PaceReasoner: Label-Free Learning to Answer Briefly from Long Reasoning
Organizations: Peking University · National University of Singapore · Shanghai Jiao Tong University · Zhongguancun Laboratory
Abstract
Extended chain-of-thought reasoning improves problem-solving performance but increases inference cost and latency, motivating more accurate answers with shorter reasoning. We introduce PaceReasoner, a label-free test-time reinforcement learning framework that uses a model's LONG responses to improve its SHORT-mode accuracy. Both modes share a trainable policy, creating a need for reliable supervision, sufficient SHORT training samples, and continued LONG responses to guide learning. PaceReasoner combines all-response and LONG-only answer consensus through progressive, confidence-weighted supervision. A behavior reward gradually shifts from encouraging SHORT exploration to rewarding agreement with LONG answers, while a ratio controller regulates the balance between modes. Across three model backbones and four benchmarks, PaceReasoner improves average SHORT accuracy over the baselines by 7.4--13.5 percentage points, with fewer than 25 reasoning tokens per response on average. Ablations support the key components, and cross-dataset results demonstrate transfer beyond the adaptation set.
Figures & tables
| Mode | Method | CSQA | GPQA-D | MATH-500 | GSM8K | Average | |||||
|---|---|---|---|---|---|---|---|---|---|---|---|
| Acc. | R-tok. | Acc. | R-tok. | Acc. | R-tok. | Acc. | R-tok. | Acc. | R-tok. | ||
| Qwen3-4B-Base | |||||||||||
| LONG | Zero-shot | 50.9 | 413.4 | 21.3 | 461.5 | 50.0 | 510.0 | 60.4 | 328.8 | 45.7 | 428.4 |
| TTRL | 80.9 | 159.1 | 31.8 | 386.3 | 79.4 | 501.2 | 93.1 | 203.0 | 71.3 | 312.4 | |
| Length-Penalty | 39.9 | 158.6 | 33.3 | 283.2 | 56.2 | 510.6 | 85.8 | 201.5 | 53.8 | 288.5 | |
| CRISP | 69.6 | 527.7 | 32.5 | 699.3 | 47.6 | 759.0 | 53.4 | 186.6 | 50.8 | 543.1 | |
| CSQA | GPQA-D | MATH-500 | GSM8K | |||||
|---|---|---|---|---|---|---|---|---|
| Supervision | Acc. | Time | Acc. | Time | Acc. | Time | Acc. | Time |
| All-response | 79.6 | 2h 24m | 32.7 | 44m | 30.2 | 2h 17m | 26.1 | 4h 25m |
| LONG-only | 78.9 | 2h 13m | 28.2 | 44m | 45.0 | 2h 22m | 36.9 | 4h 38m |
| Progressive | 80.2 | 1h 57m | 36.9 | 40m | 39.8 | 2h 9m | 39.1 | 3h 51m |
| SHORT exploration | LONG-answer matching | SHORT ratio control | Eval. mode | CSQA | GPQA-D | MATH-500 | GSM8K | ||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Acc. | R-tok. | Score | Acc. | R-tok. | Score | Acc. | R-tok. | Score | Acc. | R-tok. | Score | ||||
| No | No | No | LONG | 79.2 | 128.1 | 0.6 | 36.0 | 107.5 | 0.3 | 76.7 | 479.8 | 0.2 | 93.3 | 206.6 | 0.5 |
| Yes | No | Yes | SHORT | 79.3 | 14.7 | 5.4 | 33.3 | 31.4 | 1.1 | 40.0 | 14.5 | 2.8 | 25.8 | 19.3 | 1.3 |
| Yes | Yes | No | SHORT | 77.3 | 12.0 | 6.4 | 0.0 | 11.0 | 0.0 | 21.4 | 14.5 | 1.5 | 16.8 | 19.3 | 0.9 |
| Yes | No | No | SHORT | 80.0 | 12.0 | 6.7 | 22.7 | 31.4 | 0.7 | 24.4 | 14.5 | 1.7 | 25.2 | 19.3 | 1.3 |
| Yes | Yes | Yes | SHORT | 80.2 | 12.0 | 6.7 | 36.9 | 51.4 | 0.7 | 39.8 | 14.5 | 2.7 | 39.1 | 19.3 | 2.0 |
| Reasoning mode | Model | CSQA | GPQA-D | MATH-500 | GSM8K |
|---|---|---|---|---|---|
| LONG | Qwen3-4B (Thinking) | 56.8 | 15.2 | 65.4 | 92.3 |
| 4B-Base (PaceReasoner) | 80.8 | 36.9 | 77.2 | 93.7 | |
| Qwen3-8B (Thinking) | 82.0 | 16.7 | 61.4 | 93.6 | |
| 8B-Base (PaceReasoner) | 83.0 | 36.9 | 77.8 | 95.0 | |
| SHORT | Qwen3-4B (Non-thinking) | 76.7 | 37.8 | 18.2 | 18.4 |
| 4B-Base (PaceReasoner) | 80.2 | 36.9 | 39.8 | 39.1 |
| Eval. dataset | Training data | LONG | SHORT |
|---|---|---|---|
| GPQA-D | None (zero-shot) | 21.3 | 11.6 |
| GPQA-D | 36.9 | 36.9 | |
| CSQA | 37.4 | 34.7 | |
| MATH-500 | None (zero-shot) | 50.0 | 31.6 |
| MATH-500 | 77.2 | 39.8 | |
| GSM8K | 74.2 | 36.8 |
Appendix figures & tables10 assets
Supplementary material from the paper’s appendix.
