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</think> Doesn't Stop Reasoning: Analysis of Spurious CoT Termination

Seunghee Koh, Sungjae Choi, Minchan Kwon, Sunghyun Baek, Junmo Kim

Year: 2026Venue: arXiv preprintArea: cs.CLType: PreprintEmbeddings: 137

Intelligence

Status: succeeded | Model: Gemma-4-26B-A4B | Prompt: intel-v1 | Confidence: 94%

Last extracted: 9/4/2026, 3:40:10 AM

Summary

This paper analyzes 'spurious CoT termination' in Large Reasoning Models (LRMs) where injecting an end-of-think token (EoT) to trigger early exit fails to stop reasoning, causing the model to continue generating reasoning-like content in the answering phase. The authors hypothesize that insufficient attention to the injected EoT causes this behavior and propose Exit-token Attention Biasing (EAB) to increase attention to the EoT, which successfully reduces spurious termination and answering-phase length across multiple models and benchmarks.

Entities (14)

End-of-think token · token · 99%Spurious CoT Termination · phenomenon · 98%Exit-token Attention Biasing · method · 97%AIME · benchmark · 95%AMC · benchmark · 95%DEER · method · 95%DynaSoR · method · 95%GPQA · benchmark · 95%GSM8K · benchmark · 95%Large Reasoning Models · model-class · 95%MATH · benchmark · 95%DeepSeek-R1-Distill-Qwen · model · 90%Qwen3 · model · 90%QwQ-32B · model · 90%

Relation Signals (11)

DeepSeek-R1-Distill-Qwen → evaluatedon → AMC

confidence 95% · We evaluate on five widely used reasoning benchmarks. These are... AMC 2023 (AMC)... We study four LRMs... DeepSeek-R1-Distill-Qwen

DeepSeek-R1-Distill-Qwen → evaluatedon → AIME

confidence 95% · We evaluate on five widely used reasoning benchmarks. These are... AIME 2024 (AIME)... We study four LRMs... DeepSeek-R1-Distill-Qwen

DeepSeek-R1-Distill-Qwen → evaluatedon → GSM8K

confidence 95% · We evaluate on five widely used reasoning benchmarks. These are GSM8K... We study four LRMs... DeepSeek-R1-Distill-Qwen

DeepSeek-R1-Distill-Qwen → evaluatedon → MATH

confidence 95% · We evaluate on five widely used reasoning benchmarks. These are... MATH-500 (MATH)... We study four LRMs... DeepSeek-R1-Distill-Qwen

DeepSeek-R1-Distill-Qwen → evaluatedon → GPQA

confidence 95% · We evaluate on five widely used reasoning benchmarks. These are... GPQA-Diamond (GPQA)... We study four LRMs... DeepSeek-R1-Distill-Qwen

Exit-token Attention Biasing → reduces → Spurious CoT Termination

confidence 95% · increasing attention to the injected EoT reduces spurious CoT termination and answering-phase length.

Exit-token Attention Biasing → targets → End-of-think token

confidence 95% · we bias attention to the injected EoT during answer generation

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Abstract

Abstract:Chain-of-thought (CoT) reasoning improves large reasoning models (LRMs) on complex tasks but often produces long, redundant traces. Recent training-free early-exit methods shorten these traces by choosing an intermediate point to stop reasoning. We study one such strategy that injects an end-of-think token (EoT, </think>) at this point to trigger the reasoning-to-answering transition, and find that the injected EoT does not always induce a clean answering phase. Answering-phase generation can continue before the model regenerates another EoT, with the span preceding this regenerated EoT scaling with the reasoning tokens saved by early exit and exhibiting continued reasoning behavior. We call this spurious CoT termination, where reasoning-like generation continues into the answering phase. We hypothesize that insufficient attention to the injected EoT contributes to spurious CoT termination and probe this hypothesis with Exit-token Attention Biasing (EAB). Across four LRMs, five benchmarks, and two early-exit methods, increasing attention to the injected EoT reduces spurious CoT termination and answering-phase length. These results reveal a limitation of controlling LRMs by externally matching their explicit think-block format. Inserting the EoT token conforms to this format but does not by itself guarantee the intended reasoning-to-answering transition. Our code is available at this https URL.

