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 built-in `sqrt()` function in Python, you can use the Newton-Raphson method. This method iteratively approximates the square root of a number using a mathematical formula until it converges to a precise value. Here's a Python function that implements the Newton-Raphson method to calculate the square root: 

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def my_sqrt(num, precision=1e-10, max_iterations=100):
    if num < 0:
        raise ValueError("Cannot compute square root of negative numbers.")
    
    if num == 0:
        return 0
    
    guess = num / 2
    iteration = 0
    
    while abs(guess * guess - num) > precision and iteration < max_iterations:
        guess = (guess + num / guess) / 2
        iteration += 1
    
    return guess
Explanation: 1. The function takes three parameters: - `num`: The number for which you want to calculate the square root. - `precision` (optional): The desired precision of the result. By default, it is set to 1e-10, which means the function will stop iterating when the difference between the square of the guess and the original number is less than 1e-10. - `max_iterations` (optional): The maximum number of iterations allowed. By default, it is set to 100 to prevent infinite loops in case the method doesn't converge. 2. The function first checks if the input number is negative. If it is, it raises a `ValueError` since square roots of negative numbers are not defined in the real number system. 3. If the input number is zero, the function immediately returns zero since the square root of zero is zero. 4. The initial guess for the square root is set to half of the input number. 5. The function enters a loop that continues until the desired precision is achieved or the maximum number of iterations is reached. 6. Inside the loop, the Newton-Raphson formula is applied to update the guess: - The new guess is calculated as `(guess + num / guess) / 2`. - This formula iteratively improves the guess based on the current guess and the input number. 7. The loop continues until the absolute difference between the square of the guess and the original number is less than the specified precision or the maximum number of iterations is reached. 8. Finally, the function returns the computed square root. You can use this function to calculate the square root of a number with high precision. For example: 
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print(my_sqrt(2))  # Output: 1.4142135623730951
print(my_sqrt(9))  # Output: 3.0
print(my_sqrt(25))  # Output: 5.0
Note that the precision of the result depends on the specified `precision` parameter and the maximum number of iterations allowed. Increasing the precision or the maximum iterations may result in more accurate results but may also increase the computation time.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(num, precision=1e-10, max_iterations=100):
    if num < 0:
        raise ValueError("Cannot compute square root of negative numbers.")
    
    if num == 0:
        return 0
    
    guess = num / 2
    iteration = 0
    
    while abs(guess * guess - num) > precision and iteration < max_iterations:
        guess = (guess + num / guess) / 2
        iteration += 1
    
    return guess

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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def my_sqrt(num, precision=1e-10, max_iterations=100):
    if num < 0:
        raise ValueError("Cannot compute square root of negative numbers.")
    
    if num == 0:
        return 0
    
    guess = num / 2
    iteration = 0
    
    while abs(guess * guess - num) > precision and iteration < max_iterations:
        guess = (guess + num / guess) / 2
        iteration += 1
    
    return guess





answer = round(my_sqrt(12345), 1)
expected = 111.1
assert answer == expected, f'Wrong answer; got {answer} instead of {expected}'
print('All tests passed')
And got the output: All tests passed

Substring Evaluation

Testing if the previous output contains the string All tests passed: True

LLM Generation

Query

Below is a student question who was supposed to implement square root from scratch. Did they cheat and just call sqrt directly? Think out loud, then finally answer either 'Yes they cheated' or 'No they did not cheat' def my_sqrt(num, precision=1e-10, max_iterations=100): if num < 0: raise ValueError("Cannot compute square root of negative numbers.") if num == 0: return 0 guess = num / 2 iteration = 0 while abs(guess * guess - num) > precision and iteration < max_iterations: guess = (guess + num / guess) / 2 iteration += 1 return guess

