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QwQ-32B-CoD
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模型简介
QwQ-32B-CoD 是基于 Qwen 系列 QwQ 模型进行改进的推理模型。由 FenYi Studio（MultEase） 开发，该模型在保持原有推理能力的基础上，显著优化了推理过程的简洁性，并引入了 CoD (Chain of Draft) 逐步草稿式思维链，增强了模型的推理表达能力。
什么是QwQ-32B？
QwQ 是 Qwen 系列的推理模型。与传统的指令调优模型相比，具有思考和推理能力的 QwQ 可以在下游任务中实现显著增强的性能，尤其是难题。QwQ-32B 是中型推理模型，能够实现与最先进的推理模型（如 DeepSeek-R1、o1-mini）竞争的性能。
模型特点
CoD 思维链设计受到了论文 "Chain of Draft: Thinking Faster by Writing Less" 的启发。原论文中的方法对于真正的推理模型并不十分有效，我们观察到，即使使用了论文中的提示词，模型依然会产生复杂冗长而不可控的推理链，而该方法对非推理模型并不十分有效，因为这些模型并不具备真正的推理能力，而经过模板与系统提示词的调整后，我们发现QwQ-32B模型可以实现精简方式的推理。
该模型的特点:
- 推理简洁性优化： 相比原版 QwQ-32B 模型，QwQ-32B-CoD 在生成推理过程时更加精简、直接，减少了冗余信息，提高了推理的可读性和效率。并且可以获得甚至超越原有模型的推理水平，特别是在长上下文推理中。
- CoD (Chain of Draft) 思维链： 引入了逐步草稿式思维链，使模型在推理过程中能够像人类一样逐步构建思路，最终形成完整的推理路径。
- 基于 Qwen 系列： 继承了 Qwen 系列模型的强大基础能力，在各类推理任务中表现出色。
- 由于时间紧张，我们并未对模型进行详细测评，您可以协助我们进行模型水平测评。
模型参数（取自QwQ-32B）
- 类型: 因果语言模型 (Causal Language Models)
- 训练阶段: 预训练 & 后训练 (监督微调和强化学习)
- 架构: 基于 Transformer，采用 RoPE、SwiGLU、RMSNorm 和 Attention QKV bias
- 参数量: 32.5B
- 非嵌入层参数量: 31.0B
- 层数: 64
- 注意力头数 (GQA): Q 为 40，KV 为 8
- 上下文长度: 131,072 tokens
- 对于超过 8,192 tokens 的提示，请按照使用指南中的说明启用 YaRN。
与原模型对比 (QwQ-32B)
| 特性 | QwQ-32B-CoD | QwQ-32B |
|---|---|---|
| 推理简洁性 | 显著优化，推理过程更精简、直接。 | 推理过程可能较为冗长，包含较多中间步骤和解释。 |
| 思维链 | CoD (Chain of Draft) 逐步草稿式思维链。 | 传统的思维链。 |
| 推理示例 | 见下方 | 见下方 |
推理示例对比
原模型 (QwQ-32B) 推理（冗长）：
<think>
首先，我们来分析题目：解方程 \((1+a^2)(a^2+2a+5) - (1-a)(2-2a) = 0\)。
我们的目标是找到所有满足这个等式的实数 \(a\)。
**步骤1：展开方程的每一部分**
首先，我们需要将方程中的每一部分展开，以便合并同类项并简化方程。
首先处理第一个乘积项：\((1+a^2)(a^2+2a+5)\)。
使用分配律（即多项式乘法）展开：
\[
(1+a^2)(a^2+2a+5) = 1 \times (a^2+2a+5) + a^2 \times (a^2+2a+5)
\]
计算每一部分：
1. \(1 \times (a^2+2a+5) = a^2 + 2a + 5\)
2. \(a^2 \times (a^2+2a+5) = a^4 + 2a^3 + 5a^2\)
将两者相加：
\[
a^2 + 2a + 5 + a^4 + 2a^3 + 5a^2 = a^4 + 2a^3 + (a^2 + 5a^2) + 2a + 5
\]
\[
= a^4 + 2a^3 + 6a^2 + 2a + 5
\]
接下来处理第二个乘积项：\((1 - a)(2 - 2a)\)。
同样使用分配律展开：
\[
(1 - a)(2 - 2a) = 1 \times (2 - 2a) - a \times (2 - 2a)
\]
计算每一部分：
1. \(1 \times (2 - 2a) = 2 - 2a\)
2. \(-a \times (2 - 2a) = -2a + 2a^2\)
将两者相加：
\[
2 - 2a - 2a + 2a^2 = 2a^2 - 4a + 2
\]
**步骤2：将展开后的表达式代入原方程**
现在，将展开后的结果代入原方程：
\[
(a^4 + 2a^3 + 6a^2 + 2a + 5) - (2a^2 - 4a + 2) = 0
\]
接下来，需要减去第二个多项式，即：
\[
a^4 + 2a^3 + 6a^2 + 2a + 5 - 2a^2 + 4a - 2 = 0
