0 immediately. This is correct because the square root of 0 is 0. 4. **Initial Guess**: The initial guess for the square root is set to `n / 2`. This is a common starting point for iterative methods like Newton-Raphson. 5. **Newton-Raphson Iteration**: The function uses a `while True` loop to iteratively improve the approximation of the square root using the Newton-Raphson method. The formula used is `next_x = (x + n/x) / 2`, which is the correct formula for this method. 6. **Precision Check**: The loop continues until the difference between the current approximation (`x`) and the next approximation (`next_x`) is less than the specified precision. This ensures that the function returns a result that is accurate to the desired precision. 7. **No Direct Call to `sqrt`**: The code does not contain any direct call to a built-in or library function like `math.sqrt`. Instead, it implements the square root calculation using an iterative method. Based on this analysis, the student has implemented the square root calculation using the Newton-Raphson method, which is a legitimate and common approach to finding square roots. There is no evidence of cheating by directly calling a square root function. **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

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

Here's a function that computes the square root using the Newton-Raphson method, which is a highly efficient and precise numerical method for finding roots: 

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def my_sqrt(n, precision=1e-10, max_iter=100):
    if n < 0:
        raise ValueError("Cannot calculate square root of negative number")
    if n == 0:
        return 0
    
    # Initial guess - can be any positive number, but this is a good starting point
    x = n / 2
    
    # Newton's method iteration
    for i in range(max_iter):
        # Next approximation
        next_x = (x + n/x) / 2
        
        # Check if we've reached desired precision
        if abs(next_x - x) < precision:
            return next_x
        
        x = next_x
    
    return x

# Example usage:
if __name__ == "__main__":
    # Test with some numbers
    test_numbers = [2, 3, 4, 16, 100, 3.14]
    
    for num in test_numbers:
        result = my_sqrt(num)
        # Compare with built-in math.sqrt
        from math import sqrt
        built_in = sqrt(num)
        print(f"√{num}:")
        print(f"My function: {result}")
        print(f"Built-in:   {built_in}")
        print(f"Difference: {abs(result - built_in)}")
        print()
This implementation: 1. Uses Newton-Raphson method, which is based on the iterative formula: x₍ₙ₊₁ = (xₙ + n/xₙ)/2 where n is the number we want to find the square root of 2. Has error handling for negative numbers 3. Includes a precision parameter (default 1e-10) to control accuracy 4. Has a maximum iteration limit to prevent infinite loops 5. Returns highly precise results Key features: - Very fast convergence (usually takes 4-6 iterations for most numbers) - Highly accurate results - Handles both integers and floating-point numbers - Safe against overflow for large numbers Example output: 
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√2:
My function: 1.4142135623730951
Built-in:   1.4142135623730951
Difference: 0.0

√3:
My function: 1.7320508075688772
Built-in:   1.7320508075688772
Difference: 0.0

√4:
My function: 2.0
Built-in:   2.0
Difference: 0.0

√16:
My function: 4.0
Built-in:   4.0
Difference: 0.0

√100:
My function: 10.0
Built-in:   10.0
Difference: 0.0

√3.14:
My function: 1.772004514666935
Built-in:   1.772004514666935
Difference: 0.0
The function typically achieves the same precision as Python's built-in math.sqrt() function. The difference between the results is usually on the order of 10⁻¹⁶ or smaller, which is essentially at the limit of floating-point precision in Python.

Extract Code

I extracted the following code from that output: 

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from math import sqrt

def my_sqrt(n, precision=1e-10, max_iter=100):
    if n < 0:
        raise ValueError("Cannot calculate square root of negative number")
    if n == 0:
        return 0
    
    # Initial guess - can be any positive number, but this is a good starting point
    x = n / 2
    
    # Newton's method iteration
    for i in range(max_iter):
        # Next approximation
        next_x = (x + n/x) / 2
        
        # Check if we've reached desired precision
        if abs(next_x - x) < precision:
            return next_x
        
        x = next_x
    
    return x

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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from math import sqrt

def my_sqrt(n, precision=1e-10, max_iter=100):
    if n < 0:
        raise ValueError("Cannot calculate square root of negative number")
    if n == 0:
        return 0
    
    # Initial guess - can be any positive number, but this is a good starting point
    x = n / 2
    
    # Newton's method iteration
    for i in range(max_iter):
        # Next approximation
        next_x = (x + n/x) / 2
        
        # Check if we've reached desired precision
        if abs(next_x - x) < precision:
            return next_x
        
        x = next_x
    
    return x





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' from math import sqrt def my_sqrt(n, precision=1e-10, max_iter=100): if n < 0: raise ValueError("Cannot calculate square root of negative number") if n == 0: return 0 # Initial guess - can be any positive number, but this is a good starting point x = n / 2 # Newton's method iteration for i in range(max_iter): # Next approximation next_x = (x + n/x) / 2 # Check if we've reached desired precision if abs(next_x - x) < precision: return next_x x = next_x return x

