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 without using the built-in `sqrt()` function, we can use the Newton-Raphson method. This is an iterative method that converges quickly to the square root. Here's how it works: 1. Start with an initial guess (x0) for the square root of a number (n). 2. Improve the guess using the formula: x1 = (x0 + n/x0) / 2 3. Repeat step 2 until the difference between consecutive guesses is very small. Here's a Python function that implements this method: 

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def my_sqrt(n, epsilon=1e-15, max_iterations=1000):
    if n < 0:
        raise ValueError("Cannot compute square root of a negative number")
    if n == 0:
        return 0
    
    x = n  # Initial guess
    for _ in range(max_iterations):
        next_x = (x + n / x) / 2
        if abs(next_x - x) < epsilon:
            return next_x
        x = next_x
    
    return x  # Return the best approximation if max_iterations is reached
This function, `my_sqrt()`, takes the following parameters: - `n`: The number for which we want to calculate the square root. - `epsilon`: The desired precision (default is 1e-15, which is very precise). - `max_iterations`: The maximum number of iterations to perform (default is 1000). The function first checks for negative numbers and zero. Then it applies the Newton-Raphson method until either the desired precision is reached or the maximum number of iterations is performed. You can use this function like this: 
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print(my_sqrt(2))  # Should print something very close to 1.4142135623730951
print(my_sqrt(9))  # Should print something very close to 3.0
print(my_sqrt(0.25))  # Should print something very close to 0.5
This implementation is highly precise for most practical purposes. It will give results very close to the built-in `math.sqrt()` function. The precision can be adjusted by changing the `epsilon` value, but be aware that setting it too low might increase computation time or lead to floating-point precision issues.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(n, epsilon=1e-15, max_iterations=1000):
    if n < 0:
        raise ValueError("Cannot compute square root of a negative number")
    if n == 0:
        return 0
    
    x = n  # Initial guess
    for _ in range(max_iterations):
        next_x = (x + n / x) / 2
        if abs(next_x - x) < epsilon:
            return next_x
        x = next_x
    
    return x  # Return the best approximation if max_iterations is reached

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, max_iterations=1000):
    if n < 0:
        raise ValueError("Cannot compute square root of a negative number")
    if n == 0:
        return 0
    
    x = n  # Initial guess
    for _ in range(max_iterations):
        next_x = (x + n / x) / 2
        if abs(next_x - x) < epsilon:
            return next_x
        x = next_x
    
    return x  # Return the best approximation if max_iterations is reached





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, max_iterations=1000): if n < 0: raise ValueError("Cannot compute square root of a negative number") if n == 0: return 0 x = n # Initial guess for _ in range(max_iterations): next_x = (x + n / x) / 2 if abs(next_x - x) < epsilon: return next_x x = next_x return x # Return the best approximation if max_iterations is reached

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 good practice since the square root of a negative number is not defined in the set of real numbers. - If `n` is zero, the function returns zero immediately. This is correct because the square root of zero is zero. 2. **Initial Guess**: - The initial guess for the square root is set to `n`. This is a reasonable starting point for the iterative method. 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 `(x + n / x) / 2`. This is the core of the Babylonian method (or Heron's method) for finding square roots, which is a well-known iterative algorithm. - The function checks if the difference between `next_x` and `x` is less than `epsilon`. If it is, the function returns `next_x` as the approximate square root. - If the difference is not small enough, `x` is updated to `next_x`, and the loop continues. 4. **Return Statement**: - If the loop completes without finding an approximation within the specified tolerance (`epsilon`), the function returns the last computed value of `x`. **Conclusion**: - The student implemented the Babylonian method for computing the square root, which is a valid and well-known algorithm. - There is no direct call to the `sqrt` function from any library. **Final Answer**: No they did not cheat

Check this condition is not true:

Substring Evaluation

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

Final Answer: True

Final Answer: True

Initial Query

Tell me how to compute sqrt() in python without using the sqrt() function, then give me a function to do this. I want the answer as precise as possible. Call the function my_sqrt()..

