Batched Contextual Reinforcement: A Task-Scaling Law for Efficient Reasoning

Bangji Yang; Hongbo Ma; Jiajun Fan; Ge Liu

cs.LGApr 2, 2026

Batched Contextual Reinforcement: A Task-Scaling Law for Efficient Reasoning

Authors: Bangji Yang, Hongbo Ma, Jiajun Fan, Ge Liu

Organizations: University of Illinois Urbana-Champaign · Tsinghua University

Abstract

Large Language Models employing Chain-of-Thought reasoning achieve strong performance but suffer from excessive token consumption that inflates inference costs. Existing efficiency methods such as explicit length penalties, difficulty estimators, or multi-stage curricula either degrade reasoning quality or require complex training pipelines. We introduce Batched Contextual Reinforcement, a minimalist, single-stage training paradigm that unlocks efficient reasoning through a simple structural modification: training the model to solve N problems simultaneously within a shared context window, rewarded purely by per-instance accuracy. This formulation creates an implicit token budget that yields several key findings: (1) We identify a novel task-scaling law: as the number of concurrent problems N increases during inference, per-problem token usage decreases monotonically while accuracy degrades far more gracefully than baselines, establishing N as a controllable throughput dimension. (2) BCR challenges the traditional accuracy-efficiency trade-off by demonstrating a "free lunch" phenomenon at standard single-problem inference. Across both 1.5B and 4B model families, BCR reduces token usage by 15.8% to 62.6% while consistently maintaining or improving accuracy across five major mathematical benchmarks. (3) Qualitative analyses reveal emergent self-regulated efficiency, where models autonomously eliminate redundant metacognitive loops without explicit length supervision. (4) Crucially, we empirically demonstrate that implicit budget constraints successfully circumvent the adversarial gradients and catastrophic optimization collapse inherent to explicit length penalties, offering a highly stable, constraint-based alternative for length control. These results prove BCR practical, showing simple structural incentives unlock latent high-density reasoning in LLMs.

Figures & tables

Figure 1: A new scaling dimension: task-level inference scaling on OLYMPIAD. We vary the number of concurrent problems

N

. Left axis (bars): per-problem tokens. Right axis (lines): accuracy. The baseline (dark blue) reduces tokens as

N

grows but suffers accuracy collapse. BCR (light blue) crystallizes this efficiency:

{\sim}

60% token reduction even at

N{=}1

, with graceful accuracy degradation. This reveals a task-scaling law: more concurrent problems

\Rightarrow

more efficient reasoning .

Figure 2: Overview of BCR. We package $N$ questions into a problem group with a system instruction and shared token budget. The model generates a single completion solving all $N$ problems sequentially. Per-problem answers are extracted via a stack-based parser for accuracy verification, combined with a format reward. Training follows standard GRPO—no length penalties or auxiliary models required.

Model	AIME25		AMC23		Minerva		MATH-500		Olympiad
Model	Acc $\uparrow$	Tok $\downarrow$	Acc $\uparrow$	Tok $\downarrow$	Acc $\uparrow$	Tok $\downarrow$	Acc $\uparrow$	Tok $\downarrow$	Acc $\uparrow$	Tok $\downarrow$
General LLM
Qwen3-1.7B	40.0	15,796	85.0	8,157	57.7	5,714	90.2	4,621	67.3	9,395
Math LLM
STILL-3-1.5B	33.3	13,512	72.5	5,706	46.7	5,678	84.6	4,280	55.2	9,755
Math LLM w/ Length Control

Table 1: Main results across five benchmarks at

N{=}1

inference. We report accuracy (%) and average tokens per problem. Bold indicates best in column. BCR improves efficiency in both model families: vs. JustRL-deepseek-1.5B, token usage drops by 39.8–62.6% with accuracy gains on 3/5 benchmarks; vs. Qwen3-4B-Thinking-2507, token usage drops by 15.8–31.8% with accuracy gains on 5/5 benchmarks. This demonstrates that efficiency learned through multi-problem training transfers fully to standard single-problem evaluation.

Figure 3: Efficiency-Accuracy Pareto Frontier on Minerva. The trajectories show checkpoint evaluations during the training process. The final models (stars) demonstrate that BCR consistently pushes the Pareto frontier significantly toward higher accuracy and lower token usage for both model families.

