STAPO: Stabilizing Reinforcement Learning for LLMs by Silencing Rare Spurious Tokens
Organizations: School of Vehicle and Mobility & College of AI, Tsinghua University · Didi Voyager Labs, DiDi Autonomous Driving
Abstract
Reinforcement Learning (RL) has significantly improved large language model reasoning, but existing RL fine-tuning methods rely heavily on heuristic techniques such as entropy regularization and reweighting to maintain stability. In practice, they often suffer from late-stage performance collapse, leading to degraded reasoning quality and unstable training. We identify a key factor behind this instability: a small fraction of tokens, termed spurious tokens (around 0.01%), which contribute little to the reasoning outcome but receive disproportionately amplified gradient updates due to inheriting the full sequence-level reward. We present a unified framework for evaluating token-level optimization impacts across spurious risk, gradient norms, and entropy changes. Building on the analysis of token characteristics that severely disrupt optimization, we propose the Silencing Spurious Tokens (S2T) mechanism to efficiently suppress their gradient perturbations. Incorporating this mechanism into a group-based objective, we propose Spurious-Token-Aware Policy Optimization (STAPO), which promotes stable and effective large-scale model refinement. Across six mathematical reasoning benchmarks using Qwen 1.7B, 8B, and 14B base models, STAPO consistently demonstrates superior entropy stability and achieves an average performance improvement of 11.49% (=1.0, top-p=1.0) and 3.73% (=0.7, top-p=0.9) over GRPO, 20-Entropy, and JustRL.
Figures & tables
| Token Properties | Influence on Optimization | ||||
| Advantage | Token Prob. | Entropy | Spurious Risk | Gradient Norm | Entropy Change [ 17 ] |
| Low | Low | ||||
| Low | Medium | ||||
| Medium | Medium | ||||
| High | High | ||||
| Baseline | AIME24 | AIME25 | AMC23 | MATH500 | Minerva | Olympiad | Avg |
| RL from the Qwen3-1.7B Base Model | |||||||
| GRPO | 7.60 (10.42) | 3.33 (3.33) | 37.81 (39.84) | 65.90 (66.40) | 33.73 ( 34.83) | 31.16 (31.42) | 29.92 (31.04) |
| 20-Entropy | 11.98 (17.40) | 6.67 (11.67) | 42.66 (47.58) | 63.70 (64.90) | 21.23 (24.45) | 31.34 (32.31) | 29.60 (33.05) |
| JustRL | 9.17 (15.73) | 6.67 (9.28) | 43.20 (48.28) | 66.25 (70.10) | 30.61 (32.08) | 30.71 (32.90) | 31.10 (34.73) |
| STAPO | 18.44 ( 19.27) | 10.00 ( 13.33) | 54.22 ( 53.12) | 73.10 ( 71.80) | 28.40 (26.10) | 36.94 ( 36.80) | 36.85 ( 36.74) |
| RL from the Qwen3-8B Base Model | |||||||
| Hyperparameter | Value |
| Advantage Estimator | GRPO |
| Reward Function | DAPO |
| Probability Threshold | 0.002 |
| Entropy Percentile | 75 |
| Use KL Loss | No |
| Use Entropy Regularization | No |
| Case 1 |
| Context: The length of is , and the length of is . Since the length of is , for the minimum length of , should start immediately from the right of . The length of is the sum of the lengths of and minus the length of , so its minimum value is . Here is a Python calculation to verify the above reasoning… |
| Spurious Token Prob ( ) Adv Top-5 Distribution calculation 0.0115% 1.62 code(51.90%) | program(31.48%) | solution(11.58%) | script(3.32%) | function(0.58%) |
| Case 2 |
| Context: We are given islands with ferry lines between each pair of distinct islands, which is essentially a complete graph . Each edge of the graph is colored by one of colors, corresponding to the companies operating the ferry lines. For any closure of one company, all its lines must be broken (the corresponding edges ignored) such that the remaining graph can no longer support a Hamiltonian Cycle, meaning it’s no longer possible to visit all the islands exactly once without returning to the starting point. According to the condition, every color, therefore, must serve to connect the graph in such a way that the removal of edges of any single color makes the graph Hamiltonian non-existent. In a complete graph , we can color the edges in such a way that removing any single color results in a graph that lacks a Hamiltonian Cycle. For this, we can use a result from graph theory, informally known as "edge coloring minimum spanning trees": for nodes, we can divide the set of nodes into color classes (as far as coloring spans, and where means "the greatest integer less than "). So when we have (even) nodes, the maximal number of colors we can use is , since we can divide our nodes into colors and deleting any one will still result in a complete graph which remains Hamiltonian (since is always Hamiltonian for ). For (odd) nodes, we can divide our nodes into classes, so the discard of any one of these classes will disintegrate the graph such that a Hamiltonian Cycle will become impossible because it contains a vertex of degree 1 or a subdivision thereof. |
| Spurious Token Prob ( ) Adv Top-5 Distribution broken 0.0471% 2.47 removed(85.53%) | closed(3.32%) | deleted(2.58%) | disconnected(1.57%) | taken(0.95%) discard 0.0001% 2.47 maximal(85.89%) | maximum(9.05%) | number(1.57%) | largest(0.45%) | answer(0.45%) |
| Case 3 |
| Case 1 |
| Context: A=\frac{1}{2}\left|\text{x}_{1}(\text{y}_{2}-\text{y}_{3})+\text{x}_{2}(\text{y}_{3}-\text{y}_{{\color[rgb]{1,0,0}\textbf{2}}})+\text{x}_{3}(\text{y}_{1}-\text{y}_{2})\right| Here, s and s are the coordinates of the points. |
| Spurious Token Prob ( ) Adv Top-5 Distribution 2 0.0142% 0.54 1(99.95%) | 2(0.01%) | 3(0.01%) | 0(0.00%) | }_(0.00%) (0.50%) |
| Case 2 |
| Context: → , b=21{\color[rgb]{1,0,0}\textbf{3}} |
| Spurious Token Prob ( ) Adv Top-5 Distribution 3 0.0017% 0.94 \ (99.98%) | \n (0.01%) | 1(0.01%) | \n (0.00%) | 3(0.00%) |
| Case 3 |
| Case 1 |
| Context: To solve this problem, let’s start by using the given information about the function : 1. Since , we know that: This means . Now , substitute into the quadratic function: |
| Spurious Token Prob ( ) Adv Top-5 Distribution Now 0.0552% 0.35 2(99.74%) | Next(0.12%) | Now(0.06%) | So(0.03%) | Sub(0.02%) |
| Case 2 |
| Context: \ boxed{5 \ text{ agony}} |
| Spurious Token Prob ( ) Adv Top-5 Distribution \ 0.0015% 0.35 }\n (99.57%) | }(0.25%) | } \n\n (0.12%) | } \ (0.03%) | }. \n (0.01%) |
| Case 3 |