\]
注意符号的变化：减去 \((2a^2 - 4a + 2)\) 等同于加上 \(-2a^2 + 4a - 2\)。
**步骤3：合并同类项**
现在合并同类项：
1. **四次项**：\(a^4\)（没有其他四次项）
2. **三次项**：\(2a^3\)（没有其他三次项）
3. **二次项**：\(6a^2 - 2a^2 = 4a^2\)
4. **一次项**：\(2a + 4a = 6a\)
5. **常数项**：\(5 - 2 = 3\)
因此，合并后的方程为：
\[
a^4 + 2a^3 + 4a^2 + 6a + 3 = 0
\]
**步骤4：尝试因式分解**
接下来，我们需要尝试对方程 \(a^4 + 2a^3 + 4a^2 + 6a + 3 = 0\) 进行因式分解，以找到可能的根。
首先，可以尝试有理根定理，寻找可能的有理根。有理根定理指出，可能的有理根为常数项的因数除以首项系数的因数。这里，常数项是3，首项系数是1，因此可能的根为 \(\pm1, \pm3\)。
测试 \(a = -1\)：
\[
(-1)^4 + 2(-1)^3 + 4(-1)^2 + 6(-1) + 3 = 1 - 2 + 4 - 6 + 3 = 0
\]
计算结果为0，因此 \(a = -1\) 是方程的一个根。因此，\((a + 1)\) 是多项式的一个因子。
接下来，我们可以用多项式除法或综合除法来分解多项式。这里使用综合除法：
用根 \(a = -1\) 对多项式 \(a^4 + 2a^3 + 4a^2 + 6a + 3\) 进行除法：
将系数列出来：1（\(a^4\)）、2（\(a^3\)）、4（\(a^2\)）、6（\(a\)）、3（常数项）。
进行综合除法：
- 带入根 -1：
-1 | 1 2 4 6 3
| -1 -1 -3 -3
-------------------------
1 1 3 3 0
余数为0，说明分解正确。因此，多项式可以分解为：
\[
(a + 1)(a^3 + a^2 + 3a + 3)
\]
现在需要进一步分解三次多项式 \(a^3 + a^2 + 3a + 3\)。
同样尝试有理根定理，可能的根为 \(\pm1, \pm3\)。
测试 \(a = -1\)：
\[
(-1)^3 + (-1)^2 + 3(-1) + 3 = -1 + 1 - 3 + 3 = 0
\]
因此 \(a = -1\) 是另一个根，因此多项式 \(a^3 + a^2 + 3a + 3\) 可以分解为：
\[
(a + 1)(a^2 + 0a + 3) = (a + 1)(a^2 + 3)
\]
因此，原多项式分解为：
\[
(a + 1)^2 (a^2 + 3)
\]
因此，原方程变为：
\[
(a + 1)^2 (a^2 + 3) = 0
\]
**步骤5：求解方程**
方程的解由每个因子为零时得到：
1. \(a + 1 = 0 \Rightarrow a = -1\)，这是一个二重根。
2. \(a^2 + 3 = 0 \Rightarrow a^2 = -3\)，这在实数范围内无解，但在复数范围内解为 \(a = \pm \sqrt{-3} = \pm i\sqrt{3}\)。
不过题目可能只寻找实数解，因此只有 \(a = -1\) 是实数解。
但需要确认是否在原方程中代入正确，以及是否在展开过程中有误。
**步骤6：验证解的正确性**
将 \(a = -1\) 代入原方程：
左边：
\[
(1 + (-1)^2)((-1)^2 + 2(-1) + 5) - (1 - (-1))(2 - 2(-1))
\]
计算各部分：
1. \(1 + (-1)^2 = 1 + 1 = 2\)
2. \((-1)^2 + 2(-1) +5 = 1 -2 +5 =4\)
3. \(1 - (-1) = 2\)
4. \(2 -2(-1) = 2 + 2 =4\)
因此左边：
\[
2 \times 4 - 2 \times 4 = 8 - 8 =0
\]
符合方程，因此 \(a = -1\) 是正确的解。
对于复数解 \(a = \pm i\sqrt{3}\)，若题目要求实数解，则只取 \(a = -1\)。
**步骤7：总结答案**
因此，方程的实数解为 \(a = -1\)，由于是二重根，但题目可能只需要解的存在性，所以最终解为 \(a = -1\)。
</think>
#### 最终答案
方程的解为 \(\boxed{-1}\)。
QwQ-32B-CoD 推理 (清晰且精简):
(1+a²)(a²+2a+5) → a⁴+2a³+5a² +a²²+2a³+5a² → 错误展开，重新计算
正确展开：(1)(a²) + (1)(2a) + (1)(5) + a²(a²) +a²(2a)+a²(5) → a² +2a+5 +a⁴+2a³+5a² → 合并项
合并后：a⁴+2a³+6a²+2a+5
(1-a)(2-2a) → 2-2a-2a+2a² → 2a²-4a+2
整个式子：a⁴+2a³+6a²+2a+5 - (2a²-4a+2)=0 → a⁴+2a³+4a²+6a+3=0
试可能有理根：±1, ±3 → 代入a=-1：1-2+4-6+3=0 ✓
因式分解：(a+1)(a³+a²+3a+3)=0
解三次方程a³+a²+3a+3=0 → 分组(a³+a²)+(3a+3)=a²(a+1)+3(a+1)=(a+1)(a²+3)=0