LLM Generation

Query

Tell me how to compute sqrt() in python without using the sqrt() function, then give me a function to do this. I want the answer as precise as possible. Call the function my_sqrt()..

Output

To compute the square root without using the built-in `sqrt()` function, we can use the Newton-Raphson method, which is an iterative technique for finding roots of a function. For square roots, this method converges quickly and can provide a very precise result. Here's how the Newton-Raphson method works for square roots: 1. Start with an initial guess (x). 2. Improve the guess using the formula: x = (x + n/x) / 2, where n is the number we want to find the square root of. 3. Repeat step 2 until the desired precision is achieved. Here's a Python function that implements this method: 

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def my_sqrt(n, precision=1e-10, max_iterations=100):
    if n < 0:
        raise ValueError("Cannot compute square root of a negative number")
    if n == 0:
        return 0
    
    x = n  # Initial guess
    for _ in range(max_iterations):
        root = 0.5 * (x + n / x)  # Improved guess
        if abs(root - x) < precision:
            return root
        x = root
    
    return x  # Return the best approximation if max_iterations is reached
This function, `my_sqrt()`, takes the following parameters: - `n`: The number to find the square root of. - `precision`: The desired precision (default is 1e-10, which is very precise). - `max_iterations`: The maximum number of iterations to perform (default is 100, which is usually more than enough). The function works as follows: 1. It first checks for negative numbers and raises a `ValueError` if `n` is negative. 2. If `n` is 0, it immediately returns 0. 3. It starts with an initial guess equal to `n`. 4. It then enters a loop that improves the guess using the Newton-Raphson formula. 5. The loop continues until either the desired precision is achieved or the maximum number of iterations is reached. This implementation is very precise and efficient. It will usually converge to a very accurate result within a few iterations for most numbers. You can use this function like this: 
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print(my_sqrt(2))  # Outputs approximately 1.4142135623730951
print(my_sqrt(9))  # Outputs 3.0
print(my_sqrt(0.5))  # Outputs approximately 0.7071067811865476
This function will give results that are virtually identical to the built-in `math.sqrt()` function for most practical purposes.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(n, precision=1e-10, max_iterations=100):
    if n < 0:
        raise ValueError("Cannot compute square root of a negative number")
    if n == 0:
        return 0
    
    x = n  # Initial guess
    for _ in range(max_iterations):
        root = 0.5 * (x + n / x)  # Improved guess
        if abs(root - x) < precision:
            return root
        x = root
    
    return x  # Return the best approximation if max_iterations is reached

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=100):
    if n < 0:
        raise ValueError("Cannot compute square root of a negative number")
    if n == 0:
        return 0
    
    x = n  # Initial guess
    for _ in range(max_iterations):
        root = 0.5 * (x + n / x)  # Improved guess
        if abs(root - x) < precision:
            return root
        x = root
    
    return x  # Return the best approximation if max_iterations is reached





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=100): if n < 0: raise ValueError("Cannot compute square root of a negative number") if n == 0: return 0 x = n # Initial guess for _ in range(max_iterations): root = 0.5 * (x + n / x) # Improved guess if abs(root - x) < precision: return root x = root return x # Return the best approximation if max_iterations is reached

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 good practice since the square root of a negative number is not defined in the set of real numbers. - If `n` is zero, the function returns zero immediately. This is correct because the square root of zero is zero. 2. **Initial Guess**: - The variable `x` is initialized to `n`. This is a common initial guess for iterative methods to find square roots. 3. **Iterative Improvement**: - The function uses a `for` loop to iterate up to `max_iterations` times. - Inside the loop, it calculates a new guess `root` using the formula `0.5 * (x + n / x)`. This is the iterative step of the **Babylonian method** (also known as **Newton's method**) for finding square roots. - The function checks if the difference between the new guess `root` and the previous guess `x` is less than the specified `precision`. If so, it returns `root` as the result. - If the precision condition is not met, it updates `x` to the new guess `root` and continues the iteration. 4. **Return Statement**: - If the loop completes without finding a sufficiently precise result, the function returns the last computed value of `x`. The Babylonian method is a well-known algorithm for computing square roots and does not involve directly calling any built-in square root functions like `math.sqrt`. **Conclusion**: The student implemented the square root calculation using the Babylonian method, which is a valid and common approach. There is no evidence of directly calling a built-in `sqrt` function. **Final Answer**: No they did not cheat

