# A question about Old Pochmann method



## ilikecubing (Dec 12, 2010)

Scramble with Orange on Front,Yellow on Top

F R L2 B' U F' R' B2 L' R2 B L' D B R2 U2 R F2 L F' D B D R' U

I memorize edges first and my Buffer is UR,i use T perms to shoot to UL.

So for the scramble,my memorization went like...

For Edges

LU BR FR UB DF FU BD LB RD DL LF,so thats 11 T perms,so corners on UBR and UFR will change their positiions.

now Corners,my Buffer is UBL

BDR UBR FDL UFR new cycle I shoot to UFL DBL LFU another cylcle I shoot to DFR RFD

So Thats an odd number of swaps of the edges on UB and UL and they change thier original positions
,
So I solve corners first,After I solve all the corners,the UB and UL will change their positions...

The problem is that when I start solving the edges and I shoot to LU BR FR and so on I will be making a mess of my solve because the piece on UL has been interchanged with the piece on UB,so how deal with this kind of a problem,please help,Thanks.


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## da25centz (Dec 12, 2010)

ilikecubing said:


> Scramble with Orange on Front,Yellow on Top
> 
> F R L2 B' U F' R' B2 L' R2 B L' D B R2 U2 R F2 L F' D B D R' U
> 
> ...


 
You need to do an R perm before doing the edges


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## ilikecubing (Dec 12, 2010)

And what if there a is case in which the number of T perms required for edges is even,then in the end the corners on UBL and UFR will end up unsolved if I do R perm before the edges.


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## da25centz (Dec 12, 2010)

ilikecubing said:


> And what if there a is case in which the number of T perms required for edges is even,then in the end the corners on UBL and UFR will end up unsolved if I do R perm before the edges.


 
then dont do the R perm. Either you have an even number of swaps or you dont. You dont always need to do the R Perm, only sometimes


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## ilikecubing (Dec 12, 2010)

Are you sure that no. of algorithms required for edges and corners can both even or it can be both odd.

So there can't be a case when edges are even and corners are odd OR corners are even and edges are odd.


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## da25centz (Dec 12, 2010)

ilikecubing said:


> Are you sure that no. of algorithms required for edges and corners can both even or it can be both odd.
> 
> So there can't be a case when edges are even and corners are odd OR corners are even and edges are odd.


 
Total number of swaps = y-perms + t-perms


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## Toad (Dec 12, 2010)

ilikecubing said:


> Are you sure that no. of algorithms required for edges and corners can both even or it can be both odd.
> 
> So there can't be a case when edges are even and corners are odd OR corners are even and edges are odd.


 
If your cube is solvable, correct.

If there were to be one even and one odd this would mean a single two swap of pieces which is not possible on a 3x3 without disassembly.


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## ilikecubing (Dec 12, 2010)

k thanks guys,i think i got it


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