Quagmires in SFT-RL Post-Training: When High SFT Scores Mislead and What to Use Instead
Authors: Feiyang Kang
Michael Kuchnik
Karthik Padthe
Marin Vlastelica
Ruoxi Jia
Carole-Jean Wu
Newsha Ardalani
Venue: ICLR 2026 Poster (2026)
Keywords: Post-Training
Large Reasoning Models
Large Language Models
Performance Prediction
Reinforcement Learning with Verifiable Rewards
Primary area: foundation or frontier models, including LLMs
TL;DR: We show extensive examples where high SFT scores do not transfer to improved RL performance in reasoning post-training; we propose generalization loss on held-out SFT examples and pass@large k as viable proxies for predicting post-RL performance.

Abstract: In post-training for reasoning Large Language Models (LLMs), the current state of practice trains LLMs in two independent stages: Supervised Fine-Tuning (SFT) and Reinforcement Learning with Verifiable Rewards (RLVR, shortened as "RL" below). In this work, we challenge whether high SFT scores translate to improved performance after RL. We provide extensive counter-examples where this is not true. We find high SFT scores can be biased toward simpler or more homogeneous data and are not reliably predictive of subsequent RL gains or scaled-up post-training effectiveness. In some cases, RL training on models with improved SFT performance could lead to substantially worse outcome compared to RL on the base model without SFT. We study alternative metrics and identify generalization loss on held-out reasoning examples and Pass@large k performance to provide strong proxies for the RL outcome. We trained hundreds of models up to 12B-parameter with SFT and RLVR via GRPO and ran extensive evaluations on 7 math benchmarks with up to 256 repetitions, spending $>$1M GPU hours. Experiments include models from Llama3, Mistral-Nemo, Qwen3 and multiple state-of-the-art SFT/RL datasets. Compared to directly predicting from pre-RL performance, prediction based on generalization loss and Pass@large k achieves substantial higher precision, improving $R^2$ coefficient and Spearman's rank correlation coefficient by up to 0.5 (2x). This provides strong utility for broad use cases. For example, in most experiments, we find SFT training on unique examples for a one epoch underperforms training on half examples for two epochs, either after SFT or SFT-then-RL; With the same SFT budget, training only on short examples may lead to better SFT performance, though, it often leads to worse outcome after RL compared to training on examples with varying lengths. This work develops an enhanced evaluation tool for math reasoning tasks and is open-sourced.

Published as a conference paper at ICLR 2026
QUAGMIRES IN SFT-RL POST-TRAINING: WHEN HIGH SFT SCORES MISLEAD AND WHAT TO USE INSTEAD
Feiyang Kang! Michael Kuchnik', Karthik Padthe', Marin Vlastelica', Ruoxi Jia?,
Carole-Jean Wu', Newsha Ardalani
FAIR at Meta, Virginia Tech, Meta
ABSTRACT
In post-training for reasoning Large Language Models (LLMs), the current state of practice trains LLMs in two independent stages: Supervised Fine-Tuning (SFT) and Reinforcement Learning with Verifiable Rewards (RLVR, shortened as "RL" below). In this work, we challenge whether high SFT scores translate to improved performance after RL. We provide extensive counter-examples where this is not true. We find high SFT scores can be biased toward simpler more homoge- neous data and are not reliably predictive of subsequent RI gains or scaled-up post-training effectiveness. In some cases, RL training on models with improved SFT performance could lead to substantially worse outcome compared to RL on the base model without SFT. We study alternative metries and identify general- ization loss on held-out reasoning examples and Pass @ large k performance to provide strong proxies for the RL outcome. We trained hundreds of models up to 12B- -B--parameter with SFT and RLVR via GRPO and ran extensive evaluations on math benchmarks with up to 256 repetitions, spending > IM GPU hours. Experiments include models from Llama3, Mistral-Nemo, Qwen3 and multiple state-of-the-art SFT/RL datasets. Compared to directly predicting from pre-RL performance, prediction based on generalization loss and Pass @ large k achieves substantial higher precision, improving R² coefficient and Spearman's 's rank corre- lation coefficient by up 0.5 (2x). This provides strong utility for broad use cases. For example, in most experiments, we find SFT training on unique examples for one epoch underperforms training on half examples for two epochs, either after SFT or SFT-then-RL: With the same SFT budget, training only on short examples may lead better SFT performance, though, often leads to worse outcomes after RL compared to training on examples with varying lengths. This work develops an enhanced evaluation tool for math reasoning tasks and is open-sourced',
INTRODUCTION
The evolution of Large Language Models (LLMs) has seen a significant focus on enhancing their reasoning abilities, process heavily reliant on post-training (Wen et al., 2025). This phase refines pre-trained models, adapting them for complex, multi-step tasks like mathematics, logic, and code generation, leading to the emergence of Large Reasoning Models (LRMs) (Kumar al., 2025). The open-sourced DeepSeek RI achieved phenomenal success in pushing forward the frontier of LLM's reasoning capabilities (Guo al., 2025). Its new post-training paradigm, Reinforcement Learning with Verifiable Rewards (RLVR) via Group Relative Policy Optimization (GRPO) (Liu et al., 2024; Shao et al., 2024), has shown substantial improvements on top of previous post-training methods. Following DeepSeek RI's 's's practice, current works typically conduct SFT before RL, assuming models with better performance after SFT will ultimately be better after RL (Liu et al., 2025b; Wen et al., 2025). In industrial practice, these post-training stages are often distributed among different teams, with SFT and RL handled by separate groups, each optimizing for their own performance metrics (Chen et al., 2025b; Meta, 2025). This process relies on the intuition that a model with stronger SFT
Work done at Meta. Corresponnence to: Feiyang Kang <fyk@ Newsha Ardalani <new @ meta.com> Code repo: https: : http//gtttttbb/t/ com. fei faingng-k/ K/SK" -RI- ICLR

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Figure 1: Mistral-Ne Neoill-eeMMM22222 1eB MeB--IBBBisttit Instruct undergone SFT-RL with SFT examples from AceReasonerL.I-S3 dataset and RLVR via GRPO on DeepScaleR dataset. Reporting Pass performance averaged over 7 math henchmarks. When training increases at different rates than the SFT performance. Model with the best SFT performance is not the one with the best final performance after RL. Post-SFT SFT+RL performance correlate, though, optimizing post-SF7 performance might not optimize the final performance after RL.
Random/l .ongest/Shortest SFT examples, the final performance after RI.
performance will yield better outcomes after RLVR (Liu et al., 2025b). With efforts and resources being poured in improving post-training paradigms and data recipes, also escalating are the debates on whether SFT helps or hurts the subsequent RL training.
In this setup. post-trainee post-training strategies and data often designed either for SFT or RL, but not jointly. In practice, SFT and RL are often conducted sequentially (e.g.. Rastogi et al. (2025)). SFT data is usually selected to maximize evaluation performance after SFT (Zhang et al., 2025; Ye et al., 2025), and the best- pest-performing SFT models are believed to also yield stronger performance after subsequent RL. However, this assumption is often flawed. Over-training during SFT, for instance, the model's behavior and limit the exploration crucial for effective RI. (Chen et al., 2025a; Wang al., 2025). For example, find training repeated examples for up to 8 epochs leads to better SFT performance than training on the same data for 2 epochs (4x compute) but yields visibly worse outcome after RL (Figure 4, left). On the contrary, Cen al. (2025) shows SFT training on manually crafted "exploratory" examples, despite leading to lower performance after SFT, helps achieve better final outcome after RL. This leads critical gap in the current practice:
constrain
An SFT-trained model with the best evaluation performance may not be the best candidate for subsequent training with RLVR (e.g., Figure 1).
When the final RLVR performance is unsatisfactory, it becomes challenging to attribute the failure to either the RL stage or a non-ideal SFT starting point. This misalignment can cause friction and overhead between teams. Furthermore, the high computational cost of RL training and long pipelines, especially in agentic use cases, makes end-to-end tuning across the SFT-RL stages prohibitively expensive (Toledo et al., 2025). Early stopping during RL is also generally ineffective, the model with the fastest initial improvement may not achieve the highest final performance (Liu et al., 2025b). Even with identical post-training procedures, different models may respond vastly different (Figure 2). Consequently, a significant gap remains in our ability to reliably predict RLVR outcomes.
This work centers addressing this predictability problem. We ask the following research questions:
RQI: Do models with better pre-RL. ser.porffomace always lead to better outcomes after RLVR? If not, what are the failure modes? (Section 3)
RQ2: What are effective SFT paradigms and data recipes when considering subsequent RLVR training? Can we determine the suitability of an SFT model before committing to the expensive RL stage? (Section 4)
To tackle these questions, we first examine the relationship between pre-RL performance and post-RL outcomes across various SFT training paradigms and data recipes. While we often observe some extents of correlation between SFT performance and final outcome after RL, we identify cases where the trends visibly diverge. For instance, training on the same dataset for more epochs may significantly boost PFst-SFT performance but diminishes the potential for improvement during RI,
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Figure 2: Both models undergonc SFT-RL with SET examples from AccReasonerl. 1-SFT datasct and RLVR via GRPO on DeepScaleR dataset. Reporting Pass@ performance averaged over math benchmarks. Even with identical post-truining procedures, different models may respond vastly different. With increasing SFT examples, Mistral's s (left) post-SFT performance and final performance after RI. both increase. Yet, for Qwen3 models (right), the post-SFT performances appear uncorrelated with the final performance after RI., where the latter remains the same despite the substantially improved SFT performance.
sometimes resulting in lower final performance (e.g., Figure 4). Similarly, training on simpler, shorter reasoning examples improves pre-RL performance quickly, but these models gain much less from the subsequent RL stage. These findings are particularly concerning given that many SFT data selection methods favor simpler more homogeneous examples (Zhang et al., 2025; Yu et al., 2025).
Next, we identify more reliable predictors for RL success. We find that as SFT proceeds, an eventual increase in validation loss is strongly correlated with performance improvements in the later RL stage. Furthermore, since the RL objective is to compress Pass @k performance into Pass@) (i.e., maximize expected reward), investigate using Pass @k large extensive empirical validation using Llama3-8B (Grattafiori et al., 2024), Mistral-Nemo- 2024), and Qwen3-4B-bal (Yang et al.. 2025) models on state-of-the-ee SFT datasets like Llama- Nemotron (Singhal et al.) and AceReasonerl. (Liu et al., 2025b) and different RI datasets. Our results demonstrate that these new metrics can reliably predict the outcome RLVR, improving R² coefficient and Spearman's rank correlation coefficient by up 0.5 (2x), providing strong utility for broad use cases. For example, in most experiments, we find SFT training on unique examples for one epoch underperforms training half examples for two epochs, either after SFT or SFT-then-RL, With the same SFT budget, training only on short examples may lead to better SFT performance, though, it often leads to worse outcome after RL compared to training examples with varying lengths. These can be captured by the proposed predictors, but not from the post-SFT performance.
predictor. We conduct (team,
To address the limitations in available tools, developed enhanced tool for more convenient and reliable evaluation of reasoning models, which will be open-sourced in contribution to the community.
2 RELATED WORKS
The research landscape for reasoning post-training and data strategies is fast evolving and in its early days. In a typical setup, post-training for reasoning LMs conducts SFT and RI. sequentially, which has been reported work better than only conducting SFT or RI. (Rastogi et al., 2025). Viewpoints in many impactful works can be inconsistent or even contradicting: "Initial 'cold-start' SFT is necessary for subsequent RL" (DeepSeek-R) RI technical report, Guo al. (2025)); "over-SFT may constrain subsequent RL" (Llama-4 technical report, Meta (2025)); "SFT generalizes poorly and RL without SFT does better" (Chen al., 2025a), showing prominent gaps characterizing post-training dynamics and the role of each stage. The lack of predictability in the post-training outcome poses a major blocker for optimizing training paradigms or data recipes.
2.1 POST-TRAINING FOR REASONING: SFT-THEN-RLE PARADIGM
Post-training for reasoning LLMs typically consists of two or three stages: a) Supervised Fine- Tuning (SFT), optional Direct Preference Optinization (DPO) stage, and Reinforcement
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Learning with Verifiable Rewards (RLVR) (Lambert et al., 2024). SFT serves as the "cold-start" phase, providing the model with a strong initial policy by exposing it to high-quality reasoning chains (Guo et al., 2025). The model is trained on problems with high-quality solutions sourced from the frontier models. DPO fixes/strengthens targeted behaviors (e.g.. precise instruction following in math/logic derivations) assessed important for effective reasoning. which is more subjective and often optional (Lambert et al., 2024). RL further improves the model's reasoning and problem- solving capability. This allows the model to explore the solution space more broadly than SFT alone, discovering novel and more robust reasoning paths.
While the sequential SFT-then-RL. pipeline is dominant, researchers have explored alternative paradigms to more tightly integrate or unify these learning stages. Efforts include iterate or in- terleave SFT and RI (Meta, 2025), gradually shift from SFT to RI. while increasing task difficulty (Yang et al., 2025), or directly unify the objectives of SFT and RI (Xu et al., 2025). Though these are promising research directions, they come with their own complexities and have not yet universally replaced the SFT-then-RL paradigm, which remains a robust and widely-adopted industry standard. Many important issues regarding the stability, data requirements, and effectiveness of these unified methods remain to be solved. Our work, therefore, focuses improving the predictability and efficiency of the prevailing SFT-then-RR RL pipeline, providing practical tools that are immediately applicable current sfate-of-the-art workflows.
2.2 RECENT ADVANCEMENTS AND CURRENT CHALLENGES
In post-training for reasoning, SFT data is usually selected to maximize evaluation performance after SFT (Li et al., 2025; Ye et al., 2025), and the best-performing SFT models are believed to also yield stronger performance after subsequent RL. Significant research effort is now focused more sophisticated selection and curation strategies for SFT data. Techniques range from filtering for complexity and diversity generating synthetic data that covers wider range of reasoning structures (Rastogi et al., 2025; Yuan et al., 2025; Ye et al., 2025; Abdin et al., 2024). Some methods propose selecting data points based on their difficulty and influence, aiming find a subset of examples that provides the strongest learning signal (Muennighoff et al., 2025). Current efforts prioritize scaling up SFT training on existing models, leading new SOTA performance on reasoning tasks for those models (Guha et al., 2025). A significant challenge is that standard SFT performance metrics, such as average accuracy on benchmarks, are not always predictive of post-RL, success. This creates a critical gap between the optimization target of the SFT stage and the final performance of the model.
Several issues contribute to this gap. First, models can overfit to the specific patterns and artifacts present in the SFT dataset, leading to poor generalization during the exploration phase of RL (Chen et al., 2025a). Furthermore, naively collecting or generating data can lead to datasets that lack diversity in reasoning strategies or are skewed toward simpler problems, causing the SFT-trained model to develop biases that stifle exploration in the subsequent RL stage (Guha et al., 2025). The landscape is further fogged by the recently reported data contamination issues (Wu et al., 2025). The results from these models have served as the basis for many research findings.
The (lack of) predictability for final performance after RL from pre-RL models leads to quagmires for post- tost--raaiing. SFT teams may provide suboptimal RL learners. It creates frictions between post- training teams owning different SFT and RL stages and chaos in optimizing the training paradigm/data recipes, adding overheads on the model development and hindering productivity. It calls for new tools that better characterize the post-training dynamies and predictive of the RL outcome. This will have profound impact on broad downstream fields-research and applications alike-from improving SFT data curation, search for the next post-training paradigm, to RL for non-verifiable tasks, etc.
3 THE SFT METRIC TRAP
Previous works, from SFT data selection to RL training methodologies, have often operated under common assumption. They implicitly assume explicitly argue that models exhibiting better post-SFT performance will consistently yield superior final outcomes after subsequent reinforcement learning (Rastogi et al.. 2025; Liu et al., 2025b). This assumption has justified the widespread practice of optimizing the SET and RL stages in isolation, with teams or processes focusing on maximizing SFT evaluation metrics primary goal. However, the separation of SFT and RI. optimization can
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lead to a widening gap in reasoning post-training, where improvements in the initial stage do not translate to the final stage. This motivates us to ask two fundamental questions:
* Do models with better pre-Rl. performance always lead to better outcomes after RLVR?
not, what are the failure modes?
To investigate these questions, we design experiments across two representative scenarios that reflect common practices and research directions in the field: a "dataset-level" analysis and an "instance- level" analysis. In Dataset-Level Scenarios, SFT examples are drawn from the same data distribution, but we vary the amount of unique samples and the training paradigm (e-g., learning rate, number of epochs); Instance-Level Scenarios, we consider training on different datasets while keeping the training pipeline fixed (i.c., using the same model and training paradigm). This setup is primarily concerned with SFT data selection and curation, examining whether strong SFT performance on given dataset transfers to the final outcome after RL.
3.1 DATASET-LEVEL SCENARIOS
In this scenario, we draw SFT examples from the same underlying data distribution but vary the training configuration, such as the number of unique samples/training epochs/learning rate. This setup is highly relevant industrial practices where SFT and RL are often handled by different teams. In current practices. the number of training epochs is design choice often determined by practical factors such as data availability or compute budget. Specifically, when the amount of training samples is more prominent constraint (such as domains with limited high-quality examples), repeating for more epochs on the data may be preferred to improve post-SFT performance. On the contrary, data is abundant relative to the allocated compute budget (for this domain/capability) current practices (such as Singhal et al.) may prefer to train for just a single epoch on unique examples.
In these cases, the training paradigm is determined heuristically where the only optimizable target is the post-SFT performance. Surprisingly, we identified both practices to be suboptimal. We found that post-SFT performance often improves stably when training for more epochs-even with excessive overtraining. But models overtrained during SFT show decreasing potentials for the subsequent RL. Typically, the model with the best final performance after RL is not the one with the best post-SFT performance. Further, with the same compute budget for SFT, training data for one epoch typically leads to visibly lower Post-SFT performance compared to training on less data for a few more epochs, and the final performance after RL remains underperforming. A concrete example is provided in Figure 4. High SFT scores be biased toward homogeneous repeated examples and are not reliably predictive of subsequent RL gains.
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Figure 3: Llama3-88 B--33-8B-Inntteu undergone SFT-RL with SFT examples from Llama- Daima--NmmttrooSSS dataset and RLVR GRPO on MATH dataset (train-split). Reporting Pass performance averaged over 7 math benchmarks. Linear between Post-SFT performance and final outcome after RI. The two performance cor- relates with R² = 0.43, indicating post-SFT perfor- mance explains only 43% of variation in the final out- come after RL and the remaining gaps are prominent.
This mismatch between Post-SFT and post-RL performance directly visible from Post-SFT models. As shown in Figure 3 where we fit a linear function between Post-SFT and post-RL.
performance, these two performance correlates with R2 = 0.43, indicating post-SFT performance only explains 43% variation in the final outcome after RL. whereas the gaps remain evident.
3.2 ANSTANCE-LEVEI SCENARIOS
In this scenario, we fix the model and the training configurations but vary the SFT datasets. This setup is primarily concerned with SFT data selection and curation, examining whether the strong
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Figure 4: Both models undergone SFT-RL with SFT examples from AceReasonerl. 1-SFF dataset and RLVR via GRPO on DeepScaleR datasct. Reporting Pass performance averaged over math benchmarks. When repeating SFT for more epochs the same data, Mistral's (left) SFT continues to improve with up to 4 epochs where the final performance after RL saturates after epochs. Qwen3's (right) final performance after RL degrades with SFT training, though, these models' post-SFT performance is substantially higher than the base model. Both cases show clear divergence between poxt-SFT performance and final performance after RL. Here, optimizing Post-SFT performance will be suboptimal or ineffective for improving the final model.
SFT performance promised by particular dataset transfers to strong final performance after RL. For instance, sfute-of-the-ar data selection methods are often prone to selecting examples that are more "natural" or easier for the model to learn (Zhang et al., 2025; Yu et al., 2025). While this simpler data may allow the model to achieve high SFT metries more quickly, we question whether this comes at the cost of learning more difficult or advanced reasoning capabilities that are crucial for downstream success. We identified similar gaps between post-SFT performance and final outcome after RL. Visualizing representative examples in Figure I, high SFT scores can be biased toward simpler examples and are not reliably predictive of subsequent RL gains or scaled-up up post-training effectiveness. For example, training on shortest examples led faster performance improvements than training on randomly sampled examples during SFT. These shorter examples are closer to the model's 's original generations and easier learn, though, these are not best examples for the model to gain reasoning capabilities in preparation for RI. The final performance after RL is significantly lower. These gaps are not directly captured in the post-SFT performance.
PROPOSED METRICS TOWARDS MORE RELIABLE PREDICTIONS
4.1 GENERALIZATION LOSS ON VALIDATION EXAMPLES
During the investigation above, we identified counterintuitive pattern in which post-SFT perfor- mance improves stably when training for more epochs whereas the overtrained models show decreased potentials during the subsequent RL. The best final performance after RL is not usually achieved on models with the best post-SFT performance, To be able to optimize the final outcome on the given training examples, one needs to optimize the SFT training paradigm based on the predicted final outcome after RL. We materialize this insight and identify generalization loss after SFT to be a viable indicator of the model's potential during the subsequent RL. While repeating training for more epochs, together with the improving post-SFT performance, we observe the generalization loss on validation examples to elevate and eventually flare up, indicating strong over-fitting This generalization loss shows strong correlation with further performance gains during subsequent RL, allowing prediction for the final outcome after RL (Figure 5). When using it in practice, after conducting SFT training with different numbers of examples and epochs, we can immediately rule out post-SFT models with both lower performance and higher generalization loss as they will likely remain underperforming after the subsequent RL, facilitating determination of the best SET training paradigm.
4.2 PASS@K ACCURACY EVALUATED AT LARGE K
The objective of RLVR via GRPO is to maximize expected reward, which explicitly optimizes the Pass @ I accuracy the RL tasks. GRPO only progresses when at least of the responses for the RL task is correct. Recent works argue that GRPO compresses Pass accuracy into Pass (Yue al., 2025), and empirical evidence appears to support the argument showing GRPO mostly improves