Appendix
| Parameter | CSQA / GSM8K | GPQA-D / MATH-500 |
| Rollout batch (questions) | 256 | 64 |
| Actor update batch (question groups, configured) | 64 | 16 |
| Responses per question | 15 | 15 |
| Rollout batch (response trajectories) | 3840 | 960 |
| Actor update batch (response trajectories) | 960 | 240 |
| Training epochs | 15 | 15 |
| Ratio Controller | CSQA | GPQA | MATH-500 | GSM8K | ||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| acc | R-tok. | score | acc | R-tok. | score | acc | R-tok. | score | acc | R-tok. | score | |
| 77.3 | 12.0 | 6.4 | 0.0 | 11.0 | 0.0 | 21.4 | 14.5 | 1.5 | 16.8 | 19.3 | 0.9 | |
| 78.5 | 11.6 | 6.8 | 20.7 | 31.5 | 0.7 | 32.5 | 14.4 | 2.3 | 36.0 | 19.3 | 1.9 | |
| 79.8 | 12.0 | 6.7 | 31.2 | 31.9 | 1.0 | 27.8 | 14.5 | 1.9 | 26.0 | 19.1 | 1.4 | |
| 76.7 | 12.0 | 6.4 | 22.2 | 31.5 | 0.7 | 31.2 | 14.4 | 2.2 | 37.2 | 19.3 | 1.9 | |
| 80.2 | 12.0 | 6.7 | 36.9 | 51.4 | 0.7 | 39.8 | 14.5 | 2.7 | 39.1 | 19.3 | 2.0 | |
| Ratio Controller | CSQA | GPQA | MATH-500 | GSM8K | ||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| acc | R-tok. | score | acc | R-tok. | score | acc | R-tok. | score | acc | R-tok. | score | |
| 74.1 | 11.5 | 6.4 | 28.8 | 31.6 | 0.9 | 25.2 | 13.8 | 1.8 | 90.3 | 198.3 | 0.5 | |
| 77.2 | 12.0 | 6.4 | 27.8 | 31.4 | 0.9 | 32.3 | 13.5 | 2.4 | 89.3 | 202.4 | 0.4 | |
| 80.2 | 12.0 | 6.7 | 36.9 | 51.4 | 0.7 | 39.8 | 14.5 | 2.7 | 39.1 | 19.3 | 2.0 | |
| Behavior Bonus | CSQA | GPQA | MATH-500 | GSM8K | ||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| acc | R-tok. | score | acc | R-tok. | score | acc | R-tok. | score | acc | R-tok. | score | |
| 79.9 | 13.0 | 6.1 | 32.3 | 31.3 | 1.0 | 78.0 | 440.7 | 0.2 | 92.2 | 194.8 | 0.5 | |
| 80.2 | 12.0 | 6.7 | 36.9 | 51.4 | 0.7 | 39.8 | 14.5 | 2.7 | 39.1 | 19.3 | 2.0 | |
| 80.0 | 13.0 | 6.2 | 30.3 | 35.0 | 0.9 | 39.6 | 14.8 | 2.7 | 39.3 | 21.9 | 1.8 | |
| CSQA | GPQA | MATH-500 | GSM8K | |
|---|---|---|---|---|
| Question | Question: The man closed his eyes as the music played, what effect did the music have? (A)coma (B)enjoyable (C)soothing (D)universal (E)good or bad | Question: Astronomers are observing a planet with a radius of 1 Earth radius in a circular orbit, … What is the maximum orbital period of a second planet, with a radius of 2.5 times that of Earth, … A.37.5 B.33.5 C.12.5 D.7.5 | Question: Compute | Question: Two trains leave San Rafael at the same time. They begin traveling westward, both traveling for 80 miles. The next day, they travel northwards, covering 150 miles. What’s the distance covered by each train in the two days? |
| GT | C | B | 12 | 230 |
| Zero-shot’s outputs | The man closed his eyes as the music played, what effect did the music have? <not></not><answer> (C)soothing </answer> | To solve this problem, we need to determine the maximum orbital period of the second planet that can both transit and occult the first planet. Here’s the step-by-step reasoning: … <answer> D </answer> | To compute the expression , we can use trigonometric identities and properties. First, recall the identity: … (no answer) | The total distance covered by each train in the two days is calculated by adding the distance traveled on the first day (80 miles) and the distance traveled on the second day (150 miles). This gives us a total distance of 230 miles. |
| TTRL’s outputs | <not></not><answer> C </answer> | <not></not><answer> C </answer> | <not></not><answer> 16 </answer> | <not></not><answer> 460 miles </answer> |
| Ours | <not></not><answer> C </answer> | <not></not><answer> B.33.5 </answer> | <not></not><answer> 12 </answer> | <not></not><answer> 230 </answer> |
| CSQA | |
|---|---|
| Question | Question: A revolving door is convenient for two direction travel, but it also serves as a security measure at a what? (A)bank (B)library (C)department store (D)mall (E)new york |
| GT | A |
