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 the Newton-Raphson method. This method is iterative and can provide a very precise approximation of the square root. Here's how you can implement it: ### Explanation of the Newton-Raphson Method The Newton-Raphson method is an iterative method for finding successively better approximations to the roots (or zeroes) of a real-valued function. For the square root, the function is \( f(x) = x^2 - n \), where \( n \) is the number you want to find the square root of. The formula to update the approximation is: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] For \( f(x) = x^2 - n \), the derivative \( f'(x) \) is \( 2x \). Therefore, the update formula becomes: \[ x_{n+1} = x_n - \frac{x_n^2 - n}{2x_n} = \frac{x_n + \frac{n}{x_n}}{2} \] ### Implementation in Python Here's how you can implement this in Python: 

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def my_sqrt(n, precision=1e-10, max_iterations=1000):
    if n < 0:
        raise ValueError("Cannot compute square root of negative number.")
    if n == 0:
        return 0

    x = n  # Initial guess
    for _ in range(max_iterations):
        next_x = 0.5 * (x + n / x)
        if abs(next_x - x) < precision:
            return next_x
        x = next_x

    raise ValueError("Failed to converge within max iterations.")

# Example usage:
number = 25
print(f"The square root of {number} is approximately {my_sqrt(number)}")
### Explanation of the Code 1. **Initial Guess**: Start with an initial guess of `x = n`. 2. **Iteration**: Use the Newton-Raphson update formula to refine the guess. 3. **Convergence Check**: Stop iterating when the change in `x` is smaller than the desired precision. 4. **Error Handling**: Raise an error if the method fails to converge within a specified maximum number of iterations. ### Example Output 
1
The square root of 25 is approximately 5.0
This function will give you a very precise approximation of the square root of a number. You can adjust the `precision` and `max_iterations` parameters to control the accuracy and performance of the function.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(n, precision=1e-10, max_iterations=1000):
    if n < 0:
        raise ValueError("Cannot compute square root of negative number.")
    if n == 0:
        return 0

    x = n  # Initial guess
    for _ in range(max_iterations):
        next_x = 0.5 * (x + n / x)
        if abs(next_x - x) < precision:
            return next_x
        x = next_x

    raise ValueError("Failed to converge within max iterations.")

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):
    if n < 0:
        raise ValueError("Cannot compute square root of negative number.")
    if n == 0:
        return 0

    x = n  # Initial guess
    for _ in range(max_iterations):
        next_x = 0.5 * (x + n / x)
        if abs(next_x - x) < precision:
            return next_x
        x = next_x

    raise ValueError("Failed to converge within max iterations.")





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): if n < 0: raise ValueError("Cannot compute square root of negative number.") if n == 0: return 0 x = n # Initial guess for _ in range(max_iterations): next_x = 0.5 * (x + n / x) if abs(next_x - x) < precision: return next_x x = next_x raise ValueError("Failed to converge within max iterations.")

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. **Initial Checks**: - The function checks if `n` is negative and raises a `ValueError` if it is. This is a reasonable check for a square root function. - If `n` is zero, it returns zero immediately. This is also correct since the square root of zero is zero. 2. **Initial Guess**: - The initial guess for the square root is set to `n`. This is a common starting point for iterative methods. 3. **Iterative Method**: - The function uses a `for` loop to iterate up to `max_iterations` times. - In each iteration, it calculates `next_x` using the formula `0.5 * (x + n / x)`. This is the formula for the Newton-Raphson method (or Heron's method) for finding square roots. - It checks if the difference between `next_x` and `x` is less than the specified `precision`. If so, it returns `next_x` as the result. - If the difference is not small enough, it updates `x` to `next_x` and continues to the next iteration. 4. **Convergence Check**: - If the loop completes without finding a sufficiently precise result, it raises a `ValueError` indicating that it failed to converge. The Newton-Raphson method is a well-known algorithm for finding square roots, and the implementation here follows that method correctly. There is no direct call to the `sqrt` function from any library. **Conclusion**: The student did not cheat. They implemented the square root calculation from scratch using the Newton-Raphson 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

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-Raphson method, which is an iterative method for finding successively better approximations to the roots (or zeroes) of a real-valued function. Here's a function called `my_sqrt()` that implements this method: 

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def my_sqrt(number, precision=1e-10, max_iterations=1000):
    if number < 0:
        raise ValueError("Cannot compute the square root of a negative number.")

    if number == 0:
        return 0

    # Initial guess can be number / 2, or any other reasonable approximation
    guess = number / 2.0

    for _ in range(max_iterations):
        new_guess = 0.5 * (guess + number / guess)

        # Check if the difference is within the desired precision
        if abs(new_guess - guess) < precision:
            return new_guess

        guess = new_guess

    raise RuntimeError("Failed to converge within the specified number of iterations.")

