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| Metric | Value |
|---|---|
| Base Model | openai/gpt-oss-20b |
| Architecture | Mixture-of-Experts Transformer |
| Total Parameters | ~13.1B (pruned from 21B) |
| Original Experts per Layer | 32 |
| Pruned Experts per Layer | 19 |
| Layers | 24 |
| Top-k Routing | 4 |
| Context Length | 128K tokens |
| Attention Heads | 64 (Query), 8 (Key-Value) |
| Residual Dimension | 2880 |
| Attention Pattern | Alternating dense & sliding window (128 tokens) |
| Positional Encoding | RoPE (Rotary Position Embedding) |
| Normalization | RMSNorm |
| Precision | BF16 |
| License | Apache 2.0 |
| Specialization | Math |
1from transformers import AutoModelForCausalLM, AutoTokenizer
2import torch
3
4# Load the specialized model on CPU
5model = AutoModelForCausalLM.from_pretrained(
6 "AmanPriyanshu/gpt-oss-13.1b-specialized-math-pruned-moe-only-19-experts",
7 torch_dtype=torch.bfloat16,
8 device_map="cpu",
9 trust_remote_code=True
10)
11tokenizer = AutoTokenizer.from_pretrained("AmanPriyanshu/gpt-oss-13.1b-specialized-math-pruned-moe-only-19-experts")
12
13# Generate with the model
14messages = [
15 {"role": "user", "content": "Solve this equation: 2x + 5 = 17. Show your work step by step."}
16]
17
18inputs = tokenizer.apply_chat_template(
19 messages,
20 add_generation_prompt=True,
21 return_tensors="pt",
22 return_dict=True,
23 reasoning_effort="medium"
24)
25
26# Ensure inputs are on the same device as model
27inputs = {k: v.to(model.device) for k, v in inputs.items()}
28
29outputs = model.generate(
30 **inputs,
31 max_new_tokens=512,
32 do_sample=True,
33 temperature=0.1,
34 top_p=0.9,
35 pad_token_id=tokenizer.eos_token_id,
36 eos_token_id=tokenizer.eos_token_id
37)
38
39# Decode only the generated part
40input_length = inputs['input_ids'].shape[1]
41response_tokens = outputs[0][input_length:]
42response = tokenizer.decode(response_tokens, skip_special_tokens=True)
43print(response)1from transformers import AutoModelForCausalLM, AutoTokenizer
2import torch
3
4# Check MPS availability and load model
5device = "mps" if torch.backends.mps.is_available() else "cpu"
6
7model = AutoModelForCausalLM.from_pretrained(
8 "AmanPriyanshu/gpt-oss-13.1b-specialized-math-pruned-moe-only-19-experts",
9 torch_dtype=torch.float16, # Better MPS compatibility
10 device_map=device,
11 trust_remote_code=True,
12 low_cpu_mem_usage=True
13)
14tokenizer = AutoTokenizer.from_pretrained("AmanPriyanshu/gpt-oss-13.1b-specialized-math-pruned-moe-only-19-experts")
15
16# Generate with the model
17messages = [
18 {"role": "user", "content": "Solve this equation: 2x + 5 = 17. Show your work step by step."}
19]
20
21inputs = tokenizer.apply_chat_template(
22 messages,
23 add_generation_prompt=True,
24 return_tensors="pt",
25 return_dict=True,
26 reasoning_effort="medium"
27)
28
29# Move inputs to model device
30inputs = {k: v.to(model.device) if isinstance(v, torch.Tensor) else v for k, v in inputs.items()}
31
32# Use torch.no_grad for MPS stability
33with torch.no_grad():
34 outputs = model.generate(
35 **inputs,
36 max_new_tokens=512,
37 do_sample=True,
38 temperature=0.1,
39 top_p=0.9,
40 pad_token_id=tokenizer.eos_token_id,
41 eos_token_id=tokenizer.eos_token_id,
42 use_cache=True
43 )
44
45# Decode only the generated part
46input_length = inputs['input_ids'].shape[1]
47response_tokens = outputs[0][input_length:]
48response = tokenizer.decode(response_tokens, skip_special_tokens=True)
49print(response)1device_map="auto" # Will automatically use GPU if available
2torch_dtype=torch.bfloat16 # or torch.float16<|channel|>analysis<|message|>We need to solve linear equation. Provide step-by-step.<|end|><|start|>assistant<|channel|>final<|message|>**Equation to solve**
\[
2x + 5 = 17
\]
---
### Step 1: Isolate the term containing \(x\)
Subtract \(5\) from both sides of the equation to get the \(x\)-term alone on the left side.
