README.md

---
library_name: transformers
license: apache-2.0
language:
- ko
---
# GPT-4o Level Reasoning with Ko-R1-1.5B

<p align="center"><img src="https://cdn-uploads.huggingface.co/production/uploads/60d3e619b8448e1785bbda2a/WBhhHyqkPnY8veTTEqDip.png?scale-down-to=1024&lossless=1" alt="HRM8K" width="300" height="300" /></p>
<p align="center">| 📖 <a href="https://www.arxiv.org/abs/2501.02448" target="_blank">Paper</a> | 📝 <a href="https://www.onelineai.com/blog/ko-r1-preview" target="_blank">Blog(English)</a> | 📝 <a href="https://www.onelineai.com/blog/ko-r1-1-5b-korean" target="_blank">Blog(Korean)</a> | 🖥️ Code(Coming soon!) |</p>

## Model Overview
Ko-R1-Preview is a Korean implementation of [R1](https://huggingface.co/deepseek-ai/DeepSeek-R1). 
Similar to R1, it adapts a two-step problem-solving process (\<think\> and \<solution\>). Drawing inspiration from [Ko et al., 2025](https://www.arxiv.org/abs/2501.02448) the thinking is done in English for performance, and the solution is generated in Korean for readability.

## Performance
The model’s performance is evaluated on subsets of the HRM8K benchmark ([Ko et al., 2025](https://www.arxiv.org/abs/2501.02448)).
The evaluation files are also available [here](https://huggingface.co/OLAIR/ko-r1-1.5b-preview/tree/main/eval).

![image/png](https://cdn-uploads.huggingface.co/production/uploads/60d3e619b8448e1785bbda2a/tLKBy8-CrygkYjIrk8U9l.png)


## Inference Settings
Parameters used during evaluation:
```
params = SamplingParams(temperature=0.0, min_tokens=32, max_tokens=8192, repetition_penalty=1.1)
```


## Sample Output
Here is a sample output.

```
Q: $2^8=4^x$일 때, $x$의 값은 얼마입니까?
Response by ko-r1-preview-1.5B:
<think>
Okay, so I need to solve the equation 2^8 = 4^x and find the value of x. Let me think about how to approach this step by step.
First, I remember that when dealing with exponential equations, it's often helpful to express both sides of the equation with the same base if possible.
That way, I can set the exponents equal to each other because if a^b = c^d, then b must equal d if a and c are the same.
So, let's see if I can rewrite 4 as a power of 2. Since 4 is 2 squared, right? Because 2 times 2 is 4. So, 4 is 2^2.

So substituting that into the original equation, we have:\n\n2^8 = (2^2)^x
Now, using the exponentiation rule (a^m)^n = a^(m*n), which means multiplying the exponents. Applying that here, the right side becomes 2^(2*x).
So now the equation is:\n\n2^8 = 2^(2x)

Since the bases are the same (both are powers of 2), their exponents must be equal for the equality to hold.
Therefore, I can set the exponents equal to each other:\n\n8 = 2x\n\nTo solve for x, I just divide both sides by 2:

x = 8 / 2
x = 4

Wait, let me check that again. If x is 4, then 4^4 would be 256, and 2^8 is also 256. Yeah, that works out.
But maybe there's another way to verify?

Alternatively, I could take logarithms on both sides.
Let's try that method to confirm. Taking the natural log or common log of both sides.
Let's use natural logs for this example.\n\nTaking ln of both sides:

ln(2^8) = ln(4^x)

Using the logarithm property that ln(a^b) = b*ln(a):

8*ln(2) = x*ln(4)\n\nThen solving for x:

x = (8*ln(2)) / ln(4)

But since 4 is 2^2, ln(4) is ln(2^2) = 2*ln(2). Substituting back in:

x = (8*ln(2)) / (2*ln(2)) = 8/2 = 4

Same result. So that confirms it. Both methods give x = 4.

[skipped...]
</think>
<solution>
\\(2^8 = 4^x\\)을 풀기 위해, 우리는 양쪽을 같은 밑으로 표현합니다. \\(4\\)를 \\(2^2\\)로 다시 씁니다:

\\[
2^8 = (2^2)^x
\\]

지수 법칙 \\((a^m)^n = a^{mn}\\)을 사용하여 오른쪽을 단순화하면:

\\[\n2^8 = 2^{2x}\n\\]

밑이 같으므로 지수는 같아야 합니다:

\\[
8 = 2x
\\]

\\(x\\)를 구하면:

\\[
x = \\frac{8}{2} = 4
\\]
검증:
- \\(4^4 = 256\\)이고 \\(2^8 = 256\\)이므로 해가 맞음을 확인할 수 있습니다.
로그를 사용한 검증도 결과를 확인합니다.

따라서, \\(x\\)의 값은 \\(\\boxed{4}\\)입니다.
</solution>
```

## Citation
```
@article{ko2025understand,
  title={Understand, Solve and Translate: Bridging the Multilingual Mathematical Reasoning Gap},
  author={Ko, Hyunwoo and Son, Guijin and Choi, Dasol},
  journal={arXiv preprint arXiv:2501.02448},
  year={2025}
}
```