Check this condition is not true:

Substring Evaluation

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

Final Answer: True

Final Answer: True

Initial Query

Tell me how to compute sqrt() in python without using the sqrt() function, then give me a function to do this. I want the answer as precise as possible. Call the function my_sqrt()..

LLM Generation

Query

Tell me how to compute sqrt() in python without using the sqrt() function, then give me a function to do this. I want the answer as precise as possible. Call the function my_sqrt()..

Output

To compute the square root without using the built-in `sqrt()` function, we can use the Newton-Raphson method, which is an iterative technique for finding the roots of a function. For calculating the square root, this method converges quickly and can provide a very precise result. Here's a function that implements this method to calculate the square root: 

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def my_sqrt(x, epsilon=1e-15, max_iterations=1000):
    if x < 0:
        raise ValueError("Cannot compute square root of a negative number")
    if x == 0:
        return 0
    
    # Initial guess
    guess = x / 2
    
    for _ in range(max_iterations):
        # Newton-Raphson iteration
        next_guess = (guess + x / guess) / 2
        
        # Check for convergence
        if abs(next_guess - guess) < epsilon:
            return next_guess
        
        guess = next_guess
    
    # If we reach here, we didn't converge within max_iterations
    raise ValueError(f"Failed to converge after {max_iterations} iterations")
Let's break down this function: 1. We first check for invalid inputs (negative numbers) and the special case of 0. 2. We start with an initial guess of `x/2`. 3. We then enter a loop that will run for a maximum of `max_iterations` times. 4. In each iteration, we apply the Newton-Raphson formula: `next_guess = (guess + x / guess) / 2` 5. We check if the difference between our new guess and the previous guess is smaller than our desired precision (`epsilon`). If so, we've converged and return the result. 6. If we haven't converged, we update our guess and continue to the next iteration. 7. If we exit the loop without converging, we raise an error. The `epsilon` parameter allows you to control the precision of the result. The default value of 1e-15 will give you a very precise result in most cases. You can adjust this value if you need more or less precision. You can use this function like this: 
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print(my_sqrt(2))  # Should print something very close to 1.4142135623730951
print(my_sqrt(9))  # Should print 3.0
This implementation should give you results that are as precise as Python's floating-point arithmetic allows, which is typically equivalent to what you'd get from the built-in `math.sqrt()` function.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(x, epsilon=1e-15, max_iterations=1000):
    if x < 0:
        raise ValueError("Cannot compute square root of a negative number")
    if x == 0:
        return 0
    
    # Initial guess
    guess = x / 2
    
    for _ in range(max_iterations):
        # Newton-Raphson iteration
        next_guess = (guess + x / guess) / 2
        
        # Check for convergence
        if abs(next_guess - guess) < epsilon:
            return next_guess
        
        guess = next_guess
    
    # If we reach here, we didn't converge within max_iterations
    raise ValueError(f"Failed to converge after {max_iterations} iterations")

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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def my_sqrt(x, epsilon=1e-15, max_iterations=1000):
    if x < 0:
        raise ValueError("Cannot compute square root of a negative number")
    if x == 0:
        return 0
    
    # Initial guess
    guess = x / 2
    
    for _ in range(max_iterations):
        # Newton-Raphson iteration
        next_guess = (guess + x / guess) / 2
        
