cs.CLFeb 22, 2026

IAPO: Information-Aware Policy Optimization for Token-Efficient Reasoning

Authors: Yinhan He, Yaochen Zhu, Mingjia Shi, Wendy Zheng, Lin Su, Xiaoqing Wang, Qi Guo, Jundong Li

Organizations: University of Virginia, Charlottesville, VA, USA · LinkedIn Inc., Sunnyvale, CA, USA

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Abstract

Large language models increasingly rely on long chains of thought to improve accuracy, yet such gains come with substantial inference-time costs. We revisit token-efficient post-training and argue that existing sequence-level reward-shaping methods offer limited control over how reasoning effort is allocated across tokens. To bridge the gap, we propose IAPO, an information-theoretic post-training framework that assigns token-wise advantages based on each token's conditional mutual information (MI) with the final answer. This yields an explicit, principled mechanism for identifying informative reasoning steps and suppressing low-utility exploration. We provide a theoretical analysis showing that our IAPO can induce monotonic reductions in reasoning verbosity without harming correctness. Empirically, IAPO consistently improves reasoning accuracy while reducing reasoning length by up to 36%, outperforming existing token-efficient RL methods across various reasoning datasets. Extensive empirical evaluations demonstrate that information-aware advantage shaping is a powerful and general direction for token-efficient post-training. The code is available at https://github.com/YinhanHe123/IAPO.

Figures & tables

Figure 1: Reasoning verbosity of RL post-trained LLMs. (a) Comparison of reasoning length between LLM (DeepSeekR1-Distilled-Qwen-1.5B ( DeepSeek-AI, 2025 ) ) and a human volunteer on math problems ( Lightman et al., 2024 ) . (b) Illustration of why the reasoning generated by the LLM are unnecessarily verbose.
Figure 2: Illustration of the information-aware advantage shaping module, where si,js_{i,j} and ci,jc_{i,j} are token-wise advantages of the informativeness level and exploration adjustment of the jj th token in the ii th completion oio_{i} of the completions group {oi}i=1G\{o_{i}\}_{i=1}^{G} .
Figure 3: Illustration of early-exit-based conditional MI estimator. We highlight “ o_{{\color[rgb]{1,0,0}<}t} ” and “ o_{{\color[rgb]{1,0,0}\leq}t} ” to emphasis the inclusion and exclusion of the current examined token oto_{t} in the partial completions.
Figure 4: Illustration of the naive implementation and KV cache preloading technique in conditional MI estimation. We highlight the key, query, and values of the prompt postfix in red . We show the time complexities in the right, where KK is the length of the prompt postfix, NN is ∣oi∣|o_{i}| , LL is ∣q∣+∣oi∣|q|+|o_{i}| , and dd is the embedding dimension of the LLM. The technique is significantly faster than the naive implementation since the condition K≪LK\ll L holds.
Figure 5: Ablation Study for IAPO .
MethodP@16P@32L@16L@32R@16R@32
Dataset GSM8K ( Cobbe et al., 2021 )
Qwen2.5-0.5B-Instruct0.4261± 0.00990.5722± 0.0081148.04± 3.51149.64± 6.012.88×10−32.88\times 10^{-3}3.82×10−33.82\times 10^{-3}
+ DAPO ( Yu et al., 2025 )0.8605± 0.00060.9050± 0.0004172.95± 0.27172.53± 0.094.98×10−34.98\times 10^{-3}5.25×10−35.25\times 10^{-3}
+ GFPO ( Shrivastava et al., 2025 )0.8590± 0.00560.9052± 0.0043216.26± 0.31215.99± 0.123.97×10−33.97\times 10^{-3}4.19×10−34.19\times 10^{-3}
+ GTPO ( Tan et al., 2025 )0.8519± 0.00410.9085± 0.0028243.17± 0.26243.21± 0.323.50×10−33.50\times 10^{-3}3.74×10−33.74\times 10^{-3}
+ S-GRPO ( Lee and Tong, 2025 )0.8489± 0.00770.9017± 0.0009159.70± 0.13159.49± 0.205.32×10−3‾\underline{5.32\times 10^{-3}}5.65×10−3‾\underline{5.65\times 10^{-3}}
Table 1: Comparison of IAPO with baselines across three mathematical reasoning datasets and three LLMs in various scales. P@ kk , L@ kk , and R@ kk denote Pass@ kk , Length@ kk , and the ratio Pass@ kk /Length@ kk , respectively. The optimal R@ kk s are in bold , the runner-ups are underlined . Results are shown for k∈{16,32}k\in\{16,32\} ; for additional results with k∈{2,4,8}k\in\{2,4,8\} , please see Appendix C.2.2 .
