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. This method converges quickly and can provide a very precise approximation of the square root. Here's a function that implements this method to calculate the square root: 

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def my_sqrt(x, epsilon=1e-15):
    if x < 0:
        raise ValueError("Cannot compute square root of a negative number")
    if x == 0:
        return 0
    
    # Initial guess
    guess = x / 2
    
    while True:
        # Newton-Raphson iteration
        new_guess = (guess + x / guess) / 2
        
        # Check for convergence
        if abs(new_guess - guess) < epsilon:
            return new_guess
        
        guess = new_guess

# Example usage
number = 2
result = my_sqrt(number)
print(f"The square root of {number} is approximately {result}")
This function, `my_sqrt()`, works as follows: 1. It first checks if the input is negative (which would result in a complex number) or zero (which has a trivial square root of 0). 2. It starts with an initial guess of `x/2`. 3. It then applies the Newton-Raphson iteration: `new_guess = (guess + x / guess) / 2` 4. It checks if the difference between the new guess and the previous guess is smaller than a very small epsilon value (default is 1e-15, which provides high precision). 5. If the difference is smaller than epsilon, it returns the result. Otherwise, it continues iterating. The epsilon value can be adjusted to trade off between precision and computation time. A smaller epsilon will give a more precise result but may take more iterations. This method is very efficient and converges quickly for most inputs. The precision of the result is limited by the floating-point precision of Python, which is typically 15-17 decimal digits for doubles. You can use this function like this: 
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print(my_sqrt(2))  # Outputs something like 1.4142135623730951
print(my_sqrt(9))  # Outputs 3.0
print(my_sqrt(0.5))  # Outputs something like 0.7071067811865476
This implementation should give results that are as precise as Python's floating-point arithmetic allows, which is typically equivalent 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(x, epsilon=1e-15):
    if x < 0:
        raise ValueError("Cannot compute square root of a negative number")
    if x == 0:
        return 0
    
    # Initial guess
    guess = x / 2
    
    while True:
        # Newton-Raphson iteration
        new_guess = (guess + x / guess) / 2
        
        # Check for convergence
        if abs(new_guess - guess) < epsilon:
            return new_guess
        
        guess = 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, epsilon=1e-15):
    if x < 0:
        raise ValueError("Cannot compute square root of a negative number")
    if x == 0:
        return 0
    
    # Initial guess
    guess = x / 2
    
    while True:
        # Newton-Raphson iteration
        new_guess = (guess + x / guess) / 2
        
        # Check for convergence
        if abs(new_guess - guess) < epsilon:
            return new_guess
        
        guess = 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, epsilon=1e-15): if x < 0: raise ValueError("Cannot compute square root of a negative number") if x == 0: return 0 # Initial guess guess = x / 2 while True: # Newton-Raphson iteration new_guess = (guess + x / guess) / 2 # Check for convergence if abs(new_guess - guess) < epsilon: return new_guess guess = new_guess

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 `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 set of real numbers. - 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 the Newton-Raphson method. 3. **Newton-Raphson Iteration**: - The function enters a `while True` loop, which will continue until the square root is found within the desired precision. - Inside the loop, a new guess is calculated using the formula: `(guess + x / guess) / 2`. This is the Newton-Raphson iteration formula for finding the square root. - The function checks if the absolute difference between the new guess and the old guess is less than `epsilon`. If it is, the function returns the new guess, indicating convergence. - If not, the guess is updated to the new guess, and the loop continues. 4. **Conclusion**: - The function does not call the `sqrt` function from any library. - The implementation uses the Newton-Raphson method to iteratively find the square root of `x`. Based on this analysis, the student did not cheat and 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 without using the built-in `sqrt()` function, we can use the Newton-Raphson method, which is an iterative technique for finding roots of functions. This method is highly efficient and can provide very precise results. Here's how the Newton-Raphson method works for square roots: 1. Start with an initial guess for the square root. 2. Improve the guess using the formula: new_guess = (guess + x / guess) / 2 3. Repeat step 2 until the desired precision is reached. Here's a Python function that implements this method: 