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</think> Doesn’t Stop Reasoning: Analysis of Spurious CoT Termination Seunghee KohSungjae ChoiMinchan KwonSunghyun BaekJunmo Kim Korea Advanced Institute of Science and Technology, South Korea seunghee1215, sungjae579, kmc0207, baeksh, junmo.kim@kaist.ac.kr Abstract Chain-of-thought (CoT) reasoning improves large reasoning models (LRMs) on complex tasks but often produces long, redundant traces. Recent training-free early-exit methods shorten these traces by choosing an intermediate point to stop reasoning. We study one such strategy that injects an end-of-think token (EoT,</think>) at this point to trigger the reasoning-to-answering transition, and find that the injected EoT does not always induce a clean answering phase. Answering-phase generation can continue before the model regenerates another EoT, with the span preceding this regenerated EoT scaling with the reasoning tokens saved by early exit and exhibiting continued reasoning behavior. We call this spurious CoT termination, where reasoning-like generation continues into the answering phase. We hypothesize that insufficient attention to the injected EoT contributes to spurious CoT termination and probe this hypothesis with Exit-token Attention Biasing (EAB). Across four LRMs, five benchmarks, and two early-exit methods, increasing attention to the injected EoT reduces spurious CoT termination and answering-phase length. These results reveal a limitation of controlling LRMs by externally matching their explicit think-block format. Inserting the EoT token conforms to this format but does not by itself guarantee the intended reasoning-to- answering transition. Our code is available athttps://github.com/Seunghee-Koh/ Spurious-CoT-Termination. 1 Introduction Chain-of-thought(CoT)reasoningisa core mechanism of large reasoning models (LRMs) (DeepSeek-AI, 2025; Team, 2025), which generate extended intermediate reasoning steps before producing an answer (OpenAI, 2024; Team et al., 2025). While CoT substantially improves performance on mathematical (Mathematical The proper divisors of 12 are 1, 2, 3, 4 and 6. A proper divisor of an integer N is a positive divisor of N that is less than N. What is the sum of the proper divisors of the sum of the proper divisors of 284? <think> ... Reasoning Phase ... </think> Answering Phase. 1,090 tokens for reasoning 760 tokens for answering 2,116 tokens for reasoning 371 tokens for answering Full-CoT <think> Okay, so I need to find ... Wait a minute, ... The sum of ... ... Thus, the sum ... is \$\ 284\$. </think> <think> Okay, so I need to find ... ... So, yes, the sum of ... </think> ... Thus, the sum ... is \$\ 284\$. </think> Injection </think> Wait Figure 1: Top: CoT response format.<think>opens the reasoning phase,</think>closes it, and the answer- ing phase follows. Bottom: Illustration of Full-CoT and </think>injection. Injection can lengthen the answer- ing phase and lead to another </think>. Association of America, 2024; Project Numina, 2024; Hendrycks et al., 2021; Cobbe et al., 2021) and scientific (Rein et al., 2024) reasoning tasks, it often produces excessively long traces that increase inference costs without sufficient performance gains (Chen et al., 2025; Sui et al., 2025; Wang et al., 2025b; Wu et al., 2026; Pu et al., 2025). Training-free early-exit approaches shorten the reasoning trace by stopping reasoning early (Yang et al., 2026; Fu et al., 2026; Mao et al., 2026; Wang et al., 2026; Xiang et al., 2026). We study a strategy that injects the end-of-think token (EoT, </think>) at an intermediate point in the CoT to induce a reasoning-to-answering transition (Yang et al., 2026; Zhang et al., 2025c; Xiang et al., 2026). The EoT normally marks the end of the CoT in the explicit think-block format. However, we observe that EoT injection can arXiv:2609.03633v1 [cs.CL] 3 Sep 2026 lengthen the answering phase, with another EoT ap- pearing later in generation, as illustrated in Figure 1. We characterize this phenomenon by studying the conditions that make EoT regeneration more likely and the behavior preceding the regenerated EoT. We find that EoT regeneration becomes more fre- quent as early exit removes more of the reasoning trace. Answering phases are extended mainly by the generation before the regenerated EoT, where the model often continues reasoning-like behavior, including self-correction markers such asWait. We refer to this reasoning-like continuation as spurious CoT termination. The spurious CoT termination suggests that the injected EoT may not be sufficiently incorporated in subsequent generation. We hypothesize that sub- sequent tokens assign it insufficient attention, weak- ening its function as an internal transition signal. We probe this hypothesis with Exit-token Attention Biasing (EAB), which modulates attention to the injected EoT during answer generation. Across four reasoning models, five benchmarks, and two CoT early-exit methods, we demonstrate that EoT injection can induce spurious CoT termi- nation, where reasoning-like generation continues until another EoT appears. Increasing attention to the injected EoT via EAB reduces this behavior, supporting insufficient attention as one underly- ing mechanism. These findings show that exter- nally matching the explicit think-block format does not guarantee the intended reasoning-to-answering transition, as inserted delimiters may fail to engage the model’s internal transition mechanism. 2 Related Works 2.1 Efficient Chain-of-Thought Reasoning To reduce the cost of extended chain-of-thought rea- soning (Wei et al., 2022), prior works take several broad approaches, including post-training methods that internalize efficient reasoning behaviors (Ma et al., 2025; Munkhbat et al., 2025; Xia et al., 2025; Yu et al., 2024; Aggarwal and Welleck, 2025; Fang et al., 2025; Zhang et al., 2026) and prompt-based methods that elicit shorter CoT traces through in- structions (Xu et al., 2025; Han et al., 2025; Aytes et al., 2025; Lee et al., 2025; Chen et al., 2024). Among these strategies, decoding-time ap- proaches reduce reasoning cost by acting on gener- ated outputs or intermediate reasoning traces with- out modifying the underlying model. Recent work suppresses self-reflection tokens to prevent redun- dant CoT continuation (Wang et al., 2025a; Huang et al., 2026a), or terminates a CoT trajectory early based on answer confidence or model-internal un- certainty (Fu et al., 2026; Mao et al., 2026; Wang et al., 2026; Fu et al., 2025; Zhang et al., 2025a). DynaSoR (Fu et al., 2026) periodically probes the model for intermediate answers during reasoning and stops once these answers become consistent. DEER (Yang et al., 2026) generates trial answers at reasoning transition points and exits when the answer confidence exceeds a threshold. Some of these early-exit methods leverage EoT to force the reasoning-to-answering transi- tion (Yang et al., 2026; Zhang et al., 2025c; Xiang et al., 2026; Liu and Wang, 2025). In addition, recent studies have noted that LRMs may resume reasoning even after the thinking phase is externally terminated (Zhu et al., 2025; Zhang et al., 2025d). However, both consider prompt-level pre-filling that skips the reasoning phase entirely, whereas we study EoT injection mid-reasoning under dynamic early exit. Beyond this difference in experimental setting, we provide a finer-grained behavioral characteri- zation of EoT regeneration and a causal mechanis- tic analysis via attention intervention, and validate these findings across a broader range of model fam- ilies and reasoning benchmarks. 2.2 Attention-Based Analysis and Intervention Prior mechanistic studies of reasoning models show that CoT traces leave measurable internal signa- tures, including reasoning-relevant attention statis- tics and influential reasoning steps (Zhang et al., 2025b; Ostmeier et al., 2026; Park et al., 2025; Bogdan et al., 2025; Dutta et al., 2024; Chen et al., 2026; Choi et al., 2025). Beyond analysis, atten- tion also serves as a causal handle to control model behavior. For example, attention reweighting meth- ods (Zhang et al., 2024; Nguyen et al., 2026; Han et al., 2026) intervene directly on attention scores at inference time, while other approaches steer in- ternal activations to modify generation behavior (Li et al., 2023; Huang et al., 2026b). Building on this view of attention as an inference-time control signal, we examine whether the injected EoT is actually used by the model to induce a reasoning-to-answering transition. As a diagnostic probe, we bias attention to the injected EoT during answer generation and observe how post-exit behavior changes. 3 Preliminary 3.1 Reasoning Phase Conventions As illustrated in Figure 1, modern large reasoning models (LRMs) often use an explicit two-phase generation format, following the DeepSeek-R1 template (DeepSeek-AI, 2025): a reasoning phase delimited by the special tokens<think>(start-of- think, SoT) and</think>(end-of-think, EoT), followed by an answering phase comprising all subsequent tokens. This convention is used by the DeepSeek-R1-Distill and Qwen3 (Yang et al., 2025) reasoning models. Within the reasoning phase, these models of- ten exhibit surface markers of ongoing reasoning. In particular,Waitis commonly used as a self- reflection marker, indicating that the model is re- visiting prior reasoning (DeepSeek-AI, 2025), and has also been used in efficient-reasoning works as a signal of continued reflection (Wang et al., 2025a; Yang et al., 2026; Zhang et al., 2025c). For evaluation, we follow the official prompting guidance for the reasoning models, which recom- mends asking the model to place mathematical fi- nal answers in ...(DeepSeek-AI, 2025; Team, 2025; Yang et al., 2025). We extract the last ... as the model’s final answer. 3.2 Experimental Setup We study four LRMs spanning an order of magni- tude in scale, namely DeepSeek-R1-Distill-Qwen 1.5B and 14B (DeepSeek-AI, 2025), abbreviated as R1-Distill-1.5B and R1-Distill-14B, Qwen3- 14B (Yang et al., 2025), and QwQ-32B (Team, 2025). We evaluate on five widely used reasoning benchmarks. These are GSM8K (Cobbe et al., 2021), MATH-500 (MATH) (Hendrycks et al., 2021), AMC 2023 (AMC) (Project Numina, 2024), AIME 2024 (AIME) (Mathematical Asso- ciation of America, 2024), and GPQA-Diamond (GPQA) (Rein et al., 2024). The first four are math reasoning tasks of increasing difficulty, while GPQA-Diamond targets graduate-level scientific reasoning. Wecomparetwoearly-exitmethods, DEER (Yang et al., 2026) and DynaSoR (Fu et al., 2026), with two references, No-CoT and Full-CoT. DEER exits when an intermediate probe at aWait token is sufficiently confident, while DynaSoR probes at fixed intervals and exits when predictions become consistent. When either method selects ModelMethodMATHAMC AIMEGSM8KGPQA AVG R1-Distill-1.5B DEER1.02.56.70.01.02.2 DynaSoR1.05.06.70.00.02.5 Qwen3-14B DEER2.40.00.00.83.51.3 DynaSoR4.62.53.30.83.02.8 R1-Distill-14B DEER19.822.526.73.39.116.3 DynaSoR35.840.040.08.018.228.4 QwQ-32B DEER40.457.550.09.612.133.9 DynaSoR19.440.043.31.010.122.8 Table 1: ERR (%) across models, early-exit methods, and datasets. AVG is the row average across the five benchmarks. an exit point, we inject EoT there. No-CoT skips reasoning entirely, while Full-CoT runs without early exit. All settings otherwise share the same prompt and generation protocol, differing only in whether and when EoT is injected. Detailed configurations are provided in Section A.5. 4 Characterization of EoT Regeneration Throughout the paper, we distinguish the EoT to- ken that marks the transition point from any EoT later generated in the answering phase. We call the former the exit token, whether naturally generated or externally injected for early exit, and the latter the regenerated EoT. 4.1 Prevalence of EoT Regeneration with Answer Length Inflation We first define the EoT regeneration rate (ERR) as ERR = N regen N × 100%,(1) whereN regen is the number of EoT regenerated samples in the answering phase andNis the total number of samples. Table 1 shows ERR above 20% for QwQ-32B, consistent with prior observa- tions (Yang et al., 2026; Wang et al., 2025a). R1- Distill-14B also exceeds 15%, whereas R1-Distill- 1.5B and Qwen3-14B remain below 3%. The differ- ence between the two R1-Distill models suggests a scale effect, but the gap between the two 14B models indicates that scale alone is insufficient and training method may also matter. Next, we examine whether EoT regeneration re- lates to answer-length inflation. Table 2 reports the relative answering-phase length, defined as the me- dian answering-phase token count of samples with EoT regeneration divided by that of samples with- out EoT regeneration. EoT-regenerated samples consistently have longer answering phases, with the ratio typically exceeding two times. The effect ModelMethodMATH AMC AIME GSM8K GPQA R1-Distill-1.5B DEER2.39 2.292.23–2.34 DynaSoR2.99 1.935.02– Qwen3-14B DEER3.23–3.885.18 DynaSoR5.47 3.045.723.375.01 R1-Distill-14B DEER5.32 5.34 11.803.827.01 DynaSoR3.32 3.933.972.19 11.58 QwQ-32B DEER8.82 6.145.956.68 22.04 DynaSoR11.91 15.90 11.3414.64 13.70 Table 2: Relative answering-phase length, computed by dividing the median answering-phase token count among samples with EoT regeneration by that among samples without EoT regeneration. “–” indicates no EoT-regenerated samples.Absolute median token counts for both groups are provided in Section D.1. 0-20% 20-40%40-60%60-80%80-90% 90-100% Compression rate 0 10 20 30 40 ERR (%) Think-limit (n=23, 21.7%) Natural EoT (n=51, 0.0%) Early-exit (n=426) Figure 2: ERR by exit case and reasoning compression rate for R1-Distill-14B with DEER on MATH. For early- exit cases, bars show ERR within each compression-rate bin. Natural-EoT and think-limit cases are shown as horizontal reference lines. is particularly pronounced for QwQ-32B with Dy- naSoR, where the ratio exceeds ten times across all datasets. These results show that EoT regeneration is associated with longer answering phases, though the magnitude varies by model. 4.2 Forced Termination as a Trigger of EoT Regeneration We next test whether EoT regeneration depends on how EoT is introduced. We group samples into three exit cases. In natural EoT cases, the model generates EoT on its own. In early-exit cases, an early-exit algorithm injects EoT at a selected inter- mediate point. In think-limit cases, EoT is injected when the thinking budget is reached. For early-exit cases, we define the compression rate as the fraction by which early exit reduces the reasoning-phase length relative to its uninterrupted reasoning length. In Figure 2, we compare ERR across the three exit cases. For early-exit samples, we further group them by compression rate bin. Natural EoT cases show no ERR, whereas ERR #Token#Wait MethodDataset#Pre #PostRel. Pre Post DEER MATH8364052.0630 AMC5745651.0220 AIME4,674463 10.10220 GSM8K3962081.9020 GPQA5683931.4520 DynaSoR MATH5343711.4410 AMC1,5404943.1260 AIME2,1486283.4260 GSM8K2162240.9600 GPQA12804163.0880 Table 3: Pre/Post-regen analysis on samples with spu- rious CoT termination for R1-Distill-14B, reported as medians. Rel. = #Pre/#Post. #Wait countsWaittokens per segment. appears only when EoT is externally injected, as in early-exit and think-limit cases. Within early- exit samples, ERR remains low in the 0–20% bin but rises to about 40% in the 90–100% bin. This suggests that EoT regeneration is associated with external EoT injection and becomes more frequent with greater reasoning truncation. Takeaway 1. EoT regeneration appears un- der EoT injection and often leads to long answering-phase continuations, undermining early-exit efficiency. 5 Reasoning-Like Continuation after Exit-Token Injection We next examine behavioral patterns in EoT- regenerated answering phases. We split each an- swer phase at the last regenerated EoT. We define pre-regen as the span from the injected EoT to the last regenerated EoT and post-regen as the remain- ing span to the end of the sequence. Their token counts are denoted by #Pre and #Post, respectively. 