$N$	Model	AIME25		AMC23		Minerva
$N$	Model	Acc $\uparrow$	Tok $\downarrow$	Acc $\uparrow$	Tok $\downarrow$	Acc $\uparrow$	Tok $\downarrow$
1 $\times$	JustRL-1.5B	40.0	8,482	85.0	5,713	43.4	5,413
1 $\times$	BCR (Ours)	33.3	3,173	87.5	2,637	48.5	2,494
2 $\times$	JustRL-1.5B	23.3	7,495	75.0	4,697	33.8	4,141
2 $\times$	BCR (Ours)	30.0	2,371	83.3	1,705	37.5	1,679
3 $\times$	JustRL-1.5B	20.0	4,095	55.0	5,429	24.6	3,885

Table 2: Task-scaling law: accuracy (%) and per-problem tokens under

N{\times}

concurrent inference. BCR maintains superior efficiency across all group sizes and degrades far more gracefully than the baseline. The efficiency gap widens as

N

increases, establishing a favorable scaling relationship. See Appendix B for MATH-500 and Olympiad.

Example 1: Sudoku-Style Counting (Qwen AIME 2025) (43.2% token reduction)
Problem: Count the valid $3\times 9$ Sudoku-style fillings and write the total as $p^{a}q^{b}r^{c}s^{d}$ ; compute $pa+qb+rc+sd$ .
Ground Truth: 81
Qwen3-4B-Thinking-2507 (31,362 tokens)	BCR-Qwen3-4B (17,805 tokens)
Uses $N=9!\cdot 2^{3}\cdot(3!)^{3}$ and factors this incomplete count. Most extra tokens are spent re-describing the grid geometry and block layout before the counting argument actually starts. The actual error is structural: the trace never explicitly counts the 56 feasible middle-block assignments, so a long derivation is built on the wrong combinatorial factor. Concludes $N=2^{13}\cdot 3^{7}\cdot 5\cdot 7$ , hence $2\cdot 13+3\cdot 7+5+7=59$ .	Adds the missing structural multiplier: 56 feasible middle-block assignments. $N=9!\cdot 56\cdot(3!)^{3}\cdot(3!)^{3}$ . Thus $N=2^{16}\cdot 3^{10}\cdot 5\cdot 7^{2}$ , so $2\cdot 16+3\cdot 10+5+14=81$ .
Answer: $\boxed{59}$ $\times$	Answer: $\boxed{81}$ ✓
Example 2: Number Theory — Roots of Unity (BRUMO 2025)

Table 3: Qualitative comparison. BCR eliminates verbose metacognitive loops and missing structured logic while preserving all essential reasoning steps. Green highlights efficient reasoning; red marks verbose or failed patterns. See Appendix C for seven additional examples with detailed analysis.

Appendix figures & tables21 assets

Supplementary material from the paper’s appendix.

Appendix

$N$	MATH-500		Olympiad
$N$	acc. (%)	tokens	acc. (%)	tokens
3	90.0	2185	61.4	4052
4	89.4	2386	62.0	4594
5	89.2	2001	64.4	3873

Appendix

Table 5: Group size ablation on MATH-500 and Olympiad (extending Table 4 ). All models trained for 300 steps with the specified

N

and evaluated at

N{=}1

$N$	Model	MATH-500		Olympiad
$N$	Model	acc. (%)	tokens	acc. (%)	tokens
$1\times$	JustRL-1.5B	91.4	3099	62.1	7017
$1\times$	BCR (ours)	87.6	1868	62.9	2969
$2\times$	JustRL-1.5B	83.0	2033	47.3	5567
$2\times$	BCR (ours)	85.6	1290	52.0	1919
$3\times$	JustRL-1.5B	73.6	2544	38.7	4696

Appendix

Table 6: Task-scaling law on MATH-500 and Olympiad (extending Table 2 ). Both models evaluated under

N{\times}

concurrent inference. Bold indicates best within each

N

$B_{\max}$	AIME25		AMC23		Minerva		MATH-500		Olympiad
$B_{\max}$	acc. (%)	tokens	acc. (%)	tokens	acc. (%)	tokens	acc. (%)	tokens	acc. (%)	tokens
4096	26.7	2561	82.5	2135	46.0	2085	86.6	1524	61.1	2711
5120 (Ours)	33.3	3173	87.5	2637	48.5	2494	87.6	1868	62.9	2969
6144	30.0	3527	80.0	2701	44.9	2875	88.8	1874	63.4	3116

Appendix

Table 7: Complete token budget ablation across all five benchmarks. The 1.5B model is trained with

N{=}3

. We report accuracy (%) and average generated tokens per problem.

