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caird coinheringaahing

04:07

Well, looks like the challenge won't make it to 2018. 24th is the deadline.

4 hours later…

Mr. Xcoder

07:42

@cairdcoinheringaahing so you are saying that the next sequence won’t be solved by anyone and the challenge will die? Shame.

Wheat Wizard

It doesn't look that hard

I'm not going to do it but, it doesn't look that hard

If I could do it in Haskell I would do it but I don't want to do a hard challenge in a language I don't know.

Wheat Wizard

08:02

Eh I take it back this is pretty hard

Ok here is a pretty easy way to compute for odd n:

Start by making a edge on the hexagonal lattice a-b. No find every connected hexagonal graph of size (n-1)/2 containing that edge such that node a has order 1. Count the number of them that when mirrored vertically across a share a vertex other than a with the original graph.

This diagram from the paper demonstrates why this works.

user202729

08:55

@WheatWizard You need to make sure there is no more symmetry too.

Wheat Wizard

Oh good point

That shouldn't be very hard

user202729

Wait, mirrored vertically across the vertex [a]? How?

Wheat Wizard

Eh I was a little vague

I'll draw a diagram because words are hard

Dotted line

Actually I suppose there are only two lines and they are isomorphic

(Excuse my terrible trackpad diagrams)

user202729

But then there are cases where the original graph overlap itself, not with its reflection...

Wheat Wizard

Ah yes I suppose that is true.

Now my method goes from all odd n to all odd n < 13.

user202729

09:07

> Geometrically planar polyenoids (without overlapping vertices) are enumerated by computer
programming. Thus the numbers of geometrically nonplanar polyenoids become accessible.

Wheat Wizard

Those can probably be handled recursively.

user202729

(from the paper)

Actually I suppose that's a good idea.

Wheat Wizard

There might be some problems with that approach for example you need to pull all nonplanars for < n-1/2 of a particular but different symmetry

The author keeps saying they used code to generate these but they don't really mention anything about the code.

I'm thinking it might be easier to implement A000936

non-planarity is hard

user202729

Yes that's what I'm doing...

Wheat Wizard

Oh cool.

user202729

09:14

> Therefore we resorted to computer
programming, like Kirby [2] in order to produce these numbers,

Reference [2] links to link.springer.com/article/10.1007/BF01164203

Wheat Wizard

Yeah, I saw, but I wasn't willing to pay and I'm on break from uni so I can't get it from them.

What approach are you using for the planars?

user202729

Of course there are sites which offer papers for free...

Anyway... there is a well-known (?) method of brute-force to generate all of them.

For this I may try to generate half-trees and reflect, for better performance.

Wheat Wizard

Glad I could be of help. (If you didn't come up with it on your own that is)

How do you plan on doing even n?

I couldn't figure out how to modify the method for evens.

user202729

Instead of reflecting by a line that goes through a vertex, you reflect through the perpendicular bisector of an edge.

Wheat Wizard

Oh that's clever

I unfortunately set up my coordinate system to make the parallel reflections very cheap so it might be hard for me without a refactor.

Actually it's not that bad.

user202729

09:22

Wheat Wizard

Actually its also really easy. Wow that turned out quite nice

user202729

My coordinate system ^^

Wheat Wizard

You only have two axes?

How does it work?

user202729

Yes, why having more?

"Point" (x, y) with (x+y) %3 != 0 are points, otherwise they're vector.

Wheat Wizard

There is a really nice three axis coordinate system for triangular system.

Sorry for an even cruder drawing.

user202729

09:25

But why using more than necessary?

I don't think it's more convenient.

Wheat Wizard

It makes reflections a breeze.

There is also an very effecient method for determining if a point is on a vertex.

Add all the coordinates if the sum is zero it is at a vertex, anything else it is not.

user202729

Java 1 doesn't seem to have assert statement

Wheat Wizard

I can see the benefits of dealing with a two axis system. I think there is a give and take.

Oh you are using Java 1, god rest your soul

I'm using Haskell. If I finish it won't be a valid answer but it will still be fun.

Wheat Wizard