        # Check for convergence
        if abs(next_guess - guess) < epsilon:
            return next_guess
        
        guess = next_guess
    
    # If we reach here, we didn't converge within max_iterations
    raise ValueError(f"Failed to converge after {max_iterations} 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(x, epsilon=1e-15, max_iterations=1000): if x < 0: raise ValueError("Cannot compute square root of a negative number") if x == 0: return 0 # Initial guess guess = x / 2 for _ in range(max_iterations): # Newton-Raphson iteration next_guess = (guess + x / guess) / 2 # Check for convergence if abs(next_guess - guess) < epsilon: return next_guess guess = next_guess # If we reach here, we didn't converge within max_iterations raise ValueError(f"Failed to converge after {max_iterations} iterations")

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. **Initial Checks**: - The function checks if `x` is negative and raises a `ValueError` if it is. This is a good practice since the square root of a negative number is not defined in the real number system. - If `x` is zero, the function returns zero immediately. This is correct since the square root of zero is zero. 2. **Initial Guess**: - The initial guess for the square root is set to `x / 2`. This is a common starting point for iterative methods like Newton-Raphson. 3. **Newton-Raphson Iteration**: - The function uses a `for` loop to iterate up to `max_iterations` times. - In each iteration, it calculates `next_guess` using the formula `(guess + x / guess) / 2`. This is the Newton-Raphson method for finding square roots. - The function checks if the difference between `next_guess` and `guess` is less than `epsilon`. If it is, the function returns `next_guess` as the result. This is a convergence check to determine if the current guess is close enough to the actual square root. 4. **Non-Convergence Handling**: - If the loop completes without finding a sufficiently accurate result, the function raises a `ValueError` indicating that it failed to converge within the allowed number of iterations. **Conclusion**: - The student implemented the square root calculation using the Newton-Raphson method. - There is no direct call to the `sqrt` function from any library. - The logic and steps followed are consistent with an iterative approach to approximating the square root. **Final Answer**: No they did not cheat.

Check this condition is not true:

Substring Evaluation

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

Final Answer: True

Final Answer: True

Initial Query

Tell me how to compute sqrt() in python without using the sqrt() function, then give me a function to do this. I want the answer as precise as possible. Call the function my_sqrt()..

LLM Generation

Query

Tell me how to compute sqrt() in python without using the sqrt() function, then give me a function to do this. I want the answer as precise as possible. Call the function my_sqrt()..

Output

To compute the square root without using the built-in `sqrt()` function, we can use the Newton-Raphson method, which is an efficient iterative technique for finding roots of a function. Here's how we can implement it to calculate square roots: 1. Start with an initial guess (x0) for the square root of n. 2. Improve the guess using the formula: x1 = (x0 + n / x0) / 2 3. Repeat step 2 until the difference between successive guesses is very small. Here's a Python function that implements this method: 

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def my_sqrt(n, epsilon=1e-15, max_iterations=1000):
    if n < 0:
        raise ValueError("Cannot compute square root of a negative number")
    
    if n == 0 or n == 1:
        return n
    
    x = n  # Initial guess
    for _ in range(max_iterations):
        next_x = (x + n / x) / 2
        if abs(next_x - x) < epsilon:
            return next_x
        x = next_x
    
    return x  # Return the best approximation if max_iterations is reached
This function, `my_sqrt()`, takes the following parameters: - `n`: The number for which we want to calculate the square root. - `epsilon`: The desired precision (default is 1e-15, which is very precise). - `max_iterations`: The maximum number of iterations to perform (default is 1000). The function works as follows: 1. It first checks if the input is negative, raising a `ValueError` if it is. 2. It handles the special cases of 0 and 1, which are their own square roots. 3. It starts with an initial guess equal to `n`. 4. It then iteratively improves the guess using the Newton-Raphson formula. 5. The iteration stops when the difference between successive guesses is less than `epsilon` or when `max_iterations` is reached. This implementation is highly precise and efficient for most practical purposes. The use of `epsilon=1e-15` provides accuracy up to about 15 decimal places for most numbers, which is comparable to the built-in `math.sqrt()` function. You can use this function like this: 
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print(my_sqrt(2))  # Outputs approximately 1.4142135623730951
print(my_sqrt(9))  # Outputs 3.0
print(my_sqrt(0.5))  # Outputs approximately 0.7071067811865476
Note that for very large numbers or for cases where extreme precision is needed beyond 15-16 decimal places, you might need to adjust the `epsilon` value or consider using arbitrary-precision arithmetic libraries like `decimal` or `mpmath`.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(n, epsilon=1e-15, max_iterations=1000):
    if n < 0:
        raise ValueError("Cannot compute square root of a negative number")
    
