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 the Square Root Without sqrt() To compute the square root without using the sqrt() function, I'll implement Newton's method (also known as the Newton-Raphson method). This is an iterative algorithm that quickly converges to the square root of a number. Newton's method for square roots uses the following formula: - Start with an initial guess - Repeatedly apply: x_next = 0.5 * (x + n/x) where n is the input number - Continue until desired precision is reached Here's the implementation: 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
def my_sqrt(x, epsilon=1e-15, max_iterations=100):
    """
    Compute the square root of x using Newton's method.
    
    Args:
        x (float): The number to compute the square root of
        epsilon (float): The precision threshold
        max_iterations (int): Maximum number of iterations
        
    Returns:
        float: The square root of x with high precision
    """
    if x < 0:
        raise ValueError("Cannot compute square root of a negative number")
    
    if x == 0 or x == 1:
        return x
    
    # Initial guess
    if x > 1:
        guess = x / 2
    else:  # 0 < x < 1
        guess = x
    
    # Newton's method iteration
    for _ in range(max_iterations):
        next_guess = 0.5 * (guess + x / guess)
        if abs(next_guess - guess) < epsilon:
            return next_guess
        guess = next_guess
    
    return guess
This implementation: - Handles special cases (negative numbers, zero, and one) - Uses a smart initial guess strategy depending on whether x is greater or less than 1 - Iterates until the difference between consecutive approximations is smaller than epsilon - Has a max_iterations parameter to prevent infinite loops - Achieves high precision (epsilon=1e-15) Newton's method converges quadratically, which means the number of correct digits roughly doubles with each iteration, making it very efficient.

Extract Code

I extracted the following code from that output: 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
def my_sqrt(x, epsilon=1e-15, max_iterations=100):
    """
    Compute the square root of x using Newton's method.
    
    Args:
        x (float): The number to compute the square root of
        epsilon (float): The precision threshold
        max_iterations (int): Maximum number of iterations
        
    Returns:
        float: The square root of x with high precision
    """
    if x < 0:
        raise ValueError("Cannot compute square root of a negative number")
    
    if x == 0 or x == 1:
        return x
    
    # Initial guess
    if x > 1:
        guess = x / 2
    else:  # 0 < x < 1
        guess = x
    
    # Newton's method iteration
    for _ in range(max_iterations):
        next_guess = 0.5 * (guess + x / guess)
        if abs(next_guess - guess) < epsilon:
            return next_guess
        guess = next_guess
    
    return guess

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
def my_sqrt(x, epsilon=1e-15, max_iterations=100):
    """
    Compute the square root of x using Newton's method.
    
    Args:
        x (float): The number to compute the square root of
        epsilon (float): The precision threshold
        max_iterations (int): Maximum number of iterations
        
    Returns:
        float: The square root of x with high precision
    """
    if x < 0:
        raise ValueError("Cannot compute square root of a negative number")
    
    if x == 0 or x == 1:
        return x
    
    # Initial guess
    if x > 1:
        guess = x / 2
    else:  # 0 < x < 1
        guess = x
    
    # Newton's method iteration
    for _ in range(max_iterations):
        next_guess = 0.5 * (guess + x / guess)
        if abs(next_guess - guess) < epsilon:
            return next_guess
        guess = next_guess
    
    return 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, epsilon=1e-15, max_iterations=100): """ Compute the square root of x using Newton's method. Args: x (float): The number to compute the square root of epsilon (float): The precision threshold max_iterations (int): Maximum number of iterations Returns: float: The square root of x with high precision """ if x < 0: raise ValueError("Cannot compute square root of a negative number") if x == 0 or x == 1: return x # Initial guess if x > 1: guess = x / 2 else: # 0 < x < 1 guess = x # Newton's method iteration for _ in range(max_iterations): next_guess = 0.5 * (guess + x / guess) if abs(next_guess - guess) < epsilon: return next_guess guess = next_guess return guess

Output

To determine if the student cheated by directly calling the `sqrt` function, let's analyze the provided code: 1. **Function Purpose**: The function `my_sqrt` is designed to compute the square root of a number `x` using Newton's method, which is a well-known iterative method for finding successively better approximations to the roots (or zeroes) of a real-valued function. 2. **Initial Checks**: - The function raises a `ValueError` if `x` is negative, which is appropriate since the square root of a negative number is not defined in the set of real numbers. - It returns `x` immediately if `x` is 0 or 1, which is correct because the square root of 0 is 0 and the square root of 1 is 1. 3. **Initial Guess**: - For `x > 1`, the initial guess is set to `x / 2`. - For `0 < x < 1`, the initial guess is set to `x`. These are reasonable starting points for the Newton's method iteration. 4. **Newton's Method Iteration**: - The loop iterates up to `max_iterations` times, updating the guess using the formula: `next_guess = 0.5 * (guess + x / guess)`. - The loop breaks early if the change between successive guesses is less than `epsilon`, indicating convergence to a sufficiently precise value. 5. **Return Value**: The function returns the computed square root once the desired precision is achieved or after the maximum number of iterations. **Conclusion**: The student implemented the square root calculation using Newton's method without directly calling any built-in square root function like `math.sqrt`. The logic and structure of the code align with the typical implementation of Newton's method for finding square roots. There is no evidence of cheating by directly using 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