5.1 Answering-Phase Inflation Concentrated Before EoT Regeneration We first identify which part of the answering phase, before or after the regenerated EoT, is responsible for the length inflation. The #Token columns in Table 3 show that, across most settings, the median #Pre is greater than the median #Post, with relative length greater than one. The gap is especially large on harder benchmarks, such as DEER on AIME, where the median pre-regen span is roughly ten times as long as the median post-regen span. Figure 3: Early-exit answer spans versus uninterrupted generation for 168 EoT-regenerated early-exit samples from R1-Distill-14B with DEER across five bench- marks. Of the 178 total EoT-regenerated samples, 10 think-limit samples are excluded. Figure 3 plots #Pre and #Post against the num- ber of reasoning tokens saved by early exit, #∆Thk, and the answering-phase token count without early exit, #Ans(No Exit). Each panel reports the Pear- son correlation coefficient computed over 168 EoT- regenerated samples obtained using DEER on R1- Distill-14B across five benchmarks. It shows a clear correspondence between early- exit spans and their uninterrupted counterparts. The strongest correlations are between pre-regen length and saved reasoning tokens (ρ = 0.67), and be- tween post-regen length and uninterrupted answer- ing phase length (ρ = 0.71). This suggests that the prolonged pre-regen span primarily reflects reason- ing that would have continued without early exit, whereas the post-regen span more closely matches uninterrupted answering-phase generation. 5.2 Reasoning-Like Patterns Before the Regenerated EoT Motivated by the frequent occurrence ofWaitdur- ing the reasoning phase in Section 3.1, we examine its distribution across pre-regen and post-regen to test whether pre-regen exhibits reasoning-like be- havior. Table 3 shows thatWaitappears almost entirely in pre-regen, while post-regen has a me- dian count of 0 in every setting. The pre-regen Waitcount also increases with the pre/post length ratio (Rel.). These patterns suggest that the pre- SoTEoTEOS 0.82 0.95 Full-CoT (n=500) SoTInjected EoT Regen EoT EOS 0.80 1.02 DEER (n=99) SoTInjected EoT Regen EoT EOS 0.74 0.98 DynaSoR (n=179) Figure 4: Boxed-expression density in samples with spurious CoT termination on MATH with R1-Distill- Qwen-14B, plotted by relative position within the an- swering phase completion. Quantitative results across models and datasets are reported in Table 15. regen span behaves more like continued reasoning than a verbose answer continuation. We also examine the requested final-answer for- mat, .... In Figure 4, we plot the dis- tribution of boxed expressions for Full-CoT sam- ples and EoT-regenerated samples from early-exit methods on MATH using R1-Distill-14B. For visu- alization, we normalize each analyzed span by its average length in the dataset and plot the relative position of each boxed expression within that span. In Full-CoT, the model often generates boxed expressions immediately before EoT and<EOS>. DEER and DynaSoR select exit points based on confidence or consistency rather than whether a boxed expression appears. As a result, boxed ex- pressions rarely precede the injected EoT, whereas they frequently precede the regenerated EoT. Taken together, the quantitative and behavioral evidence suggests that EoT regeneration reflects reasoning-like continuation rather than a mere for- matting artifact. We refer to this phenomenon as spurious CoT termination. Takeaway 2. We call EoT regeneration spurious CoT termination, where reasoning- like generation continues into the answering phase rather than terminating at the injected EoT. 6Causal Probing of Exit-Token Attention To investigate a potential mechanism underlying spurious CoT termination, we use Exit-token Atten- 0.6 0.8 Acc R1-Distill-14B 0.6 0.8 Qwen3-14B 0.0 0.2 ERR 0.00 0.05 0.10 -4-2-1024 1000 200 500 2000 5000 #Ans -4-2-1024 1000 500 2000 AIMEAMCGPQAGSM8KMATH Figure 5: EAB sweep (α) with DEER on R1-Distill- Qwen-14B (left) and Qwen3-14B (right). Rows show accuracy, ERR, and answering-phase length (#Ans) vs. α. α = 0 is the base method. tion Biasing (EAB) as an inference-time diagnostic probe that modulates attention to the exit token dur- ing answer generation. In this section, we examine the causal role and token specificity of attention to the injected EoT. We evaluate EAB’s selectivity for samples with spurious CoT termination and its gen- erality across models and early-exit methods. We then compare EAB with simple control baselines. 6.1 Exit-Token Attention Biasing Letx = (x 1 ,...,x T )denote the full token se- quence consisting of the prompt and completion, with the exit token at positionI. Tokens at posi- tionsi > Iform the answering phase. We denote bys (l,h) i,j the pre-softmax attention score from query positionito key positionjat layerland headh, and byA (l,h) i,j = softmax j (s (l,h) i,: )the correspond- ing post-softmax attention weight. Under causal masking, A (l,h) i,j is defined only for j ≤ i. As a diagnostic probe, Exit-token Attention Bi- asing (EAB) adds a constantα ∈Rto the pre- softmax attention logit of the injected EoT key for each answer-phase query: ̃s (l,h) i,I = s (l,h) i,I + αfor all i > I,l,h.(2) Forj ̸= I, the logitss (l,h) i,j remain unchanged, and the modified attention weights are computed by softmax, ̃ A (l,h) i,j = softmaxj( ̃si, : (l,h) ). 0.850 0.875 0.900 MATH Acc 0.2 0.4 ERR 500 1000 #Ans 024 0.52 0.54 GPQA 024 0.0 0.2 024 400 600 800 EoTSoTOffset-1Offset-10 Figure 6: Exit-token specificity of EAB under DEER on R1-Distill-14B for MATH and GPQA. We apply EAB to the exit token (EoT), SoT, Offset-1, and Offset-10. Since EAB is applied only fori > I, it affects only answering-phase generation, leaving the pre- ceding tokens unchanged. Positiveαstrengthens attention to EoT, whereas negativeαsuppresses it. 6.2 Reduction of Spurious CoT Termination by Increasing Exit-Token Attention To investigate whether spurious CoT termination is associated with under- or over-attention to the injected EoT, we sweepαover moderate negative and positive values using DEER on R1-Distill-14B and Qwen3-14B, as shown in Figure 5. Positiveαreduces ERR and answering-phase length while largely preserving accuracy, whereas negativeαoften increases ERR or answering-phase inflation. This pattern supports a causal role for exit-token attention. Increasing it promotes the reasoning-to-answering transition, whereas sup- pressing it does not. 6.3Exit-Token Specificity of Attention Biasing To identify whether the effect of EAB is specific to the exit token, we vary where the attention bias is applied. On R1-Distill-14B with DEER, we com- pare four targets on MATH and GPQA, including the exit token, the<think>(start-of-think, SoT) token, the token immediately before the exit token (Offset-1), and the token 10 positions before the exit token (Offset-10). This comparison assesses whether biasing other positions also reduces spuri- ous CoT termination and answering-phase length. Figure 6 shows that the effect is specific to the exit token. Consistent with the prior observation, biasing attention to the exit token reduces both spu- rious CoT termination and answering-phase length Acc (%,↑)ERR (%,↓)#Ans (↓) DatasetN DEER +EAB DEER +EAB DEER +EAB Samples with spurious CoT termination AIME850.050.0100.012.55,6681,353 AMC9100.0100.0100.022.21,4851,288 GPQA1877.872.2100.011.11,489783 GSM8K4484.168.2100.04.51,042442 MATH9991.979.8100.019.21,678777 Samples without spurious CoT termination AIME2250.054.50.00.02,4081,885 AMC3180.683.90.06.51,293435 GPQA18050.652.20.00.0456256 GSM8K 1,27591.593.70.00.0233197 MATH40186.389.50.01.0627376 Table 4: Grouped comparison of EAB effects with DEER on R1-Distill-14B atα = 4, grouping samples by whether their reasoning phase spuriously terminates at α = 0. Results for the other models are in Table 16. asαincreases, while preserving accuracy. Interest- ingly, the Offset-1 token exhibits the opposite trend. It is notable that biasing the Offset-1 token, which is adjacent to the exit token and typically serves as a formatting token, increases spurious CoT termi- nation and answering-phase length. Biasing SoT or Offset-10 does not result in a clear decrease in ei- ther spurious CoT termination or answering-phase length. These results suggest that the reduction in spu- rious CoT termination is specific to increasing at- tention to the injected EoT, rather than to arbitrary preceding tokens. 6.4 Selectivity of EAB for Samples with Spurious CoT Termination To test whether EAB primarily affects samples with spurious CoT termination, we group samples by whether their reasoning phase spuriously ter- minates without bias and apply EAB to the same samples on R1-Distill-14B with DEER. We report results by group in Table 4. For samples without spurious CoT termination, EAB keeps ERR near 0%, maintains or improves accuracy, and slightly reduces answer length. For samples with spurious CoT termination, EAB re- duces ERR below 25% on all datasets and sub- stantially shortens the answering phase. EAB can reduce accuracy for these samples, as discussed further in Section 6.6. Overall, EAB selectively intervenes on samples with spurious CoT termination. This suggests that increasing attention to the injected EoT primarily helps when the token fails to serve as an effective transition signal, with little effect when the model MethodAcc (%,↑) ERR (%,↓) #Thk (↓) #Ans (↓) R1-Distill-1.5B No-CoT33.13.4–2,138 Full-CoT51.70.05,635769 DEER42.52.24,1571,059 +EAB48.00.14,157557 DynaSoR43.22.52,9471,127 +EAB49.10.02,947386 R1-Distill-14B No-CoT57.70.0–1,004 Full-CoT76.90.04,156547 DEER73.316.33,2621,252 +EAB75.13.83,262667 DynaSoR74.028.42,2841,761 +EAB70.013.02,2841,065 Qwen3-14B No-CoT67.20.0–2,903 Full-CoT83.50.05,510822 DEER82.91.43,743917 +EAB82.60.23,743832 DynaSoR73.32.92,0041,786 +EAB74.51.52,0041,297 QwQ-32B No-CoT79.760.3–4,794 Full-CoT83.10.04,868636 DEER83.233.92,7072,282 +EAB81.128.02,7072,213 DynaSoR84.222.82,0922,055 +EAB80.314.82,0921,615 Table 5: Effect of EAB averaged across MATH, AMC, AIME, GSM8K, and GPQA benchmarks. We report accuracy (Acc, %), ERR, reasoning-phase token count (#Thk), and answering-phase token count (#Ans). already transitions successfully without the bias. 6.5 Generality of the Exit-Token Attention Biasing We next test whether EAB generalizes across mod- els, benchmarks, and early-exit methods. In Table 5, we evaluate positive attention bias with DEER and DynaSoR, measuring accu- racy, ERR, reasoning-phase length (#Thk), and answering-phase length (#Ans). We useα = 2 by default andα = 4for R1-Distill-14B. For each method, we regenerate the answering phase with and without EAB from the same reasoning trace, isolating the effect of modifying attention to the exit token during answer generation. Regarding the reference methods, Full-CoT ex- hibits no spurious CoT termination, producing long reasoning phases followed by short answering phases. No-CoT, in contrast, skips the reasoning phase entirely, with all models producing substan- tially longer answering phases. Across models and benchmarks, EAB generally reduces ERR and shortens the answering phase with varying magnitude. The reduction is espe- cially visible under DynaSoR, whose base method has shorter reasoning phases (#Thk), higher ERR, and longer answering phases. The effect is most pronounced for R1-Distill- 14B, where EAB reduces ERR by more than 13 percentage points (p) and, under DEER, improves accuracy while nearly halving the answering-phase length. For R1-Distill-1.5B, EAB roughly halves the answering-phase length under both methods and improves accuracy, despite already low ERR. For QwQ-32B, EAB yields smaller ERR reduc- tions and sometimes decreases accuracy. Its No- CoT ERR exceeds 60%, whereas those of all other models remain near zero, suggesting that EoT re- generation in QwQ-32B is driven more by a learned generation pattern than by insufficient attention to the injected EoT. This indicates that some EoT re- generation arises from learned generation patterns that attention intervention does not address. 6.6 Accuracy Trade-off of EAB Building on our earlier analysis, we further discuss the accuracy trade-off of EAB. As shown in Table 4, samples with spurious CoT termination exhibit non- trivial accuracy drops in some settings, with the magnitude varying across models, datasets, and early-exit policies. Our analysis indicates that this accuracy drop mainly arises from early exit points, where early exit removes part of the reasoning that would con- tribute to the final answer. Figure 2 shows that ERR is higher when early exit removes more of the reasoning phase, while Section 5.1 shows that the pre-regen span correlates with the amount of reasoning removed by early exit. This suggests that the reasoning-like continuation associated with spu- rious CoT termination partly compensates for the removed reasoning. EAB suppresses this continua- tion by strengthening attention to the injected EoT, promoting the intended reasoning-to-answering transition. Its remaining accuracy drop therefore largely reflects the quality of the selected exit point. This interpretation also explains the variation in Table 5. The accuracy drop of EAB is larger when early exit removes more of the reasoning phase, and the amount of truncation varies across datasets and exit policies. Since DynaSoR removes more of the reasoning phase than DEER on average, its larger accuracy drops are consistent with this MethodAcc (%,↑) ERR (%,↓) #Ans (↓) Baseline (α=0)73.316.31,252 EAB (α=4)75.13.8667 Double-EoT73.34.4761 Ans-Prefix74.6 0.1444 Post-Box74.21.6136 Post-Ans-Box73.65.6118 EAB (α=−4)71.213.31,658 Block-EoT69.40.02,189 Table 6: Simple control baselines and decoding alterna- tives to EAB on R1-Distill-14B with DEER, averaged across MATH, AMC, AIME, GSM8K, and GPQA. interpretation. 6.7 Control Baselines for Exit-Token Injection EAB shows that spurious CoT termination can be reduced by increasing attention to the injected exit token. We further examine output- and prompt- level interventions that leave attention computation unchanged. Using the same DEER traces, we eval- uate Double-EoT, which inserts a second EoT after the injected EoT, and Block-EoT, which sets the </think>logit to−∞during the answering phase. Double-EoT tests whether providing an additional EoT key has a similar effect to increased attention to the exit token, whereas Block-EoT tests whether simply preventing EoT regeneration is sufficient. We also evaluate three prompt-level transition cues. Ans-Prefix prepends an explicit transition phrase before the injected EoT, while Post-Box and Post-Ans-Box insert boxed-answer templates immediately after it. Further details are provided in Section D.9. Double-EoT reduces ERR and shortens the an- swering phase, closely matching the effect of pos- itive EAB. This suggests that an additional EoT strengthens the transition cue. Block-EoT reduces ERR to 0% by design but increases answer length and decreases accuracy, with effects more severe than suppressing attention via negative EAB. Pre- venting EoT regeneration therefore eliminates the regenerated EoT itself without inducing a clean reasoning-to-answering transition. Prompt-level controls also reduce EoT regener- ation while maintaining accuracy comparable to EAB. Ans-Prefix effectively suppresses regener- ation, whereas Post-Box and Post-Ans-Box pro- duce shorter answers but depend on a task-specific boxed-answer format. These controls support the view that spurious CoT termination reflects an incomplete reasoning- to-answering transition rather than EoT regenera- tion itself. Output- and prompt-level interventions can mitigate the behavior indirectly by modify- ing the token sequence or constraining generation. EAB is complementary, as a diagnostic probe that intervenes directly on attention to the injected EoT. Takeaway 3. EAB establishes a causal ef- fect at the attention level on answering-phase behavior. Increasing attention to the injected EoT reduces spurious CoT termination and answering-phase length, supporting its role in the reasoning-to-answering transition. 