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def my_sqrt(x, epsilon=1e-15):
    if x < 0:
        raise ValueError("Cannot compute square root of a negative number")
    if x == 0:
        return 0
    
    guess = x / 2  # Initial guess
    while True:
        new_guess = (guess + x / guess) / 2
        if abs(new_guess - guess) < epsilon:
            return new_guess
        guess = new_guess
Let's break down this function: 1. We first check if the input `x` is negative, which would result in a complex number. We raise a `ValueError` in this case. 2. We handle the special case of x = 0 separately. 3. We start with an initial guess of `x / 2`. This is arbitrary, but it works well in practice. 4. We enter a loop where we keep improving our guess using the Newton-Raphson formula. 5. We continue this process until the difference between the new guess and the previous guess is smaller than our desired precision (`epsilon`). 6. The `epsilon` parameter allows you to control the precision. The default value of 1e-15 will give you a very precise result for most purposes. This function will compute the square root to a high degree of precision. The precision is controlled by the `epsilon` parameter, which you can adjust if needed. You can use this function like this: 
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print(my_sqrt(2))  # Approximately 1.4142135623730951
print(my_sqrt(9))  # 3.0
print(my_sqrt(0.5))  # Approximately 0.7071067811865476
This method is very efficient and will converge quickly for most inputs. However, for very large numbers or numbers very close to zero, you might need to adjust the initial guess or use a different method to maintain accuracy and efficiency.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(x, epsilon=1e-15):
    if x < 0:
        raise ValueError("Cannot compute square root of a negative number")
    if x == 0:
        return 0
    
    guess = x / 2  # Initial guess
    while True:
        new_guess = (guess + x / guess) / 2
        if abs(new_guess - guess) < epsilon:
            return new_guess
        guess = 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, epsilon=1e-15):
    if x < 0:
        raise ValueError("Cannot compute square root of a negative number")
    if x == 0:
        return 0
    
    guess = x / 2  # Initial guess
    while True:
        new_guess = (guess + x / guess) / 2
        if abs(new_guess - guess) < epsilon:
            return new_guess
        guess = 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, epsilon=1e-15): if x < 0: raise ValueError("Cannot compute square root of a negative number") if x == 0: return 0 guess = x / 2 # Initial guess while True: new_guess = (guess + x / guess) / 2 if abs(new_guess - guess) < epsilon: return new_guess guess = new_guess

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 x < 0:
       raise ValueError("Cannot compute square root of a negative number")
This part of the code checks if the input `x` is negative and raises an error if it is. This is a standard check for square root functions since the square root of a negative number is not defined in the realm of real numbers. 2. **Check for Zero Input**: 
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   if x == 0:
       return 0
This part handles the special case where the input is zero. The square root of zero is zero, so it returns 0 directly. 3. **Initial Guess**: 
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   guess = x / 2  # Initial guess
The code initializes a guess for the square root as half of the input value `x`. This is a common starting point for iterative methods to find square roots. 4. **Iterative Improvement**: 
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   while True:
       new_guess = (guess + x / guess) / 2
       if abs(new_guess - guess) < epsilon:
           return new_guess
       guess = new_guess
This part of the code uses an iterative method to improve the guess. Specifically, it uses the **Babylonian method** (also known as **Newton's method** for square roots). The formula `(guess + x / guess) / 2` is used to iteratively improve the guess until the difference between the new guess and the old guess is smaller than a specified tolerance `epsilon`. 5. **Conclusion**: - The code does not call any built-in square root function like `sqrt`. - It uses an iterative method to compute the square root, which is a valid and common approach to implement square root calculation from scratch. Based on the analysis, the student did not cheat. They implemented the square root calculation using an iterative method. **Final Answer: No they did not cheat**

Check this condition is not true:

Check if any of the following conditions are true:

Substring Evaluation

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

Substring Evaluation

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

Final Answer: False

Final Answer: True

Final Answer: True

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. This method converges quickly and can provide a very precise approximation of the square root. Here's a function that implements this method to calculate the square root: 

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def my_sqrt(n, epsilon=1e-15):
    if n < 0:
        raise ValueError("Cannot compute square root of a negative number")
    
    if n == 0:
        return 0
    
    # Initial guess
    x = n / 2
    
    while True:
        # Newton-Raphson iteration
        next_x = (x + n / x) / 2
        
        # Check for convergence
        if abs(next_x - x) < epsilon:
            return next_x
        
        x = next_x

# Example usage
print(my_sqrt(2))  # Should print something very close to 1.4142135623730951
Explanation of the function: 1. We first check if the input is negative, which would result in an imaginary number. We raise a `ValueError` in this case. 2. If the input is 0, we return 0 as its square root. 3. We start with an initial guess of `n/2`. 4. We then enter a loop where we apply the Newton-Raphson iteration: `x = (x + n/x) / 2` 5. We check if the difference between the new estimate and the previous one is smaller than our desired precision (epsilon). If so, we've converged and return the result. 6. If not converged, we update our estimate and continue the loop. The `epsilon` parameter allows you to control the precision. The default value of 1e-15 provides a very high precision, often matching the precision of the built-in `math.sqrt()` function for many inputs. This implementation is generally very fast and accurate. However, for very large numbers or numbers very close to zero, you might need to adjust the initial guess or use a different method to maintain accuracy and avoid potential overflow or underflow issues.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(n, epsilon=1e-15):
    if n < 0:
        raise ValueError("Cannot compute square root of a negative number")
    
