Happy Ending Problem - Numberphile

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Professor Ron Graham discusses the famed Happy Ending Problem and Ramsey Theory.
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NUMBERPHILE

Videos by Brady Haran

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I like how he described the concave/convex polygon thing using the turns, I never thought about that but it's so simple and make sense.

SquirrelASMR

Rip Ron graham . One of the greatest modern mathematicians

daddyken

Is anyone else happy that this channel exists?

OoleoleoleO

Ron Graham is a boss. His talk at ICCM 2013 was one of the best talks I've been too.

LeonhardEuler

"It's actually called the 'Happy Ending Theorem'"
Me: *pauses the video and laughs loudly*

TimJSwan

this is such a cool theorem/hypothesis, I love when there is something profound, you can say about groups of objects regardless of the configuration of those objects. Math is awesome!

tuftman

Ron Graham's office is awesome. Brady, it would be great to see more of it.

blueguitarbob

Ron Graham must have some kind of prestige to be able to hire Robert Duvall to portray him in YouTube videos.

GlorifiedTruth

Of course it is false! If you plug it in for a 1-gon, it'll say you need 1.5 points. Clearly, since you can't have half of a point, then it simply must be wrong.

...I'll show myself out.

arpyzero

This is a very simple and clever explanation of convex and concave polygons, didn't know this before. Greets

MiDnYTe

(1:42) I thought the "PROVEN" green box would take me to the video that proves it :(

fbDJLL

That's interresting. I like this problem.
This kind of problem is really in the trail of Erdös's usual problems or the problems he was keen on: a simple/straightforward combinatoric question told and often involving an asymptotic bound.

julienferte

Your video with Matt and Tony Padilla(?) on Graham's number is what drew me to Numberphile and (eventually) your other channels. Great job, Brady!

deproissant

In a recursive manner, tackling the problem backwards.

Building a convex shape with n sides (Dn lines) with Sn vertices, Dn not parallel to Dn-1 will yield a couple of things:
A set number of intersections
A set number of zones

If you "pull" a point on a segment of the n gone, into the zone, outside the n-gone, directly adjacent, only then can you get a n+1 convex n+1 gone. This point, now Sn+1 is the intersection of Dn and Dn+1. Since Dn+1 is not parallel to Dn, it will intersect all the other lines (new intersections, new zones).
This is the only way to build a convex shape. Any other point will yield something concave.
It does work with parallel lines, not at the start of course, lines must intersect. But Dn-3 can be parallel to Dn. Not with the optimum solution.

So the constellation of candidate points for the new gone is limited, and the minimum number of points is:
The sum of all the zones made up by Dn lines minus all the intersections of Dn lines minus 1 (this is the zone inside the n gone)

I'm french, I can clarify if needed.

fdutrey

Very interesting! There was some discussion about practical applications of researching mathematical stuff like this. At least one part of it involves motivation to develop computationally more efficient computers, since maths stuff like checking how many decimals of pi or how far in this problem can a computer compute gives some estimate about it. Also many of these kinda math theorems are in a way tied together, in other words even if this problem wouldn't have any easy-to-see practical applications, it's theoretically very possible that these theorems are analogous to some other math theorems, which could potentially have many many applications.. and so on. Cheers!

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