# Computing Square Root in Python Without the sqrt() Function To compute the square root of a number without using the built-in `sqrt()` function, I'll implement the Newton-Raphson method, which is both efficient and highly accurate. ## The Newton-Raphson Method for Square Roots This iterative method uses the following formula to converge on the square root: 

1
x_next = (x + n/x) / 2
Where: - `n` is the input number we want the square root of - `x` is our current guess - `x_next` is our improved guess The key insight is that this formula will rapidly converge to √n, with each iteration approximately doubling the number of correct digits. ## Implementation Here's a precise implementation of the square root function: 
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
def my_sqrt(n, epsilon=1e-15):
    """
    Compute the square root of a number using the Newton-Raphson method.
    
    Args:
        n (float): The number to find the square root of
        epsilon (float): The desired precision
        
    Returns:
        float: The square root of n with precision up to epsilon
    """
    if n < 0:
        raise ValueError("Cannot compute square root of a negative number")
    
    if n == 0:
        return 0
    
    # Initial guess
    x = max(1.0, n)
    
    # Iterate until the desired precision is reached
    while abs(x*x - n) > epsilon * max(1.0, n):
        # Newton-Raphson formula for square root
        x = 0.5 * (x + n / x)
        
    return x
## Explanation of the Implementation 1. **Input validation**: We check if n is negative, which would result in a complex number 2. **Special case**: We handle n = 0 directly 3. **Initial guess**: We start with max(1.0, n) as our initial approximation 4. **Convergence condition**: We continue iterating until x² is sufficiently close to n, with the precision scaled appropriately for the magnitude of n 5. **Newton-Raphson step**: Each iteration uses the formula (x + n/x)/2 to improve our approximation The epsilon parameter (default 1e-15) controls the precision. This implementation provides highly accurate results while ensuring numerical stability across a wide range of inputs.

Extract Code

I extracted the following code from that output: 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
def my_sqrt(n, epsilon=1e-15):
    """
    Compute the square root of a number using the Newton-Raphson method.
    
    Args:
        n (float): The number to find the square root of
        epsilon (float): The desired precision
        
    Returns:
        float: The square root of n with precision up to epsilon
    """
    if n < 0:
        raise ValueError("Cannot compute square root of a negative number")
    
    if n == 0:
        return 0
    
    # Initial guess
    x = max(1.0, n)
    
    # Iterate until the desired precision is reached
    while abs(x*x - n) > epsilon * max(1.0, n):
        # Newton-Raphson formula for square root
        x = 0.5 * (x + n / x)
        
    return x

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
def my_sqrt(n, epsilon=1e-15):
    """
    Compute the square root of a number using the Newton-Raphson method.
    
    Args:
        n (float): The number to find the square root of
        epsilon (float): The desired precision
        
    Returns:
        float: The square root of n with precision up to epsilon
    """
    if n < 0:
        raise ValueError("Cannot compute square root of a negative number")
    
    if n == 0:
        return 0
    
    # Initial guess
    x = max(1.0, n)
    
    # Iterate until the desired precision is reached
    while abs(x*x - n) > epsilon * max(1.0, n):
        # Newton-Raphson formula for square root
        x = 0.5 * (x + n / 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, epsilon=1e-15): """ Compute the square root of a number using the Newton-Raphson method. Args: n (float): The number to find the square root of epsilon (float): The desired precision Returns: float: The square root of n with precision up to epsilon """ if n < 0: raise ValueError("Cannot compute square root of a negative number") if n == 0: return 0 # Initial guess x = max(1.0, n) # Iterate until the desired precision is reached while abs(x*x - n) > epsilon * max(1.0, n): # Newton-Raphson formula for square root x = 0.5 * (x + n / x) return x

Output

Let's analyze the provided code to determine if the student implemented the square root function from scratch or if they simply called an existing function like `sqrt`. 1. **Function Definition**: The function `my_sqrt` is defined with parameters `n` and `epsilon`. The purpose is to compute the square root of `n` using 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. 2. **Handling Edge Cases**: - The function raises a `ValueError` if `n` is negative, which is appropriate since the square root of a negative number is not defined in the real number system. - If `n` is zero, the function returns zero immediately, which is correct since the square root of zero is zero. 3. **Initial Guess**: The initial guess `x` is set to `max(1.0, n)`. This is a reasonable starting point for the Newton-Raphson method, especially for positive numbers. 4. **Iterative Process**: - The while loop continues until the difference between `x*x` and `n` is within the specified precision `epsilon`. This is a typical stopping condition for iterative methods. - The update step `x = 0.5 * (x + n / x)` is the Newton-Raphson formula specifically adapted for finding square roots. This formula is derived from the general Newton-Raphson method applied to the function \( f(x) = x^2 - n \). 5. **Return Statement**: The function returns `x`, which should be the approximate square root of `n` once the loop exits. **Conclusion**: The student has implemented the square root function using the Newton-Raphson method, which is a valid and common approach to compute square roots iteratively. There is no direct call to any built-in square root function like `math.sqrt`. The logic and steps used in the function are consistent with an original implementation of the square root calculation using an iterative method. **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