7 Conclusion We investigated end-of-think token (</think>, EoT) regeneration under CoT early-exit methods from three perspectives.We studied the phe- nomenon itself, the behavior that surrounds it, and the mechanism that contributes to it. At the phenomenon level, we found that forced EoT termination often leads the model to generate another EoT later in the answering phase. Across models, benchmarks, and CoT early-exit methods, we observe regeneration with varying prevalence. At the behavior level, we showed that EoT re- generation is associated with long answering-phase continuations that remain reasoning-like before EoT regeneration, including self-correction mark- ers and boxed-before-EoT patterns. We define this reasoning-like continuation after the exit token as spurious CoT termination. At the mechanism level, we used exit-token at- tention biasing (EAB) to investigate whether insuf- ficient attention to the exit token contributes to spu- rious CoT termination. Increasing attention to the exit token reduces EoT regeneration and shortens the answering phase, especially for samples with spurious CoT termination. The token-specific and sample-selective effects of EAB support its causal contribution to whether the transition succeeds. Overall, our results suggest that the injected EoT does not always serve as a reasoning-to-answering transition signal, highlighting a limitation of EoT- injection-based reasoning control. The EAB re- sults indicate that insufficient attention during sub- sequent generation may limit the injected EoT’s effectiveness. More broadly, our findings suggest that inserting such a structural delimiter does not by itself control the transition. Its effectiveness as a state-transition signal depends on how subsequent generation attends to and incorporates it. Limitations Our analysis characterizes spurious CoT termina- tion and its relationship to exit-token attention, but it remains limited in three respects. First, our interpretation of spurious CoT termina- tion is based on indirect evidence. Because CoT be- havior is complex and the relevant training data are not fully observable, we cannot directly verify the model’s internal state after the exit token. We there- fore rely on behavioral and intervention-based evi- dence, including EoT regeneration, answer-length inflation, reasoning-like markers, boxed-before- EoT patterns, and attention-biasing effects. Second, although EAB reduces spurious CoT termination and answering-phase length, extending it beyond a causal diagnostic into a competitive efficient-reasoning method requires addressing sev- eral challenges. These include adaptive criteria for when and how strongly to apply the bias and in- ference backends that preserve efficient generation while allowing the required attention modification. Third, our analysis focuses on the injected EoT as an individual transition signal, although reason- ing termination may also depend on surrounding context. The boxed-before-EoT patterns in Full- CoT generation and the effectiveness of Ans-Prefix suggest that recurring patterns around the EoT or expressions signaling reasoning completion may in- fluence this transition. Addressing these limitations is an interesting direction for future work. Acknowledgements This work was partly supported by the Institute of Information & Communications Technology Plan- ning & Evaluation (IITP) grant funded by the Ko- rea government (MSIT) (No.RS-2025-02283048, Developing the Next-Generation General AI with Reliability, Ethics, and Adaptability, 80%) and Ba- sic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (No.RS-2025-25410835, Machine Unlearning in Continual Learning with Linearity-Based Reduction of Data Dependency for Trustworthy AI, 20%). References Pranjal Aggarwal and Sean Welleck. 2025. L1: Con- trolling how long a reasoning model thinks with re- inforcement learning. In Second Conference on Lan- guage Modeling. Simon A Aytes, Jinheon Baek, and Sung Ju Hwang. 2025. Sketch-of-thought: Efficient llm reasoning with adaptive cognitive-inspired sketching. 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For mathematical benchmarks, this follows the of- ficial prompting guidance for DeepSeek-R1 and QwQ-32B, which recommends asking the model to place the final answer in (DeepSeek- AI, 2025; Team, 2025; Yang et al., 2025). We use the same instruction for GPQA as well, al- though GPQA is not a mathematical benchmark, so that final-answer extraction is identical across all datasets. For R1-Distill and Qwen3 models, we use the standard two-phase format: the assistant response begins with<think> , the reasoning phase ends with </think> , and the answering phase follows.QwQ-32B’s chat template prepends <think> (Team, 2025). We treat its outputs un- der the same two-phase convention. Although Qwen3 officially supports/think and/no_thinkinstructions for switching between thinking and non-thinking modes, these flags are not required for single-turn thinking-mode infer- ence: Qwen3 operates in thinking mode by default, so callingapply_chat_templatewith its default arguments automatically appends<think> the prompt (Yang et al., 2025). We therefore do not use these flags in our experiments. For the No-CoT condition, we implement a zero- length reasoning phase with an explicit empty- thinking template: <think> </think> This keeps No-CoT under the same phase- segmentation convention as our early-exit outputs. A.2 Serving Engine We use two backends depending on the genera- tion phase. For reasoning phase generation and exit-token selection, which are DEER and Dyna- SoR’s early-exit decisions, we use vLLM (Kwon et al., 2023). For answering-phase generation, we feed the vLLM-produced prefix to Hugging Face transformerswith Flash Attention 2 (Dao, 2024). This split is necessary because attention interven- tions require per-step access to attention computa- tion, which vLLM does not support. To assess the effect of this backend split, Ta- ble 17 also reports the corresponding vLLM- backend results (†) for DEER and DynaSoR, along- side the transformers-backendα = 0baseline. The differences are marginal, suggesting that the main comparisons are unlikely to be substantially con- founded by the backend change. Table 9 addition- ally reports a runtime comparison across the two backends. A.3 EAB Implementation We implement the EAB by patching the attention forward pass oftransformers. For R1-Distill and QwQ-32B, we patchQwen2FlashAttention2; for Qwen3 models, we patchQwen3Attention. In both cases, the prefill step uses the original Flash Attention path, and the decode step routes through xformers.memory_efficient_attention (Lefaudeux et al., 2022), which supports arbitrary additive attention bias. The bias mask addsαto the pre-softmax attention score at the exit-token key position for all answering-phase queries, across all layers and heads. For multi-GPU setups withdevice_map="auto", the patch is applied per-instance to remain compatible with accelerate’s device-placement hooks. A.4 GPU Configuration R1-Distill-1.5B runs on a single RTX 4090D. R1- Distill-14B uses a single A6000 for thinking-phase construction, No-CoT, and Full-CoT generation, and 2 A6000s for attention interventions. QwQ- 32B uses a single A100 for all non-intervention settings; for attention interventions, we run on a single A100 by default and split a small number of OOM-prone samples across 2 A100s. A.5 Early-Exit Baseline Configurations Both DEER and DynaSoR share the following set- tings: greedy decoding (T =0, top-p=1), a max- imum total generation budget of 16,384 tokens, and a model context window 8,000 tokens larger than this budget. The reasoning phase is capped at a fixed fraction of the generation budget, fol- lowing DEER’s official recommendations (Yang et al., 2026): 0.6 for R1-Distill and QwQ-32B, and 0.8 for Qwen3. Both methods use the same probe prompt" **Final Answer** " to prompt the model for an intermediate answer, and generate up to a fixed number of new tokens per probe: 30 for MATH, since LaTeX answers tend to be long; 2 for GPQA, since the model only needs to generate a multiple-choice letter; and 20 for the other benchmarks. For confidence aggregation, we follow DEER’s recommended policy (Yang et al., 2026). We use the arithmetic mean of per-token top- 1 probabilities for R1-Distill and QwQ-32B, and the geometric mean for Qwen3, which DEER’s au- thors recommend to compensate for Qwen3’s ten- dency toward overconfident probability estimates. We also set the maximum number of probes per sample, using 10 for DEER and 20 for DynaSoR. If the maximum number of probes is reached without satisfying the exit condition, the model continues reasoning without further probes until it either gen- erates EoT naturally or reaches the reasoning-phase budget. Conversely, if the reasoning-phase budget is exhausted before the exit condition is met, rea- soning is terminated and we inject EoT to begin the answering phase. DEER. DEER triggers the early-exit probe at thought-transition tokens. Following the original paper (Yang et al., 2026), we useWaitas the stop word that segments reasoning into chunks. At each Wait, an intermediate-answer probe is induced, and exit is triggered when the probe answer’s confi- dence exceeds the threshold λ=0.95. DynaSoR. The original DynaSoR paper (Fu et al., 2026) does not specify a quantitative con- sistency threshold. We adopt the configuration de- scribed in DEER’s baseline section (Yang et al., 2026). DynaSoR periodically induces intermediate- answer probes at a fixed token interval, and exit is triggered when three consecutive probe answers are consistent. We set this interval to 256 tokens. Fur- thermore, since our analysis targets the effective- ness of injected EoT as a phase-transition signal, we inject EoT at the exit point and let the model generate an answering phase, matching the proto- col used for DEER. This allows direct comparison of spurious CoT termination behavior under a com- mon early-exit framework. A.6 Generation Budget Discrepancy As originally implemented, DEER and DynaSoR allocate only a fraction of the model’s context window to the reasoning phase, such as 60% or 80%, leaving the remainder for answer genera- tion. Our EAB intervention pipeline keeps the same reasoning-phase budget but regenerates the answering phase under a separate answering-phase limit. As a result, intervention runs can produce longer answering phases than the original budget would allow. This applies only to the DEER and DynaSoR intervention runs, not to the No-CoT and Full-CoT references. Within each intervention sweep, all conditions (α = 0, 2, 4) share the same reasoning traces and answering-phase budget, so within-sweep comparisons of accuracy, ERR, and generation length remain unaffected. A.7 Datasets We evaluate on five test splits, GSM8K (Cobbe et al., 2021), MATH-500 (Hendrycks et al., 2021), AMC 2023 (Project Numina, 2024), AIME 2024 (Mathematical Association of America, 2024), and GPQA-Diamond (Rein et al., 2024), with 1,319, 500, 40, 30, and 198 samples, respec- tively. B Additional Statements B.1 License We use publicly available models, datasets, and software libraries. DeepSeek-R1-Distill-Qwen- 1.5B and DeepSeek-R1-Distill-Qwen-14B are re- leased under the MIT License. Qwen3-14B and QwQ-32B are released under the Apache 2.0 Li- cense. For datasets, GSM8K is released under the MIT License, AMC 2023 via AI-MO/aimo- validation-amc is released under the Apache 2.0 License, and GPQA-Diamond is released under C BY 4.0. MATH-500 is distributed publicly on Hugging Face as a subset of the MATH benchmark, but we did not find an explicit license field on the dataset card. AIME 2024 consists of competition problems from MAA, and we did not find an ex- plicit open license. For software, vLLM and Hug- ging Face Transformers are released under Apache 2.0, while xFormers and FlashAttention-2 use BSD- style licenses. B.2 Artifact Use Consistent With Intended Use We use publicly released LRMs and standard reasoning benchmarks (MATH, AMC, AIME, GSM8K, GPQA) for their intended purpose of eval- uating reasoning capabilities, consistent with their original release conditions. ModelFull#ERR (%)Rel Qwen3-14BC4742.32 3.80 (1836 / 483) I263.85 0.84 (667 / 798) R1-Distill-14BC45719.47 5.33 (1138 / 214) I4323.26 6.46 (2661 / 412) QwQ-32BC47140.98 8.94 (2433 / 272) I2931.03 2.96 (3337 / 1126) Table 7: EoT regeneration rate (ERR) and relative answering-phase length on MATH with DEER, grouped by Full-CoT (Full) answer correctness (C: Correct, I: Incorrect). Relative answering-phase length (Rel) is computed in the same way as in Table 2, with abso- lute medians (with/without EoT regeneration) shown in parentheses. In the incorrect group (I) of Qwen3-14B, the relative answer length is based on a single regener- ated sample (n = 1) and thus is not representative. C Further Analyses C.1 EoT Regeneration and Answer Length by Correctness Building on the analysis in Section 4.1, we examine whether the relationship between EoT regeneration and answering-phase length can be explained by correctness. We group DEER-MATH samples us- ing two criteria. First, Full-CoT answer correctness (Table 7) serves as a proxy for problem difficulty, using a separate Full-CoT run to avoid dependence on the injected EoT. Second, DEER’s intermediate exit-probe correctness (Table 8) reflects whether the reasoning at the exit point is sufficient. Under both criteria, correctness does not fully explain the longer answering phases of EoT- regenerated samples. Grouping by Full-CoT cor- rectness, the answer-length gap persists within the correct group across all three models, and ERR re- mains broadly comparable between correct and in- correct samples without a consistent trend. Group- ing by probe correctness, probe-incorrect cases consistently show higher ERR, suggesting that EoT regeneration is more likely when reasoning at the exit point is incomplete. The relative an- swer length, however, remains comparable across probe-correct and probe-incorrect groups in R1- Distill-14B and Qwen3-14B, indicating that probe correctness alone does not account for the gap. C.2 Efficiency Analysis In Table 9, we evaluate wall-clock latency, through- put, and answer length for DEER and DEER+EAB with single-batch decoding. Latency covers only the answer-generation phase, since the prefill stage ModelProbe#ERR (%)Rel Qwen3-14BC3711.62 3.08 (1332 / 432) I2524.00 3.36 (2534 / 754) R1-Distill-14BC38219.37 6.12 (1071 / 175) I4445.45 5.17 (2154 / 417) QwQ-32BC33234.64 8.19 (1958 / 239) I11176.58 0.75 (3619 / 4834) Table 8: Same as Table 7, but grouped by the interme- diate exit probe outcome. In the probe-incorrect group of QwQ-32B, the relative answer length falls below 1 because the high ERR (76.6%) leaves only a few non- regenerated samples, whose answer lengths are excep- tionally long (median 4,834 tokens). before the injected EoT is identical across meth- ods. We report both the original DEER with the vLLM backend (DEER†) and our Transformer im- plementation with FlashAttention, as described in Section A.2. Comparing the Transformer implementations (DEER vs DEER+EAB), both follow the same at- tention path, withα = 0simply adding a zero- valued bias. Throughput is therefore comparable with and without EAB when answer lengths are similar. The vLLM and Transformer implementations (DEER † vs. DEER) differ in backend as well as the resulting answer lengths and accuracies, and are not directly comparable. vLLM achieves much higher throughput through optimizations such as PagedAttention. Even so, EAB can attain a mean latency comparable to DEER † when it shortens the answer substantially, whereas with little token reduction the latency difference largely reflects the backend gap. Overall, EAB’s runtime overhead is manageable for diagnostic use. C.3 Case Study on Reasoning-like Behavior Without Regenerating EoT Following the analysis in Table 3, we again use the occurrence of the reasoning markerWaitin the answering phase as a simple proxy for reasoning- like behavior. Table 10 summarizes all answer- ing phase samples containingWaitacross the five benchmarks. Among these, we distinguish cases that do not regenerate EoT and terminate before reaching the maximum generation budget (No EoT Regenera- tion, Before Limit), cases that continue until the budget is exhausted (No EoT Regeneration, At Limit), and cases with explicit EoT regeneration MethodMATHAMCAIMEGSM8KGPQA LatThr AnsLatThrAnsLatThrAnsLatThr AnsLatThrAns R1-Distill-14B DEER † 23.0/8.0 36.6 840 61.8/10.3 35.1 2,170 131.0/21.7 35.2 4,6096.2/4.338.7 24113.7/5.435.6488 DEER72.3/17.6 10.9 786 181.5/25.1 7.7 1,392 793.6/79.25.5 4,398 16.8/9.4 14.8 249 61.2/17.09.4578 +EAB34.7/15.7 12.8 444 41.5/22.8 11.2464261.8/60.06.0 1,574 11.5/10.5 17.0 194 26.6/17.2 10.0267 Qwen3-14B DEER † 16.4/12.4 39.3 643 22.3/22.4 38.084755.0/35.437.4 2,0556.2/5.840.3 251 34.4/21.1 39.9 1,372 DEER51.6/31.5 12.7 653 80.4/67.5 10.4836157.8/159.4 8.0 1,264 14.9/13.0 16.8 250 115.9/57.0 12.9 1,496 +EAB50.0/30.3 12.4 620 75.4/67.3 10.5792391.4/164.3 6.6 2,596 14.9/13.6 16.9 252 84.9/48.8 12.9 1,093 Table 9: Runtime comparison of DEER and DEER+EAB across benchmarks. Lat as mean/median (s), Thr in tok/s, Ans is mean #answer-phase tokens. All values are measured on a dedicated timing run. DEER † denotes the original DEER implementation served with the vLLM backend, following the notation in Table 17. All other results use our Transformer implementation. ModelMethodTotal No EoT Regeneration EoT Regen. Before Limit At Limit R1-Distill-1.5B DEER299200 DynaSoR5118312 Qwen3-14B DEER70291427 DynaSoR83321833 R1-Distill-14B DEER2191341165 DynaSoR 2821650216 QwQ-32B DEER5139036387 DynaSoR 2363940157 Table 10: Breakdown of answer-phase samples contain- ing the reasoning marker “Wait,” aggregated across all five benchmarks. Total denotes the number of Wait- containing samples. Samples without EoT regeneration are divided into those that terminate before reaching the maximum generation budget (Before Limit) and those that exhaust the budget (At Limit). (EoT Regen). The At Limit cases are truncated by the generation budget, so their termination behavior is unobservable. As shown in Table 10, the Before Limit cases range from 9 to 90 samples per model and early- exit method. Relative to the total of 2,087 eval- uation samples across the five benchmarks, these cases represent only a small fraction, at most 4.3%. This suggests that most reasoning-like continuation is accompanied by explicit EoT regeneration and is therefore captured by ERR, leaving only a small residual that ERR does not reflect. D Expanded Results Across Models and Benchmarks D.1 Full Results for Answering-phase Length Table 2 reports the median answering-phase length ratio between samples with and without EoT re- generation. Table 12 provides the underlying raw MethodExit caseMATH AMC AIME GSM8K GPQA Avg R1-Distill-1.5B DEER early-exit0.00.00.00.00.0 0.0 think-limit8.89.111.80.01.5 6.2 DynaSoR early-exit0.00.08.30.00.0 1.7 think-limit14.7 28.67.10.00.0 10.1 Qwen3-14B DEER early-exit3.00.00.00.94.5 1.7 think-limit0.00.00.00.00.0 0.0 DynaSoR early-exit4.82.94.81.53.1 3.4 think-limit0.00.00.0– 0.0 R1-Distill-14B DEER early-exit22.1 28.633.34.322.5 22.1 think-limit21.7 14.330.80.00.0 13.4 DynaSoR early-exit49.6 65.257.923.520.3 43.3 think-limit23.1 33.314.3–0.0 17.7 QwQ-32B DEER early-exit45.1 75.973.312.121.4 45.6 think-limit18.2 33.333.30.00.0 17.0 DynaSoR early-exit21.0 40.050.02.010.7 24.7 think-limit25.0 100.037.5– 54.2 Table 11: ERR (%) comparison across exit cases broken down by which mechanism triggered the end of thinking. Natural-EoT cases (the model generates EoT on its own) are omitted as they yield 0% ERR across all settings. early-exit and think-limit both force-inject</think>. AVG is the row average across the five benchmarks. medians for each (model, method, dataset, phase) cell. Across all configurations, EoT-regenerated samples require substantially more tokens in the answering phase than samples without EoT regen- eration. The gap is more pronounced on harder datasets (AIME, GPQA) and larger models (R1- Distill-14B, QwQ-32B). For the full distributional view, refer to Figure 11. 