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). It's an iterative method for finding successively better approximations to the roots (or zeroes) of a real-valued function. The formula for Newton's method is: \[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\] For the square root of a number `a`, we want to solve \(x^2 - a = 0\), so: - \(f(x) = x^2 - a\) - \(f'(x) = 2x\) The iterative formula becomes: \[x_{n+1} = x_n - \frac{x_n^2 - a}{2x_n} = \frac{1}{2} \left(x_n + \frac{a}{x_n}\right)\] This means that you start with an initial guess \(x_0\) for the square root of `a`, and then use the formula to get closer and closer to the actual square root. Here is how you could implement this in Python: 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
def my_sqrt(a, tolerance=1e-10):
    if a < 0:
        raise ValueError("Cannot compute the square root of a negative number.")
    elif a == 0:
        return 0
    else:
        x = a
        while True:
            next_x = 0.5 * (x + a / x)
            if abs(x - next_x) < 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 starts with an initial guess (`x = a`) and iteratively improves the guess until the change is less than a specified tolerance (indicating that it is sufficiently close to the actual square root). The `tolerance` parameter allows you to specify how close you want the approximation to be; a smaller value will result in a more precise answer but may require more iterations.

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

 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(a, tolerance=1e-10):
    if a < 0:
        raise ValueError("Cannot compute the square root of a negative number.")
    elif a == 0:
        return 0
    else:
        x = a
        while True:
            next_x = 0.5 * (x + a / x)
            if abs(x - next_x) < 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 3.0 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(a, tolerance=1e-10): if a < 0: raise ValueError("Cannot compute the square root of a negative number.") elif a == 0: return 0 else: x = a while True: next_x = 0.5 * (x + a / x) if abs(x - next_x) < tolerance: break x = next_x return x # Example usage: number = 9 print(f"The square root of {number} is approximately {my_sqrt(number)}")

Output

Let's analyze the provided code step by step to determine if the student directly called the `sqrt` function or implemented the square root calculation from scratch. 1. **Function Definition and Input Handling:** - The function `my_sqrt` is defined with parameters `a` and `tolerance`. - It checks if `a` is negative and raises a `ValueError` if so. - If `a` is zero, it returns 0 immediately. 2. **Initial Guess and Iterative Process:** - The initial guess for the square root is set to `a`. - A `while` loop is used to iteratively improve the guess. - The next guess `next_x` is calculated using the formula `0.5 * (x + a / x)`, which is a step in the Newton-Raphson method for finding square roots. - The loop continues until the difference between the current guess `x` and the next guess `next_x` is smaller than the specified tolerance. 3. **Return Statement:** - Once the loop condition is met, the function returns the current guess `x`. 4. **Example Usage:** - The function is called with the number 9, and the result is printed. **Thinking Out Loud:** - The Newton-Raphson method is a well-known algorithm for finding square roots, and the formula used in the code (`0.5 * (x + a / x)`) is a key part of this method. - The student did not call any built-in square root function like `math.sqrt`. - The iterative process and tolerance check are consistent with an implementation from scratch. **Conclusion:** Based on the analysis, the student implemented the square root calculation using the Newton-Raphson method and did not directly call the `sqrt` 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

To compute the square root of a number in Python without directly using the `sqrt()` function from the `math` module, you can instead use the Newton's method (also known as the Newton-Raphson method). This is an iterative algorithm that can be used to approximate the roots (or zeroes) of a real-valued function. The basic idea of Newton's method for finding the square root of a number `x` is to start with an initial guess `g` for the square root, and then improve this guess iteratively using the formula: \[ g_{\text{new}} = \frac{1}{2} \left(g + \frac{x}{g}\right) \] This formula is derived from Newton's method applied to the function \( f(y) = y^2 - x \), where \( x \) is the number you want to find the square root of, and \( y \) is an approximation of the square root. Here's a Python function that implements this method to compute the square root. It includes a loop that repeatedly applies the formula until the guess does not change significantly anymore, which indicates that it has converged to a sufficiently accurate value. 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
def my_sqrt(x, tolerance=1e-10):
    if x < 0:
        raise ValueError("Cannot compute the square root of a negative number")
    elif x == 0:
        return 0
    else:
        guess = x
        while True:
            new_guess = 0.5 * (guess + x / guess)
            if abs(new_guess - guess) < tolerance:
                return new_guess
            guess = new_guess

# Example usage:
number = 25
print(f"The square root of {number} is approximately {my_sqrt(number)}")
In this function: - `x` is the number you want to find the square root of. - `tolerance` is the precision you desire for the square root computation. The loop will continue until the change in the guess between iterations is less than this value. You can adjust the `tolerance` parameter based on your needs for precision and performance. - The function checks if `x` is negative and raises a `ValueError` if so, because the square root of a negative number is not a real number (though it could be a complex number, this function does not handle such cases). This method is quite efficient and provides a very precise approximation of the square root, suitable for most practical purposes.