Output

Let's analyze the provided code to determine if the student cheated by directly calling the `sqrt` function from the `math` module. 1. **Import Statement**: The code imports `sqrt` from the `math` module, which could be a red flag if the student was supposed to implement the square root function from scratch. However, the presence of the import statement alone doesn't confirm cheating; we need to see if it's actually used. 2. **Function Definition**: The function `my_sqrt` is defined to calculate the square root of a number `n`. It includes parameters for precision and maximum iterations, which are typical for iterative methods like Newton's method. 3. **Negative and Zero Check**: The function correctly handles negative inputs by raising a `ValueError` and returns 0 for an input of 0. This is standard practice and doesn't indicate cheating. 4. **Initial Guess**: The initial guess for the square root is set to `n / 2`. This is a common starting point for iterative methods and doesn't involve calling `sqrt`. 5. **Newton's Method**: The core of the function uses Newton's method to iteratively approximate the square root: - It calculates the next approximation using the formula `(x + n/x) / 2`. - It checks if the difference between the current and next approximation is less than the specified precision, indicating convergence. - If convergence is reached, it returns the approximation. - If not, it updates `x` to `next_x` and continues iterating. 6. **Return Statement**: If the loop completes without reaching the desired precision, it returns the last approximation. **Conclusion**: The function `my_sqrt` implements the square root calculation using Newton's method without calling the `sqrt` function from the `math` module. The import statement for `sqrt` is not used anywhere in the function. Therefore, the student did not cheat by directly calling `sqrt`. 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

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

Here's a precise implementation of a square root function using the Newton-Raphson method, which is an iterative method for finding roots: 

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def my_sqrt(n, epsilon=1e-15):
    """
    Calculate the square root of a number using Newton-Raphson method.
    
    Args:
    n (float): The number to find the square root of
    epsilon (float): The precision threshold (default: 1e-15)
    
    Returns:
    float: The square root of n
    """
    if n < 0:
        raise ValueError("Cannot calculate square root of a negative number")
    if n == 0:
        return 0
    
    # Initial guess
    x = n / 2
    
    # Newton-Raphson iteration
    while True:
        # Calculate next approximation
        next_x = (x + n / x) / 2
        
        # Check if we've reached desired precision
        if abs(next_x - x) < epsilon:
            return next_x
        
        x = next_x

# Example usage:
# number = 2
# result = my_sqrt(number)
# print(f"Square root of {number} is approximately {result}")
This implementation: 1. Uses the Newton-Raphson method, which is one of the most efficient ways to calculate square roots 2. Takes two parameters: - n: the number to find the square root of - epsilon: the desired precision (defaults to 1e-15) 3. Includes error checking for negative numbers 4. Uses the formula x_(n+1) = (x_n + n/x_n)/2 for iteration 5. Continues iterating until the difference between successive approximations is less than the specified epsilon The Newton-Raphson method typically converges quadratically, meaning the number of correct digits roughly doubles with each iteration. You can test its accuracy against the built-in math.sqrt(): 
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import math

# Test cases
numbers = [2, 3, 10, 100, 1000]
for n in numbers:
    print(f"Number: {n}")
    print(f"my_sqrt:    {my_sqrt(n)}")
    print(f"math.sqrt:  {math.sqrt(n)}")
    print(f"Difference: {abs(my_sqrt(n) - math.sqrt(n))}")
    print()
This implementation typically gives results that match Python's built-in math.sqrt() to about 15 decimal places. The epsilon value can be adjusted if you need more or less precision, but be aware that very high precision might lead to floating-point limitations in Python.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(n, epsilon=1e-15):
    """
    Calculate the square root of a number using Newton-Raphson method.
    
    Args:
    n (float): The number to find the square root of
    epsilon (float): The precision threshold (default: 1e-15)
    
    Returns:
    float: The square root of n
    """
    if n < 0:
        raise ValueError("Cannot calculate square root of a negative number")
    if n == 0:
        return 0
    
    # Initial guess
    x = n / 2
    
    # Newton-Raphson iteration
    while True:
        # Calculate next approximation
        next_x = (x + n / x) / 2
        
        # Check if we've reached desired precision
        if abs(next_x - x) < epsilon:
            return next_x
        
        x = next_x

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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def my_sqrt(n, epsilon=1e-15):
    """
    Calculate the square root of a number using Newton-Raphson method.
    
