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's method for finding the square root of a number. Here's a function called my_sqrt() that implements this method: 

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def my_sqrt(x):
    # Initial guess for the square root
    guess = x / 2
    while True:
        new_guess = 0.5 * (guess + x / guess)
        if abs(new_guess - guess) < 1e-10:  # Define precision here
            return new_guess
        guess = new_guess

# Test the function
number = 25
print(my_sqrt(number))  # Output: 5.0
This function will compute the square root of a given number using the Newton's method with a specified precision. You can adjust the precision by changing the value in the `if` condition.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(x):
    # Initial guess for the square root
    guess = x / 2
    while True:
        new_guess = 0.5 * (guess + x / guess)
        if abs(new_guess - guess) < 1e-10:  # Define precision here
            return new_guess
        guess = new_guess

# Test the function
number = 25
print(my_sqrt(number))  # Output: 5.0

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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def my_sqrt(x):
    # Initial guess for the square root
    guess = x / 2
    while True:
        new_guess = 0.5 * (guess + x / guess)
        if abs(new_guess - guess) < 1e-10:  # Define precision here
            return new_guess
        guess = new_guess

# Test the function
number = 25
print(my_sqrt(number))  # Output: 5.0





answer = round(my_sqrt(12345), 1)
expected = 111.1
assert answer == expected, f'Wrong answer; got {answer} instead of {expected}'
print('All tests passed')
And got the output: 5.0 All tests passed

Substring Evaluation

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

LLM Generation

Query

Below is a student question who was supposed to implement square root from scratch. Did they cheat and just call sqrt directly? Think out loud, then finally answer either 'Yes they cheated' or 'No they did not cheat' def my_sqrt(x): # Initial guess for the square root guess = x / 2 while True: new_guess = 0.5 * (guess + x / guess) if abs(new_guess - guess) < 1e-10: # Define precision here return new_guess guess = new_guess # Test the function number = 25 print(my_sqrt(number)) # Output: 5.0

Output

Let's analyze the provided code step by step to determine if the student directly called the `sqrt` function or implemented the square root calculation from scratch. 1. **Function Definition**: 

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   def my_sqrt(x):
The function `my_sqrt` is defined to take one argument `x`. 2. **Initial Guess**: 
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   guess = x / 2
The initial guess for the square root is set to half of `x`. 3. **Iteration Loop**: 
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   while True:
       new_guess = 0.5 * (guess + x / guess)
       if abs(new_guess - guess) < 1e-10:
           return new_guess
       guess = new_guess
- The loop runs indefinitely (`while True`). - A new guess is calculated using the formula `0.5 * (guess + x / guess)`, which is a step in the Newton-Raphson method for finding square roots. - The loop checks if the difference between the new guess and the old guess is less than a small threshold (`1e-10`). If it is, the function returns the new guess. - If not, the guess is updated to the new guess, and the loop continues. 4. **Testing the Function**: 
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   number = 25
   print(my_sqrt(number))  # Output: 5.0
The function is tested with the number 25, and the expected output is 5.0. **Thinking Out Loud**: - The student did not use the `sqrt` function from any library (e.g., `math.sqrt`). - They implemented the Newton-Raphson method, which is a well-known iterative method for finding square roots. - The code correctly follows the steps of the Newton-Raphson method: making an initial guess, iteratively improving the guess, and checking for convergence. **Conclusion**: The student implemented the square root calculation from scratch using the Newton-Raphson method. They did not directly call the `sqrt` function from any library. **Final Answer**: No they did not cheat.

Check this condition is not true:

Check if any of the following conditions are true:

Substring Evaluation

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

Substring Evaluation

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

Final Answer: False

Final Answer: True

Final Answer: True

Initial Query

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

LLM Generation

Query

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

Output

To compute the square root without using the sqrt() function in Python, you can use the Newton's method for approximating square roots. Here is a function called my_sqrt() that implements this method: 

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def my_sqrt(num):
    if num < 0:
        return "Invalid input. Cannot compute square root of a negative number."
    
    if num == 0:
        return 0
    
    x = num
    y = (x + 1) // 2
    while y < x:
        x = y
        y = (x + num // x) // 2
        
    return x
You can use this function to compute the square root of a number like this: 
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number = 25
result = my_sqrt(number)
print(f"The square root of {number} is approximately {result}")
This function will give you a precise result for the square root of a given number without using the sqrt() function in Python.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(num):
    if num < 0:
        return "Invalid input. Cannot compute square root of a negative number."
    