Figure 5: Extended Efficiency-Accuracy Pareto Frontiers. Training trajectories on AIME25, AMC23, MATH-500, and Olympiad. The arrows track intermediate model checkpoints during BCR optimization, demonstrating a continuous, stable shift toward lower token consumption and competitive accuracy. The final BCR models (stars) consistently dominate the Pareto frontier relative to baseline models of comparable or larger scale.

JustRL-DeepSeek-1.5B (7,211 tokens)	BCR-JustRL-1.5B (3,639 tokens)
Problem: Find the number of ordered pairs $(x,y)$ with $x,y\in[-100,100]$ such that $12x^{2}-xy-6y^{2}=0$ .
Ground Truth: 117
So the solutions are $x=(3y)/4$ or $x=(-2y)/3$ . But since $x$ and $y$ are integers, these expressions must result in integers. So for $x=(3y)/4$ to be integer, $3y$ must be divisible by 4, which implies that $y$ must be divisible by 4, since 3 and 4 are coprime. Similarly… Wait wait wait, wait. Wait wait, wait wait… [Extensive rechecking of calculations] Wait the final answer is $N=2016$ … Wait wait wait let me double-check…	The equation factors as $(3x+2y)(4x-3y)=0$ . Therefore, either $3x+2y=0$ or $4x-3y=0$ . For $3x+2y=0$ : $x=2k$ , $y=-3k$ , with $k\in[-33,33]$ . This gives 67 solutions. For $4x-3y=0$ : $x=3k$ , $y=4k$ , with $k\in[-25,25]$ . This gives 51 solutions. Overlap is only $(0,0)$ , so total $=67+51-1=117$ .
Answer: $\boxed{117}$ ✓	Answer: $\boxed{117}$ ✓

Appendix

Table 8: AIME 2025, Example 1: Equation Factorization (49.5% token reduction). Both models reach the correct answer via factorization, but the baseline wastes

>

3,500 tokens on self-interruption loops that revisit already-correct calculations.

JustRL-DeepSeek-1.5B (6,676 tokens)	BCR-JustRL-1.5B (4,164 tokens)
Problem: In $\triangle ABC$ , points on sides with given lengths. $M,N$ are reflections. Area of $DEGF$ is 288.
Find area of heptagon $AFNBCEM$ .
Ground Truth: 588
The shoelace formula applied to the heptagon gave $\frac{1}{2}\times 1176=588$ . So the area is 588. Wait, but let me check if this makes sense. Given that the area of triangle ABC is 588, and the heptagon AFNBCEM is covering most of it, but the calculation gives 588. Wait, but let me think again. But wait, let me check the coordinates again. Wait, when I calculated term6… Yes, that is correct. And term7… So the sum is indeed…	Now compute the sum: Term1=0, Term2= $-240$ , Term3= $+840$ , Term4= $+1176$ , Term5= $-840$ , Term6= $+240$ , Term7=0. Total sum $=0-240+840+1176-840+240+0=1176$ . Area $=\frac{1}{2}\times\|1176\|=588$ . So the area of the heptagon is 588.
Answer: $\boxed{588}$ ✓	Answer: $\boxed{588}$ ✓

Appendix

Table 9: AIME 2025, Example 2: Coordinate Geometry (37.6% token reduction). The baseline re-verifies each shoelace term individually despite having computed them correctly; BCR computes once and terminates.

JustRL-DeepSeek-1.5B (3,530 tokens)	BCR-JustRL-1.5B (2,216 tokens)
Problem: Find the sum of all positive integers $n$ such that $n+2$ divides $3(n+3)(n^{2}+9)$ .
Ground Truth: 49
Okay, so I have got this math problem here… Hmm, let us break this down step by step. First, let us write down what is given… To tackle this, maybe I can use polynomial division or factor the expression somehow. Let me try to see if I can simplify… Alternatively, since $n+2$ divides the product, by the Factor Theorem, if we substitute $n=-2$ … Wait, but actually, if $n+2$ divides the product, then substituting $n=-2$ should give zero. Let us test that. [Performs explicit polynomial long division] So the remainder is 39. Therefore… 39 must be divisible by $(n+2)$ .	The condition is that $n+2$ divides $3(n+3)(n^{2}+9)$ . Using the Remainder Theorem: substituting $n=-2$ into the numerator gives $3(1)(13)=39$ . So $n+2$ must be a divisor of 39. Divisors of 39: 1, 3, 13, 39. Since $n$ is positive, $n+2\geq 3$ , so $n\in\{1,11,37\}$ . Sum: $1+11+37=49$ .
Answer: $\boxed{49}$ ✓	Answer: $\boxed{49}$ ✓

Appendix

Table 10: AIME 2025, Example 3: Divisibility Analysis (37.2% token reduction). The baseline explores multiple strategies before finding the Remainder Theorem approach; BCR applies it directly without exploration.

