## Re: [Speed cubing group] Re: wow... just........ wow..

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• ... Sorry to answer my own question, but I guess this list would be in the billions, and so listing them may not be a good idea :-) -- Ryan Heise
Message 1 of 77 , Mar 31, 2007
Ryan Heise wrote:

> Can we create a list of all 25-move sequences that amount to the
> identity operation?

Sorry to answer my own question, but I guess this list would be in the
billions, and so listing them may not be a good idea :-)

--
Ryan Heise
http://www.ryanheise.com/cube/
• ... about ... Those ... Ah, yes, I must ve made a computation mistake. I tried again and get 12 expected cubes for 13f as well now. I did it like this: First I
Message 77 of 77 , Apr 12, 2007
--- In speedsolvingrubikscube@yahoogroups.com, h_kociemba
>
> > Which had to be expected. I think in a million random cubes,
> > three should be expected to be solvable in 13 moves or less.
Those
> > four you found with 13f, can you please find out their optimal
> > solution lengths?
>
>
> For 1 million cubes, the theoretical distribution for the *optimal*
> maneuvers should look like this:
>
> 12f: ~1
> 13f: ~12
> 14f: ~160
> 15f: ~2200
> 16f: ~29000
> 17f: ~260.000
> 18f: ~690.000
> 19f: ~30.000
> 20f: probably less than 1

Ah, yes, I must've made a computation mistake. I tried again and get
12 expected cubes for 13f as well now. I did it like this:

First I looked at the right column in this statistic:
http://cubezzz.homelinux.org/drupal/?q=node/view/68

Multiplied the last number twice more with 13 (roughly the branching
factor). Multiply with a million, divide by 4.3e19.

Yeah I know, probably there's a statistic somewhere out there about
the overall distribution, but I don't know where.

Cheers!
Stefan
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