Message #1656

From: Melinda Green <>
Subject: Re: [MC4D] Hi everyone, I’m back!
Date: Wed, 04 May 2011 00:36:05 -0700

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.


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