Initial

README.md

@@ -27,6 +27,7 @@ dataset to enhance its reasoning capabilities.
 ### TableofContents
 
 - [Usage](#usage)
+- [Comparison](#comparison)
 - [Dataset](#dataset)
 - [Training](#training)
 - [License](#licence)
@@ -64,7 +65,7 @@ model.eval()
 
 messages = [
     {"role": "system", "content": "You are a helpful coding assistant."},
-    {"role": "user", "content": "write a quick sort algorithm."}
+    {"role": "user", "content": "Please provide a Python script that checks whether a number is prime."}
 ]
 
 prompt = tokenizer.apply_chat_template(
@@ -84,6 +85,159 @@ print(tokenizer.decode(outputs[0], skip_special_tokens=True))
 
 ---
 
+### Comparison
+
+- Please pay attention to the `<think> </think>` special tokens in the [Reasoning Model](#reasoning-model). As you can
+  see, the `Reasoning Model` can now _“think”_.
+
+#### Base Model
+
+`System:` You are a helpful coding assistant.
+
+`User:` Please provide a Python script that checks whether a number is prime.
+
+`Assistant:`
+
+```python
+def check_prime(num):
+    # Check if the number is less than 2
+    if num < 2:
+        return False
+    # Check for factors from 2 to the square root of the number
+    for i in range(2, int(num**0.5) + 1):
+        if num % i == 0:
+            return False
+    # If no factors were found, the number is prime
+    return True
+
+# Example usage
+number = 7
+if check_prime(number):
+    print(f"{number} is a prime number.")
+else:
+    print(f"{number} is not a prime number.")
+```
+
+In this Python script, the function `check_prime` takes an integer as input and returns `True` if the number is prime,
+otherwise it returns `False`. The function first checks if the number is less than 2, which means it cannot be prime.
+Then, it iterates from 2 up to the square root of the number (inclusive). If any factor of the number is found within
+this range, the function immediately returns `False`, indicating that the number is not prime. If no factors are found
+after checking all possible divisors, the function concludes that the number is prime and returns `True`.
+
+The example usage demonstrates how to call the `check_prime` function with the number 7. The output will indicate
+whether 7 is a prime number or not.
+
+#### Reasoning Model
+
+`System:` You are a helpful coding assistant.
+
+`User:` Please provide a Python script that checks whether a number is prime.
+
+`Assistant:`
+`<think>`
+Okay, I need to write a Python program that checks if a given number is prime. Hmm, what's a prime number? Oh right, a
+prime number has exactly two distinct positive divisors: 1 and itself. So for a number to be prime, it must not have any
+divisors other than those two.
+
+First, let's think about the steps involved. The first thing is to determine if there are any divisors other than 1 and
+the number itself. How do I check that?
+
+Well, for a number n, starting from 2 up to sqrt(n), check if any of them divide n. If any do, then it's not prime. But
+wait, maybe even before checking up to the square root, some numbers might have factors beyond that. For example, take
+
+9. The square root is 3. So checking 2,3,4 would catch 9's factors. Wait, but 3 divides 9, so the loop from 2 to sqrt(
+   n) (which is 3 in this case) will check 2 and 3. So yes, checking up to the square root is sufficient because any
+   composite number has at least one factor less than or equal to its square root.
+
+So the plan is:
+
+1. Read an integer from input.
+2. Check if the number is less than 2. If so, it's not prime.
+3. Otherwise, check from 2 up to the square root of the number. For each i, if the number mod i is zero, return False.
+4. If none of the numbers in this range divide the number, return True.
+
+Wait, but what about 0 and 1? Well, according to the problem statement, the input could be 0 or 1, which are not primes.
+So we should handle those cases separately.
+
+Let me outline the code structure.
+
+Read the input number as an integer. Let's call it num.
+
+Handle edge cases first:
+
+- If num < 2: print "No" (since primes are >=2)
+  Else:
+  Compute sqrt_num = int(math.sqrt(num))
+  Then loop from 2 to sqrt_num inclusive. For each i, check if num % i ==0. If any such i exists, return False. Else,
+  after checking all, return True.
+
+But how to compute the square root accurately in Python? Using math.isqrt for integers, which returns the floor of the
+square root. That's better than using ** operator since it avoids floating point inaccuracies.
+
+Wait, but what if the number is a perfect square? Like 9. Then math.isqrt(9) is 3. So in that case, the loop would check
+up to 3. But when i reaches 3, 9%3 is 0, so it would return False, which is correct.
+
+So the code would look something like this:
+
+import math
+
+n = int(input())
+if n < 2:
+print("No")
+else:
+sqrt_n = math.isqrt(n)
+is_prime = True
+for i in range(2, sqrt_n +1):
+if n % i == 0:
+is_prime = False
+break
+print("Yes" if is_prime else "No")
+
+Wait, but wait, what about numbers like 25. The square root is 5. So the loop runs from 2 to 5. When i=5, 25%5 is 0, so
+returns false. Correct.
+
+This logic seems solid. Let me test some examples.
+
+Test case 1: input 7. The loop runs up to 2. 2 to 2. Check 2. 7%2 is 1, so returns true. Output Yes.
+
+Test case 2: input 6. sqrt(6) is ~2.449. So sqrt_n is 2. Loop from 2 to 2. Check 2. 6%2 is 0. So returns false. Output
+No.
+
+Another test case: 8. sqrt(8) is ~2.828. So sqrt_n is 2. Check 2. 8%2 is 0. So output No.
+
+What about a prime number like 13. sqrt(13) is ~3.605. So loop up to 3. Check 2 and 3. 13%2 is 1, 13%3 is 1. So returns
+true. Correct.
+
+Edge case: 1. Output No.
+
+Another edge case: 2. Output Yes.
+
+What about a very large number, say 999983 (a prime). The code should check up to sqrt(999983) which is around 999.9. So
+the loop runs 2 to 999. It will check all possible factors. Since it's a prime, none will divide it. So returns Yes.
+
+So the code seems to handle all cases correctly.
+
+I think that's the solution. Now, writing the code accordingly.
+`</think>`
+
+```python
+import math
+
+n = int(input())
+if n < 2:
+    print("No")
+else:
+    sqrt_n = math.isqrt(n)
+    is_prime = True
+    for i in range(2, sqrt_n + 1):
+        if n % i == 0:
+            is_prime = False
+            break
+    print("Yes" if is_prime else "No")
+```
+
+---
+
 ### Dataset
 
 - [nvidia/OpenCodeReasoning](https://huggingface.co/datasets/nvidia/OpenCodeReasoning)