| --- |
| base_model: marin-community/delphi-1e23-25Bparams-628Btokens |
| tags: [math, reasoning, grpo, reinforcement-learning, simplerl] |
| --- |
| |
| # Delphi-25B-SimpleRL-Math |
|
|
| GRPO RL fine-tune of the Marin **Delphi-25B** base model (`marin-community/delphi-1e23-25Bparams-628Btokens`, Qwen3 architecture, compute-optimal 1e23-FLOP point) using the **SimpleRL-Zoo** zero-RL recipe on hard competition math (hkust-nlp `simplelr_qwen_level3to5`, MATH levels 3-5). |
|
|
| This is the **step-160** checkpoint (20 epochs). Training reward on the hard-math distribution rose from ~4% to ~29%. |
|
|
| ## Training dynamics & generalization |
|
|
|  |
|
|
| Held-out generalization (greedy, few-shot prompt) tracked against training step: |
|
|
| | step | MATH-500 (few-shot) | MATH-500 (plain) | AIME-24 | |
| |---|---|---|---| |
| | 0 (base) | 5.2% | 1.4% | 0% | |
| | 70 | **17.4%** (peak) | 0.6% | 0% | |
| | 160 (this ckpt) | 13.6% | 2.4% | 0% | |
|
|
| Read: RL produced **real held-out gains** (MATH-500 5.2% -> ~17%), but held-out **peaks around step 70 and then overfits** while train reward keeps climbing. The ability is **few-shot-format-locked** (plain-prompt held-out stays ~1-2.5%) and AIME-24 stays at **0%** throughout. The best-generalizing checkpoint is ~step 70, not the final step. |
|
|
| ## Recipe |
| - GRPO, rule-based boxed-answer reward (`math_verify`), `kl_loss_coef=1e-3`, `rollout.n=8`, lr 5e-7, top_p 1.0. |
| - **Rollout temperature 0.5** (not the recipe default of 1.0 — a sweep showed the reward signal collapses above ~0.5 for this weak base on hard math). |
| - Seeded with a **3-shot prompt** (a base model needs demonstrations, not a bare instruction). The model expects this exact prompt format and stops at `<|im_end|>`. |
|
|
| > Note: the Delphi tokenizer is Llama-3 based, so `<|im_start|>`/`<|im_end|>` are **literal text, not special tokens**. Feed them as raw strings and **stop generation at the `<|im_end|>` string**. |
| |
| ## Prompt format (fixed 3-shot prefix, then your problem) |
| ``` |
| <|im_start|>system |
| You are a helpful assistant.<|im_end|> |
| <|im_start|>user |
| A positive integer less than 100 leaves remainder 1 when divided by 6 and remainder 4 when divided by 7. Find the sum of all such integers. |
| Please reason step by step, and put your final answer within \boxed{}.<|im_end|> |
| <|im_start|>assistant |
| We need x ≡ 1 (mod 6) and x ≡ 4 (mod 7). The numbers that are 1 mod 6 are 1, 7, 13, 19, 25, ...; the smallest of these that is also 4 mod 7 is 25, since 25 = 6·4+1 and 25 = 7·3+4. Solutions repeat every lcm(6,7) = 42, so the values below 100 are 25 and 67. Their sum is 25 + 67 = 92. |
| The final answer is \boxed{92}.<|im_end|> |
| <|im_start|>user |
| A club has 4 times as many seniors as juniors and twice as many sophomores as juniors. If there are 42 members in total and only these three groups, how many seniors are there? |
| Please reason step by step, and put your final answer within \boxed{}.<|im_end|> |
| <|im_start|>assistant |
| Let the number of juniors be x. Then the number of seniors is 4x and the number of sophomores is 2x. The total is x + 4x + 2x = 7x = 42, so x = 6. The number of seniors is 4x = 24. |
| The final answer is \boxed{24}.<|im_end|> |
| <|im_start|>user |
| A right square pyramid has base edge 10 and lateral edge 13. Find the total area of its four triangular faces. |
| Please reason step by step, and put your final answer within \boxed{}.<|im_end|> |
| <|im_start|>assistant |
| The slant height runs from the apex to the midpoint of a base edge, not to a corner. Half the base edge is 5 and the lateral edge is 13, so the slant height is √(13² − 5²) = √(169 − 25) = √144 = 12. Each triangular face has area ½ · 10 · 12 = 60, and there are four faces, so the total area is 4 · 60 = 240. |
| The final answer is \boxed{240}.<|im_end|><|im_start|>user |
| {your problem} |
| Please reason step by step, and put your final answer within \boxed{}.<|im_end|> |
| <|im_start|>assistant |
| ``` |
| |
| ## Example problems (illustrative of the reliable type; not guaranteed) |
| | Problem | Answer | |
| |---|---| |
| | What is the remainder when $2^{10}$ is divided by 7? | 2 | |
| | A right square pyramid has base edge 6 and lateral edge 5. Find the total area of its four triangular faces. | 48 | |
| | A right triangle has legs 9 and 12. What is the length of the hypotenuse? | 15 | |
| | What is 15% of 80? | 12 | |
| | A rectangle has length 8 and width 5. What is its area? | 40 | |
| |
| ## Capability notes (honest) |
| RL gave a weak base **genuine but narrow** competence, and reasoning in the correct shape (steps -> single \boxed{} answer -> clean stop). It is reliable on problems solvable by **a single operation or one known formula with small/obvious inputs**. It is **unreliable** whenever it must (a) **track/combine several quantities** or (b) **discover** something like a non-trivial factorization. It is a demonstration that the SimpleRL recipe works mechanically on a weak base; it is **not** a strong general math model. |
|
|