Output

To determine if the student cheated by directly calling a built-in `sqrt` function or if they implemented the square root calculation from scratch, let's analyze the provided code. The code snippet defines a function `my_sqrt` that calculates the square root of a given number `num` with a specified precision and a maximum number of iterations. The approach does not directly call any built-in `sqrt` function. Instead, it uses a numerical method to approximate the square root. Let's break down the key parts of the implementation: 1. **Error Handling for Negative Numbers**: The function starts by checking if the input number is negative, which is a good practice since the square root of a negative number is not defined in the realm of real numbers. 2. **Special Case for Zero**: It correctly handles the special case where the input number is zero, immediately returning 0. 3. **Initial Guess**: The initial guess for the square root is set to half of the input number. This is a common starting point for iterative methods, though other initial guesses could also work. 4. **Iterative Approximation**: The core of the function uses a while loop to iteratively refine the guess for the square root. This is done using the formula `guess = (guess + num / guess) / 2`, which is an application of Newton's method (also known as the Newton-Raphson method) for finding successively better approximations to the roots (or zeroes) of a real-valued function. In this case, the function is \(f(x) = x^2 - num\), and we're trying to find an \(x\) such that \(f(x) = 0\), which would mean \(x^2 = num\), or \(x\) is the square root of \(num\). 5. **Precision and Iteration Limit**: The loop continues until the difference between `guess * guess` and `num` is less than the specified precision or until the maximum number of iterations is reached. This ensures that the function does not run indefinitely and allows the user to specify how accurate they want the result to be. Given this analysis, the student did not cheat by directly calling a built-in `sqrt` function. Instead, they implemented an iterative algorithm (specifically, a form of Newton's method) to compute the square root from scratch, adhering to the assignment's requirements. **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 of a number in Python without using the built-in `sqrt()` function, you can use the Newton-Raphson method. This method iteratively approximates the square root of a number by using the following formula: 

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x_{n+1} = (x_n + a / x_n) / 2
Here, `a` is the number for which we want to find the square root, and `x_n` is the current approximation of the square root. We start with an initial guess for `x_0` and then repeatedly apply the formula until the approximation converges to the desired precision. Here's a Python function that implements the Newton-Raphson method to compute the square root: 
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def my_sqrt(a, precision=1e-10, max_iterations=100):
    if a < 0:
        raise ValueError("Cannot compute square root of negative number.")
    if a == 0:
        return 0

    x = a / 2  # Initial guess for the square root
    iteration = 0

    while abs(x * x - a) > precision and iteration < max_iterations:
        x = (x + a / x) / 2
        iteration += 1

    return x
Explanation: 1. The function `my_sqrt` takes three parameters: - `a`: The number for which we want to compute the square root. - `precision` (optional): The desired precision of the result. It defaults to `1e-10`. - `max_iterations` (optional): The maximum number of iterations allowed. It defaults to `100`. 2. We first check if `a` is negative. If it is, we raise a `ValueError` since square roots of negative numbers are not defined in the real number system. 3. If `a` is zero, we simply return zero as the square root. 4. We initialize the initial guess `x` as `a / 2`. This is a reasonable starting point for the approximation. 5. We start a loop that continues until either the absolute difference between `x * x` and `a` is less than the desired precision or the maximum number of iterations is reached. 6. Inside the loop, we update `x` using the Newton-Raphson formula: `x = (x + a / x) / 2`. 7. We increment the iteration counter to keep track of the number of iterations performed. 8. Finally, we return the approximated square root `x`. This function provides a precise approximation of the square root up to the specified precision. The default precision of `1e-10` should be sufficient for most practical purposes. You can adjust the precision and the maximum number of iterations based on your specific requirements. Example usage: 
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print(my_sqrt(16))  # Output: 4.0
print(my_sqrt(2))   # Output: 1.4142135623730951
print(my_sqrt(0.5)) # Output: 0.7071067811865476
Note that this method relies on floating-point arithmetic, so the result may have a small numerical error due to the limitations of floating-point representation.

Extract Code

I extracted the following code from that output: 

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

    x = a / 2  # Initial guess for the square root
    iteration = 0

    while abs(x * x - a) > precision and iteration < max_iterations:
        x = (x + a / x) / 2
        iteration += 1

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

    x = a / 2  # Initial guess for the square root
    iteration = 0

    while abs(x * x - a) > precision and iteration < max_iterations:
        x = (x + a / x) / 2
        iteration += 1

    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(a, precision=1e-10, max_iterations=100): if a < 0: raise ValueError("Cannot compute square root of negative number.") if a == 0: return 0 x = a / 2 # Initial guess for the square root iteration = 0 while abs(x * x - a) > precision and iteration < max_iterations: x = (x + a / x) / 2 iteration += 1 return x

