| | --- |
| | license: cc-by-nc-4.0 |
| | language: |
| | - en |
| | pipeline_tag: text-generation |
| | tags: |
| | - nvidia |
| | - AceMath |
| | - math |
| | - pytorch |
| | --- |
| | |
| |
|
| | ## Introduction |
| | We introduce AceMath, a family of frontier models designed for mathematical reasoning. The models in AceMath family, including AceMath-1.5B/7B/72B-Instruct and AceMath-7B/72B-RM, are <b>Improved using Qwen</b>. |
| | The AceMath-1.5B/7B/72B-Instruct models excel at solving English mathematical problems using Chain-of-Thought (CoT) reasoning, while the AceMath-7B/72B-RM models, as outcome reward models, specialize in evaluating and scoring mathematical solutions. |
| |
|
| | The AceMath-7B/72B-RM models are developed from our AceMath-7B/72B-Instruct models and trained on AceMath-RM-Training-Data using Bradley-Terry loss. The architecture employs standard sequence classification with a linear layer on top of the language model, using the final token to output a scalar score.pull |
| |
|
| | For more information about AceMath, check our [website](https://research.nvidia.com/labs/adlr/acemath/) and [paper](https://arxiv.org/abs/2412.15084). |
| |
|
| | ## All Resources |
| |
|
| | ### AceMath Instruction Models |
| | - [AceMath-1.5B-Instruct](https://huggingface.co/nvidia/AceMath-1.5B-Instruct), [AceMath-7B-Instruct](https://huggingface.co/nvidia/AceMath-7B-Instruct), [AceMath-72B-Instruct](https://huggingface.co/nvidia/AceMath-72B-Instruct) |
| |
|
| | ### AceMath Reward Models |
| | - [AceMath-7B-RM](https://huggingface.co/nvidia/AceMath-7B-RM), [AceMath-72B-RM](https://huggingface.co/nvidia/AceMath-72B-RM) |
| |
|
| | ### Evaluation & Training Data |
| | - [AceMath-RewardBench](https://huggingface.co/datasets/nvidia/AceMath-RewardBench), [AceMath-Instruct Training Data](https://huggingface.co/datasets/nvidia/AceMath-Instruct-Training-Data), [AceMath-RM Training Data](https://huggingface.co/datasets/nvidia/AceMath-RM-Training-Data) |
| |
|
| | ### General Instruction Models |
| | - [AceInstruct-1.5B](https://huggingface.co/nvidia/AceInstruct-1.5B), [AceInstruct-7B](https://huggingface.co/nvidia/AceInstruct-7B), [AceInstruct-72B](https://huggingface.co/nvidia/AceInstruct-72B) |
| |
|
| | ## Benchmark Results (AceMath-Instruct + AceMath-72B-RM) |
| |
|
| | <p align="center"> |
| | <img src="./acemath-pic.png" alt="AceMath Benchmark Results" width="800"> |
| | </p> |
| |
|
| |
|
| | We compare AceMath to leading proprietary and open-access math models in above Table. Our AceMath-7B-Instruct, largely outperforms the previous best-in-class Qwen2.5-Math-7B-Instruct (Average pass@1: 67.2 vs. 62.9) on a variety of math reasoning benchmarks, while coming close to the performance of 10× larger Qwen2.5-Math-72B-Instruct (67.2 vs. 68.2). Notably, our AceMath-72B-Instruct outperforms the state-of-the-art Qwen2.5-Math-72B-Instruct (71.8 vs. 68.2), GPT-4o (67.4) and Claude 3.5 Sonnet (65.6) by a margin. We also report the rm@8 accuracy (best of 8) achieved by our reward model, AceMath-72B-RM, which sets a new record on these reasoning benchmarks. This excludes OpenAI’s o1 model, which relies on scaled inference computation. |
| |
|
| |
|
| | ## Reward Model Benchmark Results |
| |
|
| | | Model | GSM8K | MATH500 | Minerva Math | GaoKao 2023 En | Olympiad Bench | College Math | MMLU STEM | Avg. | |
| | |---------------------------|-------|---------|--------------|----------------|-----------------|--------------|-----------|--------| |
| | | majority@8 | 96.22 | 83.11 | 41.20 | 68.21 | 42.69 | 45.01 | 78.21 | 64.95 | |
| | | Skywork-o1-Open-PRM-Qwen-2.5-7B | 96.92 | 86.64 | 41.00 | 72.34 | 46.50 | 46.30 | 74.01 | 66.24 | |
| | | Qwen2.5-Math-RM-72B | 96.61 | 86.63 | 43.60 | 73.62 | 47.21 | 47.29 | 84.24 | 68.46 | |
| | | AceMath-7B-RM (Ours) | 96.66 | 85.47 | 41.96 | 73.82 | 46.81 | 46.37 | 80.78 | 67.41 | |
| | | AceMath-72B-RM (Ours) | 97.23 | 86.72 | 45.06 | 74.69 | 49.23 | 46.79 | 87.01 | 69.53 | |
| |
|