| Zero-shot’s outputs | To determine whether to think or not, I will analyze the question and the options provided. Question: A revolving door is convenient for two-direction travel, but it also serves as a security measure at a what? Options: (A) bank (B) library (C) department store (D) mall (E) New York Reasoning: A revolving door is commonly used in places where a high volume of people move in and out, and it helps to prevent the spread of germs and maintain a sense of security. This is particularly important in financial institutions like banks, where security is a top priority. Given this information, I can conclude that the correct answer is: <answer> (A) bank </answer> |
| TTRL’s outputs | <reason> The question asks which location a revolving door serves as a security measure. A revolving door is commonly used as a security measure at banks to control access and prevent unauthorized entry. Libraries, department stores, and malls may use revolving doors for convenience, but banks are the most likely to use them for security purposes. </reason><answer> A </answer> |
| PaceReasoner ’s outputs | <reason> A revolving door is typically used as a security measure at banks, as it helps prevent tailgating (where one person follows another into a secure area) and ensures that only one person can pass through at a time. While revolving doors can also be found in other places like department stores and malls, the primary security function is most commonly associated with banks. </reason><answer> A </answer> |
| GPQA | |
|---|---|
| Question | Question: You are analyzing a small peptidic compound that has been chemically synthesized. The 1H NMR spectrum of the crude compound appears consistent with the expected molecule, with the exception of two peaks that both correspond to the same alpha-proton. These two peaks have similar chemical shifts and roughly equal integrals (together they integrate to 1H - the expected integral); based on the coupling pattern, you can rule out spin-spin coupling as an explanation for the duplicate peaks. LC-MS analysis of the crude compound at elevated temperature shows two clearly defined peaks of equal intensities. Both peaks have the same mass spectrum, which is consistent with the expected molecule. What is the most likely explanation for these observations? A. The compound is contaminated with a precursor B. The crude compound exists as a mixture of diastereoisomers C. ’Double coupling’ has occurred during an amide-bond forming reaction D. The crude compound exists as a mixture of enantiomers. You are required to first decide whether to think or not based on the given question. |
| GT | B |
| Zero-shot’s outputs | You are a helpful assistant. <reason> The question describes a small peptidic compound with a 1H NMR spectrum showing two peaks for the same alpha-proton, which are consistent in chemical shift and integral. LC-MS analysis at elevated temperature shows two peaks of equal intensities with the same mass spectrum. The absence of spin-spin coupling and the consistency with the expected molecule suggest that the issue is not due to contamination or diastereoisomers. The most plausible explanation is that the compound exists as a mixture of enantiomers, as enantiomers have identical NMR spectra and mass spectra but different optical rotations. The observation of two peaks in the NMR spectrum could be due to the presence of both enantiomers in the sample. </reason><answer> D </answer> |
| TTRL’s outputs | <reason> The question describes a scenario where a synthetic peptidic compound shows two peaks in its 1H NMR spectrum that correspond to the same alpha-proton, with similar chemical shifts and integrals. The LC-MS analysis at elevated temperature shows two peaks of equal intensity with the same mass spectrum. The key points are: 1. NMR Analysis: Two peaks for the same proton, ruling out spin-spin coupling. 2. LC-MS Analysis: Two peaks of equal intensity with the same mass spectrum, indicating identical molecular species. The most likely explanation for these observations is that the compound exists in a form where these two peaks arise from different molecular conformations or isomers. Since the peaks have similar chemical shifts and integrals, they are likely not due to contamination or precursor molecules. Double coupling is not a plausible explanation given the context. Enantiomers would typically show distinct peaks in NMR due to their chirality, but here the peaks are identical. Diastereoisomers, however, can have similar NMR spectra due to their structural similarities, especially if they are in a conformational equilibrium. </reason><answer> B. The crude compound exists as a mixture of diastereoisomers </answer> |
