Message #1660

From: schuma <>
Subject: Re: Hi everyone, I’m back!
Date: Wed, 04 May 2011 16:50:16 -0000


The official rule of WCA about scrambling can be found here:


The rule specified which program to use to generate the scramble sequences, and the length of the sequences. They shouldn’t be too long coz that’s a burden for the scramblers.

I think the most interesting case is 2x2x2. By simple brute force search, the god’s number is shown to be 11. The length of scramble sequences is random, according to the official program. The length is usually 7~9, which is less than the god’s number. Sometimes very short scramble sequence is generated from that program, like with length 5. Then the contestant will be lucky. And the record is still official. I think the reason they don’t use 11-move scrambling is because, with 9 moves most of the states (3 million out of 3.6 million states) can already be covered (see the table in So WCA feels it not necessary to use more moves.

The scrambling rule for other puzzles is more intuitive. The length is always approximately the god’s number of an estimate of it, as far as I can see. For example, the length for 3x3x3 is about 20, 21.

But I guess every one agrees that the complexity of a human solve has little to do with whether the puzzle is really fully scrambled or not, except it’s too easy. Like, nobody can take advantage from the fact that the 120-cell is scrambled by 1000 moves rather than 4000 moves.


— In, Melinda Green <melinda@…> wrote:
> I just made up the 4D pyraminx crystal as the extrapolation from the 3D
> version, but there’s no reason to go to 4D in the first place. I’m
> betting that the 3D megaminx is harder to scramble well compared with
> the pyraminx crystal.
> Hey, maybe the members interested in speedsolving can tell us how they
> solve that problem. I mean whoever runs those contests must have some
> rules regarding the minimum number of scrambling twists to use with each
> different puzzle. They must have the same dilemma that I have which is
> that they must choose a sufficiently large minimum, but not one that is
> so large as to be impractical. Anyone here know how they choose those
> numbers?
> -Melinda
> On 5/4/2011 2:28 AM, Andrey wrote:
> > 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, 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
> >
> >
> >
> > ————————————
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