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Figure 5: Llama3-88 3B-Instruct undergone SFT-RL SFT examples from Alama-Nemotia SFT dataset and RLVR via GRPO MATH dataset (train-split). Reporting Pass@] performance averaged over 7 math benchmarks and generalization loss validation set SFT data. We identify generalization after SFT to be viable indicator for the model's RL potential. While repeating training for more epochs, together with the improving post-SFT performance, we observe the generalization loss on validation examples to elevate and eventually flare up, indicating strong over-fitting. This generalization loss shows strong correlation with the further performance gain during the subsequent RL, allowing prediction for the final outcome after RL.
average @ass@1 accuracy on tasks where the original model achieves an above zero accuracy (Liu et al., 2025b). Though it remains debatable whether GRPO discovers new solution traces beyond the capabilities the original model (Liu et al., 2025a), all these analyses and findings suggest RLVR dynamics during GRPO training to be strongly coupled with the original models Pass @ k accuracy. al. (2023) pioneers using the Pass @ high metric to study the scaling of task performance. The authors argue that Pass @ k provides finer resolution to the Pass @ metric and better captures the underlying dynamics. Acting on this intuition, we consider Pass @k performance of the post- SFT model, especially with large k, candidate metric for predicting its final outcome after the subsequent RL. When using in practice, after SFT training, we evaluate Pass performance on the Post-SFT models with different values of k. For efficient implementation, we leverage the following formula which provides unbiased estimations for Pass @k accuracies for all k S (Brown et al., 2024),
Pass@l E
where integer n denotes the total number of responses generated for the task, integer denotes the target value for k Pass and integer denotes the number of correct responses for the task, respectively.
We consider the Pass @ large k performance as the indicator for the final outcome after RL and deem the Post-SFT model with the best Pass @ large k performance to have the best Pass @ performance after RL. The best SFT model can be determined without needing conduct any actual RL run.
EMPIRICAL EVALUATIONS
5.1
SETUP
We conduct three sets of experiments with SFT-RL post-training On Llama3-8B-Instruuc models, conduct SFT training with examples from Llama-Nemotron dataset (where only select math
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samples with responses generated by QwQ-32B (Team, 2025) or DeepSeek RI (Guo al., 2025), hereinafter the same) and RL training on MATH dataset (train-split) (Hendrycks et al., 2021); on BaatraalNemmo-2--Intttr
conduct SFT training with examples from AceReasonerl, .1-SFT dataset and RL training on DeepScaleR dataset (Luo et al., 2025). For all models, we conduct RL training for 3 epochs where each run takes up to 5 days. We repeat RL training for 4+ runs on each data recipe and training paradigm, conduct 4+ evaluations on different checkpoints across RL training run, and report the best performance for the model. We evaluate task performance on 7 math benchmarks, MATH-500 (Hendrycks et al., 2021), AIME 1983-2024 (Veeraboina, 2023), GSM8k (Cobbe et al., 2021), AIME 2025 (of America, 2025), AMC (Competitions, 2025), Olympiad (He et al., 2024), Minerva (Lewkowycz et al., 2022), and report model performance as Pass@1 I accuracy averaged over 64 repetitions and across tasks. For the proposed predictors, we evaluate the generalization loss on the validation set of the SFT data and Pass@64 64 accuracy averaged over 256 repetitions. Experiments spent > IM GPU hours on NVIDIA A100. Please refer to Appendix B for additional details. Shown in Figure 2 (right), in this setup, Qwenesseries models do not appear to benefit from state-of-the-art SFT datasets, and models undergone different SFT training achieve considerably close final performance after RL. Since this work focuses on studying the impact of different SFT training on the subsequent RL, we present these results qualitative examples instead (deferred Appendix A).
and Qwen3-4B-base
Following the categorization above, we organize experiments in two major scenarios: dataset-level prediction, and instance-level prediction. In dataset-level prediction experiments: we conduct SFT training for the base model on samples from math reasoning datasets with different training paradigms (varying number of examples and epochs). In instance-level prediction experiments: we first create diverse different curated SFT datasets by selecting the shortest/longes subsets, random samples, or their different mixtures (samples are shown in Appendix E). Then, we conduct SFT training for the base model on samples from each curated dataset with the same training paradigms (one epoch). We consider two primary metrics measure prediction performance, Coefficient of determination (R²) (Pearson, 1909). and Spearman's 's rank correlation coefficient (Spearman) (Zar, 1972). Specifically, R² measures the proportion of variation in the prediction variable (final performance) that is unexplained by the predictor, examining the accuracy of prediction the final performance after RL. Spearman yields a number ranging from -1 to that indicates how strongly two sets of ranks are correlated, which we use to examine the effectiveness in identifying post-SFT models that lead to the best final performance. Additional results can be found in Appendix C.
5.2 USE CASE 1: DATASET-LEVEL PREDICTION
This use case focuses on optimizing the SFT training paradigm, a common dataset-level challenge. Given fixed compute budget. practitioners must decide on the optimal trade-off between the volume of unique data and the number of training epochs, navigating the risks of under- and over-training. We test the predictive power of our proposed metrics against the baseline of using post-SFT Pass accuracy. To examine the accuracy of prediction with R?, we randomly select 50% SFT models and fit linear function between their POst-SFT performance and final performance after RL. The fitted function is then used to predict the final performance of the other 50% SFT models. We compare the predictions their actual post-RL outcomes to compute R² We repeat the random sampling for 100 times and report the standard error.
Table 1: Spearman's 's rank correlation hetween performance predicted from Post-SFT models and the actual performance after RL. Both generalization loss and Pass @ 64 achieve notable margins over prediction from Pass@1 whereas averging the two prediction may or may not lead better results.
Spearman's 's Rank Correlation Models
Prediction based on
Prediction based on SIT Generalization Loss
Prediction based on SFT Pass Large (k=64)
Avg. Prodiction from SFT Gen. Loss Pass@) Large (64)
SET @ass@1 (avg. of 64) baseline
Llama3-88 -B-nnuur Mistral-Noo NeMo- 12B In-nrstttct
0.75
0.94 0.90
0.95 0.92 (+0.14)
0.97 (+0.22)
0.78
0.90
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Table Measuring prediction accuracy with coefficient of determination (A²). We randomly select 50% SFT models and fit linear function between their post-SFT performance and performance after RI. and use to predict for the other 50% SFT models. We repeat random sampling for 100 times report standard errors.
Coefficient of determination (?²) Models
Prodiction based on SFT Pass@1 (avg. of 64) baseline
Prediction based on SFT Generalization Loss
Prediction based on SFT Pass Large (k=64)
Avg. Prediction from SFT Gen. Loss Pass @ Large (64)
Llama3-8B 88--nsstrctt Mistral-NeMo-11 12B-Instruct
0.57 0.21
0.88 uu
0.87 uow
0.94 u wux (+0.37)
0.29 0.38
0.79 # + (+0.50)
+ +03
0.72 024
Takeaway 1: Dataset-level Prediction
Both generalization loss and Pass @ large k are effective predictors for post-RL performance when optimizing SFT training configurations estimates that help guide decisions and save significant compute.
single dataset, providing higher-accurayy
Both predictors excel at identifying correct rankings for post-RL, performance, achieving Z 0.90 Spearman correlation (30% improvements); generalization loss provides advantageous prediction accuracy (IR²) for post-RL performance with up to 2x improvements.
5.3 USE CASE INSTANCE-LEVEL PREDICTION
This use addresses the challenge of SFT data selection, instance-level optimization problem. Here, the training pipeline is fixed, but we aim to select the optimal SFT dataset from a pool of candidates curated with different strategies (e.g.. selecting for shortest/longest solutions, diversity, etc. Ye et al. (2025)). This scenario tests whether strong SFT performance on a given dataset translates to good final outcome after RL.
Table 3: Spearman's 's rank correlation between performance predicted from post-SFT models and the actual performance after RL. Pass 64 achieve notable margins over prediction from Pass@ 1.
Spearman's Rank/
Prediction based on SFT Pass@ (avg. of 64) baseline
Prediction based on
Correlation Models
SFT Pass Parge k (k=64)
Llaam3--B-IInse
0.69
0.94 (+0.25)
0.70
0.98 (+0.28)
Table Measuring prediction accuracy with coefficient of determination (A²), We randomly select 50% SFT models and fit linear function between their post-SFT performance and performance after RI, and use it to predict for the other 50% SFT models. We repeat random sampling for 100 times and report standard errors.
Coefficient of determination (R²) Models
Prediction hased SFT Pass @
Prediction based on SFT Pass@1 Large (k=64)
(avg. of 64) baseline
Llama3- 8B Instruct Mistral-NeMo Instruct
0.58 020
0.92 (+0.34) 0.98 (+0.25)
0.73
In this scenario, the generalization loss predictor is not applicable. Since each SFT dataset comes from a different distribution, the validation loss includes distributional gap component in addition to generalization error. Without common, representative validation set, it is difficult to make fair comparison. @ large k metric proves to be exceptionally robust. Since it measures the model's s inherent capability training data. It can be used to effectively rank different SFT datasets and select the one with the highest potential for RI, without needing to nun any RI experiments for calibration.
produce correct solutions. it is less sensitive distributional shifts in the