# Example usage
result = my_sqrt(25)
print(f"The square root of 25 is approximately {result}")
### Explanation: 1. **Initial Guess**: Start with an initial guess. Here, `number / 2.0` is used as the initial guess. You can use other reasonable initial values as well. 2. **Newton-Raphson Iteration**: - The formula `new_guess = 0.5 * (guess + number / guess)` is derived from the Newton-Raphson method. - This formula repeatedly improves the guess by averaging the current guess and `number / guess`. 3. **Convergence Check**: - The loop continues until the difference between the new guess and the old guess is smaller than the specified precision (`precision=1e-10`). - If the loop completes the specified number of iterations (`max_iterations=1000`) without converging, a `RuntimeError` is raised. 4. **Edge Cases**: - If the input number is negative, a `ValueError` is raised. - If the input number is zero, the function immediately returns zero. You can adjust the `precision` and `max_iterations` parameters to control the accuracy and performance of the function.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(number, precision=1e-10, max_iterations=1000):
    if number < 0:
        raise ValueError("Cannot compute the square root of a negative number.")

    if number == 0:
        return 0

    # Initial guess can be number / 2, or any other reasonable approximation
    guess = number / 2.0

    for _ in range(max_iterations):
        new_guess = 0.5 * (guess + number / guess)

        # Check if the difference is within the desired precision
        if abs(new_guess - guess) < precision:
            return new_guess

        guess = new_guess

    raise RuntimeError("Failed to converge within the specified number of iterations.")

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

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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def my_sqrt(number, precision=1e-10, max_iterations=1000):
    if number < 0:
        raise ValueError("Cannot compute the square root of a negative number.")

    if number == 0:
        return 0

    # Initial guess can be number / 2, or any other reasonable approximation
    guess = number / 2.0

    for _ in range(max_iterations):
        new_guess = 0.5 * (guess + number / guess)

        # Check if the difference is within the desired precision
        if abs(new_guess - guess) < precision:
            return new_guess

        guess = new_guess

    raise RuntimeError("Failed to converge within the specified number of iterations.")

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





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(number, precision=1e-10, max_iterations=1000): if number < 0: raise ValueError("Cannot compute the square root of a negative number.") if number == 0: return 0 # Initial guess can be number / 2, or any other reasonable approximation guess = number / 2.0 for _ in range(max_iterations): new_guess = 0.5 * (guess + number / guess) # Check if the difference is within the desired precision if abs(new_guess - guess) < precision: return new_guess guess = new_guess raise RuntimeError("Failed to converge within the specified number of iterations.") # Example usage result = my_sqrt(25) print(f"The square root of 25 is approximately {result}")

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 and Input Handling**: - The function `my_sqrt` takes three parameters: `number`, `precision`, and `max_iterations`. - It checks if the input number is negative and raises a `ValueError` if so. - If the number is zero, it returns zero immediately. 2. **Initial Guess**: - The initial guess for the square root is set to `number / 2.0`. 3. **Iterative Calculation**: - The function uses a loop to iteratively improve the guess. - In each iteration, it calculates a new guess using the formula `0.5 * (guess + number / guess)`. This is known as the Babylonian method or Newton's method for finding square roots. - It checks if the difference between the new guess and the old guess is within the specified precision. If so, it returns the new guess. - If the loop completes without finding a sufficiently precise guess, it raises a `RuntimeError`. 4. **Example Usage**: - The function is called with the number 25, and the result is printed. **Thinking Out Loud**: - The student did not use the `sqrt` function from the `math` module or any other library function to directly compute the square root. - The implementation uses an iterative method (Babylonian/Newton's method) to approximate the square root, which is a valid and common approach to solving this problem from scratch. - The code includes proper handling for edge cases (negative numbers and zero) and uses a loop to refine the guess until the desired precision is achieved or the maximum number of iterations is reached. **Conclusion**: Based on the provided code, the student implemented the square root calculation from scratch using an iterative method. There is no evidence of directly calling the `sqrt` function from any library. **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 the Newton-Raphson method, which is an iterative approach to finding successively better approximations to the roots of a real-valued function. For square roots, the method can be applied as follows: \[ x_{n+1} = \frac{1}{2} \left( x_n + \frac{S}{x_n} \right) \] where \( S \) is the number for which you want to find the square root, and \( x_n \) is the current approximation. Here's a Python function called `my_sqrt()` that implements this method: 

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def my_sqrt(S, tolerance=1e-10, max_iterations=1000):
    if S < 0:
        raise ValueError("Cannot compute the square root of a negative number.")