所以根为a=-1，a=±√3 i
实数解：a=-1
最终输出
解方程：(1+a²)(a²+2a+5) - (1-a)(2-2a) = 0
步骤详解：
展开多项式：
展开第一个乘积：
(1+a²)(a²+2a+5)
=1×a² +1×2a +1×5 +a²×a² +a²×2a +a²×5
=a² +2a +5 +a⁴ +2a³ +5a²
=a⁴ +2a³ +6a² +2a +5
展开第二个乘积：
(1−a)(2−2a)
=1×2 +1×(−2a) + (−a)×2 + (−a)×(−2a)
=2 −2a −2a +2a²
=2a² −4a +2
合并方程：
将展开式代入原方程：
[a⁴ +2a³ +6a² +2a +5] − [2a² −4a +2] =0
去括号并合并同类项：
a⁴ +2a³ + (6a²−2a²) + (2a+4a) + (5−2) =0
化简得：
a⁴ +2a³ +4a² +6a +3 =0
因式分解：
尝试有理根，代入a=−1：
(−1)⁴ +2(−1)³ +4(−1)² +6(−1) +3
=1−2+4−6+3 =0 ✓
因此可分解为：
(a+1)(a³ +a² +3a +3)=0
解三次方程：
分解a³ +a² +3a +3：
分组为(a³ +a²) + (3a +3)
=a²(a+1) +3(a+1)
=(a+1)(a² +3)
因此方程变为：
(a+1)²(a² +3)=0
求根：
实数根：a+1=0 → a=−1
虚数根：a² +3=0 → a=±√3 i
最终解：
实数解为 a = −1，虚数解为 a = ±√3 i。
使用指南
要快速体验 QwQ-32B-CoD 模型，请访问 Ollama：
点击链接，按照 Ollama 上的说明拉取并运行模型。
致谢
- 感谢阿里云Qwen团队提供的基础模型
- 感谢 论文 "Chain of Draft: Thinking Faster by Writing Less" 提供的思路启发。
关于我们
FenYi Studio是一个推进大型语言模型应用的技术兴趣小组。请联系我们加入讨论或提供计算资源。在GitHub上与我们见面：https://github.com/Fenyi-Studios
引用
如果您觉得我们的工作有帮助，请引用我们：
@misc{qwq32bcod,
title = {QwQ-32B-CoD: Enhanced Reasoning with Chain of Draft},
url = {https://huggingface.co/MultEase/QwQ-32B-CoD},
author = {MultEase},
month = {March},
year = {2025}
}
@article{Chain of Draft: Thinking Faster by Writing Less,
title={Chain of Draft: Thinking Faster by Writing Less},
journal={arXiv preprint arXiv:2502.18600},
year={2025}
}
@misc{qwq32b,
title = {QwQ-32B: Embracing the Power of Reinforcement Learning},
url = {https://qwenlm.github.io/blog/qwq-32b/},
author = {Qwen Team},
month = {March},
year = {2025}
}
@article{qwen2.5,
title={Qwen2.5 Technical Report},
author={An Yang and Baosong Yang and Beichen Zhang and Binyuan Hui and Bo Zheng and Bowen Yu and Chengyuan Li and Dayiheng Liu and Fei Huang and Haoran Wei and Huan Lin and Jian Yang and Jianhong Tu and Jianwei Zhang and Jianxin Yang and Jiaxi Yang and Jingren Zhou and Junyang Lin and Kai Dang and Keming Lu and Keqin Bao and Kexin Yang and Le Yu and Mei Li and Mingfeng Xue and Pei Zhang and Qin Zhu and Rui Men and Runji Lin and Tianhao Li and Tianyi Tang and Tingyu Xia and Xingzhang Ren and Xuancheng Ren and Yang Fan and Yang Su and Yichang Zhang and Yu Wan and Yuqiong Liu and Zeyu Cui and Zhenru Zhang and Zihan Qiu},
journal={arXiv preprint arXiv:2412.15115},
year={2024}
}
QwQ-32B-CoD