JustRL-DeepSeek-1.5B (10,122 tokens)	BCR-JustRL-1.5B (3,261 tokens)
Problem: Consider the set of complex numbers $z$ satisfying $\|1+z+z^{2}\|=4$ . The maximum value of the
imaginary part of $z$ can be written as $\tfrac{\sqrt{m}}{n}$ . What is $m+n$ ?
Ground Truth: 21
… So the maximum $y$ occurs when $y=\sqrt{3}/2$ and the condition is not satisfied, so the maximum $y$ is $\sqrt{3}/2$ , and the $m+n=3+2=5$ . [Incorrect analysis: confuses constraint satisfaction] [Concludes with wrong answer]	Set $1+z+z^{2}=-4$ , giving $z^{2}+z+5=0$ . Solutions: $z=\frac{-1\pm i\sqrt{19}}{2}$ Thus $\text{Im}(z)=\pm\frac{\sqrt{19}}{2}$ , maximum is $\frac{\sqrt{19}}{2}$ . So $m=19$ , $n=2$ , thus $m+n=21$ .
Answer: $\boxed{5}$ $\times$	Answer: $\boxed{21}$ ✓

Appendix

Table 11: AMC 2023, Example 4: Complex Analysis (67.8% token reduction). BCR correct, baseline incorrect. The baseline’s verbose exploration leads to a reasoning error through constraint confusion.

JustRL-DeepSeek-1.5B (3,464 tokens)	BCR-JustRL-1.5B (1,425 tokens)
Problem: How many digits are in the base-ten representation of $8^{5}\cdot 5^{10}\cdot 15^{5}$ ?
Ground Truth: 18
Alright, so I have this math problem here… Hmm, okay. Let me break this down step by step. Starting with $8^{5}$ . I know that 8 is 2 cubed, so $8=2^{3}$ . Therefore, $8^{5}=(2^{3})^{5}$ . When you raise a power to a power, you multiply the exponents. So that would be $2^{15}$ … [Extensive explanation of basic exponent rules] So $10^{15}$ would have 16 digits. But let me confirm that. Yes, $10^{n}$ is a 1 followed by n zeros…	$8^{5}=(2^{3})^{5}=2^{15}$ $15^{5}=(3\cdot 5)^{5}=3^{5}\cdot 5^{5}$ Combined: $2^{15}\cdot 5^{15}\cdot 3^{5}=10^{15}\cdot 243$ $243\times 10^{15}$ is 243 followed by 15 zeros. Digits: $3+15=18$
Answer: $\boxed{18}$ ✓	Answer: $\boxed{18}$ ✓

Appendix

Table 12: AMC 2023, Example 5: Digit Counting (58.9% token reduction). The baseline narrates basic exponent rules that it already knows; BCR computes directly without unnecessary explanation.

JustRL-DeepSeek-1.5B (4,329 tokens)	BCR-JustRL-1.5B (2,182 tokens)
Problem: Triangle $ABC$ has side lengths in arithmetic progression, smallest side 6, and an angle of $120^{\circ}$ .
Find the area in form $m\sqrt{n}$ . What is $m+n$ ?
Ground Truth: 18
… So $d=4$ , sides are $6,10,14$ . Wait, wait, wait a second. Wait, the sides are 6, 6 + d, 6 + 2d. If d=4, then sides are 6,10,14. Let me verify if these satisfy triangle inequality… Wait, wait, wait, wait. Let me double-check the sides. Area $=\frac{1}{2}\cdot 6\cdot 10\cdot\sin(120^{\circ})=15\sqrt{3}$	Sides: $a=6$ , $b=6+d$ , $c=6+2d$ Law of Cosines with $\cos(120^{\circ})=-\frac{1}{2}$ : $(6+2d)^{2}=36+(6+d)^{2}+6(6+d)$ Solving: $3d^{2}+6d-72=0\Rightarrow d=4$ Sides: $6,10,14$ . Area $=\frac{1}{2}(6)(10)\sin 120^{\circ}=15\sqrt{3}$ So $m=15$ , $n=3$ , $m+n=18$
Answer: $\boxed{18}$ ✓	Answer: $\boxed{18}$ ✓

Appendix

Table 13: AMC 2023, Example 6: Triangle Area (49.6% token reduction). The baseline rechecks valid intermediate results with two distinct self-doubt episodes; BCR follows a linear derivation from setup to answer.