0-20% 20-40%40-60%60-80%80-90% 90-100% Compression rate 0 25 50 75 100 125 Sample count 36 102 135 92 46 15 Non EoT-regeneratedEoT-regenerated Figure 7: Sample count by reasoning compression rate on MATH with R1-Distill-Qwen-14B under DEER, split into samples with and without EoT regeneration. Together with Figure 2, this shows both the per-bin ERR and the underlying sample distribution across compres- sion rates. ModelMethodRegen MATH AMC AIME GSM8K GPQA R1-Distill-1.5B DEER Y574530589–601 N240231264238257 DynaSoR Y810670 2,105– N271347420232202 Qwen3-14B DEER Y1,593–9094,209 N494802 1,243235813 DynaSoR Y2,865 2,535 10,2299364,657 N524834 1,788278930 R1-Distill-14B DEER Y1,187 1,222 4,892620964 N223229415162138 DynaSoR Y940 1,925 2,5644321,691 N283490646197146 QwQ-32B DEER Y2,451 3,259 4,4801,0054,486 N278531753150.5203.5 DynaSoR Y3,394 6,693 6,6232,327 3,240.5 N285421584159236.5 Table 12: Median answer-phase token lengths for sam- ples with (Y) vs. without (N) EoT regeneration, across early-exit methods and datasets. “–” indicates no sam- ples with EoT regeneration. D.2 EoT Regeneration by Exit Case and Compression Rate Section 4.2 examines how EoT regeneration relates to the way the CoT is terminated. Table 11 extends this analysis to all evaluated models and bench- marks, reporting within-case ERR, with Natural- EoT cases omitted as they yield no EoT regener- ation across all settings. Figure 7 complements Figure 2 with the underlying sample distribution, showing the counts of samples with and without EoT regeneration in each bin. D.3 Pre-regen and Post-regen Analysis In Table 3, we analyze token count and Wait-token repetition frequency for the R1-Distill-14B model. We report the results for all models in Table 13. #Token#Wait MethodDatasetN#Pre #PostRel. Pre Post R1-Distill-1.5B DEERMATH51874800.3900 AMC11953340.5800 AIME2525360.1000 GPQA22004000.5000 DynaSoR MATH53495270.6600 AMC2506200.0800 AIME2 15395652.72150 Qwen3-14B DEERMATH129405461.7220 GSM8K114873691.3220 GPQA7 25809812.63120 DynaSoR MATH23 18158332.1830 AMC1 15469881.5690 AIME1 836718614.50230 GSM8K116283671.7120 GPQA6 38788144.77140 R1-Distill-14B DEERMATH998364052.0630 AMC95745651.0220 AIME8 4674463 10.10220 GSM8K443962081.9020 GPQA185683931.4520 DynaSoR MATH1795343711.4410 AMC16 15404943.1260 AIME12 21486283.4260 GSM8K 1052162240.9600 GPQA36 12804163.0880 QwQ-32B DEERMATH202 19734724.1880 AMC23 28045355.24140 AIME15 40034988.04170 GSM8K 1277442263.2930 GPQA24 39744199.49200 DynaSoR MATH97 27635445.08140 AMC16 6113583 10.49340 AIME13 5949536 11.10210 GSM8K13 21022897.2780 GPQA20 28815015.75180 Table 13: Pre/Post-regen analysis on samples with spu- rious CoT termination, reported as medians. Rel. = #Pre/#Post. #Wait counts Wait tokens per segment. D.4 Quantitative Boxed-before-EoT Statistics Across Models In Figures 4 and 8, we visualize the occurrence of . . . expressions around EoT for R1- Distill-14B, Qwen3-14B, and QwQ-32B, respec- tively. Here, we provide a quantitative, per-span, and per-dataset analysis of the boxed-before-EoT pattern across all four backbones and five bench- marks in Table 15. The metric ̄n box denotes the mean number of boxed expressions within a segment. We also use END-BOX 5 , denoting the fraction of samples whose SoTEoTEOS 1.03 1.05 Full-CoT (n=500) SoTInjected EoT Regen EoT EOS 1.001.00 DEER (n=12) SoTInjected EoT Regen EoT EOS 0.96 1.00 DynaSoR (n=23) (a) Qwen3-14B SoTEoTEOS 1.04 1.00 Full-CoT (n=500) SoTInjected EoT Regen EoT EOS 1.09 1.00 DEER (n=202) SoTInjected EoT Regen EoT EOS 1.16 0.99 DynaSoR (n=97) (b) QwQ-32B Figure 8: Boxed-expression density in spurious samples on MATH with Qwen3-14B and QwQ-32B. The density is low around the exit token and peaks immediately before the regenerated EoT, indicating that the boxed- before-EoT pattern appears near the regenerated EoT rather than the exit token. final boxed expression closes within five characters of the segment boundary. This metric serves as a proxy for the pattern in which the model writes a boxed answer immediately before ending the seg- ment. For example, a highEND-BOX 5 in the pre- regen window of spurious DEER or DynaSoR sam- ples indicates that the model has produced a boxed answer immediately before regenerating EoT. In Full-CoT, boxed expressions tend to concen- trate near the ends of the thinking and answering phases, as indicated by the highEND-BOX 5 val- ues. For samples with spurious CoT termination, this behavior instead appears prominently in the pre-regen span, which shows both high ̄n box and highEND-BOX 5 . This indicates that the model frequently produces a boxed answer immediately before regenerating EoT. The post-regen span also tends to end with a boxed expression, similar to the answering phase of Full-CoT. Overall, these results quantitatively support the boxed-before-EoT pat- tern in Section 5.2 across models, benchmarks, and early-exit methods. D.5 Selectivity of EAB for Samples with Spurious CoT Termination In Table 4, we partition samples by whether they exhibit spurious CoT termination under the unbi- ased setting (α = 0), and report group-wise results for R1-Distill-14B with EAB at α = 4. Table 16 extends this analysis to the full set of models and datasets, using DEER with EAB at α = 2andα = 4. For each group, we report ac- curacy, ERR, and answering-phase length to eval- uate whether EAB primarily affects the samples with spurious CoT termination while leaving the samples without spurious CoT termination largely intact. The results show that this trend generally extends across models, although its strength varies. EAB reduces ERR for samples with spurious CoT ter- mination under the unbiased setting, although this can be accompanied by accuracy degradation. For samples without spurious CoT termination, ERR remains close to zero, and accuracy is generally preserved, suggesting little adverse effect on this group. The effect on answering-phase length is more model-dependent, especially for samples with spurious CoT termination. D.6 Extended Sweep We extend theαsweep in Figure 5 to a wider range,α ∈ [−256, 8], using R1-Distill-14B with DEER on AMC, GPQA, and MATH, and R1- Distill-1.5B with both DEER and DynaSoR. As shown in Figure 9, large negative values ofα largely preserve accuracy but substantially increase ERR and answering-phase length. Positiveαim- proves these metrics only up to around four, after which generation becomes unstable at α = 8. D.7 Specificity of EAB Figure 10 reports the target-position ablation on Qwen3-14B with DEER, following the R1-Distill- 14B analysis in Figure 6. The results show the same qualitative pattern, with the effect specific to the exit token. D.8 Benchmark-Level EAB Results Table 17 expands the aggregate results in Table 5 to individual benchmarks, reporting EAB results atα = 0, 2, 4. It includes corresponding vLLM- backend results (†). 0.0 0.5 Acc 0.0 0.1 0.2 ERR -256-16-4-2-10248 1000 500 2000 #Ans AMCGPQAMATH (a) R1-Distill-14B (DEER) 0.0 0.5 Acc DEER 0.0 0.5 DynaSoR 0.0 0.2 0.4 ERR 0.0 0.1 0.2 -256-16-8-101248 1000 200 500 2000 5000 #Ans -256-16-8-101248 1000 500 2000 5000 AIMEAMCGPQAGSM8KMATH (b) R1-Distill-1.5B (DEER vs DynaSoR) Figure 9: EAB sweep at extendedαrange (−256to8). Panels show accuracy, ERR, and answering-phase length vs. α; colors denote benchmarks. MethodMATHAMCAIMEGSM8KGPQAAVG Acc ERR #Ans Acc ERR #Ans Acc ERR #Ans Acc ERR #Ans Acc ERR #Ans Acc ERR #Ans Baseline (α=0)87.4 19.8835 85.022.5 1,336 50.026.7 3,277 91.23.3260 53.09.1550 73.3 16.3 1,252 EAB (α=4)87.64.6455 87.5 10.0627 53.33.31,743 92.90.2205 54.01.0304 75.13.8667 Double-EoT (append) 87.64.2715 85.0 5.0832 50.010.0 1,645 91.40.3209 52.52.5404 73.34.4761 Double-EoT (inline)88.68.8747 82.5 10.0 1,271 43.36.7 2,076 92.61.3210 53.55.1660 72.16.4993 Ans-Prefix89.60.4 292 85.00.025250.00.0 1,290 92.90.0154 55.60.0230 74.60.1444 Post-Box89.01.255 85.00.05 46.76.760393.10.112 57.10.03 74.21.6136 Post-Ans-Box89.08.08285.07.5400 43.33.351 93.83.94357.13.51273.65.2118 EAB (α=−4)87.0 22.8917 80.0 20.0 2,032 50.010.0 4,452 90.04.5236 49.09.1656 71.2 13.3 1,658 Block-EoT82.80.0 2,726 75.00.0 3,117 46.70.0 3,109 90.80.0585 51.50.0 1,407 69.40.0 2,189 Table 14: Control baselines to EAB on R1-Distill-14B with DEER traces. We report accuracy (Acc, %,↑), ERR (%, ↓), and answering-phase length (#Ans,↓). Avg denotes the macro average across the five datasets. The best and second-best values in each column are shown in bold and underlined. D.9 Control Baseline Details and Extended Results This section expands on the control baselines in Section 6.7, providing implementation details and the full quantitative comparison in Table 14. Implementation.All baselines are evaluated on the same DEER traces as EAB. Our default in- jected boundary is </think> .Double- EoT (append) adds a second EoT outside this boundary,... </think> </think>, whereas Double-EoT (inline) places it after the first,... </think></think> . The main text reports Double-EoT (append), while here we compare the inline variant. Block-EoT masks the </think>logit to−∞throughout the answering phase, making EoT regeneration impossible by construction. It serves as a harder counterpart to negative-αEAB, as both reduce the EoT availabil- ity during answer generation. For the prompt-level controls, Ans-Prefix prepends the transition phrase “Based on the reasoning up until now, I will now present my final answer.” before the injected EoT. Post-Box inserts a - pression after the injected EoT, and Post-Ans-Box inserts the template**Final Answer**: after it. Double-EoT. Both variants reduce ERR and shorten the answer phase, qualitatively matching positive EAB. The append variant is closer to EAB, whereas the inline variant has a higher ERR of 0.925 0.930 0.935 MATH Acc 0.0 0.1 ERR 600 800 #Ans 024 0.50 0.55 GPQA 024 0.00 0.05 0.10 024 1000 2000 3000 EoTSoTOffset-1Offset-10 Figure 10: Specificity of EAB to the exit token under DEER on Qwen3-14B for MATH and GPQA. We ap- ply EAB to the exit token, SoT, and tokens 1 and 10 positions before the exit token, denoted Offset-1 and Offset-10. 6.4% and longer answer phases. This gap points to a formatting effect. Append preserves the de- fault boundary pattern </think> adds the second EoT after it, whereas inline stacks two </think>tokens within the boundary and is more disruptive. The direction is consistent with the attention- based explanation motivating EAB, since an extra </think>adds an EoT key that answering-phase queries can attend to. Double-EoT changes the token sequence and key-value cache, so it moves attention only indirectly and cannot isolate the ef- fect to attention as EAB does. 10 0 10 2 10 4 0 2 4 6 Count DEER AMC Non-Regen (n=39) EoT-Regen (n=1) 10 0 10 2 10 4 0 2 4 AIME Non-Regen (n=28) EoT-Regen (n=2) 10 0 10 2 10 4 0 5 10 15 20 GPQA Non-Regen (n=196) EoT-Regen (n=2) 10 0 10 2 10 4 0 100 200 300 GSM8K Non-Regen (n=1319) 10 0 10 2 10 4 0 20 40 60 80 MATH Non-Regen (n=495) EoT-Regen (n=5) 10 0 10 2 10 4 #Ans 0 2 4 6 8 Count DynaSoR Non-Regen (n=38) EoT-Regen (n=2) 10 0 10 2 10 4 #Ans 0 1 2 3 4 Non-Regen (n=28) EoT-Regen (n=2) 10 0 10 2 10 4 #Ans 0 10 20 Non-Regen (n=198) 10 0 10 2 10 4 #Ans 0 100 200 300 Non-Regen (n=1319) 10 0 10 2 10 4 #Ans 0 20 40 60 80 Non-Regen (n=495) EoT-Regen (n=5) (a) R1-Distill-1.5B 10 0 10 2 10 4 0.0 2.5 5.0 7.5 10.0 Count DEER AMC Non-Regen (n=40) 10 0 10 2 10 4 0 2 4 6 8 AIME Non-Regen (n=30) 10 0 10 2 10 4 0 10 20 30 GPQA Non-Regen (n=191) EoT-Regen (n=7) 10 0 10 2 10 4 0 100 200 GSM8K Non-Regen (n=1308) EoT-Regen (n=11) 10 0 10 2 10 4 0 20 40 60 MATH Non-Regen (n=488) EoT-Regen (n=12) 10 0 10 2 10 4 #Ans 0 2 4 6 8 Count DynaSoR Non-Regen (n=39) EoT-Regen (n=1) 10 0 10 2 10 4 #Ans 0 1 2 3 4 Non-Regen (n=29) EoT-Regen (n=1) 10 0 10 2 10 4 #Ans 0 10 20 30 40 Non-Regen (n=192) EoT-Regen (n=6) 10 0 10 2 10 4 #Ans 0 100 200 300 Non-Regen (n=1308) EoT-Regen (n=11) 10 0 10 2 10 4 #Ans 0 20 40 60 80 Non-Regen (n=477) EoT-Regen (n=23) (b) Qwen3-14B 10 0 10 2 10 4 0 2 4 Count DEER AMC Non-Regen (n=31) EoT-Regen (n=9) 10 0 10 2 10 4 0 1 2 3 AIME Non-Regen (n=22) EoT-Regen (n=8) 10 0 10 2 10 4 0 10 20 GPQA Non-Regen (n=180) EoT-Regen (n=18) 10 0 10 2 10 4 0 50 100 150 200 GSM8K Non-Regen (n=1275) EoT-Regen (n=44) 10 0 10 2 10 4 0 10 20 30 40 MATH Non-Regen (n=401) EoT-Regen (n=99) 10 0 10 2 10 4 #Ans 0 2 4 6 Count DynaSoR Non-Regen (n=24) EoT-Regen (n=16) 10 0 10 2 10 4 #Ans 0 2 4 Non-Regen (n=18) EoT-Regen (n=12) 10 0 10 2 10 4 #Ans 0 10 20 Non-Regen (n=162) EoT-Regen (n=36) 10 0 10 2 10 4 #Ans 0 100 200 300 Non-Regen (n=1214) EoT-Regen (n=105) 10 0 10 2 10 4 #Ans 0 20 40 Non-Regen (n=321) EoT-Regen (n=179) (c) R1-Distill-14B 10 0 10 2 10 4 0 2 4 6 8 Count DEER AMC Non-Regen (n=17) EoT-Regen (n=23) 10 0 10 2 10 4 0 1 2 3 4 AIME Non-Regen (n=15) EoT-Regen (n=15) 10 0 10 2 10 4 0 10 20 30 GPQA Non-Regen (n=174) EoT-Regen (n=24) 10 0 10 2 10 4 0 100 200 GSM8K Non-Regen (n=1192) EoT-Regen (n=127) 10 0 10 2 10 4 0 10 20 30 40 MATH Non-Regen (n=298) EoT-Regen (n=202) 10 0 10 2 10 4 #Ans 0 2 4 6 8 Count DynaSoR Non-Regen (n=24) EoT-Regen (n=16) 10 0 10 2 10 4 #Ans 0 1 2 3 4 Non-Regen (n=17) EoT-Regen (n=13) 10 0 10 2 10 4 #Ans 0 10 20 Non-Regen (n=178) EoT-Regen (n=20) 10 0 10 2 10 4 #Ans 0 100 200 Non-Regen (n=1306) EoT-Regen (n=13) 10 0 10 2 10 4 #Ans 0 20 40 60 80 Non-Regen (n=403) EoT-Regen (n=97) (d) QwQ-32B Figure 11: Histograms of answer-phase length (#Ans) for each model across early-exit methods (DEER, DynaSoR) and datasets. Each histogram is split into samples with and without EoT regeneration. The dashed vertical line marks the answering-phase length threshold (1000 tokens) used to trigger regeneration. ReasoningPre-regenPost-regen Dataset ̄n box END-BOX 5 ̄n box END-BOX 5 ̄n box END-BOX 5 Full-CoT MATH0.720.60–2.530.86 AMC0.550.45–1.070.90 AIME0.230.23– 30.130.77 GSM8K 0.210.06–0.950.76 GPQA0.110.11–0.210.19 DEER, spurious MATH0.000.00 1.200.601.000.60 AMC0.000.00 1.001.001.001.00 AIME0.000.00 1.001.001.001.00 GPQA0.000.00 1.001.001.001.00 DynaSoR, spurious MATH0.000.00 1.401.001.001.00 AMC0.000.00 2.501.001.001.00 AIME0.000.00 1.501.001.001.00 (a) R1-Distill-1.5B ReasoningPre-regenPost-regen Dataset ̄n box END-BOX 5 ̄n box END-BOX 5 ̄n box END-BOX 5 Full-CoT MATH1.240.87– 1.210.96 AMC1.180.90– 1.270.97 AIME0.700.50– 1.330.90 GSM8K 0.980.45– 1.070.82 GPQA0.480.47– 1.071.00 DEER, samples with spurious CoT termination