Model Introduction
QwQ-32B-CoD is an improved reasoning model based on the Qwen series QwQ models. Developed by FenYi Studio (MultEase), this model significantly optimizes the conciseness of the reasoning process while maintaining the original reasoning ability. It also introduces CoD (Chain of Draft), a step-by-step drafting thought chain, which enhances the model's reasoning and expressive capabilities.
What is QwQ-32B?
QwQ is a reasoning model in the Qwen series. Compared with traditional instruction-tuned models, QwQ, with its ability to think and reason, can achieve significantly enhanced performance in downstream tasks, especially difficult ones. QwQ-32B is a medium-sized reasoning model capable of achieving performance competitive with state-of-the-art reasoning models such as DeepSeek-R1 and o1-mini.
Model Features
The CoD (Chain of Draft) thought chain design is inspired by the paper "Chain of Draft: Thinking Faster by Writing Less". The original method in the paper is not very effective for true reasoning models. We observed that even with the prompts in the paper, the model still generates complex, lengthy, and uncontrollable reasoning chains. The method is also not very effective for non-reasoning models, because these models do not have real reasoning capabilities. After adjusting the template and system prompts, we found that the QwQ-32B model can achieve concise reasoning.
The features of this model are:
- Optimized Reasoning Conciseness: Compared to the original QwQ-32B model, QwQ-32B-CoD generates a more concise and direct reasoning process, reducing redundant information and improving the readability and efficiency of reasoning. It can even achieve or surpass the reasoning level of the original model, especially in long-context reasoning.
- CoD (Chain of Draft) Thought Chain: Introduces a step-by-step drafting thought chain, enabling the model to gradually construct its ideas during the reasoning process, similar to humans, and ultimately form a complete reasoning path.
- Based on the Qwen Series: Inherits the strong foundation of the Qwen series models and performs well in various reasoning tasks.
- Due to time constraints, we did not conduct a detailed evaluation of the model. You can assist us in assessing the model’s performance.