JustRL-DeepSeek-1.5B (32,677 tokens)	BCR-JustRL-1.5B (2,692 tokens)
Problem: What is the smallest positive integer $n$ such that $z^{n}-1$ and $(z-\sqrt{3})^{n}-1$ share a common
complex root?
Ground Truth: 12
[Model generates over 32,000 tokens] [Output degenerates into repetitive patterns:] ... 2 2 1 11 1 1 1 0, , 11 1 , (1 1 ... [No valid answer extracted]	Let $\alpha$ be a common root. Then $\alpha^{n}=1$ and $(\alpha-\sqrt{3})^{n}=1$ . If $\alpha=e^{i\pi/6}$ , then $\|\alpha\|=1$ and $\alpha-\sqrt{3}=e^{i\cdot 5\pi/6}$ . Order of $e^{i\pi/6}$ is 12. Order of $e^{i\cdot 5\pi/6}$ is also 12. Verify: $\alpha^{12}=e^{i\cdot 2\pi}=1$ ✓ $(\alpha-\sqrt{3})^{12}=e^{i\cdot 10\pi}=1$ ✓ Smallest such $n$ is 12.
Answer: None $\times$	Answer: $\boxed{12}$ ✓

Appendix

Table 14: BRUMO 2025: Roots of Unity (91.8% token reduction). The baseline generates 32K tokens and degenerates into non-mathematical character sequences; BCR identifies the key geometric insight and terminates in 2,692 tokens with the correct answer.

Qwen3-4B-Thinking-2507 (31,362 tokens)	BCR-Qwen3-4B (17,805 tokens)
Problem: Count the valid $3\times 9$ Sudoku-style fillings and write the total as $p^{a}q^{b}r^{c}s^{d}$ ; compute $pa+qb+rc+sd$ .
Ground Truth: 81
Uses $N=9!\cdot 2^{3}\cdot(3!)^{3}$ and factors this incomplete count. Most extra tokens are spent re-describing the grid geometry and block layout before the counting argument actually starts. The actual error is structural: the trace never explicitly counts the 56 feasible middle-block assignments, so a long derivation is built on the wrong combinatorial factor. Concludes $N=2^{13}\cdot 3^{7}\cdot 5\cdot 7$ , hence $2\cdot 13+3\cdot 7+5+7=59$ .	Adds the missing structural multiplier: 56 feasible middle-block assignments. $N=9!\cdot 56\cdot(3!)^{3}\cdot(3!)^{3}$ . Thus $N=2^{16}\cdot 3^{10}\cdot 5\cdot 7^{2}$ , so $2\cdot 16+3\cdot 10+5+14=81$ .
Answer: $\boxed{59}$ $\times$	Answer: $\boxed{81}$ ✓

Appendix

Table 15: Qwen AIME 2025, Example 8: Sudoku-Style Counting (43.2% token reduction). The baseline omits a key combinatorial factor; BCR restores it and obtains the correct answer.

Qwen3-4B-Thinking-2507 (22,817 tokens)	BCR-Qwen3-4B (7,921 tokens)
Problem: Evaluate $\prod_{k=4}^{63}\frac{\log_{k}(5^{k^{2}-1})}{\log_{k+1}(5^{k^{2}-4})}=\frac{m}{n}$ and find $m+n$ .
Ground Truth: 106
Long re-parsing preamble with repeated “wait/no/hold on” cycles before simplification. The token overhead comes mainly from notation interpretation loops, not from the telescoping algebra itself. Eventually derives $\prod\frac{k^{2}-1}{k^{2}-4}=\frac{31}{13}$ and $\prod\frac{\log_{k}5}{\log_{k+1}5}=3$ .	Directly applies $\log_{b}(a^{c})=c\log_{b}(a)$ and splits the product into: (i) rational telescoping $\rightarrow\frac{31}{13}$ , (ii) logarithmic telescoping $\rightarrow 3$ . Combines to $\frac{93}{13}$ and returns $\boxed{106}$ .
Answer: $\boxed{106}$ ✓	Answer: $\boxed{106}$ ✓

Appendix

Table 16: Qwen AIME 2025, Example 9: Telescoping Log Product (65.3% token reduction). Both models are correct; BCR removes prolonged setup chatter and preserves only the core derivation.

