#570 AP book of science-- Solving the proof of the "Happy Ending Problem" of Klein convex figures// math proofs by Archimedes Plutonium

[Archimedes Plutonium]

Well, I hope Harvard's Jack Murtaugh can find a more difficult problem to solve for the King of Science, rather than these overnight problems that are too easy to solve.

I got my November Scientific American yesterday and on page 80, Jack Murtaugh outlines this math problem that is unproven. Until now.

"Happily Ever After: A puzzle about dots on a page led to one of the most profound areas of modern math"

AP writes: I would dispute that opinion of "most profound", Jack. For the most profound math is that which is prominent in Physics, and convexity does not rise to that height of profound.

So, I read this article last night, slept on it and this morning I have a proof of it.

Jack ends his article by saying this much: "The discovery matched a conjecture that Erdos and Szekeres had put forth as young men. They predicted that (2^(n-2)) + 1 points is the minimum needed to force an n-sided convex polygon. The quadrilateral, pentagonal and hexagonal cases all obey this formula, but no larger shapes have been confirmed. Erdos left behind $500 cash prize to anyone who proves that it holds in general. The prize remains unclaimed, although in 2016 mathematician Andrew Suk, then at the University of Illinois Chicago, came tantalizingly close.

AP writes:: Well the huge problem throughout mathematics, and I hate to include Computer Science for CS belongs more in engineering than it belongs in science. Its name should have been Computer Engineering and never given the inflated title of a "Science". The huge problem throughout Science is people have little to no Logical brain to be in science. I say the education system throughout the world must have 2 years mandatory Logic in college or university before a degree in science is awarded. Otherwise, all these so called scientists and professors are just fumbling around in the dark.

So, I read this last night and this morning had a proof in mind.

It is really really simple and hope Jack can find a unsolved problem that can stump me for a week or month, rather than a day.

Proof of Happy Ending Problem of its (2^(n-2)) + 1 

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Statement of Problem: I am going to borrow the statement given in Wikipedia and called the Happy Ending Problem.

Proof

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I took 2 years of College logic at UC 1968-1972 and have never regretted that. For if not for logic, I would not be in science at all.

Logic tells me that this formula (2^(n-2)) + 1 is more of obfuscation than it is of help.

Logic allows me to say that such a formula is none other than Double the number subract 1, leading to this sequence.

2, 3, 5, 9, 17, 33, .....

Do you see how I double 2 subtract 1 leads to 3, then double 3 subtract 1 leads to 5, and so on.

You see, Logic always simplifies things whereas mathematicians without Logic live in a rarefied Kook atmosphere that has more of a tendency to shroud and mask simplicity and make in kookishly feverishly complex.

So I have the Sequence I need to prove. And having a logical mind I immediately question what is the sequence if I Double than add 1 rather than subtract 1.

That sequence looks like this:

1, 3, 7, 15, 31, ....

If I started with 2 instead of 1 that sequence looks like this.

2, 5, 11, 23, 47, .....

By LOGIC, a sequence of numbers has no geometrical meaning unless I compare them to a sequence with known geometrical meaning.

By Logic we pick the most meaningful geometrical sequence that relates to the Doubling -1 and the Doubling +1.  If you guessed the Fibonacci Sequence, then you get an A+ for the day.

Fibonacci Sequence is the Logarithmic Spiral, but I prefer to call it the Equiangular Spiral.

It goes like this:

1, 1, 1+1=2, 1+2=3, 2+3=5, 3+5=8, 5+8=13, .......

Using Logic, I am going to redefine what it means to be Convex, by using the Equiangular Spiral. If points and line segments connecting those points result in Not-Convex is because the points are spread apart too much to allow the Equiangular Spiral to encompass them in its spiralling out.

The Equiangular Spiral is the sequence of 1,1,2, 3, 5, 8, 13, 21, ....

The Doubling subtract 1 is 2, 3, 5, 9, 17, 33,...

The Doubling add 1 is 1, 3, 7, 15, 31, ......

The numbers in doubling stretch too far apart in going from 9 to 17 in Doubling subtract 1 and too far apart in Doubling add 1 from 7 to 15.

Stretched too far apart when we define, by logic, that the definition of Convex is when  the Equiangular Spiral encompasses every line segment connected.

AP, King of Science

#570 AP book of science--  Solving the proof of the "Happy Ending Problem" of Klein convex figures// math proofs by Archimedes Plutonium

[Archimedes Plutonium]

I am going to use Wikipedia statement of the problem.

--- quoting Wikipedia---

 Wikipedia, the free encyclopedia

"Erdős–Szekeres conjecture" redirects here. For their theorem on monotonic subsequences, see Erdős–Szekeres theorem.

The happy ending problem: every set of five points in general position contains the vertices of a convex quadrilateral

In mathematics, the "happy ending problem" (so named by Paul Erdős because it led to the marriage of mathematicians George Szekeres and Esther Klein)^[1][2][3]is the following statement:

Theorem—any set of five points in the plane in general position^[4] has a subset of four points that form the vertices of a convex quadrilateral.

This was one of the original results that led to the development of Ramsey theory.

The happy ending theorem can be proven by a simple case analysis: if four or more points are vertices of the convex hull, any four such points can be chosen. If on the other hand, the convex hull has the form of a triangle with two points inside it, the two inner points and one of the triangle sides can be chosen. See Peterson (2000) for an illustrated explanation of this proof, and Morris & Soltan (2000) for a more detailed survey of the problem.

The Erdős–Szekeres conjecture states precisely a more general relationship between the number of points in a general-position point set and its largest subset forming a convex polygon, namely that the smallest number of points for which any general position arrangement contains a convex subset of points is . It remains unproven, but less precise bounds are known.

Larger polygons

A set of eight points in general position with no convex pentagon

Erdős & Szekeres (1935) proved the following generalisation:

Theorem—for any positive integer N, any sufficiently large finite set of points in the plane in general position has a subset of N points that form the vertices of a convex polygon.

The proof appeared in the same paper that proves the Erdős–Szekeres theorem on monotonic subsequences in sequences of numbers.

Let f(N) denote the minimum M for which any set of M points in general position must contain a convex N-gon. It is known that

• f(3) = 3, trivially.
• f(4) = 5.^[5]
• f(5) = 9.^[6] A set of eight points with no convex pentagon is shown in the illustration, demonstrating that f(5) > 8; the more difficult part of the proof is to show that every set of nine points in general position contains the vertices of a convex pentagon.
• f(6) = 17.^[7]
• The value of f(N) is unknown for all N > 6. By the result of Erdős & Szekeres (1935), f(N) is known to be finite for all finite N.

On the basis of the known values of f(N) for N = 3, 4 and 5, Erdős and Szekeres conjectured in their original paper that 

A set of sixteen points in general position with no convex hexagon

They proved later, by constructing explicit examples, that^[8]In 2016 Andrew Suk^[9] showed that for N ≥ 7

Suk actually proves, for N sufficiently large, 

This was subsequently improved to:^[10]

So in my proof, I compare the Fibonacci sequence to the Happy Ending Problem.

The Doubling add 1 is 2, 5, 11, 23, 47, .....

The numbers in doubling stretch too far apart in going from 9 to 17 in Doubling subtract 1 and too far apart in Doubling add 1 from 5 to 11.