MATH0.000.00 1.170.92 1.171.00 GSM8K 0.000.00 1.000.45 1.090.82 GPQA0.000.00 1.001.00 1.141.00 DynaSoR, samples with spurious CoT termination MATH0.000.00 1.300.91 1.040.91 AMC0.000.00 1.001.00 1.001.00 AIME0.000.00 2.001.00 1.001.00 GSM8K 0.000.00 1.180.82 1.000.82 GPQA0.000.00 1.001.00 0.830.83 (b) Qwen3-14B ReasoningPre-regenPost-regen Dataset ̄n box END-BOX 5 ̄n box END-BOX 5 ̄n box END-BOX 5 Full-CoT MATH0.930.78– 1.060.87 AMC0.880.75– 1.000.95 AIME0.470.43– 1.170.83 GSM8K 0.830.24– 1.000.50 GPQA0.090.09– 0.870.87 DEER, samples with spurious CoT termination MATH0.000.00 1.060.93 1.020.93 AMC0.000.00 1.000.89 1.000.89 AIME0.000.00 0.750.75 0.880.88 GSM8K 0.000.00 0.980.43 1.000.70 GPQA0.000.00 0.940.94 1.001.00 DynaSoR, samples with spurious CoT termination MATH0.000.00 1.000.91 0.990.92 AMC0.000.00 1.000.94 1.000.94 AIME0.000.00 1.081.00 1.001.00 GSM8K 0.000.00 1.010.32 1.000.48 GPQA0.000.00 0.810.81 0.970.97 (c) R1-Distill-14B ThinkingPre-regenPost-regen Dataset ̄n box END-BOX 5 ̄n box END-BOX 5 ̄n box END-BOX 5 Full-CoT MATH1.310.87– 1.380.93 AMC1.120.80– 1.100.95 AIME0.770.47– 1.330.90 GSM8K 0.890.49– 1.010.80 GPQA0.160.16– 1.690.98 DEER, samples with spurious CoT termination MATH0.000.00 1.810.97 1.000.93 AMC0.000.00 1.480.91 1.001.00 AIME0.000.00 1.671.00 1.001.00 GSM8K 0.000.00 1.350.77 1.000.80 GPQA0.000.00 0.880.75 1.001.00 DynaSoR, samples with spurious CoT termination MATH0.000.00 2.420.99 1.000.92 AMC0.000.00 1.381.00 1.001.00 AIME0.000.00 2.001.00 0.920.92 GSM8K 0.000.00 1.620.77 1.000.77 GPQA0.000.00 1.400.95 0.950.95 (d) QwQ-32B Table 15: Quantitative analysis of boxed-before-EoT behavior across models and generation segments. We report results for the reasoning, pre-regen, and post-regen segments. For Full-CoT, the answering phase is not further segmented, and its values are reported in the post-regen columns. ̄n box is the mean number of boxed expressions per sample within the segment.END-BOX 5 is the fraction of samples whose final boxed expression ends within 5 characters of the segment boundary. Acc (%,↑)ERR (%,↓)#Ans (↓) DatasetN DEER +EAB(α=2) +EAB(α=4) DEER +EAB(α=2) +EAB(α=4) DEER +EAB(α=2) +EAB(α=4) R1-Distill-1.5B Samples with spurious CoT termination AIME20.00.00.0100.00.00.0589607590 AMC10.00.00.0100.00.00.053015517 GPQA250.050.00.0100.050.050.0600338276 MATH520.020.020.0100.020.00.0723575448 Samples without spurious CoT termination AIME2825.025.017.90.00.00.01,8841,4611,264 AMC3951.356.443.60.00.00.02,104232226 GPQA1965.110.25.60.00.03.17646231,041 GSM8K 1,31968.572.673.10.00.00.1274255247 MATH49565.979.276.80.00.00.0394275321 Qwen3-14B Samples with spurious CoT termination GPQA7100.057.157.1100.014.30.04,4342,718740 GSM8K11100.081.881.8100.00.00.01,329271247 MATH1291.775.091.7100.08.30.02,5921,831594 Samples without spurious CoT termination AIME3073.370.073.30.00.00.01,3331,5291,159 AMC4097.597.595.00.00.00.0811795804 GPQA19153.457.652.40.00.00.01,439851692 GSM8K 1,30895.595.695.30.00.10.0244252223 MATH48893.292.893.00.00.00.0593638554 R1-Distill-14B Samples with spurious CoT termination AIME850.037.550.0100.037.512.55,6688,5411,353 AMC9100.0100.0100.0100.044.422.21,4858471,288 GPQA1877.877.872.2100.038.911.11,4891,262783 GSM8K4484.172.768.2100.050.04.51,042611442 MATH9991.987.979.8100.061.619.21,6781,385777 Samples without spurious CoT termination AIME2250.054.554.50.04.50.02,4082,6221,885 AMC3180.680.683.90.03.26.51,293951435 GPQA18050.651.752.20.01.10.0456349256 GSM8K 1,27591.592.993.70.00.20.0233193197 MATH40186.390.389.50.02.51.0627455376 QwQ-32B Samples with spurious CoT termination AIME1580.073.366.7100.080.046.75,2426,5466,694 AMC23100.091.391.3100.078.352.25,2704,7655,458 GPQA2475.066.758.3100.054.225.04,4854,0924,248 GSM8K12792.191.392.1100.068.544.91,4551,3271,227 MATH20296.593.192.6100.083.257.42,9272,9922,737 Samples without spurious CoT termination AIME1553.353.353.30.06.76.74,5172,9783,379 AMC1794.194.194.10.00.00.0452513480 GPQA17461.561.559.80.01.71.18539441,085 GSM8K 1,19296.896.697.00.01.80.4188204193 MATH29889.390.390.60.03.03.7877836758 Table 16: Grouped comparison of EAB effects with DEER across all models and datasets atα = 2andα = 4, separating samples by whether their reasoning phase spuriously terminated in the unbiased (α = 0) setting. MethodMATHAMCAIMEGSM8KGPQA Acc ERRThkAns Acc ERRThkAns Acc ERRThkAns Acc ERRThkAns Acc ERRThkAns R1-Distill-1.5B No-CoT † 62.00.4– 1,121 52.55.0– 2,648 13.3 10.0– 5,543 33.10.9–268 4.50.5– 1,110 Full-CoT † 79.60.0 4,192604 72.50.0 6,111608 23.30.0 8,823 1,550 76.00.0 1,0363077.10.0 8,013775 DEER † 66.80.3 2,079428 50.00.0 4,013 1,143 26.70.0 7,308 1,594 68.10.0398276 3.50.0 6,987701 DEER65.41.0 2,079397 50.02.5 4,013 2,065 23.36.7 7,308 1,798 68.50.0398274 5.61.0 6,987763 +EAB (2) 78.60.2 2,079278 55.00.0 4,013230 23.30.0 7,308 1,404 72.60.0398255 10.60.5 6,987620 +EAB (4) 76.20.0 2,079322 42.50.0 4,013221 16.70.0 7,308 1,219 73.10.1398247 5.63.5 6,987 1,034 DynaSoR † 55.60.3 2,076647 55.04.2 3,532 1,352 26.70.0 5,984 1,949 70.40.0466307 7.10.0 2,675 1,325 DynaSoR 57.41.0 2,076536 47.55.0 3,532 1,560 33.36.7 5,984 2,396 70.80.0466298 7.10.0 2,675844 +EAB (2) 75.80.0 2,076322 55.00.0 3,532623 30.00.0 5,984359 74.80.0466279 10.10.0 2,675344 +EAB (4) 71.00.0 2,076396 45.05.0 3,532877 26.70.0 5,984891 72.70.0466262 5.60.5 2,675611 R1-Distill-14B No-CoT † 75.20.0–681 60.00.0– 1,254 26.70.0– 2,250 89.30.0–252 37.40.0–582 Full-CoT † 91.40.0 3,006449 87.50.0 4,242454 60.00.0 7,349 1,258 94.70.0 1,129215 51.00.0 5,054361 DEER † 87.8 23.9 1,664 1,128 80.0 21.4 3,446 1,920 50.0 33.3 6,207 2,794 91.45.0524231 55.6 23.8 4,470635 DEER87.4 19.8 1,664835 85.0 22.5 3,446 1,336 50.0 26.7 6,207 3,277 91.23.3524260 53.09.1 4,470550 +EAB (2) 89.8 14.2 1,664639 85.0 12.5 3,446928 50.0 13.3 6,207 4,201 92.31.9524207 54.04.5 4,470432 +EAB (4) 87.64.6 1,664455 87.5 10.0 3,446627 53.33.3 6,207 1,743 92.90.2524205 54.01.0 4,470304 DynaSoR † 86.2 48.2 1,615 1,371 85.0 60.9 2,531 2,047 53.3 63.2 4,975 2,547 91.3 23.5759361 47.5 22.0 1,539 1,701 DynaSoR 87.2 35.8 1,615 1,024 87.5 40.0 2,531 2,333 56.7 40.0 4,975 3,931 91.18.0759303 47.5 18.2 1,539 1,212 +EAB (2) 85.6 30.0 1,615984 87.5 40.0 2,531 1,958 46.7 33.3 4,975 3,522 90.24.5759268 50.0 14.6 1,539982 +EAB (4) 83.6 16.4 1,615767 82.5 20.0 2,531 1,140 50.0 16.7 4,975 2,452 88.80.5759226 44.9 11.6 1,539739 Qwen3-14B No-CoT † 86.00.0– 1,173 70.00.0– 3,540 30.00.0– 6,598 95.80.0–282 54.00.0– 2,922 Full-CoT † 94.80.0 3,801641 97.50.0 5,931846 63.30.0 10,009 1,440 96.20.0 1,339311 65.70.0 6,467870 DEER † 93.22.5 2,307667 97.50.0 5,043843 73.30.0 8,499 1,256 95.50.7651266 57.17.1 2,214 1,952 DEER93.22.4 2,307641 97.50.0 5,043811 73.30.0 8,499 1,333 95.50.8651253 55.13.5 2,214 1,545 +EAB (2) 92.40.2 2,307667 97.50.0 5,043795 70.00.0 8,499 1,529 95.50.1651252 57.60.5 2,214917 +EAB (4) 93.00.0 2,307555 95.00.0 5,043804 73.30.0 8,499 1,159 95.20.0651223 52.50.0 2,214694 DynaSoR † 90.84.8 1,236996 75.02.9 2,540 1,419 46.74.8 4,376 3,429 95.21.2850306 58.64.1 1,019 2,107 DynaSoR 90.44.6 1,236909 82.52.5 2,540 1,642 43.33.3 4,376 4,487 95.30.8850302 55.13.0 1,019 1,590 +EAB (2) 90.21.4 1,236805 82.52.5 2,540 1,194 50.03.3 4,376 3,167 95.50.1850286 54.50.0 1,019 1,034 +EAB (4) 88.40.2 1,236670 87.50.0 2,540965 40.00.0 4,376 3,160 95.50.0850263 56.60.0 1,019768 QwQ-32B No-CoT † 92.8 57.9– 3,038 92.5 80.0– 6,001 60.0 70.0– 10,427 95.9 57.2– 1,054 57.1 36.4– 3,451 Full-CoT † 94.20.0 3,354554 92.50.0 5,611600 66.70.0 8,039 1,508 96.40.0 1,226200 65.70.0 6,110319 DEER † 94.4 49.4 1,300 1,931 90.0 62.1 2,644 3,263 56.7 66.7 5,287 4,439 96.3 12.9624315 63.1 23.2 3,679 1,603 DEER92.2 40.4 1,300 1,705 97.5 57.5 2,644 3,222 66.7 50.0 5,287 4,880 96.49.6624310 63.1 12.1 3,679 1,293 +EAB (2) 91.4 35.4 1,300 1,707 92.5 45.0 2,644 2,958 63.3 43.3 5,287 4,762 96.18.2624312 62.18.1 3,679 1,326 +EAB (4) 91.4 25.4 1,300 1,558 92.5 30.0 2,644 3,342 60.0 26.7 5,287 5,037 96.54.7624293 59.64.0 3,679 1,469 DynaSoR † 93.8 24.5 1,360 1,557 90.0 31.4 2,500 2,748 63.3 50.0 4,623 4,786 96.71.6831210 63.19.6 1,145 1,810 DynaSoR 91.6 19.4 1,360 1,369 97.5 40.0 2,500 2,766 73.3 43.3 4,623 4,428 96.41.0831209 62.1 10.1 1,145 1,504 +EAB (2) 91.4 14.8 1,360 1,269 87.5 25.0 2,500 2,206 66.7 30.0 4,623 3,391 96.10.9831203 59.63.5 1,145 1,006 +EAB (4) 92.0 12.6 1,360 1,071 87.5 15.0 2,500 2,091 63.3 33.3 4,623 3,298 96.40.8831206 57.15.6 1,145982 Table 17: Per-dataset effect of EAB (transformers backend). Acc = accuracy (%,↑), ERR = EoT regeneration rate (%,↓), Thk = the number of reasoning-phase tokens (↓), Ans = the number of answering-phase tokens (↓). The number in parentheses after +EAB denotes the value ofα.†denotes vLLM backend (No-CoT † , Full-CoT † , DEER † , DynaSoR † ). E Case Studies of Correctness Changes under EAB As shown in Tables 4 and 5, applying EAB changes accuracy, ERR, and answering-phase length to- gether. As illustrated in Figures 14 and 15, EAB reduces spurious CoT termination and shortens the answering phase in many cases. We also exam- ine the samples whose correctness changes, and attach the traces before and after EAB together with the spurious CoT termination flag and the answering-phase length. Long traces are truncated for readability. We group the cases by the cause of the correct- ness change into four types. The first is when the answering phase fails to terminate and reaches the generation budget. The second is when an EoT regeneration under EAB nonetheless leads to the correct answer. The third is when the reasoning trace itself changes and the final answer changes with it. The fourth is when the reasoning is pre- served but a scoring artifact makes the correctness judgment unreliable. The first three reflect changes in model behavior, whereas the last is a measure- ment effect of the string-match criterion, so we present it separately. For each type, we present rep- resentative samples grouped by model, primarily using DEER results on MATH. E.1 Termination Failure and Budget Exhaustion The first type is when the answering phase fails to terminate and reaches the generation budget. In some cases, EAB shortens a run that reached the maximum length to only a few tokens, whereas in others the answering phase instead grows to the budget limit. When a run reaches the maxi- mum length, with or without EAB, the model some- times performs genuine reasoning and sometimes becomes trapped in a repeated “Wait, let me con- tinue” loop. Most such cases are scored as incorrect because the answering phase contains no boxed ex- pression. As this type has many cases, we sample representative ones from R1-Distill-14B and QwQ- 32B. R1-Distill-14B / DEER / MATH (IDX=240) CorrectnessSpurious CoT#Ans Without EABIncorrectNo8192 With EABCorrectNo24 [pred @ alpha=0] [pred @ alpha=4] 116 [gold] 116 [Question] In the diagram, $PT$ is parallel to $QR.$ What is the measure ,→ of $ PQR$ in degrees? [asy] draw((0,0)--(10,0),black+linewidth(1)); draw((0,0)--(10,0),MidArrow); draw((10,0)--(20,0),black+linewidth(1)); draw((0,0)--(-7,10)--(7,10)--(10,0),black+linewidth(1)); draw((-5,10)--(7,10),MidArrow); label("$x $",(-6,10),SE); label("$2x $",(7,10),SW); label("$128 $",(10,0),NE); label("$P$",(-7,10),N); label("$T$",(7,10),N); label("$R$",(10,0),S); label("$Q$",(0,0),S); [/asy] [gen @ alpha=0] - The external angle at P (x) corresponds to the internal ,→ angle at Q (angle PQR). - The external angle at T (2x) corresponds to the internal ,→ angle at R. Given that angle at R is 128 degrees, 2x = 128 => x = 64 ,→ degrees. Therefore, angle PQR = x = 64 degrees. But in the trapezoid, angle PQR + ... [truncated] ... 16 = 64 degrees. Therefore, the measure of angle PQR is 64 degrees. But wait, in the Asymptote code, angle at R is labeled as 128 ,→ degrees, which is consistent with angle R = 2x = 128. Therefore, x = 64, angle PQR = 64 degrees. Yes, that makes sense. Therefore, the measure of angle PQR is 6 [gen @ alpha=4] -parallel lines and transversals, the measure of angle PQR is ,→ 116 degrees. R1-Distill-14B / DEER / MATH (IDX=296) CorrectnessSpurious CoT#Ans Without EABCorrectYes2370 With EABIncorrectNo8192 [pred @ alpha=0] B [pred @ alpha=4] [gold] (B) [Question] Assume $0 < r < 3$. Below are five equations for $x$. Which ,→ equation has the largest solution $x$? $(A)\ 3(1 + r)^x = 7 (B)\ 3(1 + r/10) ,→ x = 7 (C)\ 3(1 + 2r)^x = 7$ $(D)\ ,→ 3(1 + r)^x = 7 (E)\ 3(1 + 1/r)^x = 7 ,→ $ [gen @ alpha=0] Wait, hold on, at r=2, E's base is 1.5, which is larger than B ,→'s 1.2. So, B is still the smallest. What about when r approaches 0? Let's say r approaches 0 from ,→ the right. A) 1 + r approaches 1 B) 1 + r/10 approaches 1 C) 1 + 2r approaches 1 D) 1 + sqrt(r) approaches 1 E) 1 + 1/r approaches in ... [truncated] ... - For $ 0 < r < 3 $, the expression $ 1 + r/10 $ (from ,→ equation B) is always the smallest base. This is because $ ,→ r/10 $ is a smaller increment compared to $ r $, $ 2r ,→ $, $ r $, or $ 1/r $. Thus, equation B has the smallest base, leading to the largest ,→ $ x $. \[ B \] [gen @ alpha=4] Wait, let me continue. Wait, let me continue. Wait, let me continue. Wait, let me continue. ... [truncated] ... inue. Wait, let me continue. Wait, QwQ-32B / DEER / MATH (IDX=232) CorrectnessSpurious CoT#Ans Without EABCorrectYes4728 With EABIncorrectNo8192 [pred @ alpha=0] 19972 [pred @ alpha=2] [gold] 19972 [Question] If $ω^1997 = 1$ and $ω ≠ 1,$ then evaluate \[ 11 + ω + 11 + ω^2 + … + \ ,→ frac11 + ω^1997.\] [gen @ alpha=0] Wait, hold on. The problem says "omega^1997 =1 and omega !=1", ,→ so omega itself is a primitive root. But the sum is from k ,→ =1 to 1997 of 1/(1 + omega^k). Since omega^1997 =1, the ,→ last term is 1/(1 + 1) = 1/2. So the sum includes all the ,→ 1997th roots of unity except 1? Wait no, actually, when k ,→ =1997, o ... [truncated] ... 1997 · 1 = 1997. \] Thus, \[ _k=0^1996 1-1 - ω^k = 1997-2. \] Multiplying both sides by $-1$ to adjust the sign in the ,→ denominator: \[ _k=0^1996 11 + ω^k = 19972. \] Therefore, the sum $S$ is: \[ 19972. \] [gen @ alpha=2] Wait, hold on. The problem says omega1997 =1 and omega !=1, so ,→ omega is a primitive 1997th root of unity. The sum is from ,→ k=1 to 1997 of 1/(1 + omega^k). But when k=1997, omega ,→ ^1997 =1, so the last term is 1/(1 +1)=1/2. So the sum ,→ includes that term. So the total terms are 1997 terms. Hmm, how do ... [truncated] ... which is undefined. Hmm, so n=2 is problematic because 1/(1 + omega) is undefined. Thus, n must be odd to avoid such terms. Since 1997 is odd, ,→ that's okay. Another test with n=5. Let me see. But maybe it's tedious, but given that the n=3 case works, and ,→ the algebra holds, I think the formula QwQ-32B / DYNASOR / MATH (IDX=80) CorrectnessSpurious CoT#Ans Without EABCorrectYes8192 With EABIncorrectNo8192 [pred @ alpha=0] 501 [pred @ alpha=2] [gold] 501 [Question] Find the number of integer values of $k$ in the closed ,→ interval $[-500,500]$ for which the equation $ (kx)=2\ ,→ log(x+2)$ has exactly one real solution. [gen @ alpha=0] So, the equation has a repeated root at x = 2. Now, check the ,→ domain conditions: 1. x > -2: 2 > -2, which is true. 2. kx > 0: 8*2 = 16 > 0, which is true. Therefore, x = 2 is a valid solution. Since it's a repeated ,→ root, there's exactly one real solution. So, k = 8 is ,→ acceptable. Therefore, the ... [truncated] ... **Domain Conditions**: - $ kx > 0 $ - $ x + 2 > 0 x > -2 $ 2. **Rewriting the Equation**: - The equation $ (kx) = 2 (x + 2)$ can be rewritten ,→ using logarithm properties as $ kx = (x + 2)^2 $. - This simplifies to the quadratic equation \( x^2 + (4 - k ,→ )x + 4 = [gen @ alpha=2] So, the equation has a repeated root at x = 2. Now, check the ,→ domain conditions: 1. x > -2: 2 > -2 is true. 2. kx > 0: k = 8, so 8*2 = 16 > 0, which is true. Therefore, x = 2 is a valid solution. Since the quadratic has ,→ a repeated root here, there is exactly one real solution. ,→ So, k = 8 is valid. ... [truncated] ... in with another perspective. Alternatively, maybe I made a miscalculation in the number of ,→ integers from -500 to -1. Let me count them: The number of integers from a to b inclusive is b -a +1. So ,→ from -500 to -1: -1 - (-500) +1 = -1 +500 +1 =500. So that's correct. So 500 ,→ numbers. Plus 1 for k=8, QwQ-32B / DYNASOR / GPQA (IDX=55) CorrectnessSpurious CoT#Ans Without EABIncorrectNo8192 With EABCorrectNo383 [pred @ alpha=0] [pred @ alpha=2] A [gold] A [Question] The |3,0,0 state in the standard notation |n,l,m ,→ of the H -atom in the non-relativistic theory decays to ,→ the state |1,0,0 via two dipole transition. The ,→ transition route and the corresponding probability are (use ,→ latex), A. |3,0,0 →|2,1,0 →|1,0,0 ... [truncated] ... ightarrow|1,0,0 and 14 C. |3,0,0 →|2,1,-1 →|1,0,0\ ,→ rangle and 14 D. →|2,1,0 →|1,0,0 ,→ and 23 Please reason step-by-step and put your choice letter without ,→ any other text with in the end. [gen @ alpha=0] The decay from |3,0,0 to |1,0,0 via two dipole transitions ,→ requires intermediate states that satisfy the selection ,→ rules Deltal = +/-1 and Deltam = 0, +/-1. 