Qwen3-4B-Thinking-2507 (32,215 tokens)	BCR-Qwen3-4B (24,754 tokens)
Problem: In $\triangle ABC$ with angles $84^{\circ},60^{\circ},36^{\circ}$ , let $D,E,F$ be side midpoints and $G,H,J$ be additional intersections
on the circumcircle of $\triangle DEF$ . Compute $\overset{\frown}{DE}+2\overset{\frown}{HJ}+3\overset{\frown}{FG}$ .
Ground Truth: 336
Recognizes the nine-point-circle setting and proposes identities such as $\overset{\frown}{DG}=72^{\circ},\ \overset{\frown}{EH}=120^{\circ},\ \overset{\frown}{FJ}=168^{\circ}$ Most extra tokens come from theorem-recall and arc-matching loops: the model keeps generating related geometric facts without consolidating them into the three arcs actually required. It gets stuck on whether these are the same arcs the problem asks for, and no boxed final answer appears.	Settles on $\overset{\frown}{DE}=72^{\circ},\ \overset{\frown}{HJ}=24^{\circ},\ \overset{\frown}{FG}=72^{\circ}$ and then computes $72+2\cdot 24+3\cdot 72=336$ .
Answer: None $\times$	Answer: $\boxed{336}$ ✓

Appendix

Table 17: Qwen AIME 2025, Example 10: Nine-Point Circle Arc Geometry (23.2% token reduction). The baseline recalls relevant geometric facts but never consolidates them into a final answer; BCR commits to a coherent arc decomposition and finishes.

Qwen3-4B-Thinking-2507 (32,577 tokens)	BCR-Qwen3-4B (10,096 tokens)
Problem: In a modified Pascal-like triangular array, find the units digit of the sum of the 2023 numbers in row 2023.
Ground Truth: 5
Starts correctly with row-sum exploration, then the trace keeps recomputing small rows instead of collapsing early to a recurrence, so token usage grows before any real progress is made. output collapses into extremely long repeated digit sequences and no valid final boxed answer is produced.	Derives $S(k)=2S(k-1)+(k-2)$ and solves it as $S(k)=2^{k}-k$ . Then computes $2^{2023}\equiv 8\pmod{10}$ and $(8-3)\bmod 10=5$ .
Answer: None $\times$	Answer: $\boxed{5}$ ✓

Appendix

Table 18: Qwen AMC 2023, Example 11: Triangular Array Sum (69.0% token reduction). The baseline degenerates without a valid answer; BCR solves the recurrence and finishes normally.

Qwen3-4B-Thinking-2507 (21,987 tokens)	BCR-Qwen3-4B (10,082 tokens)
Problem: For $P(x)=\prod_{k=1}^{10}(x-k)^{k}$ , on how many of the 11 open intervals between roots is $P(x)$ positive?
Ground Truth: 6
Narrates each interval separately, repeatedly restating which root crossings flip the sign and which do not. The token-heavy part is not the mathematics; it is the repeated interval-by-interval bookkeeping after the odd/even multiplicity invariant is already known.	Builds the sign pattern directly from multiplicity parity: odd multiplicity roots flip sign; even multiplicity roots preserve sign. Positive intervals are $(1,2),(2,3),(5,6),(6,7),(9,10),(10,\infty)$ , so the count is 6.
Answer: $\boxed{6}$ ✓	Answer: $\boxed{6}$ ✓

Appendix

Table 19: Qwen AMC 2023, Example 12: Polynomial Sign Counting (54.1% token reduction). BCR condenses interval-by-interval narration into a short parity-based sign table.

Qwen3-4B-Thinking-2507 (11,599 tokens)	BCR-Qwen3-4B (5,082 tokens)
Problem: How many complex numbers satisfy the equation $z^{5}=\overline{z}$ ?
Ground Truth: 7
Uses polar form correctly but spends many tokens re-explaining conjugates, arguments, and why the sixth roots are distinct. This is mostly pedagogical overhead: once the problem is reduced to $r=0$ or $z^{6}=1$ , the remaining narration adds cost without changing the solution.	Moves quickly to the two modulus cases: $r=0\Rightarrow z=0$ , and $r=1\Rightarrow\overline{z}=1/z$ , so $z^{6}=1$ . Thus there are $1+6=7$ solutions.
Answer: $\boxed{7}$ ✓	Answer: $\boxed{7}$ ✓

Appendix

Table 20: Qwen AMC 2023, Example 13: Complex Equation with Conjugation (56.2% token reduction). Both models are correct; BCR strips away pedagogical narration and keeps only the polar-form argument.