    if num == 0:
        return 0
    
    x = num
    y = (x + 1) // 2
    while y < x:
        x = y
        y = (x + num // x) // 2
        
    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(num):
    if num < 0:
        return "Invalid input. Cannot compute square root of a negative number."
    
    if num == 0:
        return 0
    
    x = num
    y = (x + 1) // 2
    while y < x:
        x = y
        y = (x + num // x) // 2
        
    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: Traceback (most recent call last): File "/usr/src/app/main.py", line 23, in <module> assert answer == expected, f'Wrong answer; got {answer} instead of {expected}' AssertionError: Wrong answer; got 111 instead of 111.1

Substring Evaluation

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

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): if num < 0: return "Invalid input. Cannot compute square root of a negative number." if num == 0: return 0 x = num y = (x + 1) // 2 while y < x: x = y y = (x + num // x) // 2 return x

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 first checks if the input number is negative. If so, it returns an error message. - If the input number is zero, it returns zero. These are valid initial checks and do not indicate cheating. 2. **Variable Initialization**: - The variable `x` is initialized to the input number `num`. - The variable `y` is initialized to `(x + 1) // 2`. This is a common initial guess for iterative methods to find square roots. 3. **Iterative Method**: - The code enters a `while` loop that continues as long as `y` is less than `x`. - Inside the loop, `x` is updated to `y`. - `y` is then updated to `(x + num // x) // 2`. This is a classic implementation of the Babylonian method (also known as Heron's method) for computing square roots. 4. **Return Statement**: - Finally, the function returns `x`, which should be the computed square root of the input number. The Babylonian method is a well-known algorithm for finding square roots and does not involve directly calling the `sqrt` function from any library. The student's implementation follows this method correctly. **Conclusion**: Based on the analysis, the student did not directly call the `sqrt` function and instead implemented the square root calculation using an iterative method. Therefore, the answer is: **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: False

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 sqrt() function in Python, you can use the Newton's method for finding the square root. Here's the function to calculate the square root using this method: 

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def my_sqrt(num):
    if num < 0:
        return "Square root is not defined for negative numbers"
    
    guess = num / 2  # Initial guess
    tolerance = 0.0001
    
    while abs(guess**2 - num) > tolerance:
        guess = (guess + num / guess) / 2
        
    return guess

# Test the function
number = 25
print(f"The square root of {number} is {my_sqrt(number)}")
This function `my_sqrt()` uses Newton's method to approximate the square root of a given number. You can adjust the tolerance value for more or less precision in the result.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(num):
    if num < 0:
        return "Square root is not defined for negative numbers"
    
    guess = num / 2  # Initial guess
    tolerance = 0.0001
    
    while abs(guess**2 - num) > tolerance:
        guess = (guess + num / guess) / 2
        
    return guess

# Test the function
number = 25
print(f"The square root of {number} is {my_sqrt(number)}")

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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def my_sqrt(num):
    if num < 0:
        return "Square root is not defined for negative numbers"
    
    guess = num / 2  # Initial guess
    tolerance = 0.0001
    
    while abs(guess**2 - num) > tolerance:
        guess = (guess + num / guess) / 2
        
    return guess

# Test the function
number = 25
print(f"The square root of {number} is {my_sqrt(number)}")





answer = round(my_sqrt(12345), 1)
expected = 111.1
assert answer == expected, f'Wrong answer; got {answer} instead of {expected}'
print('All tests passed')
And got the output: The square root of 25 is 5.000000000016778 All tests passed

Substring Evaluation

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

LLM Generation

Query

Below is a student question who was supposed to implement square root from scratch. Did they cheat and just call sqrt directly? Think out loud, then finally answer either 'Yes they cheated' or 'No they did not cheat' def my_sqrt(num): if num < 0: return "Square root is not defined for negative numbers" guess = num / 2 # Initial guess tolerance = 0.0001 while abs(guess**2 - num) > tolerance: guess = (guess + num / guess) / 2 return guess # Test the function number = 25 print(f"The square root of {number} is {my_sqrt(number)}")