    if n == 0 or n == 1:
        return n
    
    x = n  # Initial guess
    for _ in range(max_iterations):
        next_x = (x + n / x) / 2
        if abs(next_x - x) < epsilon:
            return next_x
        x = next_x
    
    return x  # Return the best approximation if max_iterations is reached

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, max_iterations=1000):
    if n < 0:
        raise ValueError("Cannot compute square root of a negative number")
    
    if n == 0 or n == 1:
        return n
    
    x = n  # Initial guess
    for _ in range(max_iterations):
        next_x = (x + n / x) / 2
        if abs(next_x - x) < epsilon:
            return next_x
        x = next_x
    
    return x  # Return the best approximation if max_iterations is reached





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, max_iterations=1000): if n < 0: raise ValueError("Cannot compute square root of a negative number") if n == 0 or n == 1: return n x = n # Initial guess for _ in range(max_iterations): next_x = (x + n / x) / 2 if abs(next_x - x) < epsilon: return next_x x = next_x return x # Return the best approximation if max_iterations is reached

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 Check for Negative Input**: 

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    if n < 0:
        raise ValueError("Cannot compute square root of a negative number")
This part of the code checks if the input `n` is negative and raises an error if it is. This is a standard check and does not involve calling `sqrt`. 2. **Special Cases for 0 and 1**: 
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    if n == 0 or n == 1:
        return n
This part handles the special cases where the square root of 0 is 0 and the square root of 1 is 1. Again, this does not involve calling `sqrt`. 3. **Initial Guess and Iterative Calculation**: 
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    x = n  # Initial guess
    for _ in range(max_iterations):
        next_x = (x + n / x) / 2
        if abs(next_x - x) < epsilon:
            return next_x
        x = next_x
Here, the student initializes `x` with the value of `n` and then iteratively improves the guess using the formula: \[ \text{next\_x} = \frac{x + \frac{n}{x}}{2} \] This is the essence of the **Babylonian method** (or **Newton's method**) for computing square roots. The loop continues until the difference between successive guesses is smaller than a specified tolerance (`epsilon`) or until a maximum number of iterations is reached. 4. **Return the Best Approximation**: 
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    return x  # Return the best approximation if max_iterations is reached
If the loop exits due to reaching the maximum number of iterations, the function returns the current approximation of the square root. **Conclusion**: - The student did not call the `sqrt` function from any library. - They implemented the square root calculation using an iterative method (Babylonian/Newton's method). **Final Answer**: No they did not cheat

Check this condition is not true:

Substring Evaluation

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

Final Answer: True

Final Answer: True

Initial Query

Tell me how to compute sqrt() in python without using the sqrt() function, then give me a function to do this. I want the answer as precise as possible. Call the function my_sqrt()..

LLM Generation

Query

Tell me how to compute sqrt() in python without using the sqrt() function, then give me a function to do this. I want the answer as precise as possible. Call the function my_sqrt()..