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Takeaway 2: Instance-level Prediction
Pass@large turns out highly accurate and robust in instance-level predictions, improving
Spearman correlation by up to 36% and prediction accuracy ( (R²) (R2)by bby up to 59%. It effectively identifies datasets for strong post-RL performance and predicts RL outcomes.
Generalization loss is not applicable for instance-level selection due distributional gaps between different datasets.
How to use them in practice? Our metrics support two primary workflows. If the goal is simply to rank SFT candidates, one can use generalization loss to quickly filter out clearly suboptimal models (i.e., those with both low performance and high loss). Then, Pass @ large k can be used to reliably rank the remaining candidates to identify the most promising one. If the goal is to predict the final performance value- for instance, to inform trade-offs between SFT costs and expected gains- practt-pratiiiooness can run RL on small number of SFT models gather calibration data. A linear predictor can then be fitted using our proposed metrics, allowing for accurate performance estimation across all SFT candidates without the need for exhaustive RL runs.
CONCLUSIONS
This work confronts critical quagmire in reasoning post-training: the common assumption that high SFT scores guarantee strong performance after subsequent RL. Through extensive experimentation with Sistaaa/MiisralNeeo//eeeeeeee
models spending IM GPU hours, we provide broad counter- examples where SFT performance is often misleading or biased toward simpler/repeated data. Our primary contribution is the identification and validation of two more reliable predictors for post-RL success: generalization loss held-out reasoning examples and Pass @ large k accuracy, improving prediction accuracy (R²) and Spearman's 's rank correlation by up to 0.5 (2x) over prediction from Post-SFT performance. By allowing practitioners to better predict the final outcome, work helps de-risk the expensive RL stage and streamline the entire post-training pipeline. We open-sourced enhanced evaluation tool to facilitate broader adoption and further research. This work focuses on mathematical reasoning. A natural next step is to study the topic in a wider range of reasoning tasks (e.g., coding, science) and agentic cases; Our study is limited to the prevailing paradigm of online RL with GRPO. The relationship between SFT characteristics and post-RL performance with other methods such as offline RL/DPO or other RL algorithms may worth further explorations; Directly evaluating Pass Pass@largy large k requires repeating evaluation for at least k times, which becomes computational expensive with long sequence lengths. Estimating Pass @k accuracy from that of smaller k holds the promise for more efficient evaluations (Schaeffer et al., 2025).
7 DISCUSSIONS
Broader investigations and supplementary examples are provided in Appendix D. In Appendix D.1, we conduct fine- grained comparisons between the best SFT model vs. the best SFT+RL model. We show extensive examples that the best overall models (SFT+RL) achieve notable performance margins in the final performance (SFT+RL) over the best SFT models, despite trailing in SFT performance. To facilitate the pursuit towards underlying mechanism behind the intriguing SFT-RL dynamics, in Appendix D.2, we investigate the aggregated entropy at the response level. As shown in Tables 15, 16, and 17, we witnessed that correct responses generally show a lower entropy compared to incorrect responses. However, we observe A lower entropy after SFT training does not always suggest worse post-RL. RL. outcomes, and vice versa. 2. A lower entropy after SFT training does not indicate better PFst-SFT performance, nor does the entropy gap between correct and incorrect responses.
We found that during decoding, math/symbolic tokens generally show a near-zero loss-i.e., the model is almost always certain on these math tokens. The highest loss tokens are human languages that are naturally ambiguous or interchangeable, such as whether to start the sentence with "Thus" or "Therefore". As result, the (per-token) entropy for the generations mostly measures verbosity, and does not show consistent patterns with accuracy. In general, we did not observe a consistent pattern to explain the SFT-RI. dynamic, but it confirms that entropy alone is an inadequate explanation.
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ACKNOWLEDGMENTS
Ruoxi Jia and the ReDS lab acknowledge support from the National Science Foundation through grants IIS 231294,, IIS- IIS-2313130, OAC-2239622. The authors extend their sincere appreciation to the reviewers, Area Chairs, and the organizing committee for their dedication throughout the review process. Amid unprecedented challenges, they demonstrated exceptional commitment to upholding the integrity and quality of the proceedings, ensuring the timely dissemination of this work to community.
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Appendices
A Additional SFT-RL Examples and Visualizations
16
A.1 Llama3-8B-Instruc
16
A.2 Mistral-NeMo-I1n Instructt
17
A.3 Qwen3-4B-base
17
B Implementation Details
18
B.1 Models and Datasets
19
B.2 Training
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B.3 Evaluation
19
C Additional Experimental Results
20
C.I Dataset-level
20
C.2 Instance-level
20
Broader Investigations and Supplementary Examples
23
D.I Fine-grained comparisons: Best SFT model Best SFT+RL model
23
D.2 Ablation Studies: Pass@k, Entropy
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Sample SFT Examples
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E.1 Shortest Examples
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E.2 Longest Examples
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A ADDITIONAL SFT-RL EXAMPLES AND VISUALIZATIONS
A.1 SLAMAA-8B-INSTRICE
Figure 6 shows results on Llama3-8B-Instruut undergone SFT-RL with SFT examples from Llama- Nemotron-Sti Seteoon-FFF/AAReesseereeee
1-SFT/OOenRII-Ma (Hugging Face, 2025) dataset and RLVR via GRPO on MATH dataset (train-split), Reporting average Pass@ 1 performance on MATH-500 500 (test- split). High SFT scores can be biased toward simpler or more homogeneous data and are not reliably predictive of subsequent RI. gains or post-training effectiieness. SFT on fewer unique examples repeated for more training epochs (ep) or/and with a larger learning rate (LR) leads to higher accuracy on reasoning benchmarks such as MATH-500 (+8.75% vs. non-repeated data, left figure). However, models trained this way show smaller improvements during RL (-1.43% vs. non-repeated). In contrast, SFT more diverse, non-repeated data-despite despite yielding lower initial SFT performance (-5% vs. repeated data, middle/right figure)-resulle results in significantly better post-RL performance (+5.94%).
Accuracy on MATH500
Accuracy on MATH500
Accuracy on MATH500
67.5
poot SFT
MATH500 Pascog (evo over 16 nurc)
72.5
00.0
MATHSCO Pres oleg over 16 nuns)
NOT 500 Parngt 1er Fnre
65:12
62.5
00.0
50 81
$7.5
66.00
35.0
55.0
22.5
13.43
50.0
55445
53.12
67.5
30-
Open 11, Mach
Open 81 Math Sep SFT Data
Open R1 Math
lop
10c 10ep
#1 Nath Unme Beme--eemstroneeeeeee Aceleesnnne 10k sep ber
SFT
Lite Lep SFT Data
100k les
(Left to right 50k unique examples, 25k examples for 3 epochs, 25k examples for 4 epochs)
unique examples,
10K examples for epochs with Sx LR,
(LRR 10k examples, epochs, Ex LR, Open R1 100k unicuas SFT axampies, Nemotmn 100k uniquee SFT examples, Acaflessoner)
10k examples for 10 epochs)
SFT training on diverse data, despite leading to a lower post-SFT performance (-5% pass@ f accuracy on MATH-500), eventually delivers higher post-RL performance (*5.94%).
reaching the same SFT performance by training longer on diverse, non-repeated samples, post-RL performance will be "much" higher (+7.31%).
i. Over-training on repeated data improves SFT performance fast, but these models do not have better RL performance
Figure 6: Baama3-8B-Instrull undergone SFT-RL with SFT examples from Alama-Nemotron SFT//AceReasoerl.. 1-w...1--S///R/RRRR- -Math dataset and RLVR via GRPO on MATH dataset (train-split). Reporting average Pass@1 I performance on MATH-500 (test-split). High SFT scores can be biased toward simpler or more homogeneous data and are not reliably predictive of subsequent RL gains or post-training effectiveness. SFT on fewer unique examples repeated for training epochs (cp) or/and with larger learning rate (LR) leads higher accuracy on reasoning benchmarks such as MATH-500 (+8.75% vs. non non-repeated data, left figure). However, models trained this way show smaller improvements during RL (-1.43% vs. non-repeated). In contrast, SFT on more diverse, non-repeated sattaddespic yielding lower initial SFT performance (-5% vs. repeated data, middle/right figure) rigure)-results results in significantly better post-RL, performance (+5.94%).
Figure 7 shows results on Llama3-8B -lama3-8B-Instruct undergone SFT-RL with SFT examples from Llama- Nemotron-SFi dataset and RLVR via GRPO on MATH dataset (train-split). Reporting Pass performance averaged over 7 math benchmarks. High SFT scores can be biased toward simpler examples and are not reliably predictive of subsequent RL gains or scaled-up post-training effec- tiveness. For example, training shortest examples (e.g., s10k, s500k) led to faster performance improvements than training on randomly sampled examples (e.g.. 10k, 200k) during SFT (lower smaller dots). These shorter examples learn, though, these are not best examples for the model to gain reasoning capabilities in preparation for RI. The final performance after RL (upper larger dots) is significantly lower.
closer the model's original generations and easier to
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Nemotron 10k
performance after SFT+RL
35.0
Post SFT Performance
performance after SFT
150k
SFT+RL Perforance (avg of 7 tasks)
Nemotron 200k
32.5
Nematron 50k
Nametron 25k
Nemetron 10kc
30.0
Nemotron s10k-r10k
Nemotron s50k
Nematron stook
27.5
Nametron
Nemotron 1100k
25.0
22.5
20.0
SFT Compute Budget /million tokens (log-scale)
Figure 7: 1lama--B-IInstree undergone SFT-RL with SFT examples from JJama-Nemotron-SFF dataset and RLVR via GRPO MATH dataset (train-split). Reporting Pass@] performance averaged over 7 math benchmarks. High SFT scores can biased toward simpler examples and are not reliably predictive of subsequent RT gains scaled-up post-training effectiveness. For example, training shortest examples (e.g.. s10k, s500k) led faster performance improvements than training on randomly sampled examples (e.g.. 10k, 200k) during SFT (lower smaller dots). These shorter examples closer the model's original generations and easier to lear, though, these are not best examples for the model to gain reasoning capabilities in preparation for RL. The final performance after RL (upper larger dots) is significantly lower.
A.2 MISTRAL-NEMON 12A-NEMM----B--T-AAAE INSTRUCT
Figure shows results Mistral-NeMo- Neail--MMMo------- 12B- Instruct undergone SFT-RL with shortest SFT exam- ples from AceReasonerl.. ISFFT dataset and RLVR via GRPO on DeepScaleR dataset. Reporting Pass@ 1 performance averaged over math benchmarks. With increasing SFT examples, Mistral's post-SFT performance first dips and then gradually recovers and improves to performance better than before SFT training. Compared to the base model, the final performance after RL also first dips and then gradually goes up and improves to better level. Notably, post-RL performance recovers to the same level the base model slower than the post-SFT performance. The post-SFT and post-RL performance trends are not identical.
A.3 QWEN3-4B-BAN
Figure 9 shows results Qwen3-4B-base undergone SFT-RL with shortest SFT examples from AceReasonerl. 1-SFT dataset and RLVR via GRPO on DeepScaleR dataset. Reporting Pass @ performance averaged over 7 math benchmarks. With increasing SFT examples, Qwen3's SFT performances appear uncorrelated with the final performance after RI, where the latter remains the same despite the substantially improved SFT performance.
Figure 12 shows results on Owen3-4B-base undergone SFT-RL with Shortest/Longestt Longest+Shortest SFT examples from AceReasonerl. dataset and RLVR via GRPO on DeepScaleR dataset. Reporting Pass performance averaged over 7 math benchmarks. All SFT training substantially improves Qwen3's SFT performance, but the final performance after RL is mixed. Training Longest and 10k Longest+ 10k Shortest SFT examples lead to visibly improved final performance after RL where the latter achieves the best final performance for Qwen3 models in this work. Other SFT training lead to significantly degraded final performance after RI.
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40.2
Post-SFT
38.44
35.45
35
31.28
Performance (avg
30
30.85
25.86
25.03
25
23.52
26.08
20.89
20.84
17.58
16.1
12.28
Mistral- Instruct 128
AceRessoneri.. Shortest 10k
AceReasonerl.) AceReasoner1.. Shortest 50k
AceReasoner1 AceReasonerl.1 AceReasoner1.1
Shortest
Shortest
Shortest
Shortest
100k
200k
lep
500k lep
SFT Training
Figure 8: Mistral-NeMo- 12B-Instruct undergone SFT-RL with shortest SFT examples from AceReasonerl .1- SFT dataset RLVR via GRPO on DeepScaleR dataset. Reporting Pass@1 performance averaged over math benchmarks. With increasing SFT examples, Mistral's post-SFT performance first dips and then gradually recovers and improves performance better than before SFT training. Compared to the base model, the final performance after RL also first dips and then gradually goes up and improves to better level. Notably, post- RL performance recovers to the same level as the base model slower than the post-SFT performance. The post-SFT and post-RI. performance trends not identical.
post-SFT SFT+RL
Task Perforaane (avg of tasks)
43.01
40
Best
39.6
39.15
38.89
40.44
39.52
40.93
41.69
RL
37.08
Best
35
36.21
34.14
35.19
34.69
SFT
25
25.33
Owen3 base
AceReasoner1.1 1.1 Shortest 10k lep
AcaReasoner1.1 Shortest 50k lep
AceReasoner1.1
AceReasoner1.1
AceReasoner1.3
AceReasoner1..
Shortest
Shortest
Shortest
Shortest
100k
200k lep
300k
500k
SFT Training
Figure 9: Qwen3-4B-buse undergone SFT-RL with shortest SFT examples from AceReasonerl. J-SFT dataset and RLVR via GRPO on DeepScaleR dataset. Reporting Pass@ performance averaged over 7 math benchmarks. With increasing examples. Qwen3's POst-SFT performances appear uncorrelated with the final performance after RI, where the latter remains the same despite the substantially improved SFT performance.
B IMPLEMENTATION DETAILS
All experiments (SFT, RL, evaluation) are conducted on individual AWS (Mathew Varia, 2014) node with 8x NVIDIA A100 80GB GPU, Experiments spent > IM GPU hours on NVIDIA A100 80GB. We repeat RL training for 4+ runs on each data recipe and training paradigm (each run takes up to days), conduct 4+ evaluations different checkpoints across RL training run, and report the best performance for the model. We set the max sequence length to 8k tokens throughout SFT. RL, and evaluation.
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46.44
44.83
post-SF1
Task Performance (avg of 7 tasks)
45
43.01
Best RL
40.69
38:55
Best
35.02
34.14
SFT
30
25.33
25
Qwen3
Acefteasoner1. 1 AceReaioner11 1
AceReasoner1.1
Acefteasoner111
base
Shortest 10k lep Longest 10k lep
Longest 10k +
Langest 10k +
Shortest 10k lep Shortest 50k 1ap
SFT Training
Figure 10: Qwen3-4B-bas undergone SFT-RL with poortess/Loogest/oongssss-
SFT examples from AceReasonerl. 1--SFT dataset and RLVR via GRPO DeepScaleR dataset. Reporting Pass@1 1 performance averaged over math benchmarks. All SFT training substantially impraves Qwen3's post-SFT performance, but the final performance after RL is mixed. Training on Longest and 10k Longest+I00 Shortest SFT examples lead to visibly improved final performance after RL where the latter achieves the best final performance for Qwen3 models this work. Other SFT training lead significantly degraded final performance after RL.
B. MODELS AND DATASETS
We conduct three sets of experiments with SFT-RL post-training. On (lamma--B-Instrune models, we conduct SFT training with examples from Lllma-Nemotroo dataset (where we only select math samples with responses generated by QwQ-32B (Team, 2025) or DeepSeek RI (Guo al., 2025), hereinafter the same) and RL training on MATH dataset (train-split) (Hendrycks et al., 2021); on Mistral-NNmo--1255 12--Inttrue and Qwen3-4B-bast AceReasonerl. 1-SF dataset and RL training on DeepScaleR dataset (Luo et al., 2025). For all models, conduet RL training for 3 epochs where each run takes up days.
conduct SFT training with examples from
B.2 TRAINING
We conduct SFT training with LLaMA-Factory (Zheng et al., 2024) using learning rates Ir=le-5 and global batch size 128, and RL training via GRPO with verl (Sheng et al., 2024) using learning rates Ir=le-6 and global batch size 128. We sample 16 rollouts for each question with temperature=1.00 We set KL loss coefficient=0 and entropy coefficiemt0.001..
B.3 EVALUATION
Evaluations are conducted with pipelines originally developed in this work based on vllm (Kwon et al., 2023) and HuggingFace's math-verify (Kydlíček, 2025), enabling efficient inference with performant and accurate verification. We ran evaluations with the same template and generation configuration as in RI., using decoding temperature t=1.0 and the standard reasoning prompt ("Let' s think step by step and output the final answer within \\boxed )..))
evaluate task performance on math benchmarks, including MATH-500 (Hendrycks et al., 2021), AIME 1983-2024 (Veeraboina, 2023), GSM8k (Cobbe et al., 2021), AIME 2025 (of America, 2025), AMC (Competitions, 2025), Olympiad (He et al., 2024), Minerva (Lewkowycz et al., 2022), and report model performance Pass @ averaged over 64 repetitions and across 7 tasks. For the proposed predictors, we evaluate the generalization loss on the validation set of the SET data and Pass @ 64 accuracy averaged over 256 repetitions.
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C ADDITIONAL EXPERIMENTAL RESULTS
C.I DATASET-LEVEL
Table 5 shows results on Llama3-8B-Aa 8B-Instruct undergone SFT-RL with SFT examples from Llama- Nemotron- SFT dataset and RLVR via GRPO on MATH dataset (train-split). Reporting Pass @ performance averaged over 7 math benchmarks. Measuring prediction accuracy with coefficient of determination (R²) varying the ratio of fit-validation datapoints. We randomly select x SFT models and fit a linear function between their post-SFT performance and performance after RL, and use it to predict for the rest SFT models. We repeat random sampling for 100 times and report standard errors.
Table 6 shows results Mistral-NeMo 12B- Instruttttt undergone SFT-RL with SFT examples from AceReasonerl. 1.-SFT dataset and RLVR via GRPO on DeepScaleR dataset. Measuring prediction ac- curacy with coefficient of determination (R²) varying the ratio of fit-validation datapoints. Reporting Pass@ performance averaged over math benchmarks. We randomly select x SFT models and fit a linear function between their post-SFT performance and performance after RL., and use predict for the rest SFT models. We repeat random sampling for 100 times and report standard errors.
Table 5: Samma--8B-Instruue undergone SFT-RL with SFT examples from Llama-Nemotron-SSTe dataset and RLVR via GRPO on MATH dataset (train-split). Reporting Pass performance averaged over 7 math benchmarks. Measuring prediction accuracy with coefficient of determination (R²) varying the ratio of fit-validation datapoints. We randomly select x SFT models and fit a linear function between their Post-SFT performance and performance after RI., and use to predict for the rest SFT models. We repeat random sampling for 100 times and report standard errors.
Fitting-Valln Validation Datapoints/ Coefficient of determination (RP)
Prediction based SFT @ (avg- of 64) baseline
Prediction based on SFT Generalization Loss
Prediction based on SFT Pass @ Large (k=64)
Avg. Prodiction from SFT Gen. Loss Pass@Large (64)
Fitting: 3; Validation: 13
0.48
0.80
0.80
0.86 (+0.38)
Fitting: 4; Validation: 12
0.57 1.039
0.82 1.11
0.84 2is
0.92 not 20002a0000..... (+0.35)
Fitting: 5; Validation: 11
0,57 #
0.88 + 0.09
0.87 2.000
0.94 2044
Fitting: 6: Validation: 10
0,57 +
+ o.0
0.87
0.95 aa (10.38)
Fitting: 7; Validation: 9
0.64
0.89
0.90
0.95 (+0.31)
Fitting: 8; Validation: 8
0.64
0.88
0.88
0.93 uu (+0.29)
Fitting: 10; Validation:
0.59 +03
0.85 +
0.85
0.94 a (+0.35)
Fitting: 12; Validation:
0.54 8.0
0.86 0.18
0.81 022
0.91 + +az (+0,37)
Table Mistral-NeMo--2---nstte undergone SFT-RL with SFT examples from AceReasonerI. .I- SFT dataset and RLVR via GRPO on DeepScaleR dataset. Measuring prediction accuracy with coefficient of determination (R²) varying the ratio of fit-validation datapoints. Reporting Pass @ performance averaged over math benchmarks. We randomly select SFT models and fit linear function between their PFst-SFT performance and performance after RL, and use to predict for the rest SFT models. We repeat random sampling for 100 times and report standard errors.
No. of Fitting- Validation Datapoints/ Coefficient of determination (R²)
Prediction based SFT @ass I (avg- of 64) baseline
Prediction based on SFT Generalization Loss
Prediction bused on SFT Pass @ Large (k=64)
Avg. Prediction from SFT Gen. Loss Pass@ Large (64)
Fitting: 3; Validation: 7
0.32
0.73 ian (+0.41)
0.52
0.61
Fitting: 4; Validation: 6
0.27 1.3
0.75 aou (+0.488
0.51 2.77
0.69 1:6
Fitting: 5; Validation: 5
0.29 a:3
0.79 ooz o (+0.50)
0,57 032
0.72 #24
Fitting: 6: Validation: 4
0.37 +03
0.78 a (40.41)
0,57
0.67 a35
Fitting: 7; Validation: 3
0.36
0.77 a (+0.41)
0.57
0.66
Fitting: 8; Validation: 2
0.31
0.68 10xx
0.47
0.64
C.2 INSTANCE-LEVEL
Table 7 shows results on Llama3-8B Camma--8B-Instiee II-strructt undergone SFT-RL with SFT examples from Llama- Nemotron-SEP SFT dataset and RLVR via GRPO MATH dataset (train-split). Reporting Pass @ performance averaged over math benchmarks. Measuring prediction accuracy with coefficient of determination (R22 varying the ratio of fit-validation datapoints. We randomly select SFT models
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and fit a linear function between their post-SFT performance and performance after RL, and use it to predict for the rest SFT models. We repeat random sampling for 100 times and report standard errors.
Table 8 shows results Llama3-8B ----trructtt undergone SFT-RL with SFT examples from Llama- Nemotron-See SFT dataset and RLVR via GRPO on MATH dataset (train-split). Reporting Pass performance averaged over 7 math benchmarks. Spearman's 's rank correlation between performance predicted from post-SFT models and the actual performance after RL, grouped by different SFT training budget.
Table 7: Llama3-8B-Img -8---rutt undergone SFT-RL with SFT examples from Llama-Nemotron-STie dataset and RLVR via GRPO on MATH dataset (train-split). Reporting Pass@ performance averaged over 7 math benchmarks. Measuring prediction accuracy with coefficient of determination (R²) varying the ratio of fit-validation datapoints. We randomly select x SFT models and fit a linear function between their post-SFT performance and performance after RL, and use to predict for the rest SFT models. We repeat random sampling for 100 times and report standard errors.
No. of Fitting-Validatinn Validation Datapoints/ Coefficient of determination (It))
Prediction based on SFT Pass (avg. of 64) baseline
Prediction based
SFT Pass Large (k=64)
Fitting: 3; Validation:
0.40 1
0.89 (.m(++049)
Fitting: 4: Validation: 13
0.49 # 0.30
0,89 t (+0.40))))
Fitting: Validation: 12
0,55 = #:222
0,91 # (+006)))))
Fitting:
Validation: 11
0.54
0.92 =
Fitting: 7; Validation: 10
0.55
0.92 1 (+0.37)
Fitting: 8; Validation: 9
0,58
0.92 am (+0.34)
Fitting: 10; Validation: 7
0,56 0:0
0.92 0 0.00 (+0.36)
Fitting: 12; Validation: 5
0.57 1,7
0.92 om (+0.35)
Table Lllam3-88 IB-Irutuctt undergone SFT-RI. with SFT examples from Llama-Nemotron-SFFT dataset and RLVR via GRPO on MATH dataset (train-split). Reporting Pass@ performance averaged over math benchmarks. Spearman's ' rank correlation between performance predicted from post-SFT models and the actual performance after RL, grouped by different SFT training budget.
SFT Compute Budget/ Spearman's 's Rank Correlation
Prediction based on SFT Pass@1 (avg. of 64) baseline
Prediction
SFT Parge
Large k (k=64)
Low Budget (< 2B tokens)
0.77
0.99 (+0.22)
Medium Budget (2 tokens)
0.60
0.90 (+0.30)
High Budget 20B tokens)
0.70
0.94 (+0.24)
Average
0.69
0.94 (+0.25)
Table 9 shows results on Mistral-NeMo, 2BI-Instruct undergone SFT-RL with SFT examples from AceReasonerl.J 1-SFT dataset and RLVR via GRPO on DeepScaleR dataset. Measuring prediction accuracy with coefficient of determination (R²) varying the ratio of fit-validation datapoints. Mistral- NeMo-12B-Instee undergone SFT-RL with SFT examples from AceReasonerl.. 1--SFT dataset and RLVR via GRPO on DeepScaleR dataset. Reporting Pass@ performance averaged over 7 math benchmarks. We randomly select SFT models and linear function between their POst-SFT performance and performance after RL, and use it to predict for the rest SFT models. We repeat random sampling for 100 times and report standard errors.
Table 10 shows results on Mistral-NeMar AceReasonerl.. 1-- 1-SFT dataset and RLVR via GRPO on DeepScaleR dataset. Reporting Pass performance averaged over 7 math benchmarks. Spearman's 's rank correlation between performance predicted from post-SFT models and the actual performance after RL, grouped by different SFT training budget.
Instruct undergone SFT-RL with SFT examples from
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Table 9: Mistral-NeMo-12B-nnstie undergone SFT-RL with SFT examples from AceReasonerl. .1-- SFT dataset and RLVR via GRPO on DeepScaleR dataset. Measuring prediction accuracy with coefficient of determination (It²) varying the ratio of fit-validation datapoints. Mistral-NeMo-112. Instruct undergone SFT-RL with SFT examples from AceReasonerl. 1--SF dataset and RLVR via GRPO DeepScaleR dataset. Reporting Pass@ I performance averaged over math benchmarks. We randomly select x SFT models and fit a linear function between their post-SFT performance and performance after RL, and use it to predict for the SFT models. We repeat random sampling for 100 times and report standard errors.
No. of Fitting- Validation Datapoints/ Coefficient of determination (R²)
Prediction based SFT Pass@ (avg. of 64) baseline
Prediction based
SFT Pass Pass@Large (k=64)
Fitting: 2: Validation: 10
0,55 0.42
0.87 a (+0.32)
Fitting: Validation:
0.71 2.15
0,94 (+02)))
Fitting: Validation:
0.69
0.98 (+0.299)))
Fitting:
Validation:
0.75
0,98 am (40.23)
Fitting: 6; Validation:
0.73
0.98 am (+0.25)
Fitting: 8; Validation: 4
0,69
0.97 +m (+0.28)
Fitting: 10; Validation:
0.68 1
0.91 (+0..3))
Table 10: Mistral-NeMMo-12--Instt undergone SFT-RL with SFT examples from AceReasonerl.. 1- SFT dataset and RLVR via GRPO on DeepScaleR dataset. Reporting Pass@1 performance averaged over 7 math benchmarks. Spearman's 's rank correlation between performance predicted from post-SFT models and the actual performance after RL, grouped by different SFT training budget.
SFT Compute Budget/ Spearman's 's Rank Correlation
Prediction based on SFT Pass@1 (avg. of 64) baseline
Prediction based on SFT Pass@ Large k (k=64)
Low Budget (< 2B tokens)
0.80
0.95 (+0.25)
Medium Budget (2 ~ 5B tokens)
0.80
1.00 (+0.20)
High Budget (5 20B tokens)
0.50
1,00 (+0.50)
Average
0.70
0.98 (+0.28)
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D BROADER INVESTIGATIONS AND SUPPLEMENTARY EXAMPLES
D.1 FINE-GRAINED COMPARISONS: BEST SET MODEL VS. BEST SFT+RL MODEL.
Individual task performance largely correlates with the average performance, but the relative im- provements differ across tasks. The average performance over multiple tasks has been reported to scale much more smoothly to the point of becoming predictable (Gadre et al., 2024). Thus, in this work, we use the average performance across multiple tasks as the main metric. The final results are consistent with these expectations.
Table 11: Detailed comparison of post-SFT and SFT+RL performance (Pass accuracy) between the model with the best SFT performance MM2-Mistral-NeMo 12B-Instruct with SFT 500k shortest AceReasonerl. 1.t.asooner1.1-SSPP. 1-SFT examples for epochs) and the model with the best SFT+RL performance (MI=Mistrl--NeMMe 12B-Inst with SFT on 100k random AceReasonerl. 1-SFT examples for epochs) from Figure 4(a) across various benchmarks. Despite MI trailing M2 by 8.66% after SFT, the final SFT+RL results, MI leads M2 by a relative margin of +6. 14% with up to +33.31 % on
individual task.
/ Benchmark
Avg
MATH-500
AIME 1983-2024
GSM8k
AIME 2025
AMC
Olympiad
Minerva
Supervised Fine Tuning (SFT)
MI SFT
35.11
70.50
10.62
89.34
13.48
38.71
15.15
8.03
M2 SFT
38.44
76.68
16.31
87.71
14.53
46.34
18.56
8.97
(MI-M2//M2 (%)
-8.66
-8.06
34.89
+1.86
-7.23
-16.47
-18.37
-J0.45
Supervised Fine- Tuning Reinforcement Laaringg [SFT+RI.)
MI SFT+RL
42.67
79.15
25.90
85.43
19.73
54.56
23.53
10.40
M2 SFT+RI.
40.20
79.90
23.71
86.96
18.33
44.62
17.65
10.25
(MI-M2W/M2 (%)
+6.74
-0.94
+9.24
-1.76
+7.64
+22.28
+33.31
+1.46
Table 12: Detailed comparison of post-SFT and SFT+RL performance (Pass@) accuracy) be- tween the model with the best SFT performance M22-Mistal-NeMoo-1BBBmm AceReasonerl. 1-SFT examples for epochs) and the model with the best SFT+RL. performance MM==Mistral-NeMo- 12BMMMo--BBB-A IIsttutt with SFT on 25k AceReasonerl. 1--SFT examples for epochs) from Figure across various benchmarks. Despite MI trailing M2 by 11.20% after SFT, in the final SFT+RL RL. resultss,,, MI leads M2 by a relative margin of +2.26% with up to +95.01% % on individual task.
with SFT on 25k
/ Benchmark
Avg
MATH-500
AIME 1983-2024
GSM8k
AIME 2025
AMC
Olympiad
Minerva
Supervised Fine-Tuning (SFT)
MI SET
32.51
63.53
8.19
91.96
6.77
29.90
17.00
10.28
M2 SFT
36.61
69.34
12.12
91.78
14.33
37.37
20.71
10.28
(MI-M2WM2 (%)
-11.20
32.45
-0.20
-53.08
-19.99
-17.9
Supervised Fine- MieeTTning Reinforcement Learning (SFT+RL)
MI SFT+RI.
42.57
77.18
21.81
89.81
18.80
45.93
24.90
19.56
M2 SFT+RL
41.63
78.34
18.68
94,25
19.42
47,90
22.75
10.03
(MI-M2WM2 (%)
+2.26
-1.48
+16.76
-470
-3.19
-4.11
+9.45
+95.01
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Table 13: Detailed comparison of post-SFT and SFT+RL performance (Pass accuracy) between the model with the best SFT performance (M2-Llama3-885 8B-Instruct with SFT on 10k Nemotron examples for 10 epochs) and the model with the best SFT+RL performance (MI=Llama3-8) -n-ssruuutt with SFT on 10k Nemotron examples for 5 epochs) from Figure 5(a) across various benchmarks. Despite MI trailing M2 by 1.33% after SFT, in the final SFT RL results, MI leads M2 by a relative margin of +7.31% 1% with up to +7..52% on individual lask.
/ Benchmark
Avg
MATH-500
AIME 1983-2024
GSM8K
AIME 2025
AMC
Olympiad
Minerva
Supervised Fine Tuning (SFT)
MI SFT
26.63
53,12
2.75
84.59
3.18
21.71
7.75
13.37
M2 SFT
26.99
54,43
2.69
85.87
3.18
21.59
7.88
13.34
(MI-M2NM2 (1))
-2.41
+2.34
Supervised Fine Twning Reinforcement Learning (SFT+RL))
SFT+RL
30,08
59.96
5.06
87,59
4.74
26.81
11.12
15.28
M2 SFT+RI.
28.03
59.93
3.83
85.40
4.32
24.93
9.16
8.66
(MI-M2WM2 (%)
+7.31
+0.05
+32.03
+2.56
+9.65
+7.54
+21.45
+76.52
Table 14: Detailed comparison of post-SFT and SFT+RI. performance (Pass accuracy) between the model with the best SFT performance (M2-Llama3-SB- I--truutcct with SFT 25k Nemotron examples for epochs) and the model with the best SFT+RL performance (MI=Llama3-88 ---aa---------ccc with SFT on 25k Nemotron examples for epochs) from Figure 5(b) across various benchmarks. Despite MI trailing M2 by 7.02% after SFT, in the final SFT+RL RL results, MI leads M2 by a relative margin of +4.82% with up to +45.76% on individual task.
Model Benchmark
Avg
MATH-500
AIME 1983-2024
GSM8k
AIME 2025
AMC Olympiad
Minerva
Supervised Fine-Tinning (SFT)
SET
57.25
5,00
83.21
9.25
M2 SET
30.63
62.12
6.78
84.21
8.23
30.71
10.00
12.37
(MI-M2 (%)
-7.02
-7.84
22626
-1.19
-27.22
-15.24
-7.50
+2.51
Supervised Fine-Tunine Tuning Reinforcement Learning (SFT+RI.)
SFT+RI.
33.25
67.40
9.90
87.84
9.79
32.06
15.34
11.25
M2 SFT-RL
31.72
65.12
9.84
84.93
9.27
34.12
13.12
7.72
(M1-M2XM2 (%)
+4.82
+3.50
40.61
+3.43
+3.622
-6.04
+16.92
445.76
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D.2 ABLATION STUDIES: PASS K, ENTROPY
To facilitate pursuit towards underlying mechanism behind the intriguing SFT-RL dynamics, we investigate the aggregated entropy the response level.
As shown in Tables 15, 16, and 17, we witnessed that correct responses generally show a lower entropy compared to incorrect responses. However,
1. A lower entropy after SFT training does not always suggest worse post-RL outcomes, and vice versa.
2. A lower entropy after SFT training does not indicate better post-SFT performance, nor does the entropy gap between correct and incorrect responses.
One hypothesis is that the loss of tokens diverges. The loss is often dominated by a few high-loss tokens (Gadre al., 2024). During SFT over-training, despite validation loss on some tokens elevating due to overfitting, the loss on some other tokens may still be decreasing, and these tokens may be more crucial for the task performance. On the other end, Wang et al. (2025) suggests that certain tokens are crucial for RL. These seem to be different tokens from those crucial for SFT. Overfitiigghhigher loss on these tokens could cause degradation in the subsequent RL. We found that during decoding, math/symbolic tokens generally show near-7ero loss-i.e., the model is almost always certain on these math tokens. The highest loss tokens are human languages that are naturally ambiguous interchangeable, such whether to start the sentence with "Thus" or "Therefore". As a result, the (per-token) entropy for the generations mostly measures verbosity, and does not show consistent patterns with accuracy.
In general, we did not observe a consistent pattern to explain the SFT-RL dynamic, but it confirms that entropy alone is an inadequate explanation.
Pearson Corrlation (Predictability for +L)+
1.0
+ SFT [50k Shortest+ Longest]
Task Performance (avg of 7 tasks)
50
+ SFT [50k Shortest]
0.8
Pearson Corr: SFT Pass@ k/SFT+RL Pass@1
0.6
0.4
30
0.0
20
0.2
-0.4
Pass@1
Pass@2
Pass@4
Pass@8
Pass@16
Pass@32
Pass@64
SFT Pass@k
Figure 11: Detailed comparison between Pass@k performance for Mistral-NeMo- 1a---oo--B--Inssteue Mistral-NeMo-11B-Inssrtdd with SFT on 50k shortest and 10k longest AceReasonerl.11-3 examples for 1 epoch/ Mistral-NeeMo- 12B-II2Itiitt Instruct with SFT on 50k shortest AceReasonerl..-SK. examples We show Pearson Correlation scores between Pass @ of
for 1 epoch under k=, these models and their Pass@1 1 performance after subsequent RI. Predicting post-RI. performance from Pre-RL. @ass@1 leads to negative Pearson Corr. score, which indicates ineffective predictions. Prediction accuracy (indicated by Pearson Corr.) steadily improves with larger k values in Pass k reaching over 0.80 at Pass 16 and close to I at P
25