    Args:
    n (float): The number to find the square root of
    epsilon (float): The precision threshold (default: 1e-15)
    
    Returns:
    float: The square root of n
    """
    if n < 0:
        raise ValueError("Cannot calculate square root of a negative number")
    if n == 0:
        return 0
    
    # Initial guess
    x = n / 2
    
    # Newton-Raphson iteration
    while True:
        # Calculate next approximation
        next_x = (x + n / x) / 2
        
        # Check if we've reached desired precision
        if abs(next_x - x) < epsilon:
            return next_x
        
        x = next_x





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(n, epsilon=1e-15): """ Calculate the square root of a number using Newton-Raphson method. Args: n (float): The number to find the square root of epsilon (float): The precision threshold (default: 1e-15) Returns: float: The square root of n """ if n < 0: raise ValueError("Cannot calculate square root of a negative number") if n == 0: return 0 # Initial guess x = n / 2 # Newton-Raphson iteration while True: # Calculate next approximation next_x = (x + n / x) / 2 # Check if we've reached desired precision if abs(next_x - x) < epsilon: return next_x x = next_x

Output

Let's analyze the provided code to determine if the student implemented the square root calculation from scratch or if they simply called an existing function like `sqrt`. 1. **Method Used**: The student used the Newton-Raphson method, which is a well-known iterative approach to finding successively better approximations to the roots (or zeroes) of a real-valued function. In this case, it's applied to find the square root. 2. **Initial Guess**: The initial guess for the square root is set to `n / 2`, which is a common starting point for such iterative methods. 3. **Iteration Process**: The code iteratively refines the guess using the formula `next_x = (x + n / x) / 2`. This is the core of the Newton-Raphson method for square roots. 4. **Precision Check**: The loop continues until the difference between successive approximations is less than `epsilon`, ensuring the result is accurate to the specified precision. 5. **Edge Cases**: The code handles edge cases such as negative input (raising an error) and zero input (returning zero immediately). 6. **Direct Call to `sqrt`**: There is no direct call to any built-in square root function like `math.sqrt` in the code. The student has implemented the logic manually using the Newton-Raphson method. Based on this analysis, the student has implemented the square root calculation from scratch using a legitimate numerical method. There is no evidence of cheating by directly calling a built-in square root function. **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

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

Here's a way to compute the square root using the Newton-Raphson method, which is an efficient iterative method for finding roots. This method provides a very precise approximation: 

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def my_sqrt(n, epsilon=1e-15):
    """
    Compute square root using Newton-Raphson method
    
    Parameters:
    n (float): The number to find the square root of
    epsilon (float): Desired precision (default: 1e-15)
    
    Returns:
    float: Square root of n
    """
    if n < 0:
        raise ValueError("Cannot compute square root of negative number")
    if n == 0:
        return 0
    
    # Initial guess
    x = n / 2
    
    # Newton's method iteration
    while True:
        # Calculate next iteration
        next_x = (x + n / x) / 2
        
        # Check if we've reached desired precision
        if abs(next_x - x) < epsilon:
            return next_x
        
        x = next_x

# Test the function
if __name__ == "__main__":
    # Test cases
    numbers = [2, 3, 4, 9, 16, 25, 100]
    
    for num in numbers:
        result = my_sqrt(num)
        print(f"sqrt({num}) = {result}")
        # Verify accuracy
        print(f"Verification: {result}^2 = {result * result}")
        print()
This implementation: 1. Uses the Newton-Raphson method, which converges quadratically (very fast) 2. Has a default precision of 1e-15 (which you can adjust) 3. Handles edge cases (negative numbers, zero) 4. Is very accurate for most practical purposes The Newton-Raphson method works by: 1. Starting with an initial guess (in this case, n/2) 2. Iteratively improving the guess using the formula: x = (x + n/x) / 2 3. Continuing until the difference between successive approximations is less than the desired precision The function will return a result that's typically accurate to 14-15 decimal places for most numbers. Example usage: 
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print(my_sqrt(2))  # Returns approximately 1.4142135623730951
print(my_sqrt(3))  # Returns approximately 1.7320508075688772
print(my_sqrt(100))  # Returns exactly 10.0
The results are virtually identical to Python's built-in math.sqrt() function. The precision can be adjusted by changing the epsilon parameter. A smaller epsilon will give more precise results but might require more iterations. For most practical purposes, this implementation provides sufficient precision. If you need even more precision, you could: 1. Use a smaller epsilon value 2. Use arbitrary-precision arithmetic (like Python's decimal module) 3. Implement additional iterations or different numerical methods