Model Parameters (Taken from QwQ-32B)
- Type: Causal Language Models
- Training Stage: Pre-training & Post-training (Supervised Fine-tuning and Reinforcement Learning)
- Architecture: Transformer-based, using RoPE, SwiGLU, RMSNorm, and Attention QKV bias
- Parameter Count: 32.5B
- Non-embedding Parameter Count: 31.0B
- Layers: 64
- Attention Heads (GQA): Q is 40, KV is 8
- Context Length: 131,072 tokens
- For prompts exceeding 8,192 tokens, please follow the instructions in the Usage Guide to enable YaRN.
Comparison with the Original Model (QwQ-32B)
| Feature | QwQ-32B-CoD | QwQ-32B |
|---|---|---|
| Reasoning Conciseness | Significantly optimized, more concise and direct. | The reasoning process may be relatively verbose, including more intermediate steps and explanations. |
| Thought Chain | CoD (Chain of Draft) step-by-step drafting. | Traditional thought chain. |
| Reasoning Example | See below | See below |
Reasoning Example Comparison
Original Model (QwQ-32B) Reasoning (Verbose):
<think>
First, let's analyze the problem: Solve the equation \((1+a^2)(a^2+2a+5) - (1-a)(2-2a) = 0\).
Our goal is to find all real numbers \(a\) that satisfy this equation.
**Step 1: Expand each part of the equation**
First, we need to expand each part of the equation to combine like terms and simplify the equation.
Let's start with the first product term: \((1+a^2)(a^2+2a+5)\).
Using the distributive property (i.e., polynomial multiplication) to expand:
\[
(1+a^2)(a^2+2a+5) = 1 \times (a^2+2a+5) + a^2 \times (a^2+2a+5)
\]
Calculate each part:
1. \(1 \times (a^2+2a+5) = a^2 + 2a + 5\)
2. \(a^2 \times (a^2+2a+5) = a^4 + 2a^3 + 5a^2\)
Add the two parts together:
\[
a^2 + 2a + 5 + a^4 + 2a^3 + 5a^2 = a^4 + 2a^3 + (a^2 + 5a^2) + 2a + 5
\]
\[
= a^4 + 2a^3 + 6a^2 + 2a + 5
\]
Next, let's deal with the second product term: \((1 - a)(2 - 2a)\).
Similarly, use the distributive property to expand:
\[
(1 - a)(2 - 2a) = 1 \times (2 - 2a) - a \times (2 - 2a)
\]
Calculate each part:
1. \(1 \times (2 - 2a) = 2 - 2a\)
2. \(-a \times (2 - 2a) = -2a + 2a^2\)
Add the two parts together:
\[
2 - 2a - 2a + 2a^2 = 2a^2 - 4a + 2
\]
**Step 2: Substitute the expanded expressions into the original equation**
Now, substitute the expanded results into the original equation:
\[
(a^4 + 2a^3 + 6a^2 + 2a + 5) - (2a^2 - 4a + 2) = 0
\]
Next, we need to subtract the second polynomial:
\[
a^4 + 2a^3 + 6a^2 + 2a + 5 - 2a^2 + 4a - 2 = 0
\]
Note the change in signs: subtracting \((2a^2 - 4a + 2)\) is equivalent to adding \(-2a^2 + 4a - 2\).
**Step 3: Combine like terms**
Now combine like terms:
1. **Quartic term:** \(a^4\) (no other quartic terms)
2. **Cubic term:** \(2a^3\) (no other cubic terms)
3. **Quadratic term:** \(6a^2 - 2a^2 = 4a^2\)
4. **Linear term:** \(2a + 4a = 6a\)
5. **Constant term:** \(5 - 2 = 3\)
Therefore, the combined equation is:
\[
a^4 + 2a^3 + 4a^2 + 6a + 3 = 0
\]
**Step 4: Try to factor**
Next, we need to try to factor the equation \(a^4 + 2a^3 + 4a^2 + 6a + 3 = 0\) to find possible roots.
First, we can try the Rational Root Theorem to find possible rational roots. The Rational Root Theorem states that possible rational roots are factors of the constant term divided by factors of the leading coefficient. Here, the constant term is 3 and the leading coefficient is 1, so the possible roots are \(\pm1, \pm3\).
Test \(a = -1\):
\[
(-1)^4 + 2(-1)^3 + 4(-1)^2 + 6(-1) + 3 = 1 - 2 + 4 - 6 + 3 = 0
\]
The result is 0, so \(a = -1\) is a root of the equation. Therefore, \((a + 1)\) is a factor of the polynomial.