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

 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
def my_sqrt(x, tolerance=1e-10):
    if x < 0:
        raise ValueError("Cannot compute the square root of a negative number")
    elif x == 0:
        return 0
    else:
        guess = x
        while True:
            new_guess = 0.5 * (guess + x / guess)
            if abs(new_guess - guess) < tolerance:
                return new_guess
            guess = new_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") elif x == 0: return 0 else: guess = x while True: new_guess = 0.5 * (guess + x / guess) if abs(new_guess - guess) < tolerance: return new_guess guess = new_guess # Example usage: number = 25 print(f"The square root of {number} is approximately {my_sqrt(number)}")

Output

Let's analyze the provided code step by step to determine if the student cheated by directly calling the `sqrt` function. 1. **Function Definition and Input Handling**: - The function `my_sqrt` is defined to take an input `x` and an optional `tolerance` parameter. - It first checks if `x` is negative and raises a `ValueError` if so. - If `x` is zero, it returns 0 immediately. 2. **Initial Guess**: - For positive `x`, it initializes `guess` to `x`. 3. **Iterative Improvement**: - The function enters a `while True` loop to iteratively improve the guess. - It calculates a new guess using the formula `new_guess = 0.5 * (guess + x / guess)`. This is the Newton-Raphson method for finding square roots. - It checks if the absolute difference between the new guess and the old guess is less than the specified tolerance. If so, it returns the new guess. - Otherwise, it updates `guess` to `new_guess` and continues the loop. 4. **Example Usage**: - The example usage calculates the square root of 25 and prints the result. **Thinking Out Loud**: - The student has implemented the Newton-Raphson method, which is a well-known iterative method for finding square roots. - There is no direct call to the `sqrt` function from any library. - The logic and steps in the function are consistent with an algorithmic approach to finding square roots rather than using a built-in function. **Conclusion**: Based on the provided code, the student has implemented the square root calculation from scratch using the Newton-Raphson method. There is no evidence of cheating by directly calling the `sqrt` 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

To compute the square root of a number in Python without using the `sqrt()` function, one effective method is to use the Newton-Raphson method (also known as the Newton method). This method is an iterative algorithm used for finding successively better approximations to the roots (or zeroes) of a real-valued function. Here's a basic outline of how the Newton method can be applied to find the square root of a number `x`: 1. **Initial Guess**: Start with an initial guess for the square root. A good initial guess can be `x / 2`. 2. **Iterative Formula**: Apply the Newton-Raphson formula to find a better approximation of the square root. The formula for square root calculation is: \[ new\ guess = \frac{1}{2} \left( old\ guess + \frac{x}{old\ guess} \right) \] 3. **Repeat Until Sufficiently Accurate**: Keep applying the formula iteratively, using the new guess to replace the old guess in each iteration, until the difference between successive guesses is less than a desired threshold (which represents your accuracy requirement). 4. **Stop**: When the difference between two successive guesses is less than or equal to the threshold, consider the most recent guess as the square root of `x`. Let's implement this in a Python function called `my_sqrt()`: 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
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 / 2  # Initial guess
    while True:
        new_guess = (guess + x / guess) / 2
        if abs(new_guess - guess) < tolerance:  # Check if it is within the desired tolerance
            break
        guess = new_guess
    return guess

# Example usage:
number = 25
print(f"The square root of {number} is approximately {my_sqrt(number)}")
This function starts with a guess that the square root of `x` is `x / 2`, then iteratively refines this guess using the Newton-Raphson formula. The iteration stops when the difference between successive guesses is smaller than `tolerance`, which is set to `1e-10` by default for high precision. You can adjust the `tolerance` parameter based on your precision needs.

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
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 / 2  # Initial guess
    while True:
        new_guess = (guess + x / guess) / 2
        if abs(new_guess - guess) < tolerance:  # Check if it is within the desired tolerance
            break
        guess = new_guess
    return 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: 

 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
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 / 2  # Initial guess
    while True:
        new_guess = (guess + x / guess) / 2
        if abs(new_guess - guess) < tolerance:  # Check if it is within the desired tolerance
            break
        guess = new_guess
    return 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 5.000000000016778 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 / 2 # Initial guess while True: new_guess = (guess + x / guess) / 2 if abs(new_guess - guess) < tolerance: # Check if it is within the desired tolerance break guess = new_guess return guess # Example usage: number = 25 print(f"The square root of {number} is approximately {my_sqrt(number)}")