Figure 6: Parameter Analysis for IAPO .
Figure 7: Case study: We show reasoning trajectories generated by Qwen2.5-7B-Instruct ( Team, 2024 ) post-trained by IAPO and baselines for the same query from GSM8K ( Cobbe et al., 2021 ) dataset. IAPO produces a substantially shorter yet sufficient completion by concentrating generation on high-informativeness tokens.
Appendix figures & tables22 assets

Supplementary material from the paper’s appendix.

Appendix

MethodP@16P@32L@16L@32R@16R@32
DAPO ( Yu et al., 2025 )0.7111± 0.03140.7333± 0.000025.51± 0.1225.28± 0.172.79×10−22.79\times 10^{-2}2.90×10−22.90\times 10^{-2}
GFPO ( Shrivastava et al., 2025 )0.6667± 0.00000.6667± 0.000017.71± 0.1618.02± 0.373.76×10−2\mathbf{3.76\times 10^{-2}}3.70×10−2\mathbf{3.70\times 10^{-2}}
GTPO ( Tan et al., 2025 )0.7111± 0.06290.7333± 0.0544108.91± 0.53107.16± 2.106.52×10−36.52\times 10^{-3}6.84×10−36.84\times 10^{-3}
S-GRPO ( Lee and Tong, 2025 )0.7556± 0.03140.7778± 0.031425.15± 0.2225.21± 0.143.00×10−23.00\times 10^{-2}3.09×10−23.09\times 10^{-2}
+ IAPO (ours)0.8000± 0.00000.8000± 0.000024.01± 0.2823.78± 0.023.33×10−2‾\underline{3.33\times 10^{-2}}3.36×10−2‾\underline{3.36\times 10^{-2}}
Appendix
Table 2: Comparison of IAPO with various baselines on CommonsenseQA ( Talmor et al., 2019 ) and Qwen2.5-0.5B-Instruct ( Team, 2024 ) . P@ kk , L@ kk , and R@ kk denote Pass@ kk , Length@ kk , and the ratio Pass@ kk /Length@ kk , respectively.
   DAPOGFPOGTPOS-GRPOIAPO
0.5BGSM8KP@20.6515±0.00660.6027±0.00490.6138±0.00890.6113±0.01470.6247±0.0022
L@2172.40±2.59217.02±1.56241.73±1.55159.72±0.97151.00±1.78
R@23.78×10−33.78\times 10^{-3}2.78×10−32.78\times 10^{-3}2.54×10−32.54\times 10^{-3}3.83×10−33.83\times 10^{-3}4.14×10−34.14\times 10^{-3}
Math-17kP@20.0867±0.00940.0400±0.01630.0400±0.01630.0733±0.02490.1000±0.0163
L@232.52±0.5816.86±0.5515.93±0.1917.64±0.0317.43±0.04
R@22.67×10−32.67\times 10^{-3}2.37×10−32.37\times 10^{-3}2.51×10−32.51\times 10^{-3}4.16×10−34.16\times 10^{-3}5.74×10−35.74\times 10^{-3}
Appendix
Table 3: Effectiveness and efficiency of IAPO and baselines evaluated with P@2, L@2, and R@2.
