Text Generation
Transformers
Safetensors
qwen3
Generated from Trainer
trl
sft
conversational
text-generation-inference
Instructions to use cs-552-2026-MandMP/math_model with libraries, inference providers, notebooks, and local apps. Follow these links to get started.
- Libraries
- Transformers
How to use cs-552-2026-MandMP/math_model with Transformers:
# Use a pipeline as a high-level helper from transformers import pipeline pipe = pipeline("text-generation", model="cs-552-2026-MandMP/math_model") messages = [ {"role": "user", "content": "Who are you?"}, ] pipe(messages)# Load model directly from transformers import AutoTokenizer, AutoModelForCausalLM tokenizer = AutoTokenizer.from_pretrained("cs-552-2026-MandMP/math_model") model = AutoModelForCausalLM.from_pretrained("cs-552-2026-MandMP/math_model", device_map="auto") messages = [ {"role": "user", "content": "Who are you?"}, ] inputs = tokenizer.apply_chat_template( messages, add_generation_prompt=True, tokenize=True, return_dict=True, return_tensors="pt", ).to(model.device) outputs = model.generate(**inputs, max_new_tokens=40) print(tokenizer.decode(outputs[0][inputs["input_ids"].shape[-1]:])) - Notebooks
- Google Colab
- Kaggle
- Local Apps Settings
- vLLM
How to use cs-552-2026-MandMP/math_model with vLLM:
Install from pip and serve model
# Install vLLM from pip: pip install vllm # Start the vLLM server: vllm serve "cs-552-2026-MandMP/math_model" # Call the server using curl (OpenAI-compatible API): curl -X POST "http://localhost:8000/v1/chat/completions" \ -H "Content-Type: application/json" \ --data '{ "model": "cs-552-2026-MandMP/math_model", "messages": [ { "role": "user", "content": "What is the capital of France?" } ] }'Use Docker
docker model run hf.co/cs-552-2026-MandMP/math_model
- SGLang
How to use cs-552-2026-MandMP/math_model with SGLang:
Install from pip and serve model
# Install SGLang from pip: pip install sglang # Start the SGLang server: python3 -m sglang.launch_server \ --model-path "cs-552-2026-MandMP/math_model" \ --host 0.0.0.0 \ --port 30000 # Call the server using curl (OpenAI-compatible API): curl -X POST "http://localhost:30000/v1/chat/completions" \ -H "Content-Type: application/json" \ --data '{ "model": "cs-552-2026-MandMP/math_model", "messages": [ { "role": "user", "content": "What is the capital of France?" } ] }'Use Docker images
docker run --gpus all \ --shm-size 32g \ -p 30000:30000 \ -v ~/.cache/huggingface:/root/.cache/huggingface \ --env "HF_TOKEN=<secret>" \ --ipc=host \ lmsysorg/sglang:latest \ python3 -m sglang.launch_server \ --model-path "cs-552-2026-MandMP/math_model" \ --host 0.0.0.0 \ --port 30000 # Call the server using curl (OpenAI-compatible API): curl -X POST "http://localhost:30000/v1/chat/completions" \ -H "Content-Type: application/json" \ --data '{ "model": "cs-552-2026-MandMP/math_model", "messages": [ { "role": "user", "content": "What is the capital of France?" } ] }' - Docker Model Runner
How to use cs-552-2026-MandMP/math_model with Docker Model Runner:
docker model run hf.co/cs-552-2026-MandMP/math_model
Automated MNLP evaluation report (2026-05-26)
#9
by zechen-nlp - opened
- EVAL_REPORT.md +25 -53
EVAL_REPORT.md
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- **Model repo:** [`cs-552-2026-MandMP/math_model`](https://huggingface.co/cs-552-2026-MandMP/math_model)
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- **Owner(s):** group **MandMP**
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- **Generated at:** 2026-05-
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- **Pipeline:** [mnlp-project-ci](https://github.com/eric11eca/mnlp-project-ci)
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_This PR is opened automatically by the course CI. It is **non-blocking** — you do not need to merge it. The next nightly run will refresh this file._
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| Benchmark | Accuracy | Status |
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| Math | 0.
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| Knowledge | — | not run |
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| Multilingual | — | not run |
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**Correct** (1 shown)
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- **reference**: `
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- **overall** (8/8 completions correct)
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- **extracted** (✓): `
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- **completion**:
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```text
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Okay, so I need to
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and then try to solve for x, but that would be wrong because the equation is 10^x - 10 = 9990. So if you subtract 10 from both sides, you get 10^x = 9990 - 10 = 9980, which is not 10000. So that would be a mistake.
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But I think the correct approach is adding 10 to both sides.
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So, that's another way to see it.
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I think that's
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**Final Answer**
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The
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</think>
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\boxed{
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```
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**Incorrect** (1 shown)
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- **reference**: `\frac{
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- **overall** (0/8 completions correct)
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- **extracted** (✗): `<no answer>`
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- **completion**:
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```text
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Okay, so I need to
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First, the Asymptote code draws a regular heptagon and a square sharing the side AD. The points A, B, C, D are labeled, with A and D being vertices of the square and heptagon. The square is drawn from A to C to D to F a
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point A is at 90° + 3*(360/7°) = 90° + 1080/7° = (630 + 1080)/7° = 1710/7° ≈ 244.2857°, as before.