Output

Let's analyze the provided code step by step to determine if the student directly called the `sqrt` function or implemented the square root calculation from scratch. 1. **Function Definition and Input Handling**: - The function `my_sqrt` is defined with parameters `x` and `tolerance`. - It checks if `x` is negative and raises a `ValueError` if so. - If `x` is zero, it returns zero immediately. 2. **Initial Guess**: - The initial guess for the square root is set to `x / 2`. 3. **Iterative Improvement**: - The function enters a `while True` loop to iteratively improve the guess. - A new guess is calculated using the formula `(guess + x / guess) / 2`. - The loop breaks if the difference between the new guess and the old guess is less than the specified tolerance. - Otherwise, the guess is updated to the new guess. 4. **Return Statement**: - Once the loop breaks, the function returns the final guess. 5. **Example Usage**: - The function is called with the number 25, and the result is printed. ### Analysis: - The method used in the function is known as the **Babylonian method** or **Newton's method** for finding square roots. This is a well-known iterative algorithm for approximating square roots. - There is no direct call to Python's built-in `sqrt` function from the `math` module or any other library function that computes the square root directly. ### Conclusion: The student has implemented the square root calculation from scratch using an iterative method. They did not cheat by directly calling the `sqrt` function. **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

To compute the square root of a number in Python without using the `sqrt()` function, we can use several methods, including the Newton-Raphson method (also known as the Newton method). This method is iterative and can provide a very precise square root value for any positive number. The Newton-Raphson method starts with an initial guess and iteratively refines this guess by applying a specific formula. For finding the square root of a number \(x\), the formula to get the next guess \(n_{\text{next}}\) from the current guess \(n\) is: \[n_{\text{next}} = \frac{1}{2} \left(n + \frac{x}{n}\right)\] This formula is derived from the Newton-Raphson method applied to the function \(f(n) = n^2 - x\), where we want to find the root of the function, i.e., where \(f(n) = 0\). Here is a Python function implementing this method: 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
def my_sqrt(x, tolerance=1e-10):
    if x < 0:
        raise ValueError("Cannot compute the square root of a negative number.")
    elif x == 0:
        return 0
    else:
        guess = x
        while True:
            next_guess = (guess + x / guess) / 2
            if abs(guess - next_guess) < tolerance:
                return next_guess
            guess = next_guess

# Example usage:
number = 9
print(f"The square root of {number} is approximately {my_sqrt(number)}")
This function works as follows: - It starts with an initial guess for the square root, which is the number itself (`guess = x`). While this is a simple and oftentimes not the best initial guess, it is sufficient for demonstration purposes. - It iterates, refining the guess using the given formula, until the difference between two successive guesses is smaller than a specified tolerance (indicating that the guess is sufficiently close to the actual square root). - The tolerance is set to \(1 \times 10^{-10}\) by default, which means the function iterates until the result is accurate to within 10 decimal places. You can adjust this tolerance to be stricter or more lenient according to your needs. Remember, for very large or very small numbers, floating-point precision might affect the accuracy of the result.

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
def my_sqrt(x, tolerance=1e-10):
    if x < 0:
        raise ValueError("Cannot compute the square root of a negative number.")
    elif x == 0:
        return 0
    else:
        guess = x
        while True:
            next_guess = (guess + x / guess) / 2
            if abs(guess - next_guess) < tolerance:
                return next_guess
            guess = next_guess

# 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: 

 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
def my_sqrt(x, tolerance=1e-10):
    if x < 0:
        raise ValueError("Cannot compute the square root of a negative number.")
    elif x == 0:
        return 0
    else:
        guess = x
        while True:
            next_guess = (guess + x / guess) / 2
            if abs(guess - next_guess) < tolerance:
                return next_guess
            guess = next_guess

# 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 3.0 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.") elif x == 0: return 0 else: guess = x while True: next_guess = (guess + x / guess) / 2 if abs(guess - next_guess) < tolerance: return next_guess guess = next_guess # Example usage: number = 9 print(f"The square root of {number} is approximately {my_sqrt(number)}")

Output

Let's analyze the provided code to determine if the student directly called the `sqrt` function or implemented the square root calculation from scratch. 1. **Function Definition**: - The function `my_sqrt` is defined with parameters `x` and `tolerance`. - It starts by checking if `x` is negative, raising a `ValueError` if so. - If `x` is zero, it returns 0 immediately. 2. **Square Root Calculation**: - For positive `x`, it initializes `guess` to `x`. - It then enters a `while True` loop to iteratively improve the guess. - The next guess is calculated using the formula `(guess + x / guess) / 2`, which is the iterative method known as the **Babylonian method** or **Newton's method** for finding square roots. - The loop continues until the difference between the current guess and the next guess is smaller than the specified tolerance, at which point it returns the next guess. 3. **Example Usage**: - The function is called with `number = 9`, and the result is printed. **Thinking Out Loud**: - The student did not use the `sqrt` function from any library (e.g., `math.sqrt`). - They implemented an iterative method to approximate the square root, which is a well-known algorithm for this purpose. - The code correctly handles edge cases (negative input and zero). - The iterative approach and tolerance check are consistent with a manual implementation of the square root calculation. **Conclusion**: The student implemented the square root calculation from scratch using the Babylonian 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:

