Scaling laws — estimate the power-law exponent

Background

Empirical scaling laws say test loss falls as a power law in model size:

$$L(n) \approx A\,n^{-\alpha} \quad\Longleftrightarrow\quad \log L = \log A - \alpha \log n$$

So on log–log axes the loss-vs-parameters curve is a straight line whose slope is $-\alpha$. Estimating the exponent $\alpha$ is just fitting that line.

Problem statement

Implement scaling_exponent(ns, losses) that estimates $\alpha$ from measured (n, loss) pairs via a log–log linear fit.

Input

• ns — array-like of parameter counts $n$ (all $> 0$).
• losses — array-like of corresponding test losses (all $> 0$), same length.

Output

Returns a Python float: the estimated exponent $\alpha$ (the negative slope of $\log L$ vs $\log n$).

Examples

Example 1 — a clean power law

Input:  ns = [1, 4, 16, 100], losses = [10, 5, 2.5, 1.0]
Output: 0.5

Explanation: $L = 10\,n^{-0.5}$, so the log–log slope is $-0.5$ and $\alpha = 0.5$.

Example 2

Input:  ns = [1, 2, 4, 8], losses = [2, 1, 0.5, 0.25]
Output: 1.0

Explanation: $L = 2\,n^{-1}$, slope $-1$, exponent $1$.

Constraints

• Fit a line to log(losses) vs log(ns) (natural log is fine).
• Return $\alpha = -\text{slope}$ as a float.

Notes

• A larger $\alpha$ means loss drops faster as you scale up — but the law only holds when data and compute scale alongside $n$.