Rupert's Property

88 pointsposted 2 days ago
by robinhouston
(johncarlosbaez.wordpress.com)

20 Comments

Strilanc

a day ago

Oh damn, in this year's sigbovik, Tom7 was trying to find out if shapes were Rupert or not: https://sigbovik.org/2025/proceedings.pdf#page=346

robinhouston

a day ago

I believe that the name ‘Noperthedron’ for this new polyhedron that has been proven not to be Rupert was given in homage to tom7’s coinage ‘Nopert’ in that SIGBOVIK paper.

dwrensha

2 days ago

Last month, before this result came out, the question "Is Every Convex Polyhedron Rupert?" was added as a formal Lean statement to Google's Formal Conjectures repository:

https://github.com/google-deepmind/formal-conjectures/blob/1...

I wonder how feasible it would be to formalize this new proof in Lean.

robinhouston

a day ago

Interesting. My guess is that it's not prohibitively hard, and that someone will probably do it. (There may be a technical difficulty I don't know about, though.)

David Renshaw recently gave a formal proof in Lean that the triakis tetrahedron does have Rupert's property: https://youtu.be/jDTPBdxmxKw

dwrensha

a day ago

> David Renshaw recently gave a formal proof in Lean that the triakis tetrahedron does have Rupert's property

That's me!

This result appears to be significantly harder to formalize.

Steininger and Yurkevich's proof certificate is a 2.5GB tree that partitions the state space into 18 million cells and takes 30 hours to validate in SageMath.

Formalizing the various helper lemmas in the paper does seem achievable to me, but I suspect that applying them to all of the millions of cells as part of a single Lean theorem could present some significant engineering difficulties. I think it'd be a fun challenge!

If that turns out to be infeasible, an alternate approach might be: we could write a Lean proof that the 2.5GB tree faithfully encodes the original problem, while still delegating the validation of that tree to an external SageMath process. Such a formalization would at least increase our confidence that there are no math errors in the setup. A similar approach was taken recently by Bernardo Subercaseaux et al in their recent paper where they formally verified a SAT-solver encoding for the "empty hexagon number": https://arxiv.org/abs/2403.17370

mananaysiempre

a day ago

That sounds like the current proof is too brute-force—too badly understood by humans—for humans to be able to explain it to Lean?

yorwba

a day ago

The most annoying bit might be that they use different, though equivalent, definitions of the property, so you would also need to formalize the proof of the equivalence of definitions.

karmakaze

2 days ago

Intuitively not surprising as the property doesn't hold for a sphere which can be approximated. But there's a world of difference between intuition and proof, especially on the edge.

I would hope there are others with more faces that don't have the property and this could have the fewest faces.

oersted

a day ago

Here's the Rupert in question. What a dude! Eminently impressive.

https://en.wikipedia.org/wiki/Prince_Rupert_of_the_Rhine

armoredkitten

a day ago

When I saw the title "Rupert's Property", I immediately thought of Rupert's Land which used to exist in Canada[0] (a large area around Hudson Bay, essentially). And as it turns out, it's the same guy! So apparently Canada also can be said to have Rupert's Property ;)

[0] https://en.wikipedia.org/wiki/Rupert%27s_Land