Message #1655

From: Andrey <andreyastrelin@yahoo.com>
Subject: Re: [MC4D] Hi everyone, I’m back!
Date: Wed, 04 May 2011 05:50:18 -0000

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