Output

Certainly! One common method to compute the square root without using the built-in `sqrt()` function is the Newton-Raphson method. This is an iterative method that can provide a very precise approximation of the square root. Here's how it works: 1. Start with an initial guess (x0) for the square root of a number n. 2. Improve the guess using the formula: x1 = (x0 + n / x0) / 2 3. Repeat step 2 using the new guess until the desired precision is reached. Here's a Python function that implements this method: 

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def my_sqrt(n, precision=1e-10, max_iterations=100):
    if n < 0:
        raise ValueError("Cannot compute square root of a negative number")
    if n == 0:
        return 0

    # Initial guess
    x = n / 2

    for _ in range(max_iterations):
        # Compute the next approximation
        next_x = (x + n / x) / 2
        
        # Check if we've reached the desired precision
        if abs(next_x - x) < precision:
            return next_x
        
        x = next_x

    # If we haven't converged after max_iterations, return the best approximation
    return x
This function does the following: 1. It first checks for edge cases (negative numbers and zero). 2. It sets an initial guess as half of the input number. 3. It then iteratively improves the guess using the Newton-Raphson formula. 4. The iteration continues until either the desired precision is reached or the maximum number of iterations is hit. You can use this function like this: 
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# Example usage
print(my_sqrt(2))  # Should print a number very close to 1.4142135623730951
print(my_sqrt(9))  # Should print a number very close to 3.0
print(my_sqrt(0.25))  # Should print a number very close to 0.5
The `precision` parameter allows you to control how precise you want the result to be. The default value of 1e-10 should give you a very precise result for most purposes, but you can make it even smaller if you need more precision. The `max_iterations` parameter is a safeguard to prevent infinite loops in case the method doesn't converge for some reason, although this is rare for this particular algorithm. This method is quite efficient and will converge quickly for most inputs, providing 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(n, precision=1e-10, max_iterations=100):
    if n < 0:
        raise ValueError("Cannot compute square root of a negative number")
    if n == 0:
        return 0

    # Initial guess
    x = n / 2

    for _ in range(max_iterations):
        # Compute the next approximation
        next_x = (x + n / x) / 2
        
        # Check if we've reached the desired precision
        if abs(next_x - x) < precision:
            return next_x
        
        x = next_x

    # If we haven't converged after max_iterations, return the best approximation
    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=100):
    if n < 0:
        raise ValueError("Cannot compute square root of a negative number")
    if n == 0:
        return 0

    # Initial guess
    x = n / 2

    for _ in range(max_iterations):
        # Compute the next approximation
        next_x = (x + n / x) / 2
        
        # Check if we've reached the desired precision
        if abs(next_x - x) < precision:
            return next_x
        
        x = next_x

    # If we haven't converged after max_iterations, return the best approximation
    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=100): if n < 0: raise ValueError("Cannot compute square root of a negative number") if n == 0: return 0 # Initial guess x = n / 2 for _ in range(max_iterations): # Compute the next approximation next_x = (x + n / x) / 2 # Check if we've reached the desired precision if abs(next_x - x) < precision: return next_x x = next_x # If we haven't converged after max_iterations, return the best approximation return x

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. **Initial Checks**: - The function checks if the input `n` is negative and raises a `ValueError` if it is. This is a good practice since the square root of a negative number is not defined in the set of real numbers. - If `n` is zero, the function returns zero immediately. This is correct because the square root of zero is zero. 2. **Initial Guess**: - The initial guess for the square root is set to `n / 2`. This is a common starting point for iterative methods like the Newton-Raphson method. 3. **Iterative Process**: - The function uses a `for` loop to iterate up to `max_iterations` times. - In each iteration, it calculates the next approximation using the formula `next_x = (x + n / x) / 2`. This formula is derived from the Newton-Raphson method for finding roots of equations, specifically tailored for the square root function. - The function checks if the difference between the current approximation `x` and the next approximation `next_x` is less than the specified precision. If so, it returns `next_x` as the result. - If the precision condition is not met, it updates `x` to `next_x` and continues to the next iteration. 4. **Return Statement**: - If the loop completes without meeting the precision condition, the function returns the best approximation found within the allowed number of iterations. **Conclusion**: The student implemented the square root calculation using an iterative method (Newton-Raphson method) and did not directly call the `sqrt` function from any library. The code demonstrates a clear understanding of the iterative approach to approximating the square root. **Final Answer**: No they did not cheat

Check this condition is not true:

Substring Evaluation

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

Final Answer: True

Final Answer: True