    # Initial guess can be any number, but we'll use S / 2 for simplicity.
    x = S / 2.0

    for _ in range(max_iterations):
        next_x = 0.5 * (x + S / x)
        if abs(next_x - x) < tolerance:
            return next_x
        x = next_x

    raise RuntimeError("Failed to converge within the maximum number of iterations.")

# Example usage:
print(my_sqrt(25))  # Should be very close to 5.0
print(my_sqrt(2))  # Should be very close to the square root of 2
### Explanation: 1. **Initial Guess**: The initial guess \( x \) is set to \( S / 2.0 \). This is a reasonable starting point. 2. **Iterative Refinement**: The loop runs for a maximum number of iterations (`max_iterations`). In each iteration, it calculates the next approximation using the Newton-Raphson formula. 3. **Convergence Check**: The loop checks if the difference between the current approximation and the next approximation is less than a specified tolerance (`tolerance`). If so, it returns the current approximation as the square root. 4. **Error Handling**: The function raises an error if the number of iterations exceeds `max_iterations` without converging. This function should give you a very precise approximation of the square root of a given number.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(S, tolerance=1e-10, max_iterations=1000):
    if S < 0:
        raise ValueError("Cannot compute the square root of a negative number.")

    # Initial guess can be any number, but we'll use S / 2 for simplicity.
    x = S / 2.0

    for _ in range(max_iterations):
        next_x = 0.5 * (x + S / x)
        if abs(next_x - x) < tolerance:
            return next_x
        x = next_x

    raise RuntimeError("Failed to converge within the maximum number of iterations.")

# Example usage:
print(my_sqrt(25))  # Should be very close to 5.0
print(my_sqrt(2))  # Should be very close to the square root of 2

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, max_iterations=1000):
    if S < 0:
        raise ValueError("Cannot compute the square root of a negative number.")

    # Initial guess can be any number, but we'll use S / 2 for simplicity.
    x = S / 2.0

    for _ in range(max_iterations):
        next_x = 0.5 * (x + S / x)
        if abs(next_x - x) < tolerance:
            return next_x
        x = next_x

    raise RuntimeError("Failed to converge within the maximum number of iterations.")

# Example usage:
print(my_sqrt(25))  # Should be very close to 5.0
print(my_sqrt(2))  # Should be very close to the square root of 2





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: 5.0 1.414213562373095 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, max_iterations=1000): if S < 0: raise ValueError("Cannot compute the square root of a negative number.") # Initial guess can be any number, but we'll use S / 2 for simplicity. x = S / 2.0 for _ in range(max_iterations): next_x = 0.5 * (x + S / x) if abs(next_x - x) < tolerance: return next_x x = next_x raise RuntimeError("Failed to converge within the maximum number of iterations.") # Example usage: print(my_sqrt(25)) # Should be very close to 5.0 print(my_sqrt(2)) # Should be very close to the square root of 2

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 Validation:** - The function `my_sqrt` is defined with parameters `S`, `tolerance`, and `max_iterations`. - It checks if `S` is negative and raises a `ValueError` if so. This is a good practice for handling invalid input. 2. **Initial Guess:** - The initial guess for the square root is set to `S / 2.0`. This is a common starting point for iterative methods like the Newton-Raphson method. 3. **Iterative Calculation:** - The function uses a `for` loop to iterate up to `max_iterations` times. - In each iteration, it calculates `next_x` using the formula `0.5 * (x + S / x)`. This is the core of the Newton-Raphson method for finding square roots. - It checks if the difference between `next_x` and `x` is less than the specified `tolerance`. If so, it returns `next_x` as the result. - If the difference is not within the tolerance, it updates `x` to `next_x` and continues to the next iteration. 4. **Convergence Check:** - If the loop completes without finding a sufficiently accurate result, it raises a `RuntimeError`. 5. **Example Usage:** - The function is tested with `my_sqrt(25)` and `my_sqrt(2)`, which should return values close to 5.0 and the square root of 2, respectively. **Thinking Out Loud:** - The student has implemented an iterative method (Newton-Raphson) to approximate the square root. - There is no direct call to the `sqrt` function from any library (e.g., `math.sqrt`). - The logic and steps followed in the function are consistent with a manual implementation of the square root calculation. **Conclusion:** Based on the provided code and the analysis, the student did not cheat by directly calling the `sqrt` function. They implemented the square root calculation from scratch 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