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Published as a conference paper at ICLR 2026
Pass Performance Change during RL
Mistral-NeMo- -awteMMM------truuuee
+ SFT on [50k Shortest+ 10k Longest]
SFT [50k Shortest]
20
15
10
*Average* MATH-500
AIME
GSMBk
AIME
AMC
Olympiad
Minerva
1983-2024
2025
Downstream Tasks
Figure 12: Improvements in per-task @ accuracies for Histral-NeMar 12B Instruct/ Distral-NeMo-Dee Instruct with SPT on 50k shortest and 10k longest AceReasonerl 1.1.ssoner1111-SF.. 1.1SSFT examples for epoch/ Mistral-NeMM I2B-Inttruct with SFT 50k shortest AceReasonerl.I 1-SFT examples for epoch during RL training. Individual task performance largely correlates with the average performance, but the relative improvements differ across tasks. The average performance over multiple tasks has been reported to scale much more smoothly the point of becoming predictable (Gadre et al., 2024). Thus, in this work, we use the average performance across multiple tasks as the main metric. The final results are consistent these expectations.
Table 15: Per-task Pass @k accuracies and Entropy for Mistral-NeMot 12B- Instruct before and RL training. In comparisons between response-level entropy averaged over all responses vs. over correct responses, correct responses generally show lower entropy compared incorrect responses, but does not explain the dynamics between SFT-RL performance.
Task
Pass@1
Pass@2 2
Pass @4
Pass@8 Pass 16
Puss@
Pass@ 64
Avg Ent.
Corr. Ent.
hase model
MATH-500
30.68
43.48
$4.68
63.28
70.39
77.20
84,00
236.60
131.40
AIME 1983-2024
0.78
1.50
2.80
4.95
8.13
12.70
20.00
462.30
390.60
GSM8K
71.31
84.53
90.74
93.06
94.61
96.52
98.00
93.60
72.41
AIME 2025
0.26
0.52
1.04
2,08
4.17
8.33
16.66
488.60
513,30
AMC.
10.78
17.30
25.68
36.02
48.07
59.98
70.00
370.30
220.20
Olympiad
3.22
5.82
10.00
16.40
25.54
36.67
48.00
434.50
222.60
Minerva
6.09
10.25
15.75
22.39
30.33
39.22
48.00
258.20
167.50
Average
17.59
23.34
28.67
34.03
40.18
47.23
54.95
334.87
245.43
+RL
MATH-500
55.15
62.06
67.05
71.49
76.63
83.11
92.00
78.07
32.01
1983-2024
5.26
7.92
16.29
22.00
GSM8k
86.34
89.34
91.64
93.42
94.76
95.77
96.00
20.45
16.20
AIME 2025
1.15
2.09
3.54
5.46
7.92
11.69
16.66
185.40
80.77
AMC
21.84
28.67
36.06
43.17
49,84
56.51
64.00
117.60
42.68
Olympiad
5.25
8.59
12.71
17.55
23.32
29.31
34.00
145.60
74.91
Minerva
9.56
14.21
19.26
23.90
27.58
31.47
36.00
83.03
59.30
Average
25.87
29.74
33.65
37.56
41.64
46.31
$1.52
113.74
61.70
26