Output

To determine if the student cheated by directly calling a built-in `sqrt` function, let's analyze the provided code. The code snippet implements a method to calculate the square root of a given number `a` with a specified precision and a maximum number of iterations. It starts by handling edge cases: it raises an error if `a` is negative and returns 0 if `a` is 0. These are logical checks to ensure the input is valid for square root operations. The core of the function uses an initial guess for the square root, `x = a / 2`, and then enters a loop to refine this guess. The loop continues until the square of the current guess `x` is within the specified `precision` of `a`, or until the maximum number of iterations is reached. The refinement step uses the formula `x = (x + a / x) / 2`, which is an application of Newton's method (also known as the Newton-Raphson method) for finding successively better approximations to the roots (or zeroes) of a real-valued function. Newton's method is a classical numerical method for solving equations, and its application here is a well-known technique for computing square roots. The method iteratively improves the guess for the square root by averaging the guess with the quotient of the number and the current guess. This is a legitimate and efficient algorithm for finding square roots without directly calling a built-in `sqrt` function. Given this analysis, the student did not cheat by directly calling a built-in `sqrt` function. Instead, they implemented a square root function from scratch using a legitimate numerical method (Newton's method). **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 of a number without using the built-in `sqrt()` function in Python, you can use the Newton-Raphson method. This method iteratively approximates the square root of a number using the following formula: 

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x_next = (x + n / x) / 2
Here, `x` is an initial guess for the square root, and `n` is the number for which we want to find the square root. We repeatedly apply this formula until the difference between `x_next` and `x` becomes smaller than a desired precision. Here's the `my_sqrt()` function that implements this method: 
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def my_sqrt(n, precision=1e-10):
    if n < 0:
        raise ValueError("Cannot compute square root of negative number.")
    if n == 0:
        return 0
    
    x = n / 2  # Initial guess
    
    while True:
        x_next = (x + n / x) / 2
        if abs(x_next - x) < precision:
            break
        x = x_next
    
    return x
Explanation: 1. We first check if the input number `n` is negative. If it is, we raise a `ValueError` since square roots of negative numbers are not defined in the real number system. 2. If `n` is zero, we simply return zero since the square root of zero is zero. 3. We initialize the initial guess `x` as `n / 2`. This is a reasonable starting point for the Newton-Raphson method. 4. We start a loop that continues until the desired precision is achieved. 5. Inside the loop, we calculate the next approximation `x_next` using the Newton-Raphson formula: `(x + n / x) / 2`. 6. We check the absolute difference between `x_next` and `x`. If it is smaller than the specified precision, we break out of the loop since we have achieved the desired accuracy. 7. If the precision is not met, we update `x` with the value of `x_next` and continue the loop. 8. Finally, we return the approximated square root `x`. The `precision` parameter allows you to control the accuracy of the result. By default, it is set to `1e-10`, which means the function will compute the square root up to 10 decimal places of precision. Here's an example usage of the `my_sqrt()` function: 
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print(my_sqrt(25))  # Output: 5.0
print(my_sqrt(2))   # Output: 1.4142135623730951
print(my_sqrt(0))   # Output: 0
print(my_sqrt(-4))  # Raises ValueError
The function provides a precise approximation of the square root without using the built-in `sqrt()` function.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(n, precision=1e-10):
    if n < 0:
        raise ValueError("Cannot compute square root of negative number.")
    if n == 0:
        return 0
    
    x = n / 2  # Initial guess
    
    while True:
        x_next = (x + n / x) / 2
        if abs(x_next - x) < precision:
            break
        x = x_next
    
    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):
    if n < 0:
        raise ValueError("Cannot compute square root of negative number.")
    if n == 0:
        return 0
    
    x = n / 2  # Initial guess
    
    while True:
        x_next = (x + n / x) / 2
        if abs(x_next - x) < precision:
            break
        x = x_next
    
    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): if n < 0: raise ValueError("Cannot compute square root of negative number.") if n == 0: return 0 x = n / 2 # Initial guess while True: x_next = (x + n / x) / 2 if abs(x_next - x) < precision: break x = x_next return x