| PaceReasoner ’s outputs | <reason> The question describes a situation where a synthetic peptidic compound shows two peaks in its 1H NMR spectrum that appear to be the same alpha-proton but do not couple to each other, and LC-MS analysis shows two peaks of equal intensity with identical mass spectra. The options provided are: A. Contamination with a precursor - This would likely show a different peak or pattern in the NMR or MS. B. Mixture of diastereoisomers - Diastereoisomers would have different physical properties and would show distinct peaks in NMR and MS. C. Double coupling during an amide-bond forming reaction - This would typically result in more complex splitting patterns due to the coupling constants involved. D. Mixture of enantiomers - Enantiomers would have identical physical properties and would show the same peaks in NMR and MS. </reason><answer> B. The crude compound exists as a mixture of diastereoisomers </answer> |
| MATH-500 | |
|---|---|
| Question | Question: What is the least positive integer multiple of 30 that can be written with only the digits 0 and 2? |
| GT | 2220 |
| Zero-shot’s outputs | To find the least positive integer multiple of 30 that can be written with only the digits 0 and 2, we need to consider the properties of multiples of 30. A number is a multiple of 30 if and only if it is divisible by both 2 and 15. This means the number must end in 0 (to be divisible by 2) and the sum of its digits must be divisible by 3 (to be divisible by 15). Given that the number can only contain the digits 0 and 2, let’s start by checking the smallest possible numbers: 1. Single-digit numbers: None of the single-digit numbers (0, 2) are multiples of 30. 2. Two-digit numbers: The only two-digit number that ends in 0 is 20, but 20 is not divisible by 15 (since is not divisible by 3). 3. Three-digit numbers: We need to find a number ending in 0 where the sum of the digits is divisible by 3. Let’s check: - 200: ( ) (not divisible by 3) - 220: ( ) (not divisible by 3) - 202: ( ) (not divisible by 3) - 222: ( ) (divisible by 3) Since 222 is divisible by 3 and ends in 0, it is a multiple of 30. Therefore, the least positive integer multiple of 30 that can be written with only the digits 0 and 2 is 222. <reason> The number must end in 0 to be divisible by 2 and the sum of its digits must be divisible by 3 to be divisible by 15. Checking the smallest possible numbers, we find that 222 meets both criteria. </reason><answer> 222 </answer> |
| TTRL’s outputs | <reason> To find the least positive integer multiple of 30 that can be written with only the digits 0 and 2, we need to consider the properties of multiples of 30. A number is a multiple of 30 if and only if it is a multiple of both 3 and 10. 1. Multiple of 10: A number is a multiple of 10 if its last digit is 0. Therefore, the number we are looking for must end in 0. 2. Multiple of 3: A number is a multiple of 3 if the sum of its digits is a multiple of 3. Since the only digits we can use are 0 and 2, we need to find the smallest number ending in 0 where the sum of the digits is a multiple of 3. Let’s start constructing such a number: - The smallest number ending in 0 is 20 (2 + 0 = 2, not a multiple of 3). - The next candidate is 220 (2 + 2 + 0 = 4, not a multiple of 3). - The next candidate is 200 (2 + 0 + 0 = 2, not a multiple of 3). - The next candidate is 2200 (2 + 2 + 0 + 0 = 4, not a multiple of 3). - The next candidate is 2000 (2 + 0 + 0 + 0 = 2, not a multiple of 3). - The next candidate is 2220 (2 + 2 + 2 + 0 = 6, which is a multiple of 3). Thus, 2220 is the smallest number that meets both criteria: it ends in 0 and the sum of its digits is a multiple of 3. Therefore, the least positive integer multiple of 30 that can be written with only the digits 0 and 2 is 2220. </reason><answer> 2220 </answer> |