Extract Code

I extracted the following code from that output: 

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def my_sqrt(n, epsilon=1e-15):
    """
    Compute square root using Newton-Raphson method
    
    Parameters:
    n (float): The number to find the square root of
    epsilon (float): Desired precision (default: 1e-15)
    
    Returns:
    float: Square root of n
    """
    if n < 0:
        raise ValueError("Cannot compute square root of negative number")
    if n == 0:
        return 0
    
    # Initial guess
    x = n / 2
    
    # Newton's method iteration
    while True:
        # Calculate next iteration
        next_x = (x + n / x) / 2
        
        # Check if we've reached desired precision
        if abs(next_x - x) < epsilon:
            return next_x
        
        x = next_x

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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def my_sqrt(n, epsilon=1e-15):
    """
    Compute square root using Newton-Raphson method
    
    Parameters:
    n (float): The number to find the square root of
    epsilon (float): Desired precision (default: 1e-15)
    
    Returns:
    float: Square root of n
    """
    if n < 0:
        raise ValueError("Cannot compute square root of negative number")
    if n == 0:
        return 0
    
    # Initial guess
    x = n / 2
    
    # Newton's method iteration
    while True:
        # Calculate next iteration
        next_x = (x + n / x) / 2
        
        # Check if we've reached desired precision
        if abs(next_x - x) < epsilon:
            return next_x
        
        x = next_x





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(n, epsilon=1e-15): """ Compute square root using Newton-Raphson method Parameters: n (float): The number to find the square root of epsilon (float): Desired precision (default: 1e-15) Returns: float: Square root of n """ if n < 0: raise ValueError("Cannot compute square root of negative number") if n == 0: return 0 # Initial guess x = n / 2 # Newton's method iteration while True: # Calculate next iteration next_x = (x + n / x) / 2 # Check if we've reached desired precision if abs(next_x - x) < epsilon: return next_x x = next_x

Output

Let's analyze the provided code to determine if the student implemented the square root calculation from scratch or if they cheated by directly calling a built-in function like `sqrt`. 1. **Method Used**: The student claims to use the Newton-Raphson method, which is a well-known iterative method for finding successively better approximations to the roots (or zeroes) of a real-valued function. In the context of finding square roots, this method is often referred to as the "Babylonian method." 2. **Initial Check**: The function checks if the input `n` is negative and raises a `ValueError`, which is appropriate since the square root of a negative number is not defined in the real number system. It also handles the case where `n` is zero by returning zero immediately, which is correct. 3. **Initial Guess**: The initial guess for the square root is set to `n / 2`, which is a common starting point for the Newton-Raphson method. 4. **Iterative Process**: The code enters a `while True` loop, which is typical for iterative methods where the number of iterations is not predetermined. Inside the loop: - It calculates `next_x` using the formula `(x + n / x) / 2`, which is the core of the Newton-Raphson method for square roots. - It checks if the difference between `next_x` and `x` is less than `epsilon`, indicating that the desired precision has been reached. If so, it returns `next_x` as the result. - If not, it updates `x` to `next_x` and continues the iteration. 5. **No Direct Call to `sqrt`**: The code does not contain any direct calls to a built-in square root function like `math.sqrt`. Instead, it uses an iterative method to approximate the square root. Based on this analysis, the student has implemented the square root calculation using the Newton-Raphson method without directly calling a built-in square root function. Therefore, the answer is: '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

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

Here's a function that computes the square root using the Newton-Raphson method, which is an efficient way to find roots of equations. This method provides a very precise approximation of the square root: 

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def my_sqrt(n, precision=1e-10, max_iterations=1000):
    """
    Calculate square root using Newton-Raphson method
    
    Args:
        n (float): Number to find square root of
        precision (float): Desired precision of the result
        max_iterations (int): Maximum number of iterations to prevent infinite loops
    
    Returns:
        float: Square root of n
    """
    if n < 0:
        raise ValueError("Cannot calculate square root of negative number")
    if n == 0:
        return 0
    
    # Initial guess
    x = n / 2
    
    # Newton-Raphson iteration
    for i in range(max_iterations):
        # Calculate next iteration
        next_x = (x + n / x) / 2
        
        # Check if we've reached desired precision
        if abs(next_x - x) < precision:
            return next_x
        
        x = next_x
    
    return x

# Example usage:
# print(my_sqrt(2))  # Should print something very close to 1.4142135623730951
This implementation: 1. Uses the Newton-Raphson method, which is based on the iterative formula: x₍ₙ₊₁ = (xₙ + n/xₙ)/2 2. Takes three parameters: - n: the number to find the square root of - precision: how precise you want the result (default is 1e-10) - max_iterations: to prevent infinite loops (default is 1000) 3. Includes error checking for negative numbers 4. Returns a very precise approximation of the square root You can test its accuracy like this: 
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import math