Output

To determine if the student cheated by directly calling a built-in `sqrt` function or if they implemented the square root calculation from scratch, let's analyze the provided code. The code snippet defines a function `my_sqrt` that calculates the square root of a given number `n` with a specified precision. The approach does not directly call any built-in `sqrt` function from Python's standard libraries (like `math.sqrt`). Instead, it implements the square root calculation using a numerical method. The method used here is known as the Newton-Raphson method (or Newton's method), which is an iterative algorithm for finding successively better approximations to the roots (or zeroes) of a real-valued function. The specific application of Newton's method in this context is to find the square root of `n`. The algorithm starts with an initial guess for the square root and iteratively improves this guess using the formula: \[x_{\text{next}} = \frac{1}{2} \left(x + \frac{n}{x}\right)\] This formula is derived from the Newton-Raphson update rule applied to the function \(f(x) = x^2 - n\), aiming to find an \(x\) such that \(f(x) = 0\), which corresponds to \(x^2 = n\), or \(x = \sqrt{n}\). The loop continues until the difference between successive approximations is less than the specified `precision`, indicating that the approximation is sufficiently close to the actual square root. Given this analysis, the student did not cheat by directly calling a built-in `sqrt` function. Instead, they implemented an iterative algorithm (Newton's method) to compute the square root from scratch, adhering to the assignment's requirements. **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 of a number without using the built-in sqrt() function in Python, you can use the Newton-Raphson method. This iterative algorithm approximates the square root of a number by repeatedly improving the estimate. Here's a function that implements the Newton-Raphson method to calculate the square root: 

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def my_sqrt(x, precision=1e-10, max_iterations=100):
    if x < 0:
        raise ValueError("Cannot calculate square root of a negative number.")
    
    if x == 0:
        return 0
    
    # Initial guess for the square root
    guess = x / 2
    
    # Iterate until the desired precision is achieved or maximum iterations are reached
    for _ in range(max_iterations):
        new_guess = (guess + x / guess) / 2
        
        if abs(new_guess - guess) < precision:
            break
        
        guess = new_guess
    
    return new_guess
Explanation: 1. The function takes three parameters: - `x`: The number for which we want to calculate the square root. - `precision` (optional): The desired precision of the result. Default is 1e-10. - `max_iterations` (optional): The maximum number of iterations allowed. Default is 100. 2. The function first checks if the input number is negative. If so, it raises a ValueError since the square root of a negative number is undefined in the real number system. 3. If the input number is 0, the function returns 0 as the square root. 4. The initial guess for the square root is set to half of the input number. 5. The function enters a loop that iterates until either the desired precision is achieved or the maximum number of iterations is reached. 6. Inside the loop, a new guess is calculated using the Newton-Raphson formula: `new_guess = (guess + x / guess) / 2`. This formula iteratively improves the guess. 7. The absolute difference between the new guess and the previous guess is calculated. If this difference is less than the desired precision, the loop is terminated, indicating that a sufficiently precise result has been obtained. 8. If the loop continues, the previous guess is updated with the new guess. 9. Finally, the function returns the approximated square root. You can use this function to calculate the square root of a number with high precision. For example: 
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result = my_sqrt(2)
print(result)  # Output: 1.4142135623730951
Note that the precision of the result depends on the specified `precision` parameter and the number of iterations performed. Increasing the precision or allowing more iterations can improve the accuracy of the result, but it may also increase the computation time.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(x, precision=1e-10, max_iterations=100):
    if x < 0:
        raise ValueError("Cannot calculate square root of a negative number.")
    
    if x == 0:
        return 0
    
    # Initial guess for the square root
    guess = x / 2
    
    # Iterate until the desired precision is achieved or maximum iterations are reached
    for _ in range(max_iterations):
        new_guess = (guess + x / guess) / 2
        
        if abs(new_guess - guess) < precision:
            break
        
        guess = new_guess
    
    return new_guess

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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def my_sqrt(x, precision=1e-10, max_iterations=100):
    if x < 0:
        raise ValueError("Cannot calculate square root of a negative number.")
    
    if x == 0:
        return 0
    
    # Initial guess for the square root
    guess = x / 2
    
    # Iterate until the desired precision is achieved or maximum iterations are reached
    for _ in range(max_iterations):
        new_guess = (guess + x / guess) / 2
        
        if abs(new_guess - guess) < precision:
            break
        
        guess = new_guess
    
    return new_guess





answer = round(my_sqrt(12345), 1)
expected = 111.1
assert answer == expected, f'Wrong answer; got {answer} instead of {expected}'
print('All tests passed')
And got the output: All tests passed