   DAPOGFPOGTPOS-GRPOIAPO
0.5BGSM8KP@40.7374±0.00560.7028±0.01250.7159±0.00950.7111±0.00590.7165±0.0049
L@4172.31±1.65215.23±1.60242.17±0.15159.34±0.88150.32±0.86
R@44.28×10−34.28\times 10^{-3}3.27×10−33.27\times 10^{-3}2.96×10−32.96\times 10^{-3}4.46×10−34.46\times 10^{-3}4.77×10−34.77\times 10^{-3}
Math-17kP@40.1467±0.01890.0533±0.00940.0667±0.01890.1133±0.01890.1200±0.0163
L@433.10±0.4516.60±0.2815.86±0.1117.60±0.1317.39±0.07
R@44.43×10−34.43\times 10^{-3}3.21×10−33.21\times 10^{-3}4.21×10−34.21\times 10^{-3}6.44×10−36.44\times 10^{-3}6.90×10−36.90\times 10^{-3}
Appendix
Table 4: Effectiveness and efficiency of IAPO and baselines evaluated with P@4, L@4, and R@4.
   DAPOGFPOGTPOS-GRPOIAPO
0.5BGSM8KP@80.8067±0.00190.7923±0.00870.7930±0.00590.7902±0.00530.7928±0.0054
L@8172.94±0.21216.56±1.86242.81±0.22159.81±0.35150.55±0.48
R@84.66×10−34.66\times 10^{-3}3.66×10−33.66\times 10^{-3}3.27×10−33.27\times 10^{-3}4.94×10−34.94\times 10^{-3}5.27×10−35.27\times 10^{-3}
Math-17kP@80.1867±0.00940.1600±0.02830.1133±0.03400.1400±0.03270.1800±0.0000
L@832.95±0.2216.56±0.0115.82±0.0417.49±0.1217.36±0.04
R@85.67×10−35.67\times 10^{-3}9.66×10−39.66\times 10^{-3}7.16×10−37.16\times 10^{-3}8.00×10−38.00\times 10^{-3}1.04×10−21.04\times 10^{-2}
Appendix
Table 5: Effectiveness and efficiency of IAPO and baselines evaluated with P@8, L@8, and R@8.
Figure 8: Wall-clock LLM inference time before and after IAPO training.
Figure 9: The dynamics of correctness reward during the LLM post-training process.
Figure 10: The dynamics of completion lengths during the LLM post-training process.
Figure 11: The dynamics of the ratio between the mean correctness reward and the mean completion lengths during the LLM post-training process.
Figure 12: Case study on the effectiveness of the conditional MI in identifying informative tokens. Each subplot is a reasoning completion generated by Qwen2.5-0.5B-Instruct on a question from the GSM8K dataset. (Part 1 of 13)
Figure 13: Case study on the effectiveness of the conditional MI in identifying informative tokens. Each subplot is a reasoning completion generated by Qwen2.5-0.5B-Instruct on a question from the GSM8K dataset. (Part 2 of 13)
Figure 14: Case study on the effectiveness of the conditional MI in identifying informative tokens. Each subplot is a reasoning completion generated by Qwen2.5-0.5B-Instruct on a question from the GSM8K dataset. (Part 3 of 13)
Figure 15: Case study on the effectiveness of the conditional MI in identifying informative tokens. Each subplot is a reasoning completion generated by Qwen2.5-0.5B-Instruct on a question from the GSM8K dataset. (Part 4 of 13)
Figure 16: Case study on the effectiveness of the conditional MI in identifying informative tokens. Each subplot is a reasoning completion generated by Qwen2.5-0.5B-Instruct on a question from the GSM8K dataset. (Part 5 of 13)
Figure 17: Case study on the effectiveness of the conditional MI in identifying informative tokens. Each subplot is a reasoning completion generated by Qwen2.5-0.5B-Instruct on a question from the GSM8K dataset. (Part 6 of 13)
Figure 18: Case study on the effectiveness of the conditional MI in identifying informative tokens. Each subplot is a reasoning completion generated by Qwen2.5-0.5B-Instruct on a question from the GSM8K dataset. (Part 7 of 13)
Figure 19: Case study on the effectiveness of the conditional MI in identifying informative tokens. Each subplot is a reasoning completion generated by Qwen2.5-0.5B-Instruct on a question from the GSM8K dataset. (Part 8 of 13)
