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I presume to earn this prize, one need provide a formal proof. I've never been one for formal math proofs, but I think I could formulate a brute force test...
by delinka 12y ago
I presume to earn this prize, one need provide a formal proof. I've never been one for formal math proofs, but I think I could formulate a brute force test...
- ColinWright 12y agoSo you can formulate a brute force test that works for all n? That's a bold claim. Remember, you are trying to prove that for every natural number n, once you have 2^(n-2)+1 points in a plane, no three co-linear, there is a subset of n of them that form a convex n-gon. How will you brute-force that?
- delinka 12y agoCan I? I don't know for certain. I said I think I can. Also, there's nothing that says the brute force process can't run forever (presumably it must) but I'd also suspect to generate lots of data while it runs that could indeed be used (by someone else) to write the proof.
- gambiting 12y agoMaybe,maybe not. We have generated billions of decimal numbers for PI, and we are not any closer to writing a formal proof deciding if there is a pattern to them or not. Maybe a pattern will unravel once we generate billions of billions of these numbers? We don't know. I guess my point is that brute forcing a problem is unlikely to be useful here.
- peterkelly 12y agoIn base Pi there's a quite a simple pattern.
- ColinWright 12y agoAs a starter, can you write a program to deal with just the very next case? It is currently unknown as to whether a collection of 33 points guarantees a convex 7-gon. If you can prove that, even with a computer program, then you get certainly get a paper out of it, possibly jointly with Ron Graham, and possibly getting an Erdős number of 2. That would be a start. How would you do that?
- scragg 12y agoI think brute force could be used to prove him wrong (if his conjecture is incorrect). However, it won't be enough to prove correct.