Output

Let's analyze the provided code to determine if the student directly called the `sqrt` function or implemented the square root calculation from scratch. 1. **Function Definition**: - The function `my_sqrt` is defined to take one parameter `num`. 2. **Handling Negative Numbers**: - The function checks if `num` is less than 0 and returns a message indicating that the square root is not defined for negative numbers. This is a good practice and does not involve calling `sqrt`. 3. **Initial Guess**: - The initial guess for the square root is set to `num / 2`. This is a common approach in iterative methods for finding square roots. 4. **Tolerance Level**: - A tolerance level of `0.0001` is set to determine the accuracy of the result. This is also a standard practice in numerical methods. 5. **Iterative Method**: - The function uses a `while` loop to iteratively improve the guess. The condition for the loop is `abs(guess**2 - num) > tolerance`, which ensures the loop continues until the guess squared is close enough to `num`. - Inside the loop, the guess is updated using the formula `(guess + num / guess) / 2`. This is the iterative formula for the Babylonian method (or Heron's method) for finding square roots. 6. **Return Statement**: - Once the loop exits, the function returns the final guess. 7. **Testing the Function**: - The function is tested with the number 25, and the result is printed. **Conclusion**: - The student implemented the square root calculation using the Babylonian method, which is a well-known iterative approach. - 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:

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 sqrt() function in Python, you can use the Newton's method. Here's an implementation of the my_sqrt() function using this method: 

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def my_sqrt(num):
    if num < 0:
        raise ValueError("Cannot compute square root of a negative number")
    
    guess = num / 2
    while True:
        new_guess = 0.5 * (guess + num / guess)
        if abs(new_guess - guess) < 1e-10:
            return new_guess
        guess = new_guess

# Test the function
number = 25
result = my_sqrt(number)
print(f"The square root of {number} is approximately {result}")
This function will calculate the square root of a given number using the Newton's method and return the result with a precision of 1e-10. You can adjust the precision by changing the value in the condition `abs(new_guess - guess) < 1e-10`.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(num):
    if num < 0:
        raise ValueError("Cannot compute square root of a negative number")
    
    guess = num / 2
    while True:
        new_guess = 0.5 * (guess + num / guess)
        if abs(new_guess - guess) < 1e-10:
            return new_guess
        guess = new_guess

# Test the function
number = 25
result = my_sqrt(number)
print(f"The square root of {number} is approximately {result}")

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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def my_sqrt(num):
    if num < 0:
        raise ValueError("Cannot compute square root of a negative number")
    
    guess = num / 2
    while True:
        new_guess = 0.5 * (guess + num / guess)
        if abs(new_guess - guess) < 1e-10:
            return new_guess
        guess = new_guess

# Test the function
number = 25
result = my_sqrt(number)
print(f"The square root of {number} is approximately {result}")





answer = round(my_sqrt(12345), 1)
expected = 111.1
assert answer == expected, f'Wrong answer; got {answer} instead of {expected}'
print('All tests passed')
And got the output: The square root of 25 is approximately 5.0 All tests passed

Substring Evaluation

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

LLM Generation

Query

Below is a student question who was supposed to implement square root from scratch. Did they cheat and just call sqrt directly? Think out loud, then finally answer either 'Yes they cheated' or 'No they did not cheat' def my_sqrt(num): if num < 0: raise ValueError("Cannot compute square root of a negative number") guess = num / 2 while True: new_guess = 0.5 * (guess + num / guess) if abs(new_guess - guess) < 1e-10: return new_guess guess = new_guess # Test the function number = 25 result = my_sqrt(number) print(f"The square root of {number} is approximately {result}")

Output

Let's analyze the provided code step by step to determine if the student cheated by directly calling the `sqrt` function. 1. **Function Definition**: 