Figure 20: Case study on the effectiveness of the conditional MI in identifying informative tokens. Each subplot is a reasoning completion generated by Qwen2.5-0.5B-Instruct on a question from the GSM8K dataset. (Part 9 of 13)
Figure 21: Case study on the effectiveness of the conditional MI in identifying informative tokens. Each subplot is a reasoning completion generated by Qwen2.5-0.5B-Instruct on a question from the GSM8K dataset. (Part 10 of 13)
Figure 22: Case study on the effectiveness of the conditional MI in identifying informative tokens. Each subplot is a reasoning completion generated by Qwen2.5-0.5B-Instruct on a question from the GSM8K dataset. (Part 11 of 13)
Figure 23: Case study on the effectiveness of the conditional MI in identifying informative tokens. Each subplot is a reasoning completion generated by Qwen2.5-0.5B-Instruct on a question from the GSM8K dataset. (Part 12 of 13)
Figure 24: Case study on the effectiveness of the conditional MI in identifying informative tokens. Each subplot is a reasoning completion generated by Qwen2.5-0.5B-Instruct on a question from the GSM8K dataset. (Part 13 of 13)
ProblemGFPOGTPOSGRPOLeeDAPOIAPO
Robe problemFor the robe, it requires 2 bolts of blue fiber. Since the amount of white fiber needed is half that of the blue fiber, we calculate half of 2 bolts, which is 1 bolt. Adding the blue and white fibers together, we get 2 + 1 = 3 bolts in total. Answer: 3To determine the total number of bolts needed for the robe, we first look at the amount of each type of fiber required. The problem states that a robe requires 2 bolts of blue fiber and half that much white fiber. Since half of 2 is 1, the robe will require 1 bolt of white fiber. This means the total number of bolts needed is the sum of the bolts of blue fiber and the bolts of white fiber. So, 2 (blue fiber) + 1 (white fiber) = 3 bolts in total. The robe will need a total of 3 bolts of fiber. Answer: 3First, we calculate how much white fiber is needed, which is half the amount of blue fiber. Since the robe takes 2 bolts of blue fiber, the amount of white fiber needed is 2 / 2 = 1 bolt. The total amount of fiber needed is the sum of the blue and white fibers, so 2 + 1 = 3 bolts. Answer: 3First, we need to determine how much white fiber is required. Since the amount of white fiber is half of the blue fiber, we have 2/2 = 1 bolt of white fiber. Then, we add the bolts of blue fiber and white fiber together: 2 + 1 = 3 bolts in total. Answer: 32 divided by 2 is 1, so the total is 2 + 1 = 3. Answer: 3
Duck eggs problemFirst, we calculate the total number of eggs Janet has per day which is 16. Janet consumes 3 eggs for breakfast every day, leaving 16 - 3 = 13 eggs. Then she uses 4 eggs to bake muffins for her friends, leaving 13 - 4 = 9 eggs. Since she sells the remaining eggs at the farmers’ market for 2each,shemakes9∗2 each, she makes 9 * 2 = $18 per day from selling eggs. Answer: 18To find out how much money Janet makes every day at the farmers’ market, let’s first determine how many eggs she has left after consuming some for her personal use. Janet’s ducks lay 16 eggs per day. She eats 3 eggs for breakfast, so the number of eggs left after breakfast is 16 - 3 = 13 eggs. She also uses 4 eggs to bake muffins for her friends, so the number of eggs left after baking is 13 - 4 = 9 eggs. These 9 eggs are what she sells at the farmers’ market. Since each egg is sold for 2,theamountofmoneyshemakesfromselling9eggsis9∗2=18dollars.Therefore,Janetmakes2, the amount of money she makes from selling 9 eggs is 9 * 2 = 18 