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sade_95last Monday at 4:02 AM1 replyview on HN

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.


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gus_massalast Monday at 1:25 PM

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?

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