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   def my_sqrt(num):
The function `my_sqrt` is defined to take one argument `num`. 2. **Negative Number Check**: 
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   if num < 0:
       raise ValueError("Cannot compute square root of a negative number")
The function raises a `ValueError` if the input number is negative, which is a reasonable check for a square root function. 3. **Initial Guess**: 
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   guess = num / 2
The initial guess for the square root is set to half of the input number. This is a common starting point for iterative methods. 4. **Iterative Improvement**: 
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   while True:
       new_guess = 0.5 * (guess + num / guess)
       if abs(new_guess - guess) < 1e-10:
           return new_guess
       guess = new_guess
The function uses an iterative method to improve the guess. Specifically, it uses the Newton-Raphson method (also known as the Heron's method) to approximate the square root: - `new_guess` is calculated as the average of `guess` and `num / guess`. - The loop continues until the difference between `new_guess` and `guess` is smaller than a very small threshold (`1e-10`), indicating convergence. 5. **Testing the Function**: 
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   number = 25
   result = my_sqrt(number)
   print(f"The square root of {number} is approximately {result}")
The function is tested with the number 25, and the result is printed. ### Conclusion The student implemented the square root function using the Newton-Raphson iterative method. They did not call the `sqrt` function from any library directly. The approach they used is a well-known algorithm for computing square roots and is a valid method for implementing it 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 sqrt() function in Python, you can use the Newton's method for finding roots. Here is a function called my_sqrt() that calculates the square root of a given number: 

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def my_sqrt(num):
    if num < 0:
        return "Invalid input, square root of negative number is not real"
    
    # Initial guess for the square root
    x = num / 2
    
    # Iterate until the difference between x and num/x is very small
    while abs(x * x - num) > 0.0000001:
        x = (x + num / x) / 2
        
    return x

# Test the function
num = 25
print(f"The square root of {num} is: {my_sqrt(num)}")
This my_sqrt() function uses the Newton's method to iteratively improve the estimate of the square root until it reaches a sufficiently precise value. You can adjust the precision by changing the value in the while loop condition.

Extract Code

I extracted the following code from that output: 

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def my_sqrt(num):
    if num < 0:
        return "Invalid input, square root of negative number is not real"
    
    # Initial guess for the square root
    x = num / 2
    
    # Iterate until the difference between x and num/x is very small
    while abs(x * x - num) > 0.0000001:
        x = (x + num / x) / 2
        
    return x

# Test the function
num = 25
print(f"The square root of {num} is: {my_sqrt(num)}")

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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def my_sqrt(num):
    if num < 0:
        return "Invalid input, square root of negative number is not real"
    
    # Initial guess for the square root
    x = num / 2
    
    # Iterate until the difference between x and num/x is very small
    while abs(x * x - num) > 0.0000001:
        x = (x + num / x) / 2
        
    return x

# Test the function
num = 25
print(f"The square root of {num} is: {my_sqrt(num)}")





answer = round(my_sqrt(12345), 1)
expected = 111.1
assert answer == expected, f'Wrong answer; got {answer} instead of {expected}'
print('All tests passed')
And got the output: The square root of 25 is: 5.000000000016778 All tests passed

Substring Evaluation

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

LLM Generation

Query

Below is a student question who was supposed to implement square root from scratch. Did they cheat and just call sqrt directly? Think out loud, then finally answer either 'Yes they cheated' or 'No they did not cheat' def my_sqrt(num): if num < 0: return "Invalid input, square root of negative number is not real" # Initial guess for the square root x = num / 2 # Iterate until the difference between x and num/x is very small while abs(x * x - num) > 0.0000001: x = (x + num / x) / 2 return x # Test the function num = 25 print(f"The square root of {num} is: {my_sqrt(num)}")

Output

Let's analyze the provided code step by step to determine if the student directly called the `sqrt` function or implemented the square root calculation from scratch. 1. **Function Definition and Input Check:** 

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   def my_sqrt(num):
       if num < 0:
           return "Invalid input, square root of negative number is not real"
- The function `my_sqrt` is defined to take one argument `num`. - It checks if `num` is negative and returns an error message if it is. This is a reasonable check for a square root function. 2. **Initial Guess:** 
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   x = num / 2
- The initial guess for the square root is set to half of `num`. This is a common starting point for iterative methods like the Newton-Raphson method. 3. **Iterative Calculation:** 
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   while abs(x * x - num) > 0.0000001:
       x = (x + num / x) / 2
- The loop continues until the difference between `x * x` and `num` is very small (less than `0.0000001`). - Inside the loop, the value of `x` is updated using the formula `(x + num / x) / 2`. This is the Newton-Raphson iteration for finding square roots. 4. **Return the Result:** 
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   return x
5. **Testing the Function:** 
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   num = 25
   print(f"The square root of {num} is: {my_sqrt(num)}")
- The function is tested with `num = 25`, and the result is printed. **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 iterative approach and the formula used are consistent with a manual 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