Reasoning models & test-time compute scaling

Drag the chain-of-thought length and watch accuracy climb along the curve — steeply at first, then flattening out. Every extra reasoning step still helps, but each one recovers a shrinking slice of the remaining gap to a perfect score.

Accuracy approaches, but never reaches, a ceiling as the chain of thought gets longer — a geometric decay toward $1$:

• $L$ — chain-of-thought length, in reasoning steps, spent before committing to an answer.
• $a_0$ — the single-shot accuracy at $L=0$, with no reasoning steps at all.
• $d$ — the decay factor: the fraction of the remaining gap to perfect accuracy left AFTER one more step. $d < 1$ means each step still helps, but by a shrinking amount.

1. Diminishing, not disappearing, returns

   $d^L \to 0$ as $L$ grows, so accuracy keeps climbing forever — but $d^L$ shrinks geometrically, so each additional step buys less than the one before it.

2. Compute cost is a different shape entirely

   Accuracy is concave in $L$ (bending downward), but the compute spent generating that chain of thought is linear in $L$ — a straight line, no diminishing anything.

3. A concave gain against a linear cost has an optimal stopping point

   Net utility (accuracy gained minus compute spent) rises while the marginal accuracy gain still exceeds the constant marginal cost, then falls once it doesn't — that crossing point is the compute-optimal chain-of-thought length.

Drag again, but now watch three numbers at once: accuracy climbing, compute cost climbing right alongside it in a straight line, and net utility rising, peaking, then falling. Past the peak, every extra reasoning step is spending more compute than the accuracy it buys is worth.

Base accuracy $a_0 = 0.2$, decay $d = 0.8$, cost $0.02$ compute units per reasoning step:

1. Accuracy at a few chain-of-thought lengths

   • $L=0$: $1 - 0.8 \times 0.8^0 = 1 - 0.8 \times 1 = 0.2$
   • $L=1$: $1 - 0.8 \times 0.8^1 = 1 - 0.64 = 0.36$
   • $L=4$: $0.8^4 = 0.4096$, so $1 - 0.8 \times 0.4096 = 1 - 0.32768 \approx 0.672$

   Each step's gain is smaller than the last.

2. Compute cost grows linearly

   • $L=5$ costs $0.02 \times 5 = 0.10$ units.
   • $L=10$ costs exactly twice that, $0.20$ units.

   No discount for going further.

3. Net utility peaks at L=10

   Utility $= \text{accuracy}(L) - 0.02L$ at the three candidate lengths:

   • $L=9$: $0.8^9\approx0.1342$, accuracy $=1-0.8\times0.1342\approx0.8926$, utility $\approx0.8926-0.02\times9=0.8926-0.18=0.7126$
   • $L=10$: $0.8^{10}\approx0.1074$, accuracy $=1-0.8\times0.1074\approx0.9141$, utility $\approx0.9141-0.02\times10=0.9141-0.20=0.7141$
   • $L=11$: $0.8^{11}\approx0.0859$, accuracy $=1-0.8\times0.0859\approx0.9313$, utility $\approx0.9313-0.02\times11=0.9313-0.22=0.7113$

   $L=10$'s utility ($\approx0.7141$) beats both neighbors — the compute-optimal chain-of-thought length for this cost weight, found by the same search a real inference-time compute budget would run.