Don't Think Longer, Think Wisely: Optimizing Thinking Dynamics for Large Reasoning Models

While recent success of large reasoning models (LRMs) significantly advanced LLMs' reasoning capability by optimizing the final answer accuracy using reinforcement learning, they may also drastically increase the output length due to overthinking-characterized by unnecessarily complex reasoning paths that waste computation and potentially degrade the performance. We hypothesize that such inefficiencies stem from LRMs' limited capability to dynamically select the proper modular reasoning strategies, termed thinking patterns at the right position. To investigate this hypothesis, we propose a dynamic optimization framework that segments model-generated reasoning paths into distinct thinking patterns, systematically identifying and promoting beneficial patterns that improve the answer while removing detrimental ones. Empirical analysis confirms that our optimized thinking paths yield more concise yet sufficiently informative trajectories, enhancing reasoning efficiency by reducing attention FLOPs by up to 47% while maintaining accuracy for originally correct responses. Moreover, a non-trivial portion of originally incorrect responses are transformed into correct ones, achieving a 15.6% accuracy improvement with reduced length. Motivated by the improvement brought by the optimized thinking paths, we apply a preference optimization technique supported by a pairwise dataset contrasting suboptimal and optimal reasoning paths. Experimental evaluations across multiple mathematical reasoning benchmarks reveal that our method notably reduces computational overhead while simultaneously improving reasoning accuracy, achieving up to a 12% accuracy improvement and reducing token usage from approximately 5,000 to 3,000 tokens. Q: If a and b are integers such that ab ≡ 17 mod 20, then what is the remainder when (a+10)(b+10) is divided by 20? (Answer: 17) Wait , 10a + 10b can be factored as 10 x (a + b). So, 10(a + b) mod 20. Hmm, so that term is 10 times something, and modulo 20. I remember that if you have a multiple of 10 modulo 20, it's either 0 or 10. Because 10 * even number is 0 mod 20, and 10 * odd number is 10 mod 20. So, 10(a + b) is either 0 or 10 mod 20, depending on whether (a + b) is even or odd. So, we have 17 + 10(a + b) mod 20. So, that would be: If a + b is even, then 10(a + b) ≡ 0 mod 20, so the total is 17 + 0 = 17 mod 20. If a + b is odd, then 10(a + b) ≡ 10 mod 20, so the total is 17 + 10 = 27 mod 20, which is 7. Keep Pattern 2 Ptn 2 Wait , so depending on whether a + b is even or odd, the result is either 17 or 7. … it's possible that a + b can be either even or odd, so we might have two possible remainders. Remove Pattern 3 Ptn 3 Wait, maybe I can express (a + 10)(b + 10) as ab + 10a + 10b + 100, which we already did. ... So, 10(a + b) mod 20 is either 0 or 10, depending on a + b. … Remove Pattern 11 Ptn 11 Wait , but maybe we can find the parity of a + b. Let's think about that. Given that ab ≡ 17 mod 20. So, ab is congruent to 17 mod 20. Let's think about the possible values of a and b modulo 20. Since 20 is the modulus, a and b can each be from 0 to 19. ... both a and b are odd, so their sum a + b is even. Because odd + odd = even. So, a + b is even. Therefore, 10*(a + b) mod 20 is 0, since 10even = multiple of 20. Therefore, 10(a + b) ≡ 0 mod 20. So, going back, (a + 10)(b + 10) ≡ 17 + 0 = 17 mod 20. Keep Pattern 12 Ptn 12 But hold on , earlier I thought it could be 17 or 7, but now since both a and b are odd, their sum is even, so 10(a + b) is 0 mod 20, so 17 + 0 is 17. So, the remainder is 17. Keep Pattern 13 Ptn 13 … … Wait , this is getting a bit tedious, but I think the key point is that a and b must both be odd because 17 is odd and 20 is even, so their product has to be odd, so both a and b are odd. … Ptn 37