Message #2193

From: schuma <mananself@gmail.com>
Subject: [MC4D] Re: Making a puzzle based on 11-cell
Date: Thu, 24 May 2012 18:58:22 -0000

I think it would be unfair to compare your move count with Ed’s, because after Ed solved it for the first time, I changed the way moves are counted: whole puzzle re-orientation doesn’t count any more.

It’s interesting that you mention implementing it in "3D". A natural way to represent this puzzle is to make a {3,5,3} hyperbolic honeycomb, then identify opposite faces of each icosahedron, then, to make it consistent, identify the icosahedra to which the opposite faces attach. Given your experience making the {6,3,3} honeycomb, I think you’ll be good at making an interface this way. I seriously thought about this approach, but I really don’t know how to draw the 3D hyperbolic space, especially like your {6,3,3}, where the view is "what you see from inside of the space". By the way I think that would be the only way to code the 57-cell.

In my applet, I do have a 3D model but in an Euclidean space, basically trying to simulate the hyperbolic space locally. I put a red icosahedron at the origin. Then for each of its ten visible faces, I place an icosahedron on top of it. Then I have to move the surrounding icosahedra outside for a distance, just because in Euclidean space, they don’t fit in. This is my 3D model of the 11 cells. And in this model, all the rotations on the central cell are implemented as a normal 3D rotation. Rotation on other cells are hard. I first somehow move the cell-to-twist to the central position and rotate it, then move it back to its original position. My code for this part is dirty and not elegant at all. If you simulate a hyperbolic space, all the rotations on all the cells will be well defined hyperbolic rotation. That’s what a mathematician likes.

My 2D painting of the puzzle is related to my 3D model. The central red cell is just what an icosahedron looks like normally. The cyan cell is the one in front of the red cell, attaching it on the front triangular face. In the 2D painting I just place it under the red cell so that we can see the red one. I’m drawing the BACK side of the cyan cell, rather than the front side. The reason is that in this way, both face-stickers have a vertex on top and an edge at the bottom, so they look like together. If I draw the front side of cyan cell as usual, the result is like this:

http://people.bu.edu/nanma/ElevenCell/ElevenCell2.html

the face on the cyan cell that directly faces the red cell would be invisible, the identified face that’s visible is upside down. I don’t feel that’s intuitive.

All the other cells are treated similarly. The orientation of them are just as in the 3D model, and I’m drawing the back sides. The green, magenta and brown cells look more symmetrical because they attach the red cell on the three neighboring faces of the center one, which are more symmetrical than the other six farther faces. I use perspective drawing for all the cells to distort them a little bit so that the faces close to the boundaries of the icosahedra looks larger.

So this is how I made it. Please feel free to make the 11-cell in your favorite way or even the 57-cell!

Nan

— In 4D_Cubing@yahoogroups.com, "Andrey" <andreyastrelin@…> wrote:
>
> My first result for 1-scramble was 4 twists :)
>
>
> — In 4D_Cubing@yahoogroups.com, "Eduard Baumann" <baumann@> wrote:
> >
> > I’m proud. I did the 1-scramble!
> > 17 moves, 00:04:33
> >
> > ;-)
> > Ed
> >
>