Tag Archives: conjecture

Midpoint Polygon Conjecture is false

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Contrary to my expectations, my previous post turned out to be like  Popular-­Lonely primes and Decimal Problem, i.e. I discovered nothing new.

My conjecture is false. Following counterexample is given on pp. 234 of this paper:

poly-counter

Counterexample of the conjecture, taken from: Berlekamp, E. R., E. N. Gilbert, and F. W. Sinden. “A Polygon Problem.” The American Mathematical Monthly 72, no. 3 (1965): 233-41. doi:10.2307/2313689.

As pointed out by uncombed_coconut, the correct theorem is:

Theorem (Berlekamp-Gilbert-Sinden, 1965). For almost all simple polygons there exist a smallest natural number k such that after k iterations of midpoint polygon procedure, we obtain a convex polygon.

The proof of this theorem is very interesting. Till now I thought that proving euclidean geometry theorems using complex numbers was an overkill. But using an N-tuple of complex numbers to represent the vertices of a closed polygon (in given order),  \mathbf{z} = (z_1,\ldots , z_N),  we can restate the problem in terms of eigenvectors (referred to as eigenpolygons) and eigenvalues.  Following are the crucial facts used is the proof:

  • An arbitrary N-gon (need not be simple) can be written as a sum of regular N-gons i.e. the eigenvalues are distinct.
  • The coefficient of k^{th} eigenvector (when N-gon is written as linear combination of eigenpolygons) is the centroid of the polygon obtained by “winding” \mathbf{z} k times.
  • All vertices of the midpoint polygons (obtained by repeating the midpoint polygon procedure infinitely many times) converge to the centroid.
  • The sum of two convex components of \mathbf{z} is a polygon. This polygon is the affine image of a regular convex N-gon whose all vertices lie on an ellipse. (as pointed out by Nikhil)
  • A necessary and sufficient condition for \mathbf{z} to have a convex midpoint polygon (after some finite iterations of the midpoint polygon procedure) is that the ellipse circumscribing the sum of two convex components of \mathbf{z} is nondegenerate. (The degenerate form of an ellipse is a point. )

For a nice outline of the proof, please refer to the comment by uncombed_coconut on previous post.

Since I didn’t know that this is a well studied problem (and that too by a well known mathematician!) I was trying to prove it on my own. Though I didn’t make much progress, but I discovered some interesting theorems which I will share in my future posts.

Midpoint Polygon Conjecture

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This is going to be yet another ambitious post like New Diagonal Contribution Theorem, and Cross Diagonal Cover Problem.

Let’s begin with following definitions from Wikipedia:

A convex polygon is a simple polygon (not self-intersecting) in which no line segment between two points on the boundary ever goes outside the polygon.

A concave polygon is a simple polygon (not self-intersecting) which has at least one reflex interior angle – that is, an angle with a measure that is between 180 degrees and 360 degrees exclusive.

We know that 3 sided polygons (a.k.a triangles) are always convex.  So, they are not very interesting. Let’s define the following procedure:

Midpoint Polygon Procedure: Given a n-sided simple polygon, join the consecutive midpoints of the sides to generate another n-sided simple polygon.

Here are few examples:

concave-convex

Red polygon is the original polygon, green polygon is the one generated by joining midpoints.

We observe that if the original polygon was a concave polygon then the polygon generated by joining midpoints need not be a convex polygon. So, let’s try to observe what happens when we apply the midpoint polygon procedure iteratively on the 4th case above (i.e. when we didn’t get a convex polygon):

big-gon

In fourth iteration of the described procedure, we get the convex polygon.

Based on this observation, I want to make following conjecture:

Midpoint Polygon Conjecture: For every simple polygon there exists a smallest natural number k such that after k iterations of midpoint polygon procedure, we obtain a convex polygon.

If the given polygon is convex, then there is nothing to prove since k=1, but I have no idea about how to prove it for concave polygons. I tried to Google for the answer, but couldn’t find anything similar. So, if you know the proof or counterexample of this conjecture, please let me know.

Cross Diagonal Cover – III

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I found many counterexamples to my conjecture, like

  • for Case 2 in 7 by 5 grid we have 12, 11 by 5 grid we have 19 and 15 by 5 grid we have 26 filled squares
  • for Case 3 in 4 by 10 grid we have 10 filled squares

In a comment by grant93jr   following part of of my initial question, to determine the number of squares in m\times n grid that can be covered by crosses (x) by following Cross Diagonal Cover Algorithm, was proved:

Given m>n, whenever n-1 divides m-1 we have m filled squares.

Examples of such grids are: 3 by 5, 4 by 7, 4 by 10, ….

Also in that comment it has been suggested that the algorithm terminates after k(m-1) = l(n-1) steps where k, l are natural numbers. It is certainly true, but I haven’t been able to use this idea for counting number of filled squares when n-1 does not divide m-1 since in that case some squares are filled twice.

There is another unexplained observation:

Why no filled square has more than 2 crosses?

Frustrated by so many unanswered questions I started colouring squares, so that number of times I visited a square is not visible. Since I really don’t care about how many times I visited  given square while counting the number of filled squares this may help in understanding the underlying symmetry.

New Doc 21_1

Replacing cross (x) by any colour and applying Cross Diagonal Cover Algorithm

After observing above diagrams I suspect that my algorithm leads to a non-deterministic cellular automata.

So, the question which still remains to be answered is:

How many filled squares are there when  m>n and n-1 does not divide m-1 ?

Examples of such grids are: 7 by 5, 11 by 5, 15 by 5 …

 

Cross Diagonal Cover – II

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I will continue the previous post. Now the question to  be answered is that:

What percentage (or how many) of given m\times n grid can be covered by following Cross Diagonal Cover Algorithm?

I did some calculations, for example:

 

New Doc 16_1

As per my calculations, I have following conjecture:

The following 3 cases are possible:

  1. If m = n, then m squares will be covered.
  2. If m>n and both m,n are odd, then m squares will be covered.
  3. If at least one of m,n is even then \frac{mn}{2} squares will be covered.

As you can see that Case – 1 is quite trivial. I tried to prove other two cases for one week, but failed, so decided to post it as conjecture.