

The Happy Ending Problem [video] - ColinWright
http://www.numberphile.com/videos/happy_ending.html

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ilzmastr
Reminds me of VC dimension kind of stuff for describing binary classifiers.

That first example w/ 5 points necessitating a convex 4 sided polygon is like
the VC dimension proof that an axis aligned rectangle can never correctly
classify 5 points, since you can never arrange 5 points w/o one of them being
in the body of a rectangular region surrounding the other 4 (and that interior
point's label therefore can't be arbitrary and still correctly classified).
more:
[https://www.cs.princeton.edu/courses/archive/spring13/cos511...](https://www.cs.princeton.edu/courses/archive/spring13/cos511/scribe_notes/0221.pdf)

Wonder if the 2^(n-2) + 1 rule is somewhere in the VC dim lit

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agentultra
I love graphs!

For those who found this interesting or inspiring check out "Graph Theory" by
_W. T. Tutte_ \-- it's an exceptional book for those scarred by mathematics in
elementary or secondary school and want a refresher to show them how cool it
is.

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ColinWright
This isn't actually graph theory, because the position in 2D space matters
here. In graph theory the actual placement of the vertices is irrelevant - the
only thing that matters is which ones are joined.

But I share your enthusiasm - my PhD is in graph theory.

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agentultra
Right, it's a geometrical problem... my suggestion is a bit of a tangent.

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willvarfar
It kind of feels - in my mathematical nativity - that there must be a general
way to reduce all higher n down to the proof given for n=4 rather like Euler
solved the Königsberg problem.

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delinka
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...

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ColinWright
So 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?

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delinka
_Can_ 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.

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gambiting
Maybe,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.

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peterkelly
In base Pi there's a quite a simple pattern.

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zackmorris
Does anyone know the answer to this in three dimensions? I'm curious how many
points it takes to always be able to find a convex tetrahedron.

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ColinWright
You'll need to be more careful about what you're asking for, because 4 points
in 3D always gives a convex tetrahedron, as there are no non-convex
tetrahedra. The convex hull of four points (no three colinear) is a
tetrahedron.

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dec0dedab0de
if you put all the points in a straight line wouldn't that invalidate any n?

edit: just saw the bit where no more than 3 can be in a straight line

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willvarfar
No more than two may be on any line.

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ivancamilov
I admit I was thinking of a different happy ending.