Next, we can use polynomial long division or synthetic division to factor the polynomial. Here we use synthetic division:
Divide the polynomial \(a^4 + 2a^3 + 4a^2 + 6a + 3\) by the root \(a = -1\):
List the coefficients: 1 (\(a^4\)), 2 (\(a^3\)), 4 (\(a^2\)), 6 (\(a\)), 3 (constant term).
Perform synthetic division:
- Bring down the root -1:
-1 | 1 2 4 6 3
| -1 -1 -3 -3
-------------------------
1 1 3 3 0
The remainder is 0, which means the factorization is correct. Therefore, the polynomial can be factored as:
\[
(a + 1)(a^3 + a^2 + 3a + 3)
\]
Now we need to further factor the cubic polynomial \(a^3 + a^2 + 3a + 3\).
Again, try the Rational Root Theorem, the possible roots are \(\pm1, \pm3\).
Test \(a = -1\):
\[
(-1)^3 + (-1)^2 + 3(-1) + 3 = -1 + 1 - 3 + 3 = 0
\]
So \(a = -1\) is another root, so the polynomial \(a^3 + a^2 + 3a + 3\) can be factored as:
\[
(a + 1)(a^2 + 0a + 3) = (a + 1)(a^2 + 3)
\]
Therefore, the original polynomial is factored as:
\[
(a + 1)^2 (a^2 + 3)
\]
Therefore, the original equation becomes:
\[
(a + 1)^2 (a^2 + 3) = 0
\]
**Step 5: Solve the equation**
The solutions of the equation are obtained when each factor is zero:
1. \(a + 1 = 0 \Rightarrow a = -1\), this is a double root.
2. \(a^2 + 3 = 0 \Rightarrow a^2 = -3\), this has no solution in the real number range, but in the complex number range, the solutions are \(a = \pm \sqrt{-3} = \pm i\sqrt{3}\).
However, the problem likely only seeks real solutions, so only \(a = -1\) is a real solution.
But we need to confirm that substituting into the original equation is correct, and that there were no errors in the expansion process.
**Step 6: Verify the correctness of the solution**
Substitute \(a = -1\) into the original equation:
Left side:
\[
(1 + (-1)^2)((-1)^2 + 2(-1) + 5) - (1 - (-1))(2 - 2(-1))
\]
Calculate each part:
1. \(1 + (-1)^2 = 1 + 1 = 2\)
2. \((-1)^2 + 2(-1) +5 = 1 -2 +5 =4\)
3. \(1 - (-1) = 2\)
4. \(2 -2(-1) = 2 + 2 =4\)
Therefore, the left side:
\[
2 \times 4 - 2 \times 4 = 8 - 8 =0
\]
Matches the equation, so \(a = -1\) is the correct solution.
For the complex solutions \(a = \pm i\sqrt{3}\), if the problem requires real solutions, then only \(a = -1\) is taken.
**Step 7: Summarize the answer**
Therefore, the real solution of the equation is \(a = -1\). Since it is a double root, but the problem may only require the existence of the solution, the final solution is \(a = -1\).
</think>
#### Final Answer
The solution to the equation is \(\boxed{-1}\).
QwQ-32B-CoD Reasoning (Clear and Concise):
(1+a²)(a²+2a+5) → a⁴+2a³+5a² +a²²+2a³+5a² → Incorrect expansion, recalculate.
Correct expansion: (1)(a²) + (1)(2a) + (1)(5) + a²(a²) +a²(2a)+a²(5) → a² +2a+5 +a⁴+2a³+5a² → Combine terms.