Qwen3-4B-Thinking-2507 (32,645 tokens)	BCR-Qwen3-4B (11,007 tokens)
Problem: In a 54-card deck with two jokers, find the expected number of draws until at least one Ace, one King, and one Queen have appeared.
Ground Truth: $\frac{737}{39}$
Large state and permutation enumeration expands to 32K tokens and never yields a valid boxed answer. The token explosion is caused by explicit order/case enumeration over first-hit events, which grows much faster than the final inclusion-exclusion formula actually requires.	Introduces first-hit positions for progressively larger target sets and applies $\mathbb{E}[T]=3\mathbb{E}[M]-3\mathbb{E}[L]+\mathbb{E}[R]$ . With $\mathbb{E}[M]=\frac{55}{5}$ , $\mathbb{E}[L]=\frac{55}{9}$ , and $\mathbb{E}[R]=\frac{55}{13}$ , $\mathbb{E}[T]=\frac{737}{39}$ .
Answer: None $\times$	Answer: $\boxed{\frac{737}{39}}$ ✓

Appendix

Table 21: Qwen BRUMO 2025, Example 14: Card-Draw Stopping Time (66.3% token reduction). The baseline stalls in state enumeration; BCR switches to a first-hit formulation and reaches the exact expectation.

Qwen3-4B-Thinking-2507 (26,146 tokens)	BCR-Qwen3-4B (14,139 tokens)
Problem: Each guest has exactly 12 enemies and “the friend of an enemy is an enemy.” Find the sum of all possible values of $n$ .
Ground Truth: 100
Builds graph-theoretic constraints but concludes only $n=13$ and $n=24$ , so the final answer is $37$ . Most extra tokens come from testing local degree bounds and candidate sizes one by one instead of identifying the global clique decomposition early.	Represents the friend graph as a disjoint union of $k$ equal cliques of size $s$ : $n=ks$ and $n-s=12$ , so $n=\frac{12k}{k-1}=12+\frac{12}{k-1}$ . Since $k-1$ must divide 12, the feasible values are $13,14,15,16,18,24$ , whose sum is $100$ .
Answer: $\boxed{37}$ $\times$	Answer: $\boxed{100}$ ✓

Appendix

Table 22: Qwen BRUMO 2025, Example 15: Party Friend/Enemy Graph (45.9% token reduction). BCR uses clique decomposition to enumerate all feasible values of

n

; the baseline undercounts the solution set.

Qwen3-4B-Thinking-2507 (28,830 tokens)	BCR-Qwen3-4B (11,648 tokens)
Problem: Four bears hold positive integers summing to 17 and make sequential public statements; determine the product of their four numbers.
Ground Truth: 160
Eventually identifies the valid tuple $(2,5,2,8)$ but spends many tokens repeatedly rebuilding the remaining candidate sets before testing Druno’s uniqueness condition. The token-heavy step is repeated candidate-set reconstruction: the same small family of tuples is regenerated several times instead of being frozen into one compact table.	After deducing $C=2$ , quickly tabulates the surviving possibilities: $(2,5,8)$ , $(2,8,5)$ , $(5,5,5)$ , $(5,8,2)$ , $(8,5,2)$ and observes that only $D=8$ is unique, yielding $(A,B,C,D)=(2,5,2,8)$ and product 160.
Answer: $\boxed{160}$ ✓	Answer: $\boxed{160}$ ✓

Appendix

Table 23: Qwen BRUMO 2025, Example 16: Four-Bear Deduction Puzzle (59.6% token reduction). Both models are correct; BCR organizes the surviving candidate tuples into a compact uniqueness table much earlier.

Ex.	Source	Base Tok	BCR Tok	Reduction	Dominant Mechanism
1	AIME 2025	7,211	3,639	49.5%	Metacognitive loop elimination
2	AIME 2025	6,676	4,164	37.6%	Redundant verification removal
3	AIME 2025	3,530	2,216	37.2%	Strategy exploration reduction
4	AMC 2023	10,122	3,261	67.8%	Error prevention via conciseness
5	AMC 2023	3,464	1,425	58.9%	Pedagogical narration removal
6	AMC 2023	4,329	2,182	49.6%	Redundant verification removal

Appendix

Table 24: Summary of qualitative examples across three benchmarks and four compression mechanisms. Reductions measured as percentage decrease in token usage relative to the baseline.

Efficient Language Model Reasoning

Efficient language model reasoning studies how to improve problem-solving accuracy while reducing inference, memory, latency, and training costs.