dollars. Therefore, Janet makes 18 every day at the farmers’ market. Answer: 18Janet’s ducks lay 16 eggs per day. She eats 3 eggs every morning for breakfast, so the number of eggs used for breakfast is 3. She also uses 4 eggs per day to bake muffins for her friends, so the number of eggs used for baking is 4. The total number of eggs used every day is 3 + 4 = 7 eggs. The number of eggs left to sell at the farmers’ market is 16 - 7 = 9 eggs. She sells the remaining eggs at $2 per egg, so the amount she makes from selling the eggs is 9 * 2 = 18 dollars. Answer: 18Janet’s ducks lay 16 eggs per day. She eats 3 eggs every morning for breakfast and uses 4 eggs every day to bake muffins for her friends. In total, she uses 3 + 4 = 7 eggs every day. The number of eggs left to sell at the farmers’ market is 16 - 7 = 9 eggs. She sells each egg for 2,sotheamountshemakeseverydayatthefarmers’marketis9∗2, so the amount she makes every day at the farmers’ market is 9 * 2 = $18. Answer: 18Janet gets 16 - 3 - 4 = 9 eggs per day. At $2 per egg, she makes 9 * 2 = 18 dollars. Answer: 18
Sprints problemTo find out how many total meters James runs in a week, we first determine how many meters he runs in one session. Since he runs 3 sprints each session and each sprint is 60 meters, he runs 3 * 60 = 180 meters per session. Since he does this 3 times a week, the total meters he runs in a week is 180 * 3 = 540 meters. Answer: 540Let’s break down the problem step by step. First, we need to find out how many meters James runs in a single sprint session. Since each sprint is 60 meters and he does 3 sprints per session, the total distance he runs in one session is 3 * 60 = 180 meters. Next, we need to find out how many meters he runs in a week. Since he does these sprint sessions 3 times a week, the total distance he runs in a week is 180 * 3 = 540 meters. Therefore, James runs 540 meters in a week. Answer: 540First, we calculate the number of meters James runs in one session. He runs 3 sprints and each sprint is 60 meters, so in one session, he runs 3 * 60 = 180 meters. Then we calculate the total for the week. He runs 3 times a week, so the total meters he runs in a week is 180 * 3 = 540 meters. Answer: 540First, we calculate the total distance James runs in one sprint session. Since he runs 60 meters each sprint and does 3 sprints per session, the distance covered in one session is 60 * 3 = 180 meters. Next, since he does 3 sessions a week, the total distance he runs in a week is 180 * 3 = 540 meters. Answer: 540Each sprint is 60 meters, and he does 3 sprints 3 times a week, so 3 * 60 * 3 = 540 meters. Answer: 540
Glasses problemTo solve this, we need to calculate the total cost of 16 glasses, where every second glass is at 60% of the price. First, let’s determine the cost of one pair of glasses: the first glass is 5andthesecondis605 and the second is 60% of 5, which is 3.Thetotalforonepairis3. The total for one pair is 5 + 3=3 = 8. Since Kylar is buying 16 glasses, which form 8 pairs, we multiply the cost of one pair by 8: 8∗8=8 * 8 = 64. Answer: 64To determine the total cost of 16 glasses, we need to consider the pricing scheme. The first glass is priced at 5,andeverysecondglassispricedat605, and every second glass is priced at 60% of 5, which is 5∗0.6=5 * 0.6 = 3. Therefore, the cost of a pair (1 + 1) of glasses is 5+5 + 3 = 8.SinceKylarisbuying16glasses,wecanimaginethesequenceinpairs,andsince16isanevennumber,thereareexactly8pairs(because16/2=8).So,for8pairs,thetotalcostis8∗8=64.Hence,Kylarneedstopay8. Since Kylar is buying 