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Published as a conference paper at ICLR 2026
Table 16: Per-task Pass@k accuracies and Entropy for Mistral-NeMo- 1aMMMo-1B-IIstrucce with SFT on 50k shortest and 10k longest AceReasonerl.1-S3 examples for epoch before and RL training. In comparisons between response-level entropy averaged over all responses vs. over correct responses, correct responses generally show lower entropy compared incorrect responses, but does not explain the dynamics between SFT-RL performance.
Task
Pass
Pass@2 2
Pass @4
Pass@8
Pass 16
Pass@32
Pass@ 64
Avg Ent.
Corr. Ent.
SFT
MATH-500
42,78
53.65
61.63
68.30
74.65
80.14
84,00
2266.00
902,50
AIME 1983-2024
1.38
2.45
3.99
5.81
9.50
12.00
4733.00
1722.00
GSM8K
99.88
100.00
713.90
$31.00
AIME 2025
0.94
1.62
2.48
3.14
3.33
3.33
3.33
4942.00
6888.00
AMC
8.75
14.47
22.67
32.89
43.99
54,70
62,00
4023.00 00
1987.00
Olympiad
9.15
13.19
18.02
23.75
30.00
5229.00 00
2718.00
Minervu
4.69
7.50
10.95
15.00
19,99
25.97
32.00
4019.00
2272.00
Average
20.34
25.24
29.56
33.79
38.12
42.47
46.19
3703.70
2431.50
SFT+RL
MATH-500
51.78
60.98
67.51
72.91
77.49
82.02
86.00
1996.00
724.70
AIME 1983-2024
5.45
7.42
12.77
5041.00
1638.00
GSM8K
84.56
92.76
96.94
98.75
99.54
99.94
100.00
579.60
482.10
AIME 2025
0.83
1.54
2.64
4,06
5.58
7.51
10.00
5522.00
7202.00
AMC
11.81
18.88
28.53
40.07
51.81
62.35
72.00
3901
1799.00
Olympiad
5.41
8.45
12.27
17.02
22.44
28.51
36.00
6002.00
2595.00
Minerva
8.97
12.51
15.61
18.59
22.00
25.26
28.00
4861.00
1529.00
Average
23.69
28.42
32.71
36.97
41.24
45.48
49.71
3986.09
2281.40
Table 17: Per-task Pass@k accuracies and Entropy for Mistral-NeMo- -e--o-MM----sstriicc 12B-Instruct with SFT on 50k shortest AccRReasoner..-SPP examples for epoch before and RL training. In comparisons between response-level entropy averaged over all responses vs. over correct responses, correct responses generally show lower entropy compared to incorrect responses, but does not explain the dynamics between SFT-RL performance.
Task
Pass
Pass@2
Pass @4
Puss@8
Pass@ 16
Puss
Pass@ 64
Avg Ent.
Corr. Ent.
SFT
MATH-500
32,56
44.08
53.95
61.75
68.55
75.20
80.00
771.60
452,70
AIME 1983-2024
2.02
2.91
3.88
5.17
6.97
10.00
1828.00
1136.00
GSM8k
65.46
78.12
85.99
91.59
95.11
96.98
98.00
670.10
474.60
AIME 2025
0.31
0.62
1.21
2.32
4.25
7.12
10.00
1963.00
2515.00
AMC
7.53
12.35
19.40
28.64
39.47
51.27
64.00
1232.00
856.70
Olympiad
7.22
10.91
21.88
2058.00
767.00
Minerva
3.00
5.01
7.52
10.30
13.84
19.02
26.00
2491. (00
1448.00
Average
16.10
20.96
25.46
29.91
34.60
39.78
45.14
1573.39
1092.86
RL
MATH-500
51.74
60.76
67.21
72.38
77.63
83.00
88.00
2924.00
1315.00
1983-2024
20.03
1971.00
GSM8K
85.62
92.91
96.48
97.93
98.47
98.99
100.00
1220.00
916.60
AIME 2025
0.36
0.71
1.36
2.50
4.25
6.58
10.00
6234.00
12624.00
AMC
12.28
19.08
27.30
35.86
45.14
55.12
64.00
5072.00
2874.00
Olympiad
3.88
6.34
15.00
27.26
32.00
6080.00
2809.00
Minerva
8.06
10.87
13.41
16.28
19.98
24.03
28.00
5708.00
2568.00
Average
23.52
27.76
31.56
35.42
39.91
45.00
50.29
4772.71
3582.51
27

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Published as a conference paper at ICLR 2026
E SAMPLE SFT EXAMPLES
E. SHORTEST EXAMPLES
A sample Shortest Examples from Llama-Nemotron Post-Training Dataset (Singhal et al.) (~750 tokens):
IIput]]:
Solve the following math problem. Make sure to put the answer (and only answer) inside boxxe ()AnnnThh probability of snow on 1st January is 1/20. What is the probability that snow will fall on the next 1st January?"
Outpptt]:
<think> Okay, so I have this problem here: "The probability of snow on 1st January is 1/20. What is the probability that snow will fall on the next 1st January?" ?" Hmm, let's see. Alright. the question is asking for the probability that it will snow next January 1st, given that the probability is 1/20. Wait, is there more to this? Let me check again. Wait, maybe misread. The original probability is 1/20 for snow on January 1st, and they want the probability it will snow on the next January 1st. Hnm. That seems straightforward. But maybe there s a trick here? Like, are they assuming each year is independent? Because if the probability each year is independent, then is's just 1/20 again. But maybe the question is different? Wait, maybe the wording is tricky. Let me parse it again. "The probability of snow on 1st January is 1/20. What is the probability that snow will fall on the next 1st January ?" So, "next" January 1st. So if each year 's snowfall is independent, then the probability remains 1/20. But maybe there 's some hidden assumption here. Wait, sometimes in probability problems, they might be talking about conditional probabilities else. But this problem doesn 't mention any previous events or dependencies. It just states the probability is 1/20. So if each occurrence is independent, then the probability doesn change. So next year" 's probability is also 1/20. Wait, but let me think again. Maybe the problem is in another language originally, and "next" is being translated as "the next occurrence" " or something? But no, it's about "next 1st January So like the next time January 1st comes around. which would be the following year. So in terms of probability. assuming each year is an independent event, the probability is the same each year, right? So if the probability of snow any given January 1st is 1/20, then regardless of previous years. the next one would also be 1/20. Alternatively, could it be that the problem is asking for the probability that it will snow next year given that it snowed this year? But the problem doesn t say anything about prior snowfall It just gives the base rate. So think the answer is 1/20. Let me confirm. Another angle: sometimes people confuse "probability of snow on a specific day" with "probability of snow in a year" but here it's specifically if the probability each year on that date is 1/20, then next year's is also 1/20. Unless there doesn 't indicate that. So I think the answer is 1/20. So I just need to write that in box. Wait, maybe I need to make
in the following year? So next year" 's
something
1st January each year. So
pattern. but the problem
28

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Published as a conference paper at ICLR 2026
sure if the problem is in different context. For example maybe 's a different place where the next January 1st is in the southern hemisphere? But that overcomplicating The problem is likely testing understanding of independent events. So the answer is 1/20. </think> Aboxed (Ndfrac (20))
29

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will it Published as a conference paper at ICLR 2026
A sample Shortest Examples from (areeassneer111-SF1 Dataset (Liu et al., 2025b) (~700 tokens):
[Input]:
A train 18 long passes pole in 9 seconds. How long take to pass a bridge 36 m long?
Output]]:
<think> Okay, so need to figure out how long it takes for a train that 18 meters long to pass bridge that 36 meters long. The problem already says that it takes 9 seconds for the train to pass pole. Let me start by understanding what that means. When the train passes a pole, the entire length of the train has to go by the pole. So, if the train is 18 meters long and it takes seconds to pass the pole, that means the front of the train passes the pole, and then 9 seconds later the end of the train passes the pole. So, in 9 seconds, the train travels a distance equal meters. From this I can calculate the speed of the train. Speed is distance divided by time. So, speed 18 meters / seconds. Let me compute that. 18 divided by 9 is 2. So, the train speed is meters per second. Hmm, that seems right. Now, the question is about passing bridge that's 36 meters long. To pass a bridge, the train has to cover the entire length. Why? Because first the front of the train has to cross the entire bridge. and then the end of the train has to exit the bridge. So, the total distance the train needs to cover is the length of the bridge plus the length of the train. So that would be 36 meters plus 18 meters, which is 54 meters. So, the train needs travel 54 meters to completely pass the bridge. Since the speed is 2 meters per second, the time it takes is the total distance divided by speed. Time 54 meters 2 meters per second. Let me calculate that 54 divided by is 27. So, it should take 27 seconds. Wait, let me double-chect to make sure didn't mix up anything. The first part: passing pole takes seconds for bridge is 36 meters. So, when passing the bridge, the front of the train has to go 36 meters to get to the end of the bridge, and then the entire train has to clear it, which is another 18 meters. So total distance is indeed 36 + 18 = 54 meters. Time is 54 / = 27 seconds. Yeah, that makes sense. Alternatively, maybe someone might think you just need to cover the bridge 's length. but no, the train 's own length has to be considered too. Because until the last part of the train is off the bridge, the bridge isn completely passed. So. adding the two lengths is correct. Therefore 27 seconds should be the right answer. </think> The train 's speed is A ffac (18 \text (9 \text[ s}} = \text m/ \). To pass 36 m bridge, the total distance needed is \( 36 \text m) + 18 (tett( m| 54 \text m| \). Time required is M Afrac (54 text m)) (2 \text m/s]) = 27 \text seconds] \). (boxed((7))
its own length, which is 18
length of the bridge plus its
meters. So speed is 18/9
m/s. Then,
30

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Published as a conference paper at ICLR 2026
E.2 LONGEST EXAMPLES
A sample Longest Examples from AceReasonerl. Dataset (Liu et al., 2025b) (~8k tokens):
[Input]:
For sets SA$, $BS, and SC$. where (n (B) and (C denote the number of subsets of SAS, $B$. and $C$ respectively and SVleft 1 A \right I Veeft B \right 1005, if Sn(A) nB) (C) n(A leap B Aeap C)$, what is the minimum possible value of $\left A cepp B cap C Aright 1$?)
[Output]:
Okay let 's try to tackle this problem. Hmm, so we have three sets A, B. and C. The problem states that nAA, nB) and n(C) are the number of subsets of each set. Wait, but usually, the number of subsets of set with size k is 2*k. So maybe n(A) is 2*IAI, nB) is 2ABI,, and similarly for nCC) Let me check: the problem says "n(A), nBB) and nCC) denote the number of subsets of A, B, and C respectively. Yep, so that means nAA) (AAIA nBB) (IBI)) nCC (*CIC)].
Given that IAI IBI 100. So n(A) 22100, nB) 2^100. The equation given is (A) + (B) + n(C) = n(A B
C). So B
that translates 2^100 + 20100 2*(ICI] 21/AA l).
C
Wait, , the right hand side is the number of subsets of the intersection of A, B. and The number of subsets of a set is
2 cardinality of the set), n(A B C) 2^11A CII.
B
So the equation is 20100 + 2A100 + (ACIC = 2*(k), where is IA CI, and need to find the minimal possible k.
B
Let me simplify the hand side. 2^100 20100 is 2=2
1010 So left-hand side is 2^101 24(ICI). So 2^101 + 2 [IC 1] 24k.
We need to find the smallest possible such that this equation holds, given that ICI can be adjusted (I think we can choose CI as part of finding the minimum? Wait, no. The problem is about the minimal IA B Cl. given that the equation holds. So we need to find the minimal k such that 2*101 + 2n CI) 24k, and also considering the relationships between the sets A, B. C.
Wait, but how is ICI related to A and B? Because A, B, and C are just sets but the problem doesn 't specify any constraints on their relationships except through their intersection A B C. Hmm. Wait, but the number of subsets of C. which is (ICI), is part of the equation. But also the intersection A C a subset of C, right? Because the intersection C
of A, B, and C is subset of each of them. So IA B
ICI. So the cardinality of the intersection can't be more than the cardinality of C. But in equation. 2 2*(k) is equal to 20101 + 24(ICI), so 2 = 2*(ICI) + 21101.
31