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

Certainly! One way to compute the square root of a number without using the `sqrt()` function is to use the Newton-Raphson method. This method iteratively improves an initial guess until it converges to the square root. Here's a Python function called `my_sqrt()` that implements this method: 

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def my_sqrt(x, tolerance=1e-10, max_iterations=1000):
    if x < 0:
        raise ValueError("Cannot compute the square root of a negative number.")

    if x == 0:
        return 0

    # Initial guess can be x / 2 (or any reasonable number)
    guess = x / 2.0
    iteration = 0

    while iteration < max_iterations:
        better_guess = (guess + x / guess) / 2.0
        if abs(better_guess - guess) < tolerance:
            return better_guess
        guess = better_guess
        iteration += 1

    raise ValueError("Failed to converge within the maximum number of iterations.")

# Example usage:
print(my_sqrt(25))  # Should print 5.0
print(my_sqrt(2))   # Should print a value close to 1.41421356237
### Explanation: 1. **Initial Guess**: The initial guess is set to `x / 2.0`. This is a simple heuristic to start the iteration process. 2. **Iterative Improvement**: The Newton-Raphson formula for square roots is `better_guess = (guess + x / guess) / 2.0`. This formula is applied iteratively to improve the guess. 3. **Convergence Check**: The loop continues until the difference between the current guess and the better guess is less than a specified `tolerance`. 4. **Error Handling**: The function raises an error if the input is negative or if the algorithm fails to converge within a specified number of iterations (`max_iterations`). This approach ensures that the result is as precise as possible 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, max_iterations=1000):
    if x < 0:
        raise ValueError("Cannot compute the square root of a negative number.")

    if x == 0:
        return 0

    # Initial guess can be x / 2 (or any reasonable number)
    guess = x / 2.0
    iteration = 0

    while iteration < max_iterations:
        better_guess = (guess + x / guess) / 2.0
        if abs(better_guess - guess) < tolerance:
            return better_guess
        guess = better_guess
        iteration += 1

    raise ValueError("Failed to converge within the maximum number of iterations.")

# Example usage:
print(my_sqrt(25))  # Should print 5.0
print(my_sqrt(2))   # Should print a value 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(x, tolerance=1e-10, max_iterations=1000):
    if x < 0:
        raise ValueError("Cannot compute the square root of a negative number.")

    if x == 0:
        return 0

    # Initial guess can be x / 2 (or any reasonable number)
    guess = x / 2.0
    iteration = 0

    while iteration < max_iterations:
        better_guess = (guess + x / guess) / 2.0
        if abs(better_guess - guess) < tolerance:
            return better_guess
        guess = better_guess
        iteration += 1

    raise ValueError("Failed to converge within the maximum number of iterations.")

# Example usage:
print(my_sqrt(25))  # Should print 5.0
print(my_sqrt(2))   # Should print a value 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: 5.0 1.414213562373095 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, max_iterations=1000): if x < 0: raise ValueError("Cannot compute the square root of a negative number.") if x == 0: return 0 # Initial guess can be x / 2 (or any reasonable number) guess = x / 2.0 iteration = 0 while iteration < max_iterations: better_guess = (guess + x / guess) / 2.0 if abs(better_guess - guess) < tolerance: return better_guess guess = better_guess iteration += 1 raise ValueError("Failed to converge within the maximum number of iterations.") # Example usage: print(my_sqrt(25)) # Should print 5.0 print(my_sqrt(2)) # Should print a value close to 1.41421356237

