# Message #1657

From: Andrey <andreyastrelin@yahoo.com>

Subject: Re: [MC4D] Hi everyone, I’m back!

Date: Wed, 04 May 2011 09:28:19 -0000

Melinda,

May be, you are right. I don’t know the structure of 4D pyraminx crystal, but I made an experiment on 120-cell.

Take one 2C piece. A random twist moves one of its cells with probability 1/60, so in 2000-twist sequence it will be moved about 33 times. It can take one of 1440 positions+orientations, and it’s not difficult to find a distribution of its states after a number of twists.

I can say that after 33 moves touching the piece its probabilities will be distributed from 0.869/1440 (positions on the opposite cell) to 1.133/1440 (on the same cell where it started). For me it’s random enough. After 67 moves (=4000 twists of the puzzle) it will vary from 0.9957/1440 to 1.0043/1440.

But after 1000-twist scramble there are 4 times more chance of the piece to be in the starting position than to appear at the opposite cell: 1.75/1440 vs 0.42/1440.

So 2000-twist scramble is good, 4000-twist is very good, but 1000-twist is not enough if somebody wants really good scrambling :)

Andrey

— In 4D_Cubing@yahoogroups.com, Melinda Green <melinda@…> wrote:

>

> Well, if Andrey is backing away from this problem, then it must be more

> problematic than I first supposed! ;-)

>

> I don’t quite accept the idea that the number of puzzle states is a good

> proxy for the number of scrambling twists needed to fully scramble.

> Consider the 4D megaminx (AKA 120 Cell) compared with the 4D pyraminx

> crystal. It seems as if they might both have roughly the same number of

> possible states, but I feel that the first will be much harder to

> scramble than the second because it is far less internally connected.

> It’s as if in the first case, a given piece must stumble it’s way all

> the way around a roughly spherical surface whereas the in the second

> case, pieces can move "through" the sphere and quickly find themselves

> just about anywhere.

>

> -Melinda

>

> On 5/3/2011 10:50 PM, Andrey wrote:

> > Hi all,

> > If we estimate number of twists for fully scrambled puzzle by its number of states, the simplest formula may be like this:

> >

> > N=log(number_of_colors)/log(number_of_possible_twists)*number_of_stickers.

> >

> > Here we count all "chemically" possible paintings of the puzzle. Their number is not much more than number of the possible states (for 120-cell it’s close to the square of the number of states). So we have some reserve for "Brownian motion" in the graph of possible states, and it should be enough for practical purposes. For 7^5 it gives 66000 twists, but for smaller puzzles results are more acceptable:

> >

> > 3^4: 80

> > 4^4: 180

> > 5^4: 350

> > 3^5: 360

> > 120-cell: 4000

> > {3}x{3}, 3 layers: 64

> >

> > But in my new program I’ll use something more simple: N=1000 :)

> >

> > Andrey

>