    if n == 0:
        return 0
    
    # Initial guess
    x = n / 2
    
    while True:
        # Newton-Raphson iteration
        next_x = (x + n / x) / 2
        
        # Check for convergence
        if abs(next_x - x) < epsilon:
            return next_x
        
        x = next_x

# Example usage
print(my_sqrt(2))  # Should print something very close to 1.4142135623730951

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):
    if n < 0:
        raise ValueError("Cannot compute square root of a negative number")
    
    if n == 0:
        return 0
    
    # Initial guess
    x = n / 2
    
    while True:
        # Newton-Raphson iteration
        next_x = (x + n / x) / 2
        
        # Check for convergence
        if abs(next_x - x) < epsilon:
            return next_x
        
        x = next_x

# Example usage
print(my_sqrt(2))  # Should print something very close to 1.4142135623730951





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: 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(n, epsilon=1e-15): if n < 0: raise ValueError("Cannot compute square root of a negative number") if n == 0: return 0 # Initial guess x = n / 2 while True: # Newton-Raphson iteration next_x = (x + n / x) / 2 # Check for convergence if abs(next_x - x) < epsilon: return next_x x = next_x # Example usage print(my_sqrt(2)) # Should print something very close to 1.4142135623730951

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` is defined with parameters `n` and `epsilon`. - It checks if `n` is negative and raises a `ValueError` if so. - If `n` is zero, it returns zero immediately. 2. **Initial Guess**: - The initial guess for the square root is set to `n / 2`. 3. **Newton-Raphson Iteration**: - The code enters a `while True` loop to perform the Newton-Raphson iteration. - In each iteration, it calculates `next_x` using the formula `(x + n / x) / 2`. - It checks for convergence by comparing the absolute difference `abs(next_x - x)` with `epsilon`. - If the difference is less than `epsilon`, it returns `next_x`. - Otherwise, it updates `x` to `next_x` and continues the loop. 4. **Example Usage**: - The function is called with `my_sqrt(2)`, which should return a value close to the square root of 2. The Newton-Raphson method is a well-known algorithm for finding successively better approximations to the roots (or zeroes) of a real-valued function. In this case, it is used to approximate the square root of `n`. **Key Points**: - The student did not use the `sqrt` function from any library. - They implemented the square root calculation using the Newton-Raphson method. - The code includes proper handling for edge cases (negative input and zero). **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 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. This method is efficient and can provide a very precise result. Here's a function that implements the Newton-Raphson method to calculate the square root: 

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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 = 0.5 * (x + n / x)  # Newton-Raphson formula for sqrt
        if abs(next_x - x) < epsilon:  # Check for convergence
            return next_x
        x = next_x
    
    # If we haven't converged after max_iterations, return the best approximation
    return x
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). Here's how the function works: 1. It first checks if the input is negative, raising a `ValueError` if it is. 2. If the input is 0, it returns 0 immediately. 3. It starts with an initial guess equal to the input number. 4. It then iteratively applies the Newton-Raphson formula: x_{n+1} = 0.5 * (x_n + n / x_n) 5. It checks for convergence by comparing the difference between the new approximation and the previous one with the epsilon value. 6. If the difference is smaller than epsilon, it returns the result. 7. If it hasn't converged after the maximum number of iterations, it returns the best approximation found. This implementation can provide very precise results. The default epsilon of 1e-15 will give you about 15 decimal places of precision for most numbers. You can use this function like this: 
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print(my_sqrt(2))  # Output: 1.4142135623730951
print(my_sqrt(9))  # Output: 3.0
print(my_sqrt(0.5))  # Output: 0.7071067811865476
This method is generally very fast and accurate. However, for very large numbers or numbers very close to zero, you might need to adjust the epsilon or max_iterations parameters for optimal performance and accuracy.

