## Re: [Speed cubing group] Petrus- breakdown # of moves

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• It depends what you mean. If you re solving for speed, you ll use more moves than if you think about every step. If you mean the average for a perfect solution
Message 1 of 1 , Dec 1, 2004
It depends what you mean. If you're solving for speed, you'll use more
moves than if you think about every step.

If you mean the average for a perfect solution treating each step in
isolation, I would guess it's something like this:

S1: 5-6
S2: 5 or so
S3: 5
S4: Hardest to say... 10-12 maybe

If you're doing the final layer strictly as the individual steps I
think you'll get S5: 6, S6: 9, S7: 9

That adds up to about 50. I expect anyone who does it seriously to
learn more than those simple steps, and you'll get fewer moves the
better you get. In my new final layer approach I get under 15 moves for
the final layer, which would give you 41.

An other sample is my 13 solutions on my example page which had these
averages:

* S1+S2: 10.2 turns
* S3: 5.6 turns
* S4: 11.3 turns
* S5-7: 15.4 turns
* Total: 42.5 turns

This method is much less explored than Fridrich, so there isn't nearly
as much info around.

/Lars

On Nov 30, 2004, at 9:59, Brent Morgan wrote:

>
>
> Hi everyone,
> First off, I don't use Petrus...Which is why i need help ;).
> I was trying to find if there was a table somewhere that someone
> has, of the Petrus method broken down into steps and the range
> [min,max] (or even the average) number of moves of each step of the
> Petrus solution... I went to Lars Petrus' site and I couldn't fine
> one...
> But like, how Lars V. calculated the possiblities of the number of
> moves in forming the cross for the Frdirich method, is kind of what
> I'm looking for, or trying to figure out how to find (for Petrus
> method). Can someone help me? Lars? :). thanks
> peace out
> -bm
>
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>
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