Output

Let's analyze the provided code to determine if the student cheated by directly calling the `sqrt` function. 1. **Function Definition**: The function `my_sqrt` is defined with parameters `x`, `tolerance`, and `max_iterations`. 2. **Negative Input Handling**: 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 real number system. 3. **Zero Input Handling**: The function returns 0 if `x` is 0, which is correct since the square root of 0 is 0. 4. **Initial Guess**: The initial guess for the square root is set to `x / 2.0`. This is a reasonable starting point for the iterative method. 5. **Iteration Loop**: The function uses a `while` loop to iteratively improve the guess using the formula `(guess + x / guess) / 2.0`. This is the Newton-Raphson method for finding square roots. 6. **Convergence Check**: The loop checks if the absolute difference between the current guess and the next guess is less than the specified tolerance. If so, it returns the better guess. 7. **Iteration Limit**: The function raises a `ValueError` if it fails to converge within the maximum number of iterations. The function does not call any built-in square root functions like `math.sqrt`. Instead, it implements the Newton-Raphson method to approximate the square root. The example usage at the end demonstrates the function's correctness by computing the square roots of 25 and 2. Based on this analysis, the student did not cheat. They implemented the square root calculation from scratch using an iterative method. **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, you can use various algorithms. One common method is the Newton-Raphson method, which is an iterative approach that converges quickly to the square root. Here's how you can implement the Newton-Raphson method to compute the square root: 1. Start with an initial guess for the square root. A common choice is to start with half the number (as long as the number is positive). 2. Iteratively improve the guess using the formula: \[ \text{new\_guess} = \frac{1}{2} \left( \text{old\_guess} + \frac{\text{number}}{\text{old\_guess}} \right) \] 3. Repeat the process until the difference between the old guess and the new guess is very small (within a certain tolerance). Here's the Python function `my_sqrt()` that implements this method: 

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def my_sqrt(number, tolerance=1e-10):
    if number < 0:
        raise ValueError("Cannot compute the square root of a negative number.")

    if number == 0:
        return 0

    guess = number / 2.0
    while True:
        new_guess = 0.5 * (guess + number / guess)
        if abs(new_guess - guess) < tolerance:
            return new_guess
        guess = new_guess

# Example usage:
# result = my_sqrt(25)
# print(result)  # Output should be close to 5.0
### Explanation: - **Initial Guess**: We start with an initial guess of `number / 2.0`. - **Newton-Raphson Iteration**: We repeatedly update the guess using the formula `0.5 * (guess + number / guess)`. - **Tolerance**: We stop the iteration when the difference between the old guess and the new guess is less than the specified tolerance (`1e-10` by default). - **Edge Cases**: The function handles the case where the input is zero and raises an error for negative inputs. This function should give you a very precise approximation of the square root.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(number, tolerance=1e-10):
    if number < 0:
        raise ValueError("Cannot compute the square root of a negative number.")

    if number == 0:
        return 0

    guess = number / 2.0
    while True:
        new_guess = 0.5 * (guess + number / guess)
        if abs(new_guess - guess) < tolerance:
            return new_guess
        guess = new_guess

# Example usage:
# result = my_sqrt(25)
# print(result)  # Output should be close to 5.0

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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def my_sqrt(number, tolerance=1e-10):
    if number < 0:
        raise ValueError("Cannot compute the square root of a negative number.")

    if number == 0:
        return 0

    guess = number / 2.0
    while True:
        new_guess = 0.5 * (guess + number / guess)
        if abs(new_guess - guess) < tolerance:
            return new_guess
        guess = new_guess

# Example usage:
# result = my_sqrt(25)
# print(result)  # Output should be close to 5.0





answer = round(my_sqrt(12345), 1)
expected = 111.1
assert answer == expected, f'Wrong answer; got {answer} instead of {expected}'
print('All tests passed')
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(number, tolerance=1e-10): if number < 0: raise ValueError("Cannot compute the square root of a negative number.") if number == 0: return 0 guess = number / 2.0 while True: new_guess = 0.5 * (guess + number / guess) if abs(new_guess - guess) < tolerance: return new_guess guess = new_guess # Example usage: # result = my_sqrt(25) # print(result) # Output should be close to 5.0

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 `number` and `tolerance`. - It includes a check for negative numbers and raises a `ValueError` if the input is negative. - It handles the special case where the input number is 0 by returning 0 immediately. 2. **Initial Guess**: - The initial guess for the square root is set to `number / 2.0`. 3. **Iterative Improvement**: - The function uses a `while True` loop to iteratively improve the guess. - The new guess is calculated using the formula `0.5 * (guess + number / guess)`, which is a step in the Newton-Raphson method for finding square roots. - The loop continues until the difference between the new guess and the old guess is less than the specified tolerance. 4. **Return Statement**: - Once the tolerance condition is met, the function returns the new guess as the square root. 5. **Example Usage**: - The example usage shows how to call the function and print the result. ### Conclusion: The student implemented the square root calculation using the Newton-Raphson method, which is a well-known iterative approach to finding square roots. There is no direct call to the `sqrt` function from any library. The logic and steps are consistent with an algorithmic approach to computing square roots from scratch. **Final Answer**: No they did not cheat

Check this condition is not true:

Check if any of the following conditions are true:

Substring Evaluation

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

Substring Evaluation

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

Final Answer: False

Final Answer: True

Final Answer: True