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 = 0.5 * (x + n / x)  # Newton-Raphson formula for sqrt
        if abs(next_x - x) < epsilon:  # Check for convergence
            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, 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 = 0.5 * (x + n / x)  # Newton-Raphson formula for sqrt
        if abs(next_x - x) < epsilon:  # Check for convergence
            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, 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 = 0.5 * (x + n / x) # Newton-Raphson formula for sqrt if abs(next_x - x) < epsilon: # Check for convergence 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 cheated by directly calling the `sqrt` function. 1. **Function Definition and Parameters**: - The function `my_sqrt` is defined with parameters `n`, `epsilon`, and `max_iterations`. - `epsilon` is a small value to determine the precision of the result. - `max_iterations` is the maximum number of iterations allowed for the approximation. 2. **Initial Checks**: - The function raises a `ValueError` if `n` is negative, which is appropriate since the square root of a negative number is not defined in the real number system. - If `n` is zero, the function immediately returns 0, which is correct since the square root of 0 is 0. 3. **Initial Guess**: - The initial guess `x` is set to `n`. 4. **Newton-Raphson Iteration**: - The function uses the Newton-Raphson method to approximate the square root. - The formula used is `next_x = 0.5 * (x + n / x)`, which is the correct iterative formula for finding square roots. - The loop runs for a maximum of `max_iterations` times, updating `x` to `next_x` in each iteration. - The convergence check `if abs(next_x - x) < epsilon` ensures that the function returns once the approximation is within the desired precision. 5. **Return Value**: - If the loop completes without converging, the function returns the best approximation found. **Conclusion**: - The student implemented the Newton-Raphson method for computing the square root. - There is no direct call to the `sqrt` function from any library. - The logic and implementation are consistent with the requirements to compute the square root 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

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. Here's how we can implement it for square root calculation: 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 difference between consecutive guesses is very small (i.e., we've reached the desired precision). Here's a Python function that implements this method: 

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def my_sqrt(n, epsilon=1e-15):
    if n < 0:
        raise ValueError("Cannot compute square root of a negative number")
    if n == 0:
        return 0
    
    x = n  # Initial guess
    while True:
        root = (x + n / x) / 2
        if abs(root - x) < epsilon:
            return root
        x = root

# Example usage:
print(my_sqrt(2))  # Should print something very close to 1.4142135623730951
In this implementation: - `n` is the number we want to find the square root of. - `epsilon` is the desired precision (default is 1e-15, which is very precise). - We start with an initial guess of `x = n`. - We iterate using the Newton-Raphson formula until the difference between consecutive guesses is less than `epsilon`. - We handle special cases like negative numbers (raising a `ValueError`) and zero. This function will give a very precise approximation of the square root. The precision can be adjusted by changing the `epsilon` value. A smaller epsilon will give more precise results but may require more iterations. Note that for very large numbers, you might need to adjust the initial guess or use a different method to avoid potential overflow issues. This implementation works well for most common use cases.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(n, epsilon=1e-15):
    if n < 0:
        raise ValueError("Cannot compute square root of a negative number")
    if n == 0:
        return 0
    
    x = n  # Initial guess
    while True:
        root = (x + n / x) / 2
        if abs(root - x) < epsilon:
            return root
        x = root

# Example usage:
print(my_sqrt(2))  # Should print something very close to 1.4142135623730951

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):
    if n < 0:
        raise ValueError("Cannot compute square root of a negative number")
    if n == 0:
        return 0
    
    x = n  # Initial guess
    while True:
        root = (x + n / x) / 2
        if abs(root - x) < epsilon:
            return root
        x = root

# Example usage:
print(my_sqrt(2))  # Should print something very close to 1.4142135623730951





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: 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(n, epsilon=1e-15): if n < 0: raise ValueError("Cannot compute square root of a negative number") if n == 0: return 0 x = n # Initial guess while True: root = (x + n / x) / 2 if abs(root - x) < epsilon: return root x = root # Example usage: print(my_sqrt(2)) # Should print something very close to 1.4142135623730951

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 to take a number `n` and an optional parameter `epsilon` for precision. 2. **Handling Negative Numbers**: - The function raises a `ValueError` if `n` is negative, which is a reasonable check since the square root of a negative number is not defined in the real number system. 3. **Handling Zero**: - If `n` is zero, the function returns zero immediately. This is correct since the square root of zero is zero. 4. **Initial Guess**: - The initial guess for the square root is set to `n`. 5. **Iterative Approach**: - The function uses a `while True` loop to iteratively improve the guess. - The new guess `root` is calculated using the formula `(x + n / x) / 2`, which is the Newton-Raphson method for finding square roots. - The loop continues until the difference between the new guess `root` and the previous guess `x` is less than `epsilon`. 6. **Returning the Result**: - Once the condition `abs(root - x) < epsilon` is met, the function returns the computed square root. 7. **Example Usage**: - The example usage `print(my_sqrt(2))` is provided, which should print a value close to the square root of 2. **Conclusion**: - The student implemented the square root function using the Newton-Raphson method, which is a well-known iterative approach for finding square roots. - There is no direct call to the `sqrt` function from any library. - The logic and steps followed in the function are consistent with an independent implementation of the square root calculation. **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