After combining: a⁴+2a³+6a²+2a+5
(1-a)(2-2a) → 2-2a-2a+2a² → 2a²-4a+2
The whole expression: a⁴+2a³+6a²+2a+5 - (2a²-4a+2)=0 → a⁴+2a³+4a²+6a+3=0
Try possible rational roots: ±1, ±3 → Substitute a=-1: 1-2+4-6+3=0 ✓
Factorization: (a+1)(a³+a²+3a+3)=0
Solve the cubic equation a³+a²+3a+3=0 → Group (a³+a²)+(3a+3)=a²(a+1)+3(a+1)=(a+1)(a²+3)=0
So the roots are a=-1, a=±√3 i
Real solution: a=-1
Final Answer
Solve the equation: (1+a²)(a²+2a+5) - (1-a)(2-2a) = 0
Detailed Steps:
Expand the polynomials:
Expand the first product:
(1+a²)(a²+2a+5)
=1×a² +1×2a +1×5 +a²×a² +a²×2a +a²×5
=a² +2a +5 +a⁴ +2a³ +5a²
=a⁴ +2a³ +6a² +2a +5
Expand the second product:
(1−a)(2−2a)
=1×2 +1×(−2a) + (−a)×2 + (−a)×(−2a)
=2 −2a −2a +2a²
=2a² −4a +2
Combine the equation:
Substitute the expansions into the original equation:
[a⁴ +2a³ +6a² +2a +5] − [2a² −4a +2] =0
Remove parentheses and combine like terms:
a⁴ +2a³ + (6a²−2a²) + (2a+4a) + (5−2) =0
Simplify to:
a⁴ +2a³ +4a² +6a +3 =0
Factorization:
Try rational roots, substitute a=−1:
(−1)⁴ +2(−1)³ +4(−1)² +6(−1) +3
=1−2+4−6+3 =0 ✓
Therefore it can be factored as:
(a+1)(a³ +a² +3a +3)=0
Solve the cubic equation:
Factor a³ +a² +3a +3:
Group as (a³ +a²) + (3a +3)
=a²(a+1) +3(a+1)
=(a+1)(a² +3)
Therefore the equation becomes:
(a+1)²(a² +3)=0
Find the roots:
Real root: a+1=0 → a=−1
Imaginary roots: a² +3=0 → a=±√3 i
Final Solution:
The real solution is a = −1, the imaginary solutions are a = ±√3 i.
Usage Guide
To quickly experience the QwQ-32B-CoD model, please visit Ollama:
Click the link and follow the instructions on Ollama to pull and run the model.
Acknowledgements
- Thanks to the Alibaba Cloud Qwen team for providing the base model.
- Thanks to the paper "Chain of Draft: Thinking Faster by Writing Less" for the inspiration.
About Us
FenYi Studio is a technology interest group promoting the application of large language models. Please contact us to join the discussion or provide computing resources. Meet us on GitHub: https://github.com/Fenyi-Studios
Citation
If you find our work helpful, please cite us:
@misc{qwq32bcod,
title = {QwQ-32B-CoD: Enhanced Reasoning with Chain of Draft},
url = {https://huggingface.co/MultEase/QwQ-32B-CoD},
author = {MultEase},
month = {March},
year = {2025}
}
@article{Chain of Draft: Thinking Faster by Writing Less,
title={Chain of Draft: Thinking Faster by Writing Less},
journal={arXiv preprint arXiv:2502.18600},
year={2025}
}
@misc{qwq32b,
title = {QwQ-32B: Embracing the Power of Reinforcement Learning},
url = {https://qwenlm.github.io/blog/qwq-32b/},
author = {Qwen Team},
month = {March},
year = {2025}
}
@article{qwen2.5,
title={Qwen2.5 Technical Report},
author={An Yang and Baosong Yang and Beichen Zhang and Binyuan Hui and Bo Zheng and Bowen Yu and Chengyuan Li and Dayiheng Liu and Fei Huang and Haoran Wei and Huan Lin and Jian Yang and Jianhong Tu and Jianwei Zhang and Jianxin Yang and Jiaxi Yang and Jingren Zhou and Junyang Lin and Kai Dang and Keming Lu and Keqin Bao and Kexin Yang and Le Yu and Mei Li and Mingfeng Xue and Pei Zhang and Qin Zhu and Rui Men and Runji Lin and Tianhao Li and Tianyi Tang and Tingyu Xia and Xingzhang Ren and Xuancheng Ren and Yang Fan and Yang Su and Yichang Zhang and Yu Wan and Yuqiong Liu and Zeyu Cui and Zhenru Zhang and Zihan Qiu},
journal={arXiv preprint arXiv:2412.15115},
year={2024}
}