| | *Reward model evaluation on [AceMath-RewardBench](https://huggingface.co/datasets/nvidia/AceMath-RewardBench). The average results (rm@8) of reward models on math benchmarks, randomly sample 8 responses from 64 candidates with 100 random seeds. Response candidates are generated from a pool of 8 LLMs. |
| | |
| | ## How to use |
| | ```python |
| | import torch |
| | from transformers import AutoModelForSequenceClassification, AutoTokenizer |
| | |
| | model_name = "nvidia/AceMath-7B-RM" # Path to the model |
| | device = "auto" # the device to load the model onto |
| | |
| | model = AutoModelForSequenceClassification.from_pretrained( |
| | model_name, |
| | device_map=device, |
| | num_labels=1, |
| | torch_dtype=torch.bfloat16, |
| | trust_remote_code=True, |
| | ).eval() |
| | |
| | tokenizer = AutoTokenizer.from_pretrained(model_name, trust_remote_code=True) |
| | |
| | |
| | prompt = """Compute |
| | \[\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}},\]where $F_n$ denotes the $n$th Fibonacci number, so $F_0 = 0$ and $F_1 = 1.$""" |
| | |
| | resp_1 = """To compute \[\sum_{n=1}^\infty \frac{F_{n+1}}{F_n F_{n+2}},\] where \( F_n \) is the \(n\)-th Fibonacci number, we will use the recurrence relation \( F_{n+2} = F_{n+1} + F_n \) and properties of Fibonacci numbers.\n\n---\n\n### Step 1: Simplify the General Term\n\nThe given term is\n\[\frac{F_{n+1}}{F_n F_{n+2}}.\]\n\nUsing the Fibonacci recurrence relation \( F_{n+2} = F_{n+1} + F_n \), we rewrite \( \frac{1}{F_{n+2}} \) as\n\[\frac{1}{F_{n+2}} = \frac{1}{F_{n+1} + F_n}.\]\n\nThus, the term becomes\n\[\frac{F_{n+1}}{F_n F_{n+2}} = \frac{F_{n+1}}{F_n (F_{n+1} + F_n)}.\]\n\nNow, separate the fraction into partial fractions:\n\[\frac{F_{n+1}}{F_n (F_{n+1} + F_n)} = \frac{1}{F_n} - \frac{1}{F_{n+1}}.\]\n\n---\n\n### Step 2: Write the Series\n\nThe series becomes\n\[\sum_{n=1}^\infty \frac{F_{n+1}}{F_n F_{n+2}} = \sum_{n=1}^\infty \left( \frac{1}{F_n} - \frac{1}{F_{n+1}} \right).\]\n\nThis is a **telescoping series**, where most terms cancel out. Writing the first few terms explicitly:\n\[\sum_{n=1}^\infty \left( \frac{1}{F_n} - \frac{1}{F_{n+1}} \right) = \left( \frac{1}{F_1} - \frac{1}{F_2} \right) + \left( \frac{1}{F_2} - \frac{1}{F_3} \right) + \left( \frac{1}{F_3} - \frac{1}{F_4} \right) + \cdots.\]\n\nAll intermediate terms cancel, leaving:\n\[\frac{1}{F_1} - \lim_{n \to \infty} \frac{1}{F_{n+1}}.\]\n\n---\n\n### Step 3: Evaluate the Remaining Terms\n\n- The first Fibonacci number \( F_1 = 1 \), so \( \frac{1}{F_1} = 1 \).\n- As \( n \to \infty \), \( F_{n+1} \to \infty \), so \( \frac{1}{F_{n+1}} \to 0 \).\n\nThus, the sum is:\n\[\sum_{n=1}^\infty \frac{F_{n+1}}{F_n F_{n+2}} = 1 - 0 = 1.\]\n\n---\n\n### Final Answer:\n\n\[\boxed{1}\]""" |
| | |
| | chat = [ |
| | {"role": "system", "content": "Please reason step by step, and check your final answer within \\boxed{}."}, |
| | {"role": "user", "content": prompt}, |
| | {"role": "assistant", "content": resp_1} |
| | ] |
| | |
| | |
| | conversation_str = tokenizer.apply_chat_template( |
| | chat, |
| | tokenize=False, |
| | add_generation_prompt=False |
| | ) |
| | |
| | input_ids = tokenizer.encode( |
| | conversation_str, |
| | return_tensors="pt", |
| | add_special_tokens=False |
| | ).to(model.device) |
| | |
| | outputs = model(input_ids=input_ids) |
| | print(outputs[0][0]) |
| | ``` |
| | |
| | |
| | ## Correspondence to |
| | Zihan Liu (zihanl@nvidia.com), Yang Chen (yachen@nvidia.com), Wei Ping (wping@nvidia.com) |
| | |
| | |
| | ## Citation |
| | If you find our work helpful, we’d appreciate it if you could cite us. |
| | <pre> |
| | @article{acemath2024, |
| | title={AceMath: Advancing Frontier Math Reasoning with Post-Training and Reward Modeling}, |
| | author={Liu, Zihan and Chen, Yang and Shoeybi, Mohammad and Catanzaro, Bryan and Ping, Wei}, |
| | journal={arXiv preprint}, |
| | year={2024} |
| | } |
| | </pre> |
| | |
| | |
| | ## License |
| | All models in the AceMath family are for non-commercial use only, subject to [Terms of Use](https://openai.com/policies/row-terms-of-use/) of the data generated by OpenAI. We put the AceMath models under the license of [Creative Commons Attribution: Non-Commercial 4.0 International](https://spdx.org/licenses/CC-BY-NC-4.0). |
| | |