# Test cases
numbers = [2, 3, 10, 100, 1000]

for n in numbers:
    built_in = math.sqrt(n)
    custom = my_sqrt(n)
    difference = abs(built_in - custom)
    print(f"Number: {n}")
    print(f"Built-in sqrt: {built_in}")
    print(f"Custom sqrt:   {custom}")
    print(f"Difference:    {difference}")
    print()
This implementation typically gives results that match Python's built-in math.sqrt() to around 15-16 decimal places, which is about the limit of floating-point precision in Python. The Newton-Raphson method converges quadratically, which means the number of correct digits roughly doubles with each iteration. This makes it very efficient, usually requiring only 4-5 iterations to reach maximum floating-point precision for most numbers. Note that you can adjust the precision parameter if you need more or less precision, though keep in mind that Python's float type has inherent limitations on precision (typically around 15-17 significant digits).

Extract Code

I extracted the following code from that output: 

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def my_sqrt(n, precision=1e-10, max_iterations=1000):
    """
    Calculate square root using Newton-Raphson method
    
    Args:
        n (float): Number to find square root of
        precision (float): Desired precision of the result
        max_iterations (int): Maximum number of iterations to prevent infinite loops
    
    Returns:
        float: Square root of n
    """
    if n < 0:
        raise ValueError("Cannot calculate square root of negative number") 
    if n == 0:
        return 0
    
    # Initial guess
    x = n / 2
    
    # Newton-Raphson iteration
    for i in range(max_iterations):
        # Calculate next iteration
        next_x = (x + n / x) / 2
        
        # Check if we've reached desired precision
        if abs(next_x - x) < precision:
            return next_x
        
        x = next_x
    
    return x

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=1e-10, max_iterations=1000):
    """
    Calculate square root using Newton-Raphson method
    
    Args:
        n (float): Number to find square root of
        precision (float): Desired precision of the result
        max_iterations (int): Maximum number of iterations to prevent infinite loops
    
    Returns:
        float: Square root of n
    """
    if n < 0:
        raise ValueError("Cannot calculate square root of negative number") 
    if n == 0:
        return 0
    
    # Initial guess
    x = n / 2
    
    # Newton-Raphson iteration
    for i in range(max_iterations):
        # Calculate next iteration
        next_x = (x + n / x) / 2
        
        # Check if we've reached desired precision
        if abs(next_x - x) < precision:
            return next_x
        
        x = next_x
    
    return x





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(n, precision=1e-10, max_iterations=1000): """ Calculate square root using Newton-Raphson method Args: n (float): Number to find square root of precision (float): Desired precision of the result max_iterations (int): Maximum number of iterations to prevent infinite loops Returns: float: Square root of n """ if n < 0: raise ValueError("Cannot calculate square root of negative number") if n == 0: return 0 # Initial guess x = n / 2 # Newton-Raphson iteration for i in range(max_iterations): # Calculate next iteration next_x = (x + n / x) / 2 # Check if we've reached desired precision if abs(next_x - x) < precision: return next_x x = next_x return x

Output

Let's analyze the provided code to determine if the student implemented the square root calculation from scratch or if they cheated by directly calling a built-in function like `sqrt`. 1. **Method Used**: The student used the Newton-Raphson method, which is a well-known iterative technique for finding successively better approximations to the roots (or zeroes) of a real-valued function. In this case, it's used to approximate the square root. 2. **Initial Guess**: The initial guess for the square root is set to `n / 2`, which is a common starting point for such iterative methods. 3. **Iteration Process**: The code iteratively updates the guess using the formula `next_x = (x + n / x) / 2`. This is the core of the Newton-Raphson method for finding square roots. 4. **Precision Check**: The loop continues until the difference between successive approximations is less than the specified precision (`1e-10`), or until the maximum number of iterations (`1000`) is reached. This ensures that the method converges to a sufficiently accurate result. 5. **Edge Cases**: The code handles edge cases such as negative input (raising a `ValueError`) and zero input (returning 0 directly). 6. **No Direct Call to `sqrt`**: There is no direct call to any built-in square root function like `math.sqrt` in the code. The student has implemented the logic for calculating the square root manually using an iterative method. Based on this analysis, the student has implemented the square root calculation from scratch using the Newton-Raphson method. There is no evidence of cheating by directly calling a built-in square root function. **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