16 glasses, we can imagine the sequence in pairs, and since 16 is an even number, there are exactly 8 pairs (because 16 / 2 = 8). So, for 8 pairs, the total cost is 8 * 8 = 64. Hence, Kylar needs to pay 64 for 16 glasses. Answer: 64First, we need to determine the cost for one pair of glasses (two glasses). The first glass costs 5andthesecondglasscosts605 and the second glass costs 60% of 5, which is 5∗0.6=5 * 0.6 = 3. So the total cost for one pair of glasses is 5+5 + 3 = 8.Kylarwantstobuy16glasses,whichis8pairsofglassessince16/2=8.Thetotalcostfor8pairsofglassesat8. Kylar wants to buy 16 glasses, which is 8 pairs of glasses since 16 / 2 = 8. The total cost for 8 pairs of glasses at 8 each is 8 * 8=8 = 64. Answer: 64First, let’s calculate the cost of one pair of glasses, where the second glass costs 60% of 5.So,thecostofthesecondglassis0.60∗5. So, the cost of the second glass is 0.60 * 5 = 3.Therefore,thecostofonepairofglassesis3. Therefore, the cost of one pair of glasses is 5 + 3=3 = 8. Since Kylar wants to buy 16 glasses, and he gets a discounted second glass for every pair, he will have 8 pairs of glasses. The total cost will be 8 * 8=8 = 64. Answer: 64Every second pair of glasses costs 0.6 * 5 = $3. For 16 glasses, we can group them into 8 pairs, so 8 * (5 + 3) = 8 * 8 = 64. Answer: 64
Chicken feed problemFirst, we need to calculate how many cups of feed Wendi provides to each chicken in the morning and afternoon. In the morning, she gives 15 cups, and in the afternoon, she gives 25 cups. Since she feeds each chicken 15/20 = 0.75 cups in the morning, and 25/20 = 1.25 cups in the afternoon, the total amount each chicken gets in these two meals is 0.75 + 1.25 = 2 cups. Since each chicken should get 3 cups of feed per day, the final meal must make up the difference: 3 - 2 = 1 cup of feed per chicken. With 20 chickens, Wendi needs to provide 20 * 1 = 20 cups of feed in the final meal. Answer: 20Wendi gives her chickens 15 cups of feed in the morning and 25 cups of feed in the afternoon. First, we need to determine how many cups of feed each chicken gets per day. Since there are three feedings a day and each chicken should get 3 cups of feed per day, the total feed required for all 20 chickens per day is 20 * 3 = 60 cups. In the first two feedings (morning and afternoon), Wendi has already provided 15 + 25 = 40 cups of feed. To find out how much feed is needed for the final feeding, we subtract the amount already given from the total daily requirement: 60 - 40 = 20 cups. Therefore, Wendi needs to give her chickens 20 cups of feed in the final feeding of the day. Answer: 20First, we calculate the total amount of feed given to the chickens in a day. Wendi gives 15 cups in the morning and 25 cups in the afternoon, so the total for these two meals is 15 + 25 = 40 cups. Each chicken is given 3 cups of feed per day. Since there are three meals, the total daily feed needed is 3 * 20 * 3 = 180 cups. We already know the morning and afternoon feed totals, so the feed for the final meal is 180 - 40 = 140 cups. Answer: 140First, we need to calculate the total amount of feed that each chicken consumes daily. Since each chicken gets 3 cups of feed and there are 20 chickens, the daily requirement is 20 * 3 = 60 cups of feed. Wendi gives her chickens 15 cups in the morning and 25 cups in the afternoon, so the total feed given in these two meals is 15 + 25 = 40 cups. To find out how much feed is needed in the final meal, we subtract: 60 - 40 = 20 cups. Answer: 20First, calculate the total daily feed required: 20 * 3 = 60 cups. Then, the remaining feed is 60 - (15 + 25) = 60 - 40 = 20 cups. Answer: 20