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Let me denote θ = 360/7°, so that point A is at angle 90° + 3θ, and point F is at angle 90° + 4θ.
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So, coordinates of point A: (cos(90° + 3θ), sin(90° + 3θ))
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Similarly, coordinates of point F: (cos(90° + 4θ), sin(90° + 4θ))
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But cos(90° + x) = -sin(x), and sin(90° + x) = cos(x).
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Therefore, coordinates of A: (-sin(3θ), cos(3θ))
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Coordinates of F: (-sin(4θ), cos(4θ))
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Similarly, coordinates of point C is A + dir(90)*(F - A).
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dir(90) is (0,1), so adding that vector to point A.
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```
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- **Model repo:** [`cs-552-2026-MandMP/math_model`](https://huggingface.co/cs-552-2026-MandMP/math_model)
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- **Owner(s):** group **MandMP**
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- **Generated at:** 2026-05-26T12:29:10+00:00 (UTC)
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- **Pipeline:** [mnlp-project-ci](https://github.com/eric11eca/mnlp-project-ci)
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_This PR is opened automatically by the course CI. It is **non-blocking** — you do not need to merge it. The next nightly run will refresh this file._
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| Benchmark | Accuracy | Status |
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| Math | 0.2200 | ok |
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| Knowledge | — | not run |
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| Multilingual | — | not run |
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| Safety | — | not run |
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**Correct** (1 shown)
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- **reference**: `1600`
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- **overall** (8/8 completions correct)
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- **extracted** (✓): `1600`
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- **completion**:
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```text
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Okay, so I need to figure out what this expression equals: 441 plus 2 times 21 times 19 plus 361. Let me write that down again to make sure I have it right:
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$441 + 2 \times 21 \times 19 + 361 = x$
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Hmm, let me break this down step by step. First, maybe I should calculate each part separately and then add them up. Let me start with the multiplication part because there's a 2 times 21 times 19 insi
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o recognize this expression without calculating each term separately? Let me think. The original expression is 441 + 2×21×19 + 361. Wait, 441 and 361 are both squares. 21 squared is 441, right? 21×21 is 441. And 19 squared is 361. Oh! Wait, that's a key insight. So the expression is 21² + 2×21×19 + 19². Which is a perfect square. Specifically, it's (a + b)² where a = 21 and b = 19. Let me check that.
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(a + b)² = a² + 2ab + b². Exactly! So that's exactly the expression given. Therefore, x = (21 + 19)². Let me compute that. 21 + 19 is 40. Then 40 squared is 1600. There we go! So that's a much quicker way. Instead of computing each term separately, recognizing the pattern as a perfect square. That makes sense. So x is 1600.
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But just to be thorough, let me confirm once more. If I expand (21 + 19)^2, it's 21^2 + 2*21*19 + 19^2. Which is exactly the original expression. Therefore, substituting 21 + 19 = 40, so 40 squared is 1600. Yep, that's correct.
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Therefore, the answer is 1600. I think that's solid. I don't see any mistakes in my reasoning. Both methods gave me the same result, so I feel confident about this.
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**Final Answer**
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The value of $x$ is \boxed{1600}.
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</think>
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\boxed{1600}
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```
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**Incorrect** (1 shown)
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- **reference**: `$\frac{1}{3}$`
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- **overall** (0/8 completions correct)
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- **extracted** (✗): `<no answer>`
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- **completion**:
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```text
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Okay, so I need to solve this geometry problem about a tetrahedron and some points related to it. Let me try to parse the problem step by step.
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First, we have a tetrahedron $ T_1T_2T_3T_4 $ in 3D space. There's a point $ C $ such that its distance to all four vertices is equal. So, $ CT_1 = CT_2 = CT_3 = CT_4 $. That means $ C $ is equidistant from all four vertices of the tetrahedron.
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Then, th
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,-1,1), C(-1,1,-1), D(1,-1,-1). Let me find the equation of the plane through these three points.
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First, compute two vectors in the plane: vector BC = (0,2,-2) as before, and vector BD = (2,0,-2). Then, the normal vector is (-4, -4, -4) as before. Therefore, the equation of the plane is -4(x +1) -4(y +1) -4(z -1) = 0? Wait, no. Wait, the plane passes through point B(-1,-1,1). So the equation is -4(x - (-1)) -4(y - (-1)) -4(z - 1) = 0? Wait, no. Wait, the normal vector is (-4, -4, -4), so the plane equation is -4(x - x0) -4(y - y0) -4(z - z0) = 0. Using point B(-1,-1,1):
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-4(x +1) -4(y +1) -4(z -1) = 0 => -4x -4 -4y -4 -4z +4 = 0 => -4x -4y -4z -4 = 0 => divide by -4: x + y + z + 1 = 0. So the plane equation is x + y + z = -1.
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Now, the line $ T_1K $ is the line from A(1,1,1) to K(0,0,0). Parametric equations: x = 1 - t, y = 1 - t, z = 1 - t, where t ranges from 0 to 1. We need to find where this line intersects the plane x + y + z = -1. Substitute x = 1 - t, y = 1 - t, z = 1 - t into the plane equation:
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(1 - t) + (1 - t) + (1 - t) = -1 => 3 - 3t = -1 => 3t = 4 => t = 4/3.
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Therefore, the intersection point X is at t = 4/3: x = 1 - 4/3 = -1/3, y = -1/3, z = -1/3. So X is (-1/3, -1
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```
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