\[
\begin{aligned}
2x + 5 &= 17 \\
2x + 5 - 5 &= 17 - 5 \\
2x &= 12
\endaligned
\]
---
### Step 2: Solve for \(x\)
Divide both sides by the coefficient of \(x\), which is \(2\).
\[
\beginaligned
\frac{2x}{2} &= \frac{12}{2} \\
x &= 6
\endaligned
\]
---
### Step 3: Verify the solution
Plug \(x = 6\) back into the original equation:
\[
2(6) + 5 = 12 + 5 = 17
\]
The left side equals the right side, so the solution is correct.
---
\[
\boxed{x = 6}
\]<|return|><|channel|>analysis<|message|>Need explain theorem, formula, usage.<|end|><|start|>assistant<|channel|>final<|message|>**The Pythagorean Theorem**
> In a right‑angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) equals the sum of the squares of the lengths of the other two sides.
Mathematically:
\[
a^{2} + b^{2} = c^{2}
\]
where
* \(a\) and \(b\) are the legs (the two sides that form the right angle),
* \(c\) is the hypotenuse.
---
## How it’s used
| **Scenario** | **What you need** | **How the theorem helps** |
|--------------|-------------------|---------------------------|
| **Finding a missing side** | One leg and the hypotenuse, or both legs | Rearsolve the equation:
\(c = \sqrt{a^{2}+b^{2}}\) or \(a = \sqrt{c^{2}-b^{2}}\) |
| **Checking if a triangle is right‑angled** | All three side lengths | Verify if \(a^{2}+b^{2}=c^{2}\) (within rounding error) |
| **Geometry & construction** | Designing right‑angled shapes, ladders, bridges | Ensures structural angles are 90° |
| **Trigonometry** | Relating sine, cosine, and tangent | \( \sin^{2}\theta + \cos^{2}\theta = 1\) comes from the theorem |
| **Distance in coordinate geometry** | Distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) | \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\) |
| **Physics** | Projectile motion, velocity vectors, etc. | Decompose vectors into perpendicular components |
---
### Quick Examples
1. **Right triangle with legs 3 m and 4 m**
\(c = \sqrt{3^{2}+4^{2}} = \sqrt{9+16} = \sqrt{25} = 5\) m.
2. **Triangle with hypotenuse 10 m and one leg 6 m**
\(b = \sqrt{10^{2}-6^{2}} = \sqrt{100-36} =<|channel|>analysis<|message|>Need area = πr^2 = π*49 ≈ 153.938.<|end|><|start|>assistant<|channel|>final<|message|>The area \(A\) of a circle is given by
\[
A = \pi r^{2}
\]
where \(r\) is the radius.
For a radius of \(7\) m:
\[
A = \pi (7\ \text{m})^{2}
= \pi \times 49\ \text{m}^{2}
\approx 3.14159 \times 49\ \text{m}^{2}
\approx 153.938\ \text{m}^{2}
\]
So the circle has an area of about **\(154\ \text{m}^{2}\)** (rounded to the nearest square meter).<|return|>1@misc{priyanshu2025gptoss,
2 title={{GPT-OSS MoE Expert Fingerprinting: Analyzing Expert Activation Patterns in Mixture of Experts Models}},
3 author={Priyanshu, Aman and Vijay, Supriti},
4 year={2025},
5 howpublished={\url{https://amanpriyanshu.github.io/GPT-OSS-MoE-ExpertFingerprinting/}},
6 note={Interactive analysis tool for expert activation patterns in MoE architectures}
7}