Lemon tree problemTo determine how many years it will take for Carlos to start earning money from his lemon tree, we first need to calculate the net income each year. Each year, the tree grows 7 lemons, and he sells them for 1.5each,resultinginarevenueof1.5 each, resulting in a revenue of 7\times 1.5=10.5dollars.Thecosttowaterandfeedthetreeeachyearisdollars. The cost to water and feed the tree each year is3. Therefore, the net income each year is 10.5−3=7.510.5-3=7.5 dollars. The initial cost to plant the tree is 90.Tofindouthowmanyyearsitwilltaketorecoverthiscost,wedivide:90. To find out how many years it will take to recover this cost, we divide: 90/7.5=12$ . Thus, it will take 12 years. Answer: 12Let’s break down the problem step by step. First, the initial cost of planting the lemon tree is 90.Next,eachyear,thetreewillgrow7lemons,whichCarloscansellat90. Next, each year, the tree will grow 7 lemons, which Carlos can sell at 1.5 per lemon. The cost to water and feed the tree each year is 3.TodeterminehowmanyyearsitwilltakebeforeCarlosstartsearningmoney,weneedtofindthepointwherehistotalearningsfromsellinglemonsexceedtheinitialplantingcostplustheyearlycosts.Theprofiteachyearfromsellinglemonsis7∗1.5=10.5dollars.Eachyear,thenetprofit(aftersubtractingtheyearlycost)is10.5−3=7.5dollars.Now,weneedtofindouthowmanyyearsitwilltakeforthetotalnetprofittoexceedtheinitialplantingcostof3. To determine how many years it will take before Carlos starts earning money, we need to find the point where his total earnings from selling lemons exceed the initial planting cost plus the yearly costs. The profit each year from selling lemons is 7 * 1.5 = 10.5 dollars. Each year, the net profit (after subtracting the yearly cost) is 10.5 - 3 = 7.5 dollars. Now, we need to find out how many years it will take for the total net profit to exceed the initial planting cost of 90. We can set up the equation: 7.5 * n = 90, where n is the number of years. Solving for n, we get n = 90 / 7.5 = 12. Therefore, it will take 12 years for Carlos to start earning money on the lemon tree. Answer: 12The tree costs 90toplant.Eachyearthetreeproduces7lemonsat90 to plant. Each year the tree produces 7 lemons at 1.5 each, so revenue is 7×1.5=10.57\times 1.5=10.5 dollars. Annual cost is 3.Netprofitperyearis3. Net profit per year is 10.5-3=7.5dollars.Yearstobreakeven:dollars. Years to break even:90/7.5=12$ years. Answer: 12To determine how many years it will take for Carlos to start earning money on the lemon tree, we need to calculate the costs and income over time. Let’s denote the number of years by nn . Annual income from lemons: 7×1.5=10.57\times 1.5=10.5 dollars. Annual cost: 3.Netannualprofit:3. Net annual profit: 7.5n-90.Wewantthesmallest. We want the smallestnsuchthatnetprofitispositive:such that net profit is positive:7.5n-90>0.Solving:. Solving:7.5n>90,thus, thusn>12.Since. Sincen$ must be a whole number, we round up to 13. Therefore, it will take 13 years. Answer: 13First, we calculate the net profit: 7 * 1.5 - 3 = 10.5 - 3 = 7.5. To break even, we need 90 / 7.5 = 12 years. Answer: 12
Appendix
Table 6: Comparison of completions across different methods
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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 ModelsEfficient Language Model ReasoningQuadratic ProgrammingEfficient Language Model InferenceLatent Space OptimizationLarge Reasoning Models
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