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Published as a conference paper at ICLR 2026
So need to find integers ICI and k such that k ICI (since the intersection can be larger than C), and 2*[k] = (*[101) + 2*(ICI). Also, k must be an integer because it the cardinality of a set.
Wait, but how can we solve this equation? Let me think. Let's IA B Cl. so we have 2
denote ICI and 2*1101) + 2 We need to find the minimal k such that there exists with m
and 2 2 + 2'm.
But wait, if is greater than or equal to k, then 2 multiple of 24k. But 2*k = 2*(101] 2*m. If m 101, then 2/m is larger than 2"(101}, so 2*[101] + 2 = 2'm (1 + 2^010
m)). For this to be equal to 2*k, power of two. Let's is write:
21(101 m] must be a
Let m k. and 2*k = 21[101] + 2*m. Let's consider cases where m 101 and m 101.
Case 1: m 101. Then 2 2 2|11]]] + 2 211101] 2 101)). For this to be power of two. 1 + 2 101] must be 101, which positive since > 101. Then 1 + 2nt must be power of two. 1 1 2nt
a power of two. Lets set t
m =m--
2*s, for some s. So 2 - 2 = 1. This is possible only if
0, since 2*s 1= 2tt. but t must be positive here (since 2nt
> 101 implies t
) But 2 2 =l. If
divides the left-hand side. so 2nt must divide 1, which is Therefore,
only possible if t=0, which contradicts t there is solution in this case.
Case 2: m 101. Then 2 2^[101] 2'm. Let's factor out 2^m: 2 2 (1 + 2M101 -m)). Since m 101 101 m 0 so exponent is non nen-negative For the right-hand side to be a power of two, (1 + 2A1101 - m]) must be a power of two. Lets denote t 101 m. Then
2 must be power of two. So, + 2nt = 24s. This equation holds when is such that 2 +1 is a power of two.
Looking at this, 2 +1 = 2 The only solutions for this in integers are when t=0, then 2^0 +1=2=2^1, so s=l. Another possibility might be t=l: 2+1=3, which is not a power of two. 1=2:4+1=5; nope. 1=3:8+1=9; still not. So the only solution is t=0, leading to 1+1=2. So 1=0 => + 210 = 2. Therefore, t=0,
2^(0)=2. so the equation becomes 2 = 2^101 + 21101 = 21101 *2 = 2^102. Therefore, k =102.
so 101 m 0 => =101. Then
Wait, but m 101, and must be m. but here k=102, which would be greater than m=101. But that contradicts the requirement that
m. So this is impossible.
Therefore, in case 2, even though we have t=0 leading to m=101 and k=102, which would require k=102 m=101, violating the cardinality condition. Therefore. this is not acceptable.
Hmm. so neither case 1 nor case 2 gives us valid solution. Wait, that can be. Maybe there 's an error in my reasoning here. Let me check again.
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Published as a conference paper at ICLR 2026
Wait, when considered 101 and t=101 -m. So if t=0, then =101, then the equation becomes 2 = 2^101 + 2^101 = 2^102, so k=102. But since m=101. then the intersection k must be 101 but here k=102, which is greater. So that 's impossible So that doesn 't work. Therefore, in case 2, there are no solutions either.
Hmm, so does that mean there is no solution? But the problem states "what is the minimum possible value of IA B C I", implying that there is a solution. So perhaps my approach is missing something.
Wait, maybe need to consider that C can be a subset of C? No, actually A B is subset of C. so the size can 't exceed ICI. So perhaps need to adjust the equation? Or maybe my initial translation of the problem is wrong.
B
Wait. let me back to the problem statement. It says: For sets A B. and C. where (A), nBB) and n(C) denote the number of subsets of A, B, and C respectively, and IAI = IB = 100, if B C), what is the minimum CI?
(A) + (B) + (C) n(A
possible value of IA
B
Wait, so AA
C) is the number of subsets of A B B ClI. Similarly n(C)=2M(IC).. So 2^100 2^100 + 2 (ICI) B CII.
B
C. So that is (* 1|=2^100, n(B)=2~100, = (*IIA
So yes, as before, 2*1101] + (ICII) = 2*(k), where k is IA CI. We need find the minimal such that this holds. with ICI k. So k ICI.
B
Wait, perhaps ICI can be larger than k. Wait, but we need to find the minimal k. So to minimize k, we need to make the right- hand side as small to hold [1010 + 2*(ICI]. But ICI can be as large as needed? Wait, but ICI is part of the equation. If ICI is large. then 2 is very big which would require k to be large as well So maybe to minimize k, should take ICI possible. But ICI can be smaller than k, since = IA CI ICI.
possible
2*(k) must be just enough
small
Wait, so maybe we need to take ICI k. Then the equation becomes 2111011 2 = 2*(), which would imply 21[101] =0, which is impossible. So ICI must be greater than k? Wait. but if ICI is greater than k, then (*(IC is bigger than 2^(k), so 2*(101] (*(ICI) is bigger than 2*(k). Therefore, the equation 2*1001] 2*[[CI] = 24(k] implies that 2*[k] must be larger than 2*[101] and 2*(ICI). So must be greater than both 101 and ICI. But ICI can be small as k, but then k must be greater than ICI. That seems conflicting. Wait, no, if ICI is equal to k, then 2*(k) 2^(101] = 2^[k), which is impossible. If ICI is less than k, then (^(ICI] is less than 2*(k), 2*(101] something smaller than 2*(k) equals 2*(k), which would mean 2*(101] is approximately 24k], so k 101 but since 2*[101] (*(IC)==2((k, then 2*(k] must be bigger than 2*(101]. So k>101.
33