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

To compute the square root of a number without using the `sqrt()` function in Python, you can use various algorithms. One common and efficient method is the Newton-Raphson method (also known as the Newton method). This method iterates over guesses to get closer to the actual square root value. It uses the formula: \[ x_{n+1} = \frac{1}{2} \left( x_n + \frac{a}{x_n} \right) \] where \( a \) is the number you want to find the square root of, and \( x_n \) is the current guess. You iteratively improve your guess using this formula until the guess is as precise as you need it to be. Here's how you could implement this in a Python function: 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
def my_sqrt(a, epsilon=1e-10):
    if a < 0:
        raise ValueError("Cannot compute the square root of a negative number.")
    if a == 0:
        return 0
    x = a
    while True:
        # Newton-Raphson iteration
        next_x = 0.5 * (x + a / x)
        # Check for convergence
        if abs(x - next_x) < epsilon:
            break
        x = next_x
    return x

# Example usage:
print(my_sqrt(9))  # Should print a number very close to 3
print(my_sqrt(2))  # Should print a number very close to 1.41421356237
This function starts with an initial guess for the square root (in this case, the value of `a` itself, though you could choose a more sophisticated starting point if you have one). It iterates, adjusting the guess using the Newton-Raphson formula until the change between iterations is smaller than a tiny `epsilon` value, which ensures precision. The parameter `epsilon` is used to determine how precise the result should be. You can adjust this value to be smaller if you need more precision, but note that due to the limitations of floating-point arithmetic, there's a practical limit to the precision you can achieve.

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
def my_sqrt(a, epsilon=1e-10):
    if a < 0:
        raise ValueError("Cannot compute the square root of a negative number.")
    if a == 0:
        return 0
    x = a
    while True:
        # Newton-Raphson iteration
        next_x = 0.5 * (x + a / x)
        # Check for convergence
        if abs(x - next_x) < epsilon:
            break
        x = next_x
    return x

# Example usage:
print(my_sqrt(9))  # Should print a number very close to 3
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: 

 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
def my_sqrt(a, epsilon=1e-10):
    if a < 0:
        raise ValueError("Cannot compute the square root of a negative number.")
    if a == 0:
        return 0
    x = a
    while True:
        # Newton-Raphson iteration
        next_x = 0.5 * (x + a / x)
        # Check for convergence
        if abs(x - next_x) < epsilon:
            break
        x = next_x
    return x

# Example usage:
print(my_sqrt(9))  # Should print a number very close to 3
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: 3.0 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(a, epsilon=1e-10): if a < 0: raise ValueError("Cannot compute the square root of a negative number.") if a == 0: return 0 x = a while True: # Newton-Raphson iteration next_x = 0.5 * (x + a / x) # Check for convergence if abs(x - next_x) < epsilon: break x = next_x return x # Example usage: print(my_sqrt(9)) # Should print a number very close to 3 print(my_sqrt(2)) # Should print a number very close to 1.41421356237

Output

Let's analyze the provided code to determine if the student directly called the `sqrt` function or implemented the square root calculation from scratch. 1. **Function Definition**: - The function `my_sqrt` is defined with parameters `a` (the number to find the square root of) and `epsilon` (the tolerance for convergence). 2. **Handling Special Cases**: - The function raises a `ValueError` if `a` is negative, which is appropriate since the square root of a negative number is not defined in the real number system. - If `a` is zero, the function returns 0 immediately, which is correct since the square root of 0 is 0. 3. **Initial Guess**: - The initial guess for the square root, `x`, is set to `a`. 4. **Newton-Raphson Iteration**: - The function uses a `while True` loop to iteratively improve the guess `x`. - The next guess, `next_x`, is calculated using the Newton-Raphson formula: `next_x = 0.5 * (x + a / x)`. - The loop checks for convergence by comparing the absolute difference between `x` and `next_x` to `epsilon`. If the difference is smaller than `epsilon`, the loop breaks, indicating convergence. 5. **Return Value**: - Once the loop breaks, the function returns the current value of `x`, which should be a good approximation of the square root of `a`. 6. **Example Usage**: - The function is tested with `my_sqrt(9)` and `my_sqrt(2)`, which should return values close to 3 and approximately 1.41421356237, respectively. **Conclusion**: - The student implemented the square root calculation using 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. - There is no direct call to the `sqrt` function from any library (such as `math.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