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Mar 29, 2026

Avoiding Overthinking and Underthinking: Curriculum-Aware Budget Scheduling for LLMs

Scaling test-time compute via extended reasoning has become a key paradigm for improving the capabilities of large language models (LLMs). However, existing approaches optimize reasoning under fixed or uniformly sampled token budgets, ignoring the fundamental mismatch between problem difficulty and allocated compute. This leads to overthinking on easy problems and underthinking on hard ones, resulting in suboptimal token efficiency across diverse reasoning scenarios. In this paper, we propose Budget-Adaptive Curriculum Reasoning (BCAE), a unified framework that jointly optimizes reasoning quality and token efficiency through three synergistic components: (1) a budget-conditioned unified policy that embeds the token budget as a continuous conditioning signal, eliminating the need for decoupled thinking and summarization strategies; (2) a curriculum-aware budget scheduler that adaptively shifts the training budget distribution from easy to hard problems based on real-time learning progress; and (3) a truncation-aware dense reward mechanism that provides fine-grained credit assignment at intermediate reasoning steps via process-level verification. We further introduce Budget-Conditioned Advantage Estimation (BCAE), a novel variance reduction technique that conditions the advantage baseline on the sampled budget, yielding more stable policy gradients. Experiments on mathematical reasoning benchmarks (MATH, GSM8K, AIME, and Minerva Math) demonstrate that BACR consistently outperforms other strong baselines across all token budgets, achieving up to 8.3% accuracy improvement under tight budgets while reducing average token consumption by 34% compared to unconstrained reasoning.

Amirul Rahman, Aisha Karim, Kenji Nakamura +1

University of Malaya

Efficient Language Model Reasoning Inference-Time Compute Allocation Overthinking in Language Models RL for Language Model Reasoning Numerical Reasoning in Language Models

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Sep 28, 2026

Efficient Reasoning via Constrained Optimization in Latent Space

Large Reasoning Models (LRMs) have shown remarkable reasoning capabilities, yet they still suffer from overthinking, generating redundant reasoning steps which incur substantial token consumption. Existing methods, such as suppressing reflective keywords or forcing shorter reasoning lengths, attempt to mitigate this issue but inevitably truncate necessary steps and induce underthinking, thereby compromising performance. To address this dilemma, we investigate the latent representations and observe that efficient reasoning steps naturally cluster into a concentrated region in latent space, while those deviating from this region tend to produce verbose sequences. To leverage this, we keep reasoning focused within this region via a quadratic program which projects deviating hidden states back into the region. Then we propose a novel training-free framework to achieve efficient reasoning that reduces token generation costs without sacrificing performance. Extensive experiments conducted on four models ranging from 1.5B to 14B, and across six benchmarks in math reasoning, coding, and scientific QA, validate the effectiveness of our method, up to a 12.1% improvement in accuracy while reducing generated tokens by 11.8% to 52.8%. Codes are available at https://github.com/hzn18/Opt4Reasoning.

Zhinan Hou, Xingchen Li, Keyou You

Tsinghua University

Overthinking in Language Models Efficient Language Model Reasoning Quadratic Programming Efficient Language Model Inference Latent Space Optimization Large Reasoning Models

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Aug 30, 2026

To Think or Not to Think: Allocating Reasoning Where It Helps

Reinforcement learning (RL) has proven effective in enhancing the reasoning performance of large language models (LLMs), particularly in complex mathematical and programming tasks. However, this capability comes with systematic length misallocation, in which models devote excessive reasoning to simple questions while terminating prematurely on harder ones, degrading inference efficiency with negligible accuracy improvement. Many length-adaptive methods mitigate this issue by allocating token budgets according to question difficulty, under the implicit assumption that harder questions benefit monotonically from extended reasoning. In contrast, we find that the effect of reasoning length on accuracy is concentrated on partially solvable questions. Our further analysis reveals that explicit length rewards can produce unintended training dynamics. Motivated by these findings, we propose CARE---Contrastive Accuracy Reward Estimation---which compares the beneficial length adjustment per question from online sampled responses and applies adaptive length rewards within Group Relative Policy Optimization, with no extra hyperparameters or additional inference cost. Experiments across multiple reasoning benchmarks demonstrate that our method improves Pass@1 by up to

4\%

while simultaneously reducing reasoning length by

37\%

, achieving higher token efficiency. Code will be available upon the acceptance of this paper.

Zhengdong He, Yunfan Zhou, Jianguo Yao +2

Shanghai Key Laboratory of Scalable Computing and Systems · School of Computer Science, Shanghai Jiao Tong University

Efficient Language Model Reasoning RL for Language Model Reasoning

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Abstract

Efficient Language Model Reasoning

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Avoiding Overthinking and Underthinking: Curriculum-Aware Budget Scheduling for LLMs

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To Think or Not to Think: Allocating Reasoning Where It Helps