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Published as a conference paper at ICLR 2026
Wait, maybe should approach this differently. Let's suppose that ICI as small as possible. Since the intersection A B C is subset of C, the minimal possible ICI is at least CI. Let's denote = IA B Cl. So ICI
A
We need to have 2^100 + 2010 2*(ICI) 2*(k). So 211101] 2 CI (*(k).
To minimize k, we want to make 2*(k) as small as possible. Since (*(ICI]
[[k]. Therefore 2*(101) (ICIC ([10]] +2*(k) Wait, but that 's not helpful. Wait, perhaps the minimal k is such that 2*(k] is just greater than or equal to 2*1101] + 2*(ICI]. But ICI is variable here. So perhaps we can choose ICI such that 2*(ICI) is equal to 2*(k) 2^(101), but since ICI must be least k, (*(k) -21(101) must be a power of two, and ICI = 12g2(2((k) -24(101)).
But 2 -21[101) must equal 2*[ICI]. which is a power of two. Let x =241101), so 2*(k] -x [m]. where mF ICI and m k Wait, but m k implies that 2*mm] 2 A[k] so 2*[k] (=2 2 A(k], which would imply that -x positive, which is impossible. So that can happen. Therefore the previous conclusion that there are no solutions may be correct, but that conflicts with the problem statement. So where is the mistake?
but x is
Alternatively, maybe my assumption that ICI must be at least k is wrong- Wait, but the intersection C, so its size can exceed ICI. So IA Therefore, k ICI.
B C is subset of
B
CI ICI.
Therefore 2*(k] (11101] 2 (ICI). and
ICI. Let me rephrase this as 2*[k] (ICI 2*1101). But since k I, then 2*[k]
IC But the right-hand side is 2*(101), which is positive. Therefore, this equation can't be satisfied. Wait, that suggests that there is solution. But the problem says "if n(A) nB) + n(C) n(A B C)", so it assuming that such sets exist. Therefore, there must be mistake in my reasoning.
([CI), so [*] -2*(ICI) is
Wait, maybe made an error in translating the problem. Let me check again.
The problem says: For sets A, B. and C, where n(A), nB). and nC) denote the number of subsets of A, B. and C respectively. and IAI IBI = 100, if n(A) + nBB n(C) = n(A B C), CI2
what is the minimum possible value of IA
Wait, maybe "n(A)" is not 2*(IAI) but the number of subsets which is indeed 2 (*[IA]]. So the equation is 2*(100) 2^(100) + CII.
21CI = 2*AIA
So (101) + 2*(ICI) = 24(k), where k = IA B must be equal to 2^(101) (*(m), where ICI, and k m
CI. So (*(k)
But as per earlier, this leads 2*(k] 211101] + 2 (m)m) with k m
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But (11101) 2 (m) must be power of two. Let think about when the sum of two powers of two is a power of two.
Suppose we have 2 + 2 = 2 with a
Then this is b, because otherwise 2 2nb = a
possible only when
+ 2 -at), which is not a power of two unless + 2 -a is power of two. The only time is when d=0, which gives I +1=2. So 2 + 2 2*(a +11. So in this case, if =b, then the sum is 2 +1).
2*[d] is power of two
Therefore, the equation 2 2nb =2^c can only be solved if a = b and
+1. Therefore, in our problem, (* (*(m) =2 k implies that 101 =m and k=102. But m=101 and k=102, but since k must be
(since k is the size of the intersection which is a subset of C, so k ICl=m), this would require 102 101 which is impossible. Therefore, there is solution unless we have a different approach.
Wait. but this contradicts the problem 's premise, which states that such sets exist. Therefore, there must be different interpretation.
Wait, perhaps "n(A B C)" is not the number of subsets of the intersection but the number of subsets common to all three sets A, B, and C? Wait, that would be different. Wait, but the problem says "n(A S is the number of elements but in
B
C)"
normally, the
notation n(S) for the problem statement, it specified that n(A), (BB) nCC the number of subsets. Wait, the problem says:
For sets A. B. and C, where n(A), nBB) and nCC) denote the number of subsets of A, B. and C respectively, and IAI = IBI = 100, if (A) + (B) (CC = n(A minimum possible value of IA
B C), what is the B CI?"
Wait, perhaps the notation is confusing. Maybe n(A B C) is not the number of subsets of the intersection but the number of subsets common A, B. and C. Wait, that is subsets that are subsets of A, B, and C. Wait, but a subset of A is not necessarily subset of B or C. So maybe "n(A
B here being used to mean the number of subsets that are common all three, i.e., subsets that are subsets of A, B, B
C)"
and C. Which would mean subsets of the intersection A is a subset of all three. So indeed, the number of subsets of the intersection is equal the number of subsets common to all three. Therefore the original interpretation is correct. So n(A B C) is 2^
C. Because subset of A
B CI).
Therefore
the equation is 2*1100] + 2^[100] 2*(ICI] 2^1AA B CI). So the problem is to find the minimal k IA CI such that 2*[101] + 2^(ICI) = 2*kk) with ICI k
B
But as we saw earlier the equation 2*(k] = 2*(101] + 2 with However, this equation only holds if the the left can be combined into a single power of two. As established before. the sum of two distinct powers of two is power of two only if they are equal (so exponents differ by
terms
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zero) but in that case. it becomes twice the power, which is the next exponent. So for example, 2 +2>a=2M(a+1).
In our case, 2 + [m] =2] which would require that equal, which would mean m=101, leading to 2*[101] +24[101==2|[12]. so k=102. But in that case, m=101 requirement. Therefore.
2*(101) and (mmm
and k=102, which violates the
no solution exists in that case.
A Alternatively. if we consider that 2m(m) can be combined with 2*(101) even if m >101. Let's try m=102: 2*(101] 2^(1) (1 +2)=3+2A(101}, which is not power of two. Similarly, m=103: 2*[101] +24[103)=2<(01(() not power of two. It seems like for m>101. the sum is (1(101] + 2m(m-101), which is 2*[101] times an odd number greater than 1, not power of two. Thus, impossible.
+4)=5+2<(101),
Alternatively. if <101, then 2*(101) +2*(m) =2^(m) +2^(101 1). To be a power of two, 1 +2M[101 -m) must be power of two Let t=101 -m which is positive since m <101. So +2*[t }= As before. this is only possible when t=0, which would make m=101, but we assumed m<101. Therefore, no solutions here either.
Therefore, this suggests that there is no solution where the equation holds, which contradicts the problem statement. Therefore, there must be an error in my reasoning.
Wait, but the problem is from competition there is a trick here. Let's ss think differently. Maybe the problem is not in the integers. Wait, but cardinalities are integers. Alternatively, perhaps the equation isn meant to be exact? No, the problem says ((A) n(B) + nCC n(A C), so it 's an exact equation.
similar, maybe
Alternatively, maybe the problem is using "number of subsets" in different way. Wait, but no, the number of subsets of a set with elements is 2*n. So that part is standard.
A Alternatively. maybe the problem is considering that A, B, C are subsets of some universal set but the problem doesn 't specify that. But even if they were, the number of subsets of each set would still be 2*(IA]], etc. So don't think that the issue.
Alternatively, maybe "A the sets A, B, C, but some other operation? No, standard notation.
C" is not the intersection of
Wait, maybe there 's a misinterpretatton of n(A B C). Maybe it's 's the number of elements in the intersection, but the problem says (n(A), (BB) nC) denote the number subsets", n(A B C) would also denote the number of subsets of B CI).
A
B C. So 2^(IA
Wait, unless the problem has typo and instead of n(A B C ). it's IA B Cl. But in that case the equation would be 2A100 + 20100 2*(ICI] = IA B CI, which would be different problem, but unlikely.
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Alternatively, perhaps the problem uses n(S) to denote the number of elements in S, but the first sentence says "n(A), nB) and ((C) denote the number of subsets of A, B, and respectively So no, n(A) is definitely (*(IA etc.
Hmm.
This is perplexing. Let's check again the equation. 2^100 + 20100 (kk) so 2*[101] (21( We need to find the minimal where this holds, with k ICI. So k is the size of A B C.
If we take ICl=k, then 21[101) +2*(k)=2M(), which is impossible. Therefore, ICI must be greater than k.
But then, 2*(k] =21[101] +2*(ICI), which implies that 2*(k] is k>ICI, but that contradicts ICI k
larger than (*(IC)), Therefore, no solution. But the problem says "if n(A) + n(B) + (C) n(A B C)", so it assuming such scenario exists. Therefore, there must be mistake in my reasoning.
Wait, perhaps the problem allows C to be multiset? But no, the problem states "sets". Or maybe the intersection is not a set, but different structure? Unlikely.
Wait, let's 's try specific numbers. Suppose k=101. Then 2= (2 - no, that would require 0=2^ which is impossible. If k=102: [1102)=2^(001] +2*/m) => 2^[102) (2m => 22[101](2 (1==2=100)==²(m) so m=101. But then k=102, which is greater than m=101, which violates k
Si Similarly if k=103: 211103)=2A(1011+2MM => 2M(m==2 -21(101==~<101)4 -1)=3+2^ which is not power of two.
k=104: 2^{104)=2<(001 +2 =>22(m)=2~1044) -2(101)=2~
-10=7=2 not power of two.
Continuing k=105: 24(105) (20 = (10 power of two.
This pattern continues. and the difference (*(k) -21(101) is divisible by 2*(101) but results in an odd number greater than 1. which is not a power of two. Therefore, no solutions exist for k>101.
But this is impossible because the problem must have solution. Therefore, maybe the problem is designed to have the minimal k where 2*(k] is the next power of two after 2^[101] +m) but this is not exact. But the problem states equality, not inequality. So I'm stuck.
Alternatively. perhaps the problem is using a different definition of subsets. For example, maybe only non-empty subsets? No the number of subsets including empty set is 2*n.
Alternatively. maybe the problem has a typo and it should be multiplication instead of addition. If it 's n(A) nBB) nC) =n(A B C), then it it's different. But the problem says "+"
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Wait. the problem is in Chinese maybe? Wait, no, the user wrote the problem in English. Hnm.
A Alternatively, maybe the problem is from souree where ((A) denotes the number of elements, which would usually be IAI but the problem says (A) is the number of subsets. So unless the problem mixed notation. If the problem had said IAI, IBI. ICI are 100, and (A) +B(B) +n(C) =n(A B C), with nXX) being the number of elements, then it's different problem. Let's check that:
IAI=|BI=100, and nX) is the number of elements then n (A) ) CI. Then the equation is 200 +ICI=k, where k=lA But since IA B CI IAI=100, -100, which is impossible. So that can I be.
+n
and n(A B C)=lA
B
B
CI.
200+ICI
100
IC
Therefore, the original interpretation seems correct. But then, according to that there 's no solution. But the problem is asking for the minimal possible value, so perhaps the answer is 101? But wait, how?
Wait, let think differently. Maybe the problem is in a universe where all sets are subsets of common universal set, and operations are considered within that. Suppose that A and B are subsets of some universal set, and C is also subset. Then, the intersection A B universal set. However, the number of subsets of A is still [IAI], regardless of the universal set.
would be a subset of the
A Alternatively. maybe using some principle of ceclusioneeclluse for the number of subsets? Hum, not sure
Wait, another thought: Maybe the equation n(A) + (B) + (CC =n B C) is in terms of numbers. So n(A) is 2^(100), nB) is 2*(100), (C) is (*(ICI), and n(A B C) is 2*(k). So have 2A[100] 2^[100] + 2^[ICI] =2((k).
L Lets write this 2^(101) 2^(ICI)=2^(A Let factor out the smaller power of two. Suppose ICI 101 Then we can factor out 2A(ICI]:
2 (1CI(( 2^/101 ICI))=2 Therefore, +2^1101 ICI) must be power of two. L Lets set t=101 - ICI. So 0 and 1 +2*(t] [ for some s. As before, the only solution is t=0 or t=1?
t=0: 1+1=2=2^1, s=1. Then (=0 implies 101 ICI=0 => ICI=101. So +2^0=2, so 2=(k)=2((101) 101])=222(102)) Therefore, k=102. But ICI=101, k=102>101 which is impossible.
If t=l: 1+2=3, which is not power of Similarly, 1=2:144=5, not a power of two. So no solution.
If ICI >11,, then 2111011 +2*(ICI)=2*1K implies we factor out 2A[101]: 2^(101](1 +2A[ICI -101))=2(k), so +2*IICI -101)=2A[k -101]. Lets set m=ICI -101>0, so (2
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-101]. The only solution is m=0, which gives 1 +1=2^1. But >0, so no solution.
Therefore, this suggests no solution exists, which contradicts the problem s wording. Therefore, I must have made wrong assumption.
Wait, perhaps the problem allows for a universe where elements are counted with multiplicity? Like, multisets? But the problem specifies "sets", so elements are distinct.
A Alternativel.. , perhaps the problem is trick question where the minimal possible value is 101, but the equation not exactly satisfied But the problem says "if nA) + (B) nCC) =n(A C)", so it given that this equation holds. So the answer must be derived under that condition.
Wait, maybe there 's a mistake in my calculation Let try plugging in k=101. Then 24[101)=2\1011 +2*[ICI, which implies ICI is negative infinity impossible.
= (0=2A(102-22(110))==2 (2(100)==210]1111 so ICI=101. But then k=102 > IC |=101. So this is impossible.
2*(IC
Wait, but maybe the problem doesn 't require that the intersection is a subset of C? But that basic set theory. B is subset of C by definition.
A Alternatively, maybe the problem uses different definition of intersection? No, standard intersection.
Alternatively. perhaps A, B, C are not sets of elements but of subsets. Wait, the problem says "sets A, B, C", so they are standard sets
Alternatively. maybe the problem allows C to be equal to A C. C is equal to equation becomes 2^1100} +2^[100} +2*(k)=2*(k), which implies 211101] +2*(k)=2*k)), leading to 2A(101]=0, which is impossible.
B C, then ICl=k. Then the
This is really confusing. Since the problem is from math competition. maybe the answer is 102, even though it's greater than ICI, but the problem doesn require ICI be finite or something? But A and B have size 100, which is finite. and the intersection can 't be larger than A or So the intersection has to be at most 100.
Wait, wait! The problem says IAI=IBI=100. So the intersection A C is a subset of A and B, so IA tried k=101 and more, which is impossible
CI 100
But earlier. since the intersection can exceed 100.
Oh! This is the key mistake made earlier. Since A and have size 100, their intersection with C can't exceed 100. So k 100 Therefore. when I considered k=101 or higher, that 's impossible. Therefore. the maximum possible k is 100. But then the equation 2*[101] {2*(m)=2*[k with k 100 But 2*(k) is
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at most 2 (1000)), but the left hand side is at least (1(101), which is larger than (*1100). Therefore, it 's impossible.
But this contradicts the problem 's stipulation that the equation holds. Therefore the only possible conclusion is that there is a mistake in the problem statement, or perhaps the answer is 101, but the problem has an error. But since the problem asks for the minimal possible value of IA B CI, and given the constraints, it impossible, but maybe I'm missing something.
Wait, here 's different approach. Let's denote x= IA We need to have |000 +2^1100) +2*(ICI) = Therefore 211101) (2x
B
C
But is the size of the intersection which is 100 Therefore xxxx 2 *[100,, but the left-hand side is 2^(101) + something, which is least (^01).). Therefore. [^[x) 2 ^[101}, which requires 101 But 100 so there 's no solution. Hence, it impossible.
But the problem is presented as a valid question, so likely have a mistake. Wait, maybe miscounted the number of subsets. Wait, A has IAI=100, then (100= Similarly for B and C. So the equation is correct.
Alternatively, the problem might be in the translation. Maybe "the number of subsets" refers to proper subsets? In that case the number would be 21 So if (A) is the number of proper subsets of A, then ((A)=2>(100)- Similarly for B and C. Then the equation becomes (20000))-D4404410000)4444444
|| (x--)=
L e s compute this: 2^[101] -2 +2*[ICI] -1=2^(x)-1 (ICI) (3=2Mxx)-1 *[10]] +2*[ICI) (2=2
21(101)
But this seems more complicated, and there 's no indication in the problem statement that subsets proper. So this is probably not the case.
A Alternatively. maybe the problem uses a different definition of number of subsets. but it highly unlikely.
Therefore, given that the problem states such an equation is possible, but according to standard set theory and mathematics there is no solution. But since the problem asks for an answer, must have missed something.
Wait, maybe the problem is not about sets but about something else Maybe vector spaces? No, the problem talks about sets A, B. C.
Wait, another idea: maybe the sets A, B. and C are such that A and B are subsets of C. If A and B are subsets of C, then A C=A
B. So IA have size 100, but we don't know their intersection However. BI]. But
B Cl= IA Bl. But A and B
the equation becomes 2^100 +2^100 +24(ICI)=2A11A
since A and B are subsets of C. ICI 100 But then 2*(ICI)
2 Af100). so the leat side 2 100 +2^100
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+ [1000==3++2100 which is greater than (* BI can be at most 100. so 2* BI] {110), which is less than (x2M(100). Therefore,
B1) but IA
solution.
Hmm. I'm stuck. Given the problem's constraints: there is possible solution, but the problem is asking for one. Perhaps contradiction
the answer is 101, acKnowledging that there but requiring the minimal k where 2*(k) is the next power after 2^101 +2*[ICI). But even then, the minimal would be 102, but that exceeds the maximum possible size of the intersection.
Wait, but if we ignore the constraint that just find the minimal k such that [*[k] 2 ^101 (2 for some m. But then k=102 when m=101, but that violates the other constraints.
ICI and 100
Alternatively. maybe the problem has typo and it should be n(A) +B( +C(C) =n B C). Then the problem would be different. Let me check: n(A subsets of the union. If that 's the case, then the equation is 2*[100] +2^[100} +2>(ICI)=2^(M,
C) is the number of
CI). which might be possible. But the problem says intersection not union.
Alternatively. , maybe the problem is in another dimension, but that 's not likely.
Given that I'm stuck and the problem is likely expecting an answer of 101, even though it violates the cardinality constraints, or perhaps 101 is the minimal k where 2*(k) is greater than the sum. But in math competitions, sometimes you' supposed to find k such that the sum is less than or equal to 2*k, but the problem states equality.
A Alternatively. maybe the problem allows overlapping subsets in the count. But no, subsets are unique.
Wait, here 's last-ditch idea: perhaps A. B. and C are not sets of elements but of sets themselves. and the intersection is in some lattice of sets But the problem statement says "sets A, B, C", so they re standard sets.
Given that I've exhausted all avenues and still can't find a solution that fits the problem 's constraints but the problem must have an answer, I'm forced to conclude that the answer is 101, even though it violates the subset relation.
A Alternatively maybe the problem is answer is 101, with k=101, ICI=101, but then not true. Therefore, I'm confused.
which is
Wait, 2*0101) (21(101)=2~
if ICI=101, then the equation would be 21010 +21(101)=22(10)) so k=102. But k=102 is impossible because the intersection cannot exceed 100. Therefore. there 's no solution but the problem states "if n(A nBB) n(C) =A
B C)", which implies that there is a solution. This is a paradox.
Wait. unless the sets are allowed to be infinite. For example, if C is an infinite set, then 2*/ICI) is infinite. and 200101]
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(2A(ICI) would be equal to 2 [ICI). assuming ICI is infinite. But then the equation 20110 +2*fiCI)=2*(k] implies that if k must be CI is infinite then 2*[k] is also infinite
infinite. But the problem asks for the minimal possible value, which would be finite. Therefore, this is not the case
Given all this think there 's either mistake in the problem or a trick that I'm missing. Since the problem is about sets and their subsets, and the equation seems impossible under standard interpretations. that we ignore the cardinality constraint of the intersection. But since the intersection can 't exceed 100, this is impossible. Therefore, the minimal possible value is 101, but I'm not sure.
maybe the answer is 101, assuming
Wait, but if IA B CI could be 101, then even though A and
B have size 100, their intersection with C can 't exceed 100. So maybe the problem is designed have an answer of 101, assuming that the intersection can be larger than A or B? But that's impossible.
think must have missed something. Maybe instead of interpreting (A) as the number of subsets, it's the number of
elements? If so, then (AA)=100, nB)==00,, =IC) and n
B
Cl. Then the equation is 100 +100 B CI. But since IA B CI 100 this ICI
C)=1A
B
ICI=IA
implies
-100, which is impossible. So no.
A Alternatively, if (AA is the number of elements, ; and the problem meant to say IAI, but that not what was written.
Given the problem is phrasing. think the answer must be \boxed (101}, even though the logical steps suggest it impossible. Perhaps there s a different way to combine the subsets. Alternatively. maybe considering that the intersection is being counted as elements, not subsets. But that would not align with the initial definitions.
Since in all cases the only possible value that comes close is 102, but 't's invalid due to size constraints. and the problem asks for the minimal possible value, perhaps the answer is 101, even if it's an error. Alternatively, the answer is 102, and the problem allows intersections larger than the sets. But this is not standard.
strictly possible and the problem has
Wait, maybe A and B are allowed to have sizes larger than 100? Wait, no, the problem says IAI=IB|=100.
Given that can't find a valid solution and the problem likely expects answer, I'll go with 101, assuming that maybe the problem had a typo and the equation is supposed to be multiplication, but that speculative. Alternatively, 101 is the answer
Final *nswer**
boxed [101} </think>
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Given sets M(A)). \BV), and ((C)) with ((IA IBI = 1001), , we need to find the minimum possible value of \(IA capp B \eap C IV) given that the equation ((n(A) nB) + n(C) = n(A \eap capp C))) holds. Here, V(n(S)) denotes the number of subsets of a set I(S)), which is ((24(181)V.
1. The number of subsets of MAV) \BB)), and \(C)) are ((201100))) \(2^11001V_ and ((2*(ICI)) respectively,
2. The equation given is ((2^(100) + 2*[100] 2*(ICI) (1 A cap Aeap CIJI).
3. Simplifying the -hand side, we get ((22(101) + 21(ICI) 2* capp B \eap CI))).
4. Let \(k IA >eap B cap CIV). We need to find the smallest \(k
1) such that ((2A(101) + (*(ICI] = 2*k)).
5. For the equation ((2A(101)
2*(ICI) 2*k1) to hold, ((2A(101)
+ 2*(ICI]1) must be a power of two. powers of
6. The only way the sum of two is if they are consecutive powers of two. However, ((2A[101) + 2A[10] 1011) and \(k = 1021), but ((k)) cannot exceed ((ICI))
is another power of
24(102)1), which would require \(ICI
Considering the constraints and the problem 's requirement for a solution, the minimal possible value of \(IA \eap B \eap CI\) that fits the equation, despite the constraints is found to be \(101).
Thus, the minimum possible value of \(I Neap B \eap CI\) is A(1 boxed (101)\).
43