1. **First Transition (3,0,0 -> intermediate):** - Initial l = 0, so intermediate l must be 1 (Deltal = +1). - Possible m values: 0, +1 ... [truncated] ... ser must choose one, but the options are separate, I'l choose ,→ B and C. But since I can only select one, perhaps the ,→ question intended to have both, but the options are ,→ separate. Alternatively, maybe the question's answer is D, but D's ,→ probability is 2/3. Wait, option D is written as "->|2,1,0-> [gen @ alpha=2] The decay from |3,0,0 to |1,0,0 via two dipole transitions ,→ must follow the selection rules Deltal = +/-1 and Deltam = ,→ 0, +/-1 for each step. **First Transition (3,0,0 -> intermediate state):** - Initial l = 0 => intermediate l must be 1 (Deltal = +1). - Possible m values: 0, +/-1 (Deltam = 0, +/-1 ... [truncated] ... is the product divided by the normalization. However, the ,→ problem states the total probability is given as fractions. ,→ Among the options, **A** and **D** have the correct path. ,→ However, **D** is missing the initial state in its notation ,→ . **A** is correctly formatted with probability 1/3. A E.2 Correct Answer Despite EoT Regeneration For this type, we select cases where the run with spurious CoT termination is correct, while the run without it is incorrect. These cases illustrate that changes in spurious CoT termination do not always translate into corresponding changes in correctness. R1-Distill-14B / DEER / MATH (IDX=180) CorrectnessSpurious CoT#Ans Without EABIncorrectNo8192 With EABCorrectYes956 [pred @ alpha=0] [pred @ alpha=4] 3 [gold] 3 [Question] How many positive two-digit integers are factors of both 100 ,→ and 150? [gen @ alpha=0] Wait, hold on. 10 is a two-digit number, 25 is also two-digit, ,→ and 50 is two-digit as well. So, that's three numbers. But ,→ wait, 50 is a two-digit number? Hmm, 50 is written as ,→'50', which is two digits, right? So, yes, 50 is a two- ,→ digit number. But hold on, the question says "positive two-digit int ... [truncated] ... rs as from 10 to 99, but 50 is definitely within that range. So, yeah, the answer is three. But wait, let me think again. Maybe I should consider that 50 ,→ is a two-digit number, so it's included. Yes, 50 is a two-digit number, so it's included. So, the answer is three. Wait, but I'm a bit unsure [gen @ alpha=4] Wait, hold on, I almost forgot. The two-digit factors of 50 ,→ are 10, 25, and 50. But wait, 50 is a two-digit number, ,→ right? It's 5 and 0, so yes, it's a two-digit number. So, ,→ 10, 25, and 50 are all two-digit factors of 50. Therefore, ,→ these are also the two-digit numbers that are factors of ,→ both 100 a ... [truncated] ... So, the total number of positive two-digit integers that are ,→ factors of both 100 and 150 is 3. **Final Answer** The number of positive two-digit integers that are factors of ,→ both 100 and 150 is 3. </think> To determine how many positive two-digit integers are factors ,→ of both 100 and 150, we start by finding the greatest ,→ common divisor (GCD) of 100 and 150. First, we perform the prime factorization of each number: - The prime factorization of 100 is $2^2 × 5^2$. - The prime factorization of 150 is $2 × 3 × 5^2$. The common prime factors are 2 and 5. Taking the lowest powers ,→ of these common prime factors, we get: - For 2: the lowest power is $2^1$. - For 5: the lowest power is $5^2$. Thus, the GCD is: \[ 2^1 × 5^2 = 2 × 25 = 50 \] Next, we list the factors of 50, which are 1, 2, 5, 10, 25, ,→ and 50. Among these, the two-digit factors are 10, 25, and ,→ 50. Therefore, the number of positive two-digit integers that are ,→ factors of both 100 and 150 is $ 3$. QwQ-32B / DEER / MATH (IDX=328) CorrectnessSpurious CoT#Ans Without EABIncorrectNo8192 With EABCorrectYes6345 [pred @ alpha=0] [pred @ alpha=2] 3 [gold] 3 [Question] Let $f(x)= (- 58 )^x $ be ,→ a function that is defined for all values of $x$ in $[0,\ ,→ infty)$ such that $f(x)$ is a real number. How many ,→ distinct values exist in the range of $f(x)$? [gen @ alpha=0] Wait, perhaps I misread the problem. Let me check again. The ,→ function is f(x) = floor[ (-5/8)^x ], defined for all x in ,→ [0, inf) where f(x) is real. Hmm. Since the base is ,→ negative, (-5/8)^x is real only when x is an integer or a ,→ fraction with an odd denominator. But since x is in [0, inf ,→ ), which in ... [truncated] ... denominator 2 is even, so (-5/8)^1/2 is imaginary. Therefore, excluded. x=1/3 is allowed, gives floor -1. x=2/3 is allowed, gives floor 0. x=3/5 is allowed, gives floor -1. x=4/5 is allowed, gives floor 0. Therefore, all non-integer x in the domain give either -1 or ,→ 0. Therefore, the [gen @ alpha=2] Wait, hold on. Maybe I misread the problem. Let me check again ,→ . The function is f(x) = floor[ (-5/8)^x ], defined for all ,→ x in [0, inf) where it's real. Hmm. Since the base is ,→ negative, (-5/8)^x is real only when x is an integer or a ,→ fraction with an odd denominator. But since x is in [0, inf ,→ ), whic ... [truncated] ... ion**: - The function $ f(x) $ takes the value 1 at $ x = 0 $, ,→ -1 for all odd $ x ≥ 1 $, and 0 for all even $ x \ ,→ geq 2 $. - Therefore, the distinct values in the range of $ f(x) $ ,→ are $ \ -1, 0, 1 \ $. The number of distinct values in the range of $ f(x) $ is ,→ $ 3$. E.3 Genuine Reasoning and Option-Selection Flips In these cases, applying EAB alters the reasoning trace itself rather than only facilitating the transi- tion to the answering phase, and correctness flips as a result. On MATH, this appears as a change in the arithmetic or geometric derivation, while on GPQA it appears more often as a change in the selected option. As this type has many cases, we sample representative ones from R1-Distill-14B and QwQ-32B. R1-Distill-14B / DEER / MATH (IDX=43) CorrectnessSpurious CoT#Ans Without EABCorrectYes4379 With EABIncorrectNo677 [pred @ alpha=0] 70 2 [pred @ alpha=4] 10 65 [gold] 70 2 [Question] For $0 ≤ x ≤ 40$ and $0 ≤ y ≤ 50,$ find the minimum ,→ value of \[ x^2 + 400 + y^2 + 900 + x^2 + y^2 - 80x ,→ - 100y + 4100.\] [gen @ alpha=0] Wait, so now we have: \[ x^2 + 400 + y^2 + 900 + (x - 40)^2 + (y - ,→ 50)^2 \] Which is the sum of distances from P to A, Q to B', and Q to C ,→ . But I still don't see how to connect these. Wait, maybe I can think of it as a path from A to P to Q to C, ,→ where P is on the x-axis and ... [truncated] ... = 20 2\) 2. $ 30^2 + 900 = 1800 = 30 2$ 3. $ (20 - 40)^2 + (30 - 50)^2 = 800 = 20 ,→ 2$ Summing these distances, we get: \[ 20 2 + 30 2 + 20 2 = 70 2 \] Thus, the minimum value of the given expression is $ ,→ 70 2$. [gen @ alpha=4] Wait, so I have this problem here where I need to find the ,→ minimum value of the expression: \[ x^2 + 400 + y^2 + 900 + x^2 + y^2 - 80x - ,→ 100y + 4100 \] for $0 ≤ x ≤ 40$ and $0 ≤ y ≤ 50$. Hmm, that ,→ looks a bit complicated, but maybe I can break it down. First, ... [truncated] ... nimal total distance. Calculating the distance between $(0, -20)$ and $(-40, 50) ,→ $: \[ (-40 - 0)^2 + (50 - (-20))^2 = (-40)^2 + (70)^2 ,→ = 1600 + 4900 = 6500 = 10 65 \] Therefore, the minimum value of the given expression is $10\ ,→ sqrt65$. 10 65 R1-Distill-14B / DEER / MATH (IDX=80) CorrectnessSpurious CoT#Ans Without EABIncorrectNo602 With EABCorrectNo418 [pred @ alpha=0] 500 [pred @ alpha=4] 501 [gold] 501 [Question] Find the number of integer values of $k$ in the closed ,→ interval $[-500,500]$ for which the equation $ (kx)=2\ ,→ log(x+2)$ has exactly one real solution. [gen @ alpha=0] To determine the number of integer values of $ k $ in the ,→ interval $[-500, 500]$ for which the equation $ (kx) ,→ = 2 (x + 2) $ has exactly one real solution, we ,→ analyze the equation step by step. 1. **Simplify the Equation**: Using logarithm properties, the equation simplifies to: ... [truncated] ... id. - Including $ k = -500 $ and $ k = 500 $, but $ k = ,→ -500 $ is invalid as shown, so we exclude it. Thus, the total number of valid integer values of $ k $ is ,→ 499 (negative integers from -499 to -1) plus 1 (for $ k = ,→ 8 $), resulting in 500 valid values. **Final Answer**: 500 [gen @ alpha=4] To determine the number of integer values of $ k $ in the ,→ interval $[-500, 500]$ for which the equation $ (kx) ,→ = 2 (x + 2)$ has exactly one real solution, we start by ,→ simplifying the given equation using logarithmic ,→ properties. The equation can be rewritten as: \[ (kx) = ((x + 2 ... [truncated] ... th roots are positive, leading to two valid solutions, which ,→ is not desired. Thus, the valid integer values of $ k $ are all negative ,→ integers in $[-500, 500]$ and $ k = 8 $. The number of ,→ negative integers is 500, and adding $ k = 8 $ gives a ,→ total of 501 integer values. \[ 501 \] R1-Distill-14B / DEER / MATH (IDX=120) CorrectnessSpurious CoT#Ans Without EABCorrectYes3917 With EABIncorrectYes1430 [pred @ alpha=0] even [pred @ alpha=4] odd [gold] even [Question] Let $f(x)$ be an odd function, and let $g(x)$ be an even ,→ function. Is $f(f(g(f(g(f(x))))))$ even, odd, or neither? Enter "odd", "even", or "neither". [gen @ alpha=0] Wait, no, not necessarily. Let me correct that. If I have g(f ,→ (-x)), since f is odd, f(-x) = -f(x). Then, since g is even ,→ , g(-f(x)) = g(f(x)). So, g(f(-x)) = g(f(x)). Therefore, ,→ the composition g(f(x)) is even because replacing x with -x ,→ doesn't change the value. So, g(f(x)) is even. But wait, in th ... [truncated] ... = x^3 \) (odd) and $ g(x) = x^2 $ (even), the composition ,→ results in an even function. - Using $ f(x) = (x) $ (odd) and $ g(x) = (x) $ ,→ (even), the composition results in an even function. 4. **Conclusion**: - The composition $ f(f(g(f(g(f(x)))))) $ is even. \[ even \] [gen @ alpha=4] Wait, let me try to figure this out step by step. So, I have ,→ the function $ f(f(g(f(g(f(x)))))) $. Let me break it ,→ down from the inside out. 1. **Innermost Function**: Start with $ f(x) $. Since $ f ,→ $ is odd, $ f(-x) = -f(x) $. 2. **Next Layer**: Apply $ g $ to $ f(x) $, so we have $ ... [truncated] ... f(g(f(g(f(-x))))) = f(g(f(g(f(x))))) $. This implies $ f(g( ,→ f(g(f(x))))) $ is even. However, upon re-evaluating the final step, we realize that ,→ applying an odd function to an even function results in an ,→ odd function. Therefore, the composition $ f(f(g(f(g(f(x)) ,→ )))) $ is odd. \[ odd \] QwQ-32B / DYNASOR / MATH (IDX=302) CorrectnessSpurious CoT#Ans Without EABIncorrectNo440 With EABCorrectNo546 [pred @ alpha=0] 135 [pred @ alpha=2] 500 [gold] 5 [Question] When converting $ 3111111$ to a decimal, the decimal ,→ turns out to be a repeating decimal. How many digits repeat ,→ in this repeating decimal? For example, if you get the repeating decimal $0. ,→ 123,$ then your answer should be $3,$ and if you get $0 ,→ .436 7,$ your answer should be $1.$ [gen @ alpha=0] To determine the number of repeating digits in the decimal ,→ expansion of $ 3111111$, we first confirm that ,→ the fraction is in its simplest form. Since 31 does not ,→ divide 11111 (as shown by division with remainder), the ,→ fraction cannot be simplified further. Next, we need to find the perio ... [truncated] ... \). - The period of $ 1271$ is 135 because $10^135 \ ,→ equiv 1 271$ (the smallest such exponent). The LCM of 5 and 135 is 135. Therefore, the period of $ ,→ 3111111$ is also 135, as multiplying by 31 (which is ,→ coprime to 11111) does not change the period. \[ 135 \] [gen @ alpha=2] To determine the number of repeating digits in the decimal ,→ expansion of $ 3111111$, we first note that the ,→ fraction is in its simplest form since 31 and 11111 share ,→ no common factors other than 1. The length of the repeating ,→ period of a fraction $ ab$ is given by the ,→ smallest pos ... [truncated] ... isfying the condition is 500. This is done by checking powers ,→ of 10 modulo 11111 for exponents that are divisors of 11110 ,→ until the congruence holds. The smallest such $k$ is ,→ found to be 500. Thus, the number of repeating digits in the decimal expansion ,→ of $ 3111111$ is $ 500$. QwQ-32B / DEER / MATH (IDX=166) CorrectnessSpurious CoT#Ans Without EABCorrectYes7270 With EABIncorrectNo907 [pred @ alpha=0] 2707 [pred @ alpha=2] 1807 [gold] 2707 degrees [Question] A square and a regular heptagon are coplanar and share a ,→ common side $ AD$, as shown. What is the degree ,→ measure of angle $BAC$? Express your answer as a common ,→ fraction. [asy] for(int i=0; i <=7; ++i) draw(dir(360*i/7+90)--dir(360*(i+1)/7+90)); pair A = dir(360*3/7+90); pair F = dir(360*4/7+90); pair C = A+dir(90)*(F-A); pair D = C+F-A; pair B = dir(360*2/7+90); draw(A--C--D--F); label("$A$",A,S); label("$B$",B,W); label("$C$",C,SE); label("$D$",F,S); [/asy] [gen @ alpha=0] Okay, let me try to approach this step by step. First, since ,→ both the square and the regular heptagon share the side AD, ,→ we need to figure out the positions of points A, B, C, D ,→ in such a way that AD is a side of both polygons. The angle ,→ we need is angle BAC, so points B, A, and C are key here. Fir ... [truncated] ... ween $ BA $ and $ AC $ is the ,→ internal angle at $ A $ minus 90 degrees. - Therefore, $ BAC = 900 7 - 90 ,→ = 900 - 630 7 = 270 7 $ ,→ . Thus, the degree measure of angle $ BAC $ is $ \ ,→ dfrac2707 $. [gen @ alpha=2] Okay, let's tackle this problem step by step. We need to find ,→ the measure of angle $ BAC $ where a square and a regular ,→ heptagon share a common side $ AD $. First, let's understand the structure. A regular heptagon has ,→ seven equal sides and angles, each internal angle being $ ,→ ... [truncated] ... these vectors using the dot product formula: \[ θ = AB · \ ,→ overrightarrowAC| AB| | ,→ AC| \] - After calculations, the angle \( BAC $ is found to be $ \ ,→ frac180 7 $. **Final Answer** \[ 1807 \] E.4 Scoring Artifacts In these cases, there is no parsable answer in a boxed expression, while the value appearing in the free-form text may be either correct or incorrect. A related situation arises when the model includes a unit in the answer that the parser fails to match, even though the prompt neither requires nor pro- hibits including a unit. These cases primarily re- flect final-answer formatting and limitations of the checker rather than the model’s reasoning capabil- ity, so we treat them as scoring artifacts rather than changes in model behavior. We sample representa- tive ones from R1-Distill-14B. R1-Distill-14B / DEER / MATH (IDX=24) CorrectnessSpurious CoT#Ans Without EABIncorrectNo573 With EABCorrectNo154 [pred @ alpha=0] [pred @ alpha=4] 10 [gold] 10 [Question] A worker receives an annual wage of $\$20,000$, which he ,→ always deposits into a savings account at the end of the ,→ year. By the end of the third year (when he makes the third ,→ deposit), he wants to have at least $\$66,200$ in the ,→ account to finance the purchase of a house. What is the ,→ minimal compound interest rate that the savings account ,→ must provide? Express your answer as a percentage, but do ,→ not include the percent sign. [gen @ alpha=0] The minimal compound interest rate required is 10%. **Step-by-Step Explanation:** 1. **Identify the Deposits and Their Timeframes:** - The worker makes three deposits of $20,000 each at the ,→ end of each year. - The first deposit earns interest for 2 years. - The second deposit earns intere ... [truncated] ... vant positive root is $ x = 1.1 $, leading to $ r = 0.1 $ ,→ or 10%. 6. **Verify the Solution:** - Calculate the future values with $ r = 0.1 $: - $ FV_1 = 24,200 $ - $ FV_2 = 22,000 $ - $ FV_3 = 20,000 $ - Total = $66,200, which meets the requirement. **Answer:** 10% [gen @ alpha=4] The minimal compound interest rate required is 10%. To determine the minimal compound interest rate needed for the ,→ worker to accumulate at least $66,200 by the end of the ,→ third year, we analyze the future value of each annual ,→ deposit. The worker makes three deposits of $20,000 each, ,→ with each depos ... [truncated] ... pound interest, we calculate the future value of each deposit ,→ and sum them up. Setting this total equal to $66,200, we ,→ solve for the interest rate $ r $. Through algebraic ,→ manipulation and solving the resulting quadratic equation, ,→ we find that the minimal interest rate required is 10%. 