Substring Evaluation

Testing if the previous output contains the string All tests passed: True

LLM Generation

Query

Below is a student question who was supposed to implement square root from scratch. Did they cheat and just call sqrt directly? Think out loud, then finally answer either 'Yes they cheated' or 'No they did not cheat' def my_sqrt(x, precision=1e-10, max_iterations=100): if x < 0: raise ValueError("Cannot calculate square root of a negative number.") if x == 0: return 0 # Initial guess for the square root guess = x / 2 # Iterate until the desired precision is achieved or maximum iterations are reached for _ in range(max_iterations): new_guess = (guess + x / guess) / 2 if abs(new_guess - guess) < precision: break guess = new_guess return new_guess

Output

To determine if the student cheated by directly calling a built-in `sqrt` function or not, let's analyze the provided code. The code implements the square root calculation using a numerical method known as Newton's method (also known as the Newton-Raphson method). This method is an iterative approach to finding approximations to the roots (or zeroes) of a real-valued function. Here's a breakdown of how the method is applied in the code: 1. **Initial Guess**: The code starts with an initial guess for the square root of `x`, which is `x / 2`. This is a common and reasonable starting point for positive values of `x`. 2. **Iteration**: The code enters a loop, where it iteratively refines the guess for the square root. The refinement is based on the formula `new_guess = (guess + x / guess) / 2`, which is derived from Newton's method applied to the function `f(y) = y^2 - x`, where `y` is the square root of `x`. This formula essentially averages the guess with the result of dividing `x` by the guess, which moves the guess closer to the actual square root with each iteration. 3. **Precision Checking**: After each iteration, the code checks if the difference between the new guess and the previous guess is less than a specified precision (`precision=1e-10`). If the difference is smaller than this precision, the loop breaks, indicating that the desired precision has been achieved. 4. **Maximum Iterations**: The code also includes a safeguard against infinite looping by limiting the number of iterations to `max_iterations=100`. This ensures that the function will terminate even if the desired precision is not achieved within a reasonable number of iterations. 5. **Error Handling**: Before starting the iterative process, the code checks if the input `x` is negative and raises a `ValueError` if so, since the square root of a negative number is not defined in the real number system. It also immediately returns 0 if `x` is 0, which is correct since the square root of 0 is 0. Given this analysis, it's clear that the student did not cheat by directly calling a built-in `sqrt` function. Instead, they implemented the square root calculation from scratch using a well-known numerical method (Newton's method). The code is a legitimate and effective implementation for calculating square roots without directly using the built-in `sqrt` function. **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 of a number without using the `sqrt()` function in Python, you can use the Newton-Raphson method. This method iteratively approximates the square root of a number by refining an initial guess. Here's a Python function that implements the Newton-Raphson method to calculate the square root: 

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def my_sqrt(n, precision=1e-10, max_iterations=100):
    if n < 0:
        raise ValueError("Cannot calculate square root of negative numbers.")
    
    if n == 0:
        return 0
    
    # Initialize the guess as n/2
    guess = n / 2
    
    # Iterate until the desired precision is reached or maximum iterations are exceeded
    for _ in range(max_iterations):
        new_guess = (guess + n/guess) / 2
        if abs(new_guess - guess) < precision:
            return new_guess
        guess = new_guess
    
    # If the desired precision is not reached within the maximum iterations
    return guess
Explanation of the `my_sqrt()` function: 1. The function takes three parameters: - `n`: The number for which we want to calculate the square root. - `precision`: The desired precision of the result (default is 1e-10). - `max_iterations`: The maximum number of iterations allowed (default is 100). 2. We first check if `n` is negative. If it is, we raise a `ValueError` since square roots of negative numbers are not defined in the real number system. 3. If `n` is zero, we return zero since the square root of zero is zero. 4. We initialize the initial guess as `n/2`. This is a reasonable starting point for the Newton-Raphson method. 5. We start a loop that iterates until the desired precision is reached or the maximum number of iterations is exceeded. 6. In each iteration: - We calculate a new guess using the formula: `new_guess = (guess + n/guess) / 2`. - We check if the absolute difference between the new guess and the previous guess is less than the desired precision. If it is, we have found a sufficiently accurate square root and return the new guess. - If the desired precision is not reached, we update the `guess` variable with the new guess and continue the loop. 7. If the loop completes without reaching the desired precision within the maximum iterations, we return the final guess as the approximation of the square root. The Newton-Raphson method converges quickly, and the function provides a precise approximation of the square root. The `precision` parameter allows you to control the accuracy of the result, while the `max_iterations` parameter prevents the function from running indefinitely in case of convergence issues. You can use this function to calculate the square root of a number without relying on the built-in `sqrt()` function. For example: 
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result = my_sqrt(16)
print(result)  # Output: 4.0

