[sade_95]

Good questions — I ran all three: Semi-minor axis: I added b as a second feature (b = a·sqrt(1−e²), so corr(a,b) = 1.00000 to 5 decimals; Mercury is the only planet where they differ by more than 2%). It still picked a and ignored b completely: T = 164.78·a·sqrt(a), R² = 0.99999998. What saves it on such a nasty collinear decoy is the constant fitting: the law is exact in a and only almost-exact in b, so Levenberg-Marquardt makes that 2% Mercury error decisive. Kepler's 6 planets: works fine, and it's actually nicer — it returned pow(a, 1.500812) explicitly. Six clean points on a power law is plenty. Noise: this is where I have to be honest. With 1% gaussian noise on T the fit is still R² = 0.9996 and a·sqrt(a) is still in there, but wrapped in junk (exp(tanh(...))). At 5% the clean form is gone — it returns a bounded exp(−a²) family that fits well but isn't the law. So fit quality degrades gracefully, symbolic recovery doesn't. Making that part noise-robust is pretty much the open frontier of the whole field, not just of my tool.

[gus_massa]

Remember to use two enters

to get a new paragraph here.

> b = a·sqrt(1−e²), so corr(a,b) = 1.00000 to 5 decimals

Isn't e different for each planet?

> Mercury is the only planet where they differ by more than 2%

I remember something about Mars been the planet with the most eccentric elipse

> *So fit quality degrades gracefully, symbolic recovery doesn't. Making that part noise-robust is pretty much the open frontier of the whole field, not just of my tool.

Nice. It's a hard problem. Which heuristic are you using to pick the "best" formula?

[sade_95]

Thanks for the formatting tip, noted.

Isn't e different for each planet?

Yes, each planet got its own e (Mercury 0.206, Venus 0.007, Earth 0.017...). The correlation still comes out at 1.00000 because all eccentricities are small while a spans 0.39 to 30 AU — a ≤2% per-planet wobble is invisible to Pearson across two decades of range. That's exactly what makes it a nasty decoy: almost collinear, but not quite.

Mars been the planet with the most eccentric ellipse

Close — Mercury is actually the most eccentric (0.206), Mars is second (0.093). Funny enough, Mars is the famous one precisely because Kepler derived his laws fighting with Tycho's Mars data: its ellipse was just eccentric enough to kill every circular fit. If Tycho had handed him Venus data instead, we might have waited a while longer.

Which heuristic are you using to pick the "best" formula?

Hold-out validation during evolution, then the old CART-style one-standard-error rule: pick the smallest formula whose validation error is within ~1 SE of the best one. Constants get refit with Levenberg-Marquardt before that comparison, so small forms compete at their best. It also returns the full accuracy-vs-size Pareto front so you can override the choice. And to connect it to your noise question: under noise that 1-SE band is exactly where wrong-but-simpler formulas sneak in and tie the true one — the two problems are really the same problem.