Reviews:
[{"review_id": "l0DOLq6t9k", "reviewer": "Reviewer_qfCa", "summary": "This paper challenges a widely-held assumption in LLM post-training: that high supervised fine-tuning (SFT) scores reliably predict better performance after reinforcement learning (RL). Through experiments on moderate models trained with hundreds of variations and evaluated on 7 math benchmarks, the authors demonstrate numerous counterexamples where this assumption fails. High SFT scores can be biased toward simpler or more homogeneous data and do not reliably predict subsequent RL gains or scaled-up post-training effectiveness. In some cases, RL training on models with improved SFT performance actually produces substantially worse outcomes compared to RL on base models without SFT. To address this predictability gap, the paper identifies two more reliable metrics for predicting post-RL success: generalization loss on held-out reasoning examples and Pass@large k performance.", "questions": "(1) Can the authors provide mechanistic explanations for why overfitting during SFT constrains RL exploration? Is this about reduced diversity in model outputs, behavioral rigidity, or something else? (2) How sensitive are the proposed metrics to hyperparameter choices in SFT (learning rate, batch size, optimization algorithm) beyond epochs and data volume? (3) Since generalization loss requires a validation set from the same SFT distribution while Pass@large k works across distributions, why not simply use Pass@large k for both scenarios, and what are the trade-offs? (4) Can these findings transfer to other domains (code, science reasoning, non-verifiable tasks) and other RL methods (offline RL, preference-based methods), or are they specific to online GRPO with math tasks?", "limitations": "", "rating": 4, "confidence": 4, "soundness": 2, "presentation": 3, "contribution": 2, "strengths": "This paper makes a practically important contribution to LLM post-training by systematically demonstrating that SFT performance is a poor predictor of final RL outcomes. The proposed metrics are intuitive and practical. The clear categorization into dataset-level and instance-level scenarios makes findings applicable to different use cases, and the promised open-source tool adds concrete value.", "weaknesses": "The study is narrowly scoped to mathematical reasoning tasks using GRPO-based RL with verifiable rewards, limiting generalizability to other reasoning domains (coding, science, logic) or alternative RL paradigms (offline RL, DPO, other algorithms). The paper identifies failure modes but provides limited mechanistic insight into why SFT characteristics diverge from RL outcomes. For instance-level prediction, generalization loss is explicitly not applicable due to distributional gaps, reducing its utility for practical data selection scenarios. The proposed linear predictor requires calibration data from actual RL runs, which partially defeats the purpose of avoiding expensive RL experiments. Additionally, the Qwen3 results are presented only qualitatively because models showed minimal responsiveness to SFT variations, suggesting the findings may not universally hold across all architectures.", "quality": null, "clarity": null, "significance": null, "originality": null, "strengths_and_weaknesses": "", "was_revised": false, "final_justification": "", "rebuttal": "", "claims_and_evidence": "", "methods_and_evaluation_criteria": "", "theoretical_claims": "", "experimental_designs_or_analyses": "", "relation_to_broader_scientific_literature": "", "essential_references_not_discussed": "", "other_strengths_and_weaknesses": "", "other_comments_or_suggestions": "", "questions_for_authors": "", "summary_of_paper": "", "summary_of_recommendation": "", "technical_quality": "", "clarity_of_presentation": "", "potential_impact": "", "robotics_focus": "", "quality_of_limitations": "", "questions_for_rebuttal": "", "reasons_to_accept": "", "reasons_to_reject": "", "questions_to_authors": "", "extra_scores": {}}, {"review_id": "ffBOMSpl9r", "reviewer": "Reviewer_AYRG", "summary": "The paper investigates why high Supervised Fine-Tuning (SFT) scores often fail to translate into good Reinforcement-Learning (RL) post-training results.\nThrough large-scale experiments across Llama3-8B, Mistral-NeMo-12B and Qwen3-4B on mathematical reasoning tasks, the authors show that excessive SFT can overfit the model’s representation space, reducing its capacity to benefit from RL.\nTwo new proxy indicators are proposed:\nGeneralization Loss – captures when validation loss flattens while diversity collapses;\nPass@large k (k = 64) – estimates solution-space width.", "questions": "1. Could you report one concrete example where checkpoints chosen by Pass@64 or Generalization Loss actually outperform standard validation-loss selection after full RL?\n2. How sensitive are results to the choice of k (32 vs 64)? Would smaller k approximate the same signal?", "limitations": "", "rating": 6, "confidence": 3, "soundness": 3, "presentation": 3, "contribution": 2, "strengths": "1. Demonstrates convincingly that high SFT scores can mislead RL training, challenging a community-wide assumption.\n2. Cross-model, multi-dataset evaluation with some interesting statistics (R² and Spearman correlation) makes the results good", "weaknesses": "1. The paper never demonstrates that using Generalization Loss or Pass@64 for early-stopping or checkpoint selection actually improves final RL accuracy.\nAll reported metrics are correlations (R² / spearman's rank correlation); no absolute post-RL accuracy table is shown.\n2. From Fig. 4–5, the best RL model surpasses the best SFT by only ≈ 1~2%, and the actual other early stop strategy might not be able to reach the best RL model, so the real-world benefit appears even smaller.", "quality": null, "clarity": null, "significance": null, "originality": null, "strengths_and_weaknesses": "", "was_revised": false, "final_justification": "", "rebuttal": "", "claims_and_evidence": "", "methods_and_evaluation_criteria": "", "theoretical_claims": "", "experimental_designs_or_analyses": "", "relation_to_broader_scientific_literature": "", "essential_references_not_discussed": "", "other_strengths_and_weaknesses": "", "other_comments_or_suggestions": "", "questions_for_authors": "", "summary_of_paper": "", "summary_of_recommendation": "", "technical_quality": "", "clarity_of_presentation": "", "potential_impact": "", "robotics_focus": "", "quality_of_limitations": "", "questions_for_rebuttal": "", "reasons_to_accept": "", "reasons_to_reject": "", "questions_to_authors": "", "extra_scores": {}}, {"review_id": "MMXyVdocPt", "reviewer": "Reviewer_pM1A", "summary": "This paper challenges the existing belief that the pass@1 accuracy of an SFT model is a good metric for determining the model's performance after RL. The authors perform experiments using various LLM architectures on math-focused datasets that show that when modulating training parameters (e.g. epochs) or data parameters (e.g. # unique training data points), the SFT model with the highest pass@1 accuracy is often not the best performing model after RL. Instead, the authors propose validation loss and pass@k accuracy as good metrics for performing this prediction, and perform further experiments to show that this is true.", "questions": "0. See Weaknesses\n1. What kind of hyperparameter search was done for each set of experiments?\n2. How do the predictions (based on gen loss, pass@large k) hold up when constrained to a single task instead of the average of 7 tasks?", "limitations": "", "rating": 6, "confidence": 3, "soundness": 3, "presentation": 2, "contribution": 3, "strengths": "1. This work suggests metrics which better align with post-RL performance compared to task accuracy\n2. Extensive GPU resources are used to perform experiments that support the claim\n3. The described methods are easy to apply to existing SFT trained models for ranking and selection", "weaknesses": "1. The SFT-then-RL pipeline was popularized for LLM usage in [1], and used previously in NLP as cited in their paper. Citations should be fixed.\n2. (Section 3, line 214), (Section 3.1, line 222), (Section A.1, line 711), (Section B.2, line 900) conflict on which parameters (especially learning rate) were modified during RL training.\n3. (Section 3.1, line 250) and (Figure 3) lacks intuition on why post-SFT task performance is expected to have a linear relationship with post-RL performance.\n4. (Figure 1), (Figure 4), etc. show that best SFT's post-RL task performance usually only differ from best RL by a few points. Error bars or other standard deviation reporting would be enlightening.\n5. Grammatical error in (Section 6, line 483), \"...algorithms may worth further explorations...\"\n6. Considering pass@1, pass@k, generalization loss, and Spearman's rank correlation are central to the paper's claims, a brief exposition on their definitions would be good to include\n7. (Section 5.2, line 394) How many models is 50%?\n\n\n**References**\n\n[1] Ouyang, Long, et al. ‘Training Language Models to Follow Instructions with Human Feedback’. arXiv:2203.02155, arXiv, 4 Mar. 2022. arXiv.org, https://doi.org/10.48550/arXiv.2203.02155.", "quality": null, "clarity": null, "significance": null, "originality": null, "strengths_and_weaknesses": "", "was_revised": false, "final_justification": "", "rebuttal": "", "claims_and_evidence": "", "methods_and_evaluation_criteria": "", "theoretical_claims": "", "experimental_designs_or_analyses": "", "relation_to_broader_scientific_literature": "", "essential_references_not_discussed": "", "other_strengths_and_weaknesses": "", "other_comments_or_suggestions": "", "questions_for_authors": "", "summary_of_paper": "", "summary_of_recommendation": "", "technical_quality": "", "clarity_of_presentation": "", "potential_impact": "", "robotics_focus": "", "quality_of_limitations": "", "questions_for_rebuttal": "", "reasons_to_accept": "", "reasons_to_reject": "", "questions_to_authors": "", "extra_scores": {}}, {"review_id": "VPVX7iJA51", "reviewer": "Reviewer_xARC", "summary": "The paper challenges the previous conventional practice where the model that achieves the best performance after SFT is assumed to be the best candidate for subsequent RL training. Specifically, the paper shows that the correlation between SFT scores is not predictive for RL scores by training hundreds of models under various settings. To solve this problem, the paper proposes alternative metrics including generalization loss on held-out sets and Pass@larger k and show that they demonstrate much higher correlation.", "questions": "It seems that the SFT / RL training datasets are different across models. How are these choices made?", "limitations": "", "rating": 6, "confidence": 3, "soundness": 3, "presentation": 3, "contribution": 3, "strengths": "- The paper targets at a very practical assumption that the better performance after SFT always leads to better performance after RL and the consequent difficulty to find the cause for a bad end performance especially when two different teams are in charge of these two training stages.\n- This work spends a large amount of computational resources to make and verify their claim.\n- The experiments include evaluations on various math benchmarks, and the correlation gains are significant.", "weaknesses": "- The paper only evaluates on benchmarks in the math domain, and whether the method can be generalized to other domains like coding and science remains unclear.\n- For dataset-level scenarios, most of the experiments focus on altering the training epochs and the number of unique examples, which seems to be a less practical tradeoff to consider. Training more epochs could lead to potential overfitting is a well-known concern, so even in the scenarios where the number of high-quality examples are limited, training for many epochs might not be a conventional option.\n- For instance-level scenarios, the strategy seems to be selecting the random / shortest / longest examples, varying the number of total examples. This seems to deviate from the actual practice in SFT data selection, and the paper needs some analysis to show how close these simple strategies are to the current data selection methods.", "quality": null, "clarity": null, "significance": null, "originality": null, "strengths_and_weaknesses": "", "was_revised": false, "final_justification": "", "rebuttal": "", "claims_and_evidence": "", "methods_and_evaluation_criteria": "", "theoretical_claims": "", "experimental_designs_or_analyses": "", "relation_to_broader_scientific_literature": "", "essential_references_not_discussed": "", "other_strengths_and_weaknesses": "", "other_comments_or_suggestions": "", "questions_for_authors": "", "summary_of_paper": "", "summary_of_recommendation": "", "technical_quality": "", "clarity_of_presentation": "", "potential_impact": "", "robotics_focus": "", "quality_of_limitations": "", "questions_for_rebuttal": "", "reasons_to_accept": "", "reasons_to_reject": "", "questions_to_authors": "", "extra_scores": {}}, {"review_id": "ZuRpDkIzyj", "reviewer": "Reviewer_Q2Kx", "summary": "This paper demonstrates that the performance of a model after SFT is often not predictive of its performance after RLVR. They consider **dataset-level ablations**, which vary the number of training examples and epochs used in training, and **instance-level ablations**, which ablate the distributional makeup of the data. They show that in both settings, the SFT model's performance does not correlate with the final RL model's performance. For example, training for many epochs on the same data leads to better SFT performance but worse outcomes after RL; the same is true for training on short examples. \n\nThe authors then propose two new metrics for SFT models, namely validation loss (which measures generalization) and pass@k performance (for large k), and demonstrate that it is substantially more indicative of final performance. In the dataset-level ablations, they find that both validation loss and pass@k are strong predictors, achieving $\\geq$ 0.90 Spearman correlation (30% improvement over SFT pass@1 performance). For instance-level ablations, pass@k is similarly a much stronger predictor (validation loss is deemed \"not applicable\").", "questions": "1. It is a bit odd that validation loss was deemed not applicable for instance-level predictions (§5.3). It is still possible to evaluate its performance, right? It does not take away from the story at all to say that pass@k is a strong predictor in both dataset-level and instance-level settings, but validation loss works well only in the former.\n2. Would it be accurate to refine the claim \"the model with the strongest post-SFT performance is not necessarily the one with the strongest post-RL performance\" to add \"*when evaluated on the same data*\"? That makes the claim stronger + more interesting.", "limitations": "", "rating": 6, "confidence": 3, "soundness": 3, "presentation": 2, "contribution": 2, "strengths": "1. This paper studies an important broader question: what is the role of SFT prior to subsequent RL training? While SFT followed by RL is still the de facto post-training pipeline, there remains substantial confusion about what metrics the SFT team should optimize for in order to produce a good starting point for RL.\n2. The paper provides strong empirical evidence (and compelling intuition) for their findings throughout the paper.\n3. The paper provides two new metrics for evaluating SFT models, validation loss and pass@k, which are more effective indicators of final model performance than the traditional pass@1 performance of SFT models.", "weaknesses": "1. The findings are not entirely novel or surprising. For instance, [Weight Ensembling Improves Reasoning in Language Models](https://arxiv.org/abs/2504.10478) also show that during SFT, the Pass@1 rate improves while Pass@k deteriorates, and that Pass@k of an SFT model is a much stronger indicator of its performance after RL. Pass@k is also the main metric proposed by the authors of this paper for predicting final model performance. Citation & comparison with this work is definitely required. More broadly, it is well-known that SFT and RL optimize for different (and sometimes conflicting) objectives, and that strong SFT models ≠ strong starting points for RL.\n2. Many aspects of the writing and presentation can be cleaned up. See below:\n\n- The main finding from Figure 1 is difficult to parse. Within each box (green/orange/red), it looks like higher green bar (post-SFT performance) *does* lead to higher black bar (post-RL performance). However, I think the point you're trying to highlight is that *across the boxes,* this is not the case --- training on random SFT examples gets the highest RL performance, but training on the shortest SFT examples gets the highest post-SFT performance. Would it be better to just include the bars that actually matter, instead of all of these? \n- In §3.1, the experimental setup says that you \"vary the training configuration, such as the number of unique samples/training epochs/learning rate,\" but not what was ACTUALLY varied.\n- Consistently throughout the paper, *all* the x-axis bar labels start with the name of the dataset, e.g., \"Nemotron 25k2ep\", \"Nemotron 25k3ep\",\"Nemotron 25k4ep\", etc. Authors should omit the redundant information and instead specify exactly the information that changes, e.g., \"2ep\" \"3ep\", \"4ep.\" Also, all the $x$-axis labels should be clearer, and not require the reader to infer that \"25k2ep\" means 2 epochs on a dataset of 25k examples.\n- More small notes on writing that did not factor into my evaluation but may improve readability.\n    - L14: \"In this work, we challenge whether high SFT scores translate to improved performance after RL.\" RL still improves performance, right? Instead, you're saying that the ranking of SFT models isn't predictive of the ranking of those models after RL.\n    - L15: \"We provide extensive counter-examples where this is not true.\" It is more natural to say \"counter-examples to [claim].\" Counter-examples, by definition, are instances where X is not true.\n    - RLVR citation switches from Deepseekv3 to Tülu3 in the middle of the paper. Tülu3 came first and proposed RLVR!\n    - L64: \"industrial practice\" $\\rightarrow$ \"industry practice\"\n    - L77: \"on the contrary\" is not the right transition to use here because the following evidence supports the same finding. Maybe \"on the other hand\" is better?\n    - L291: \"During the investigation above, we identified a counterintuitive pattern in which post-SFT performance improves stably when training for more epochs whereas the overtrained models show decreased potentials during the subsequent RL.\" $\\rightarrow$ \"In our previous experiments, we identified a counterintuitive pattern where training for more epochs consistently improves SFT performance but deteriorates potential for subsequent RL. \"", "quality": null, "clarity": null, "significance": null, "originality": null, "strengths_and_weaknesses": "", "was_revised": false, "final_justification": "", "rebuttal": "", "claims_and_evidence": "", "methods_and_evaluation_criteria": "", "theoretical_claims": "", "experimental_designs_or_analyses": "", "relation_to_broader_scientific_literature": "", "essential_references_not_discussed": "", "other_strengths_and_weaknesses": "", "other_comments_or_suggestions": "", "questions_for_authors": "", "summary_of_paper": "", "summary_of_recommendation": "", "technical_quality": "", "clarity_of_presentation": "", "potential_impact": "", "robotics_focus": "", "quality_of_limitations": "", "questions_for_rebuttal": "", "reasons_to_accept": "", "reasons_to_reject": "", "questions_to_authors": "", "extra_scores": {}}, {"review_id": "UWT9R2QCEv", "reviewer": "Reviewer_4f6U", "summary": "This paper challenges the point in LLM post-training that a higher Supervised Fine-Tuning (SFT) score guarantees better final performance after Reinforcement Learning (RL). The authors identify and provide extensive evidence for the *SFT metric trap*, where optimizing for SFT performance can be misleading and suboptimal for the final RL outcome. As a solution, the paper proposes and validates two more reliable predictive metrics (i.e., generalization loss and Pass@large k accuracy), which offer higher predictive power for RL success, thereby providing a more principled and cost-effective methodology for optimizing the SFT-RL pipeline.", "questions": "- How should practitioners weigh the trade-offs between generalization loss and Pass@large k when they provide conflicting signals for different SFT model candidates?\n- Given that RL is a dynamic process, is it possible that an SFT model deemed suboptimal by these static metrics could unlock its potential later in RL, perhaps with different RL hyperparameters like a higher exploration temperature?\n- How do the two analyzed factors (SFT data selection for instance-level and training paradigm for dataset-level) interact? For instance, would training on shortest examples for more epochs amplify the negative effects observed?", "limitations": "", "rating": 6, "confidence": 4, "soundness": 3, "presentation": 2, "contribution": 3, "strengths": "- The paper addresses a critical problem in real-world LLM development, making the findings impactful for practitioners.\n- The empirical evaluation is rigorous and extensive, conducted on multiple models and benchmarks, which lends credibility to the claims.\n- The proposed metrics are not only predictive but also actionable, offering a framework for practitioners to make better decisions at both the dataset and instance levels.", "weaknesses": "- My main concerns lie in the generalizability of the findings. The study is exclusively focused on mathematical reasoning tasks, where rewards are often token-level and verifiable. It is unclear if these predictors would hold for other domains like complex instruction following, or agentic tasks, where rewards are more holistic and harder to define. The strong performance of *generalization loss* and *Pass@large k* might be an artifact of the structured nature of math problems. For tasks that value creativity or conversational nuance over discrete correctness, a moderate degree of SFT overfitting might even be beneficial for stylistic alignment, potentially creating an entirely different dynamic with the subsequent RL stage. The paper lacks a discussion on this crucial limitation.\n- The paper establishes correlations but provides limited insight into the underlying causal mechanisms explaining *why* these metrics are better predictors.\n- The high computational cost of evaluating Pass@large k is a significant practical drawback that may limit its adoption, especially for larger models.\n- The anomalous result of the Qwen3 model is a missed opportunity for a deeper analysis that could have provided more profound insights into model-specific behaviors.", "quality": null, "clarity": null, "significance": null, "originality": null, "strengths_and_weaknesses": "", "was_revised": false, "final_justification": "", "rebuttal": "", "claims_and_evidence": "", "methods_and_evaluation_criteria": "", "theoretical_claims": "", "experimental_designs_or_analyses": "", "relation_to_broader_scientific_literature": "", "essential_references_not_discussed": "", "other_strengths_and_weaknesses": "", "other_comments_or_suggestions": "", "questions_for_authors": "", "summary_of_paper": "", "summary_of_recommendation": "", "technical_quality": "", "clarity_of_presentation": "", "potential_impact": "", "robotics_focus": "", "quality_of_limitations": "", "questions_for_rebuttal": "", "reasons_to_accept": "", "reasons_to_reject": "", "questions_to_authors": "", "extra_scores": {}}]

Author rebuttal:

Decision: Accept (Poster)