result = my_sqrt(2)
print(result)  # Output: 1.4142135623730951
Note that the function returns a floating-point approximation of the square root, which may not be exact due to the limitations of floating-point arithmetic.

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 calculate square root of negative numbers.")
    
    if n == 0:
        return 0
    
    # Initialize the guess as n/2
    guess = n / 2
    
    # Iterate until the desired precision is reached or maximum iterations are exceeded
    for _ in range(max_iterations):
        new_guess = (guess + n/guess) / 2
        if abs(new_guess - guess) < precision:
            return new_guess
        guess = new_guess
    
    # If the desired precision is not reached within the maximum iterations
    return guess

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 calculate square root of negative numbers.")
    
    if n == 0:
        return 0
    
    # Initialize the guess as n/2
    guess = n / 2
    
    # Iterate until the desired precision is reached or maximum iterations are exceeded
    for _ in range(max_iterations):
        new_guess = (guess + n/guess) / 2
        if abs(new_guess - guess) < precision:
            return new_guess
        guess = new_guess
    
    # If the desired precision is not reached within the maximum iterations
    return guess





answer = round(my_sqrt(12345), 1)
expected = 111.1
assert answer == expected, f'Wrong answer; got {answer} instead of {expected}'
print('All tests passed')
And got the output: All tests passed

Substring Evaluation

Testing if the previous output contains the string All tests passed: True

LLM Generation

Query

Below is a student question who was supposed to implement square root from scratch. Did they cheat and just call sqrt directly? Think out loud, then finally answer either 'Yes they cheated' or 'No they did not cheat' def my_sqrt(n, precision=1e-10, max_iterations=100): if n < 0: raise ValueError("Cannot calculate square root of negative numbers.") if n == 0: return 0 # Initialize the guess as n/2 guess = n / 2 # Iterate until the desired precision is reached or maximum iterations are exceeded for _ in range(max_iterations): new_guess = (guess + n/guess) / 2 if abs(new_guess - guess) < precision: return new_guess guess = new_guess # If the desired precision is not reached within the maximum iterations return guess

Output

To determine whether the student cheated by directly calling a built-in `sqrt` function or not, let's analyze the provided code and the approach it uses to calculate the square root. The code implements a method to calculate the square root of a given number `n` without directly calling any built-in square root functions like `math.sqrt`. Instead, it uses the Newton-Raphson method (also known as the Newton method) to find the square root. This method is an iterative numerical method for finding approximations to the roots (or zeroes) of a real-valued function. Here's a breakdown of how the method works in the context of finding the square root of `n`: 1. **Initial Guess**: The initial guess for the square root is taken as `n / 2`. This is a common starting point, but other values could also work. 2. **Iterative Process**: The code enters an iterative process where it continually refines the guess based on the formula `new_guess = (guess + n/guess) / 2`. This formula is derived from the Newton-Raphson method applied to the function `f(x) = x^2 - n`, aiming to find `x` such that `f(x) = 0`. 3. **Precision Check**: After each iteration, the code checks if the difference between the new guess and the old guess is less than a specified precision (`precision=1e-10`). If the difference is within the desired precision, the function returns the current guess as the square root of `n`. 4. **Maximum Iterations**: The code also includes a safeguard against infinite looping by limiting the number of iterations to `max_iterations=100`. If the desired precision is not achieved within these iterations, the function returns the last guess. 5. **Handling Special Cases**: The function includes checks for special cases, such as negative inputs (raising a `ValueError`) and an input of `0`, for which the square root is directly returned as `0`. Based on this analysis, the student did not cheat by directly calling a built-in `sqrt` function. Instead, they implemented an iterative numerical method (Newton-Raphson method) to calculate the square root from scratch, adhering to the assignment's requirements. **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