10 R1-Distill-14B / DEER / MATH (IDX=169) CorrectnessSpurious CoT#Ans Without EABCorrectNo274 With EABIncorrectNo183 [pred @ alpha=0] 12 [pred @ alpha=4] [gold] 12 [Question] What is the number of square units in the area of a triangle ,→ whose sides measure 5, 5 and 6 units? [gen @ alpha=0] To find the area of a triangle with sides measuring 5, 5, and ,→ 6 units, we can use Heron's formula. Here's a step-by-step ,→ explanation: 1. **Identify the sides of the triangle:** - Let $ a = 5 $ units - Let $ b = 5 $ units - Let $ c = 6 $ units 2. **Calculate the semi-perimeter ($ s ... [truncated] ... s \] 3. **Apply Heron's formula to find the area (\( A $):** \[ A = s(s - a)(s - b)(s - c) = 8(8 - 5)(8 - 5)(8 ,→ - 6) = 8 × 3 × 3 × 2 = 144 = ,→ 12 square units \] **Final Answer:** The area of the triangle is $ 12$ ,→ square units. [gen @ alpha=4] To find the area of a triangle with sides measuring 5, 5, and ,→ 6 units, we can use Heron's formula. 1. **Calculate the semi-perimeter (s):** \[ s = 5 + 5 + 62 = 162 = 8 \] 2. **Apply Heron's formula:** \[ Area = s(s - a)(s - b)(s - c) = 8(8 - ,→ 5)(8 - 5)(8 - 6) = 8 × 3 × 3 × 2 = \ ,→ sqrt144 = 12 \] **The area of the triangle is 12 square units.** R1-Distill-14B / DEER / MATH (IDX=79) CorrectnessSpurious CoT#Ans Without EABIncorrectNo115 With EABCorrectNo126 [pred @ alpha=0] [pred @ alpha=4] 2 [gold] 2 [Question] For what value of $x$ is $2^3·3^x=72$? [gen @ alpha=0] To solve for $ x $ in the equation $ 2^3 · 3^x = 72 $: 1. Simplify $ 2^3 $ to get 8. 2. The equation becomes $ 8 · 3^x = 72 $. 3. Divide both sides by 8: $ 3^x = 9 $. 4. Recognize that $ 9 = 3^2 $, so $ x = 2 $. **Answer:** $ x = 2 $ [gen @ alpha=4] To solve the equation $2^3 · 3^x = 72$, we start by ,→ simplifying $2^3$ to 8. This gives us: \[ 8 · 3^x = 72 \] Next, we isolate $3^x$ by dividing both sides by 8: \[ 3^x = 728 = 9 \] Recognizing that $9$ is $3^2$, we find: \[ x = 2 \] **Answer:** $ 2$ R1-Distill-14B / DEER / GSM8K (IDX=221) CorrectnessSpurious CoT#Ans Without EABIncorrectNo290 With EABCorrectNo249 [pred @ alpha=0] 90\ liters [pred @ alpha=4] 90 liters [gold] 90 [Question] A water tank is filled with 120 liters of water. Celine used ,→ 90 liters of water from the tank to water her small garden. ,→ She was then able to collect rainwater that is twice as ,→ much as what was left. How many liters of water are in the ,→ tank now? [gen @ alpha=0] Let's break down the problem step by step: 1. **Initial Amount of Water in the Tank:** - The tank is initially filled with **120 liters** of water ,→ . 2. **Water Used by Celine:** - Celine used **90 liters** of water to water her garden. - **Remaining water in the tank:** \[ 120\ ... [truncated] ... xtliters \] 4. **Total Water in the Tank Now:** - **Remaining water after use:** 30 liters - **Added rainwater:** 60 liters - **Total water now:** \[ 30\ liters + 60\ liters = 90\ ,→ liters \] **Final Answer:** $ 90\ liters$ [gen @ alpha=4] Let's break down the problem step by step to find out how many ,→ liters of water are in the tank now. **Step 1: Determine the initial amount of water in the tank.** - The tank initially contains **120 liters** of water. **Step 2: Calculate the remaining water after Celine used some ,→ for her garden.** ... [truncated] ... r - Rainwater collected = 2 x **30 liters** = **60 liters** **Step 4: Determine the total amount of water in the tank now ,→ .** - Total water now = Remaining water + Rainwater collected - Total water now = **30 liters** + **60 liters** = **90 ,→ liters** **Final Answer:** \[ 90 liters \] F Full Spurious CoT Termination Traces Figure 1 presents a shortened example comparing Full-CoT generation with early exit by EoT injec- tion. Here, we provide the corresponding full traces without truncation. We report the Full-CoT trace, the DEER trace, and the traces obtained after ap- plying EAB withα = 2andα = 4to the same DEER exit point in Figures 12 to 15. These ex- amples illustrate how spurious CoT termination can lead to an extended answering phase, and how increasing attention to the exit token shortens the answering phase while preserving the transition to a final answer. The proper divisors of 12 are 1, 2, 3, 4 and 6. A proper divisor of an integer $N$ is a positive divisor of $N$ that is less than $N$. What is the sum of the proper divisors of the sum of the proper divisors of 284? Full-CoT <think> , soIneedtofindthesumof theproperdivisorsof thesumof theproperdivisorsof 284. Hmm, that soundsabitcomplicated, butIthinkIcanbreakitdownstepbystep. , letmemakesureIunderstandwhataproper divisoris. Aproperdivisorof anintegerNisapositivedivisorof NthatislessthanN. So, forexample, theproperdivisorsof 12 are1, 2, 3, 4, and 6, asgivenintheproblem. Theirsumis1 + 2 + 3 + 4 + 6 = 16. So, ifIweretofindthesumof theproper divisorsof 12, itwouldbe16. , applyingthattotheproblem, Ineedtodotwomainsteps: 1. Findthesumof the properdivisorsof 284. 2. Then, findthesumof theproperdivisorsof thatresult. , let'sstartwiththefirststep: finding thesumof theproperdivisorsof 284. divisors, including1 and thenumberitself, we canusetheformularelatedtoprimefactorization. The formulaisthatifanumberNcanbeexpressedasaproductof primes raisedtotheirrespectivepowers, likeN= p1^a1 * p2^a2 * ... * pn^an, thenthesumof alldivisorsof Nis(1 + p1 + p1^2 + ... + p1^a1) * (1 + p2 + p2^2 + ... + p2^a2) * ... * (1 + pn+ pn^2 + ... + pn^an). , we'llsubtractNitselffromthissum. , first, Ineedtofactorize284 intoitsprimefactors. 284 by smallprimes: 284 ÷2 = 142. So, 2 isaprimefactor. 142 ÷2 = 71. So, 2 again. 71 isaprimenumberbecauseit's notdivisibleby2, 3, 5, 7, or11. Letmecheck: 71 ÷2 = 35.5, notinteger. 71 ÷3 ≈ 23.666, notinteger. 71 ÷5 = 14.2, notinteger. 71 ÷7 ≈ 10.142, notinteger. 71 ÷11 ≈ 6.454, notinteger. 71 islessthan11^2 (whichis 121), wedon'tneedtocheckfurther. So, 71 isprime. , theprimefactorizationof 284 is2^2 * 71^1. , usingtheformulaforthesumof divisors: divisors= (1 + 2 + 2^2) * (1 + 71) = (1 + 2 + 4) * (1 + 71) = (7) * (72) = 504. , that'sthesumof alldivisors, including284 itself. Sinceweneedthesumof properdivisors, wesubtract284 fromthistotal. , sumof properdivisorsof 284 = 504 -284 = 220. , sothesumof theproperdivisorsof 284 is 220. , movingontothesecondstep: findingthesumof theproperdivisorsof 220. , Ineedtofactorize220 intoitsprimefactors. 220 ÷2 = 110 110 ÷2 = 55 55 ÷5 = 11 11 isaprimenumber. , theprime factorizationof 220 is2^2 * 5^1 * 11^1. divisorsformula: divisors= (1 + 2 + 2^2) * (1 + 5) * (1 + 11) = (1 + 2 + 4) * (1 + 5) * (1 + 11) = (7) * (6) * (12). : 7 * 6 = 42 42 * 12 = 504 , thesumof alldivisorsof 220 is504. Subtracting220 togetthesumof properdivisors: properdivisorsof 220 = 504 -220 = 284. , that'sinteresting. The sumof theproperdivisorsof 284 is220, and thesumof theproper divisorsof 220 is284. So, theyformwhat'scalledanamicablepair. Irememberhearingaboutamicablenumbersbefore; they aretwodifferentnumberswhereeachisthesumof theproperdivisorsof theother. , goingbacktothe problem. The questionisaskingforthesumof theproperdivisorsof thesumof theproperdivisorsof 284. So, thesumof the properdivisorsof 284 is220, and thenthesumof theproperdivisorsof 220 is284. , theanswershouldbe 284. 'tmakeanymistakesinmycalculations, letmeverifyeachstep. , factorizing 284: 284 ÷2 = 142 142 ÷2 = 71 71 isprime. So, 284 = 2^2 * 71. divisors: (1 + 2 + 4) * (1 + 71) = 7 * 72 = 504. properdivisors: 504 -284 = 220. Thatseemscorrect. , factorizing220: 220 ÷2 = 110 110 ÷2 = 55 55 ÷5 = 11 11 isprime. So, 220 = 2^2 * 5 * 11. divisors: (1 + 2 + 4) * (1 + 5) * (1 + 11) = 7 * 6 * 12. 7 * 6 = 42; 42 * 12 = 504. properdivisors: 504 -220 = 284. Correct. , yes, thesumof theproper divisorsof 220 isindeed284. , thefinalansweris284. -check, maybeIcanlistoutallthe properdivisorsof 284 and addthemuptomakesureIget220. 284: 1, 2, 4, 71, 142. , isthatall? Letmesee. 284 is2^2 * 71, soitsdivisorsare1, 2, 4, 71, 142, 284. , properdivisorsare1, 2, 4, 71, 142. themup: 1 + 2 + 4 + 71 + 142. 1 + 2 = 3; 3 + 4 = 7; 7 + 71 = 78; 78 + 142 = 220. Yes, that'scorrect. , for220, let'slistitsproperdivisors. 220 is2^2 * 5 * 11. , itsdivisorsare1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110, 220. divisorsare1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110. : 1 + 2 = 3; 3 + 4 = 7; 7 + 5 = 12; 12 + 10 = 22; 22 + 11 = 33; 33 + 20 = 53; 53 + 22 = 75; 75 + 44 = 119; 119 + 55 = 174; 174 + 110 = 284. , thataddsupto284. So, that's correct. , Icanbeconfidentthatthesumof theproperdivisorsof thesumof theproperdivisorsof 284 is 284. **FinalAnswer** theproperdivisorsof thesumof theproperdivisorsof 284 is\ 284. </think> "Tofindthesumof theproperdivisorsof thesumof theproperdivisorsof 284, weproceedasfollows: 1. **Findthesum of theproperdivisorsof 284:** -The primefactorizationof 284 is\$2^2 \× 71\$. -The sumof alldivisorsof 284 is calculatedusingtheformulaforthesumof divisors: \\[ (1 + 2 + 4) \× (1 + 71) = 7 \× 72 = 504 \\] - Subtracting284 fromthissumgivesthesumof theproperdivisors: \\[ 504 -284 = 220 \\] 2. **Findthesumof the properdivisorsof 220:** -The primefactorizationof 220 is\$2^2 \× 5 \× 11\$. -The sumof alldivisorsof 220 iscalculatedusingtheformulaforthesumof divisors: \\[ (1 + 2 + 4) \× (1 + 5) \× (1 + 11) = 7 \× 6 \× 12 = 504 \\] -Subtracting220 fromthissumgivesthesumof theproperdivisors: \\[ 504 -220 = 284 \\] , thesumof theproperdivisorsof thesumof theproperdivisorsof 284 is\$\ 284\$. 2116 tokens 371 tokens Figure 12: Full-CoT trace for the example shown in Figure 1. The proper divisors of 12 are 1, 2, 3, 4 and 6. A proper divisor of an integer $N$ is a positive divisor of $N$ that is less than $N$. What is the sum of the proper divisors of the sum of the proper divisors of 284? DEER <think> , soIneedtofindthesumof theproperdivisorsof thesumof theproperdivisorsof 284. Hmm, that soundsabitcomplicated, butIthinkIcanbreakitdownstepbystep. , letmemakesureIunderstandwhataproper divisoris. Aproperdivisorof anintegerNisapositivedivisorof NthatislessthanN. So, forexample, theproperdivisorsof 12 are1, 2, 3, 4, and 6, asgivenintheproblem. Theirsumis1 + 2 + 3 + 4 + 6 = 16. So, ifIweretofindthesumof theproper divisorsof 12, itwouldbe16. , applyingthattotheproblem, Ineedtodotwomainsteps: 1. Findthesumof the properdivisorsof 284. 2. Then, findthesumof theproperdivisorsof thatresult. , let'sstartwiththefirststep: finding thesumof theproperdivisorsof 284. divisors, including1 and thenumberitself, we canusetheformularelatedtoprimefactorization. The formulaisthatifanumberNcanbeexpressedasaproductof primes raisedtotheirrespectivepowers, likeN= p1^a1 * p2^a2 * ... * pn^an, thenthesumof alldivisorsof Nis(1 + p1 + p1^2 + ... + p1^a1) * (1 + p2 + p2^2 + ... + p2^a2) * ... * (1 + pn+ pn^2 + ... + pn^an). , we'llsubtractNitselffromthissum. , first, Ineedtofactorize284 intoitsprimefactors. 284 by smallprimes: 284 ÷2 = 142. So, 2 isaprimefactor. 142 ÷2 = 71. So, 2 again. 71 isaprimenumberbecauseit's notdivisibleby2, 3, 5, 7, or11. Letmecheck: 71 ÷2 = 35.5 → notinteger. 71 ÷3 ≈ 23.666... → notinteger. 71 ÷ 5 = 14.2 → notinteger. 71 ÷7 ≈ 10.142... → notinteger. 71 ÷11 ≈ 6.454... → notinteger. 71 islessthan 121 (11 squared), wedon'tneedtocheckhigherprimes. So, 71 isprime. , theprimefactorizationof 284 is2^2 * 71^1. , usingtheformulaforthesumof divisors: divisors= (1 + 2 + 2^2) * (1 + 71) = (1 + 2 + 4) * (1 + 71) = (7) * (72) = 504. , that'sthesumof alldivisors, including284 itself. Sinceweneedthesumof properdivisors, wesubtract284 fromthistotal. , sumof properdivisorsof 284 = 504 -284 = 220. , sothesumof theproper divisorsof 284 is220. , movingontothesecondstep: findingthesumof theproperdivisorsof 220. , Ineed tofactorize220 intoitsprimefactors. 'sdothat: 220 ÷2 = 110. 110 ÷2 = 55. 55 ÷5 = 11. 11 isaprime number. , theprimefactorizationof 220 is2^2 * 5^1 * 11^1. divisorsformula: divisors = (1 + 2 + 2^2) * (1 + 5) * (1 + 11) = (1 + 2 + 4) * (1 + 5) * (1 + 11) = (7) * (6) * (12). : 7 * 6 = 42. 42 * 12 = 504. , thesumof alldivisorsof 220 is504. Butagain, weneedthesumof properdivisors, sowe subtract220. properdivisorsof 220 = 504 -220 = 284. "Waitasecond, that'sinteresting. The sumof theproperdivisorsof 284 is220, and thesumof theproperdivisorsof 220 is 284. Thatmeans284 and 220 areamicablenumbers. Amicablenumbersarepairswhereeachnumberisthesumof theproper divisorsof theother. So, that'sanicelittlefactoidhere. , goingbacktotheproblem, thequestionwasaskingforthe sumof theproperdivisorsof thesumof theproperdivisorsof 284. So, thesumof theproperdivisorsof 284 is220, and then thesumof theproperdivisorsof 220 is284. So, thefinalansweris284. -checkmycalculationsto makesureIdidn'tmakeanymistakes. , for284: : 2^2 * 71^1. divisors: (1 + 2 + 4) * (1 + 71) = 7 * 72 = 504. properdivisors: 504 -284 = 220. Thatseemscorrect. , for220: : 2^2 * 5^1 * 11^1. divisors: (1 + 2 + 4) * (1 + 5) * (1 + 11) = 7 * 6 * 12. 7 * 6 = 42; 42 * 12 = 504. Correct. of properdivisors: 504 -220 = 284. Correct. , yes, thecalculationsseemaccurate. Therefore, thesumof theproper divisorsof thesumof theproperdivisorsof 284 is284. **FinalAnswer** theproperdivisorsof thesumof theproperdivisorsof 284 is\ 284. </think> findthesumof theproperdivisorsof thesumof theproper divisorsof 284, wefollowthesesteps: 1. **Findthesumof theproperdivisorsof 284:** -Primefactorizationof 284: \$284 = 2^2 \× 71\$ -Sumof divisorsformula: \$(1 + 2 + 4) \× (1 + 71) = 7 \× 72 = 504\$ -Sumof properdivisors: \$504 -284 = 220\$ 2. **Findthesumof theproperdivisorsof 220:** -Primefactorizationof 220: \$220 = 2^2 \× 5 \× 11\$ -Sumof divisorsformula: \$(1 + 2 + 4) \× (1 + 5) \× (1 + 11) = 7 \× 6 \× 12 = 504\$ -Sumof properdivisors: \$504 -220 = 284\$ , thesumof theproperdivisorsof thesumof the properdivisorsof 284 is\$\ 284\$. 1090 tokens 760 tokens </think> " Figure 13: DEER trace for the same example shown in Figure 1. The injected exit token induces spurious CoT termination. The proper divisors of 12 are 1, 2, 3, 4 and 6. A proper divisor of an integer $N$ is a positive divisor of $N$ that is less than $N$. What is the sum of the proper divisors of the sum of the proper divisors of 284? DEER + EAB(휶=2) Same reasoning trace as DEER Wait, holdonasecond. So, thesumof theproperdivisorsof 284 is220, and thesumof theproperdivisorsof 220 is284. Thatseemsinterestingbecause284 and 220 areknownasamicablenumbers. Theyformanamicablepairwhereeachisthe sumof theproperdivisorsof theother. So, inthiscase, thesumof theproperdivisorsof 284 is220, and thesumof theproper divisorsof 220 is284. Therefore, theanswertotheproblemis284. -checkmycalculationstomakesure Ididn'tmakeamistake. , for284: : 2^2 * 71^1. divisors: (1 + 2 + 4) * (1 + 71) = 7 * 72 = 504. properdivisors: 504 -284 = 220. Thatseemscorrect. , for220: : 2^2 * 5^1 * 11^1. divisors: (1 + 2 + 4) * (1 + 5) * (1 + 11) = 7 * 6 * 12. 7 * 6 = 42; 42 * 12 = 504. proper divisors: 504 -220 = 284. Thatalsoseemscorrect. , yes, thesumof theproperdivisorsof thesumof theproperdivisors of 284 isindeed284. **FinalAnswer** theproperdivisorsof thesumof theproperdivisorsof 284 is \ 284. </think> findthesumof theproperdivisorsof thesumof theproperdivisorsof 284, wefollow thesesteps: 1. **Findthesumof theproperdivisorsof 284:** -Primefactorizationof 284: \$284 = 2^2 \× 71\$ -Sumof divisorsformula: \$(1 + 2 + 4) \× (1 + 71) = 7 \× 72 = 504\$ -Sumof properdivisors: \$504 - 284 = 220\$ 2. **Findthesumof theproperdivisorsof 220:** -Primefactorizationof 220: \$220 = 2^2 \× 5 \× 11\$ -Sumof divisorsformula: \$(1 + 2 + 4) \× (1 + 5) \× (1 + 11) = 7 \× 6 \× 12 = 504\$ - Sumof properdivisors: \$504 -220 = 284\$ , thesumof theproperdivisorsof thesumof theproperdivisorsof 284 is\$\ 284\$. 1090 tokens 719 tokens </think> " Figure 14: EAB trace for the same DEER exit point withα = 2. Increasing attention to the injected exit token shortens the answering phase relative to the unbiased DEER trace. The proper divisors of 12 are 1, 2, 3, 4 and 6. A proper divisor of an integer $N$ is a positive divisor of $N$ that is less than $N$. What is the sum of the proper divisors of the sum of the proper divisors of 284? DEER + EAB(휶=4) Same reasoning trace as DEER 1090 tokens </think> " Wait, that'sinteresting. So, thesumof theproperdivisorsof 284 is220, and thesumof theproperdivisorsof 220 is284. That seemslikeaspecialrelationshipbetweenthetwonumbers. Ithinktheymightbecalledamicablenumbersorsomethinglike that. Butanyway, IthinkI'vefollowedthestepscorrectly. , torecap: 1. Findthesumof theproperdivisorsof 284: -Primefactorizationof 284: \$2^2 \× 71\$ -Sumof divisors: \$(1 + 2 + 4) \× (1 + 71) = 7 \× 72 = 504\$ -Sumof properdivisors: \$504 -284 = 220\$ 2. Findthesumof theproperdivisorsof 220: -Prime factorizationof 220: \$2^2 \× 5 \× 11\$ -Sumof divisors: \$(1 + 2 + 4) \× (1 + 5) \× (1 + 11) = 7 \× 6 \× 12 = 504\$ -Sumof properdivisors: \$504 -220 = 284\$ , thesumof theproperdivisorsof the sumof theproperdivisorsof 284 is284. **FinalAnswer** theproperdivisorsof thesumof theproper divisorsof 284 is\ 284. 369 tokens Figure 15: EAB trace for the same DEER exit point withα = 4. A stronger EAB further reduces answering-phase length in this example.

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