| PaceReasoner ’s outputs | <reason> To find the least positive integer multiple of 30 that can be written with only the digits 0 and 2, we need to consider the properties of multiples of 30. A number is a multiple of 30 if and only if it is a multiple of both 3 and 10. 1. Multiple of 10: A number is a multiple of 10 if its last digit is 0. Therefore, the number we are looking for must end in 0. 2. Multiple of 3: A number is a multiple of 3 if the sum of its digits is a multiple of 3. Since the only digits we can use are 0 and 2, the sum of the digits must be a multiple of 3. Given that the number must end in 0, the possible forms are 2220, 220, 20, etc. We need to check the smallest ones first: - 20: Sum of digits = 2 + 0 = 2 (not a multiple of 3) - 220: Sum of digits = 2 + 2 + 0 = 4 (not a multiple of 3) - 2220: Sum of digits = 2 + 2 + 2 + 0 = 6 (a multiple of 3) Since 2220 is the smallest number that meets both criteria (ends in 0 and the sum of its digits is a multiple of 3), it is the least positive integer multiple of 30 that can be written with only the digits 0 and 2. </reason><answer> 2220 </answer> |
| GSM8K | |
|---|---|
| Question | Question: I have 10 liters of orange drink that are two-thirds water and I wish to add it to 15 liters of pineapple drink that is three-fifths water. But as I pour it, I spill one liter of the orange drink. How much water is in the remaining 24 liters? |
| GT | 15 |
| Zero-shot’s outputs | You are a helpful assistant. <reason> First, let’s determine the amount of water in the original 10 liters of orange drink. Since it is two-thirds water, the amount of water in the orange drink is (2/3) * 10 = 6.67 liters. Next, we need to account for the spill. Since one liter of the orange drink is spilled, we subtract that from the total amount of water in the orange drink: 6.67 - 1 = 5.67 liters. Now, let’s calculate the amount of water in the 15 liters of pineapple drink. Since it is three-fifths water, the amount of water in the pineapple drink is (3/5) * 15 = 9 liters. Finally, we add the remaining water from the orange drink and the water from the pineapple drink to find the total amount of water in the remaining 24 liters: 5.67 + 9 = 14.67 liters. Therefore, there are 14.67 liters of water in the remaining 24 liters. </reason><answer> 14.67 </answer> |
| TTRL’s outputs | <reason> First, let’s determine the amount of water in the original 10 liters of orange drink. Since it is two-thirds water, the amount of water in the orange drink is: . Next, let’s determine the amount of water in the 15 liters of pineapple drink. Since it is three-fifths water, the amount of water in the pineapple drink is: . Since you spill 1 liter of the orange drink, the amount of orange drink left is: . Now, let’s calculate the amount of water in the remaining 9 liters of orange drink: . Finally, the total amount of water in the combined mixture is the sum of the water from the remaining orange drink and the pineapple drink: . So, the total amount of water in the remaining 24 liters is 15 liters. </reason><answer> 15 </answer> |
| PaceReasoner ’s outputs | <reason> First, let’s determine the amount of water in the original 10 liters of orange drink. Since the orange drink is two-thirds water, the amount of water in it is: . Next, let’s determine the amount of water in the 15 liters of pineapple drink. Since the pineapple drink is three-fifths water, the amount of water in it is: . Since you spill 1 liter of the orange drink, the amount of orange drink left is: . Now, let’s calculate the amount of water in the remaining 9 liters of orange drink: . The total amount of water in the combined mixture is the sum of the water in the remaining orange drink and the water in the pineapple drink: . The total volume of the mixture after spilling is: . Therefore, the amount of water in the remaining 24 liters is: . </reason><answer> 15 </answer> |
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