# Serious parity problem....



## Sora (Sep 26, 2009)

I using old stefan pochman method(solve edge follow by corner).There are 13 edges target to solve(odd number),but unfortunately it not happens any parity case...While another case,i had 12 edge target(12 wrong piece,2 cycle) BUT it happens parity problem.

I heard many tutorial had been told that odd number target will happen parity.But for me,sometimes even number of target also happen parity..Any1 can help??

Thanks for any reply


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## Novriil (Sep 26, 2009)

Sora said:


> I using old stefan pochman method(solve edge follow by corner).There are 13 edges target to solve(odd number),but unfortunately it not happens any parity case...While another case,i had 12 edge target(12 wrong piece,2 cycle) BUT it happens parity problem.
> 
> I heard many tutorial had been told that odd number target will happen parity.But for me,sometimes even number of target also happen parity..Any1 can help??
> 
> Thanks for any reply



I don't know but do you use old pochmann for regular solving? What do you do when cycle ends? Rotate the cube? or break into a new cycle? When you rotate then it might be that you will have a parity on 13.. I can't think of any other reasons right now.


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## Sora (Sep 26, 2009)

i break it into new cycle when cycle ends...So,it is the part of old stefan pochman method??Lols ,i think i more confusing now.I follow excatly the video tutorial of badmephisto


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## MatsBergsten (Sep 26, 2009)

Sora said:


> i break it into new cycle when cycle ends...So,it is the part of old stefan pochman method??Lols ,i think i more confusing now



When you break into a new cycle there will be an extra edge. If there is only one cycle of all twelve edges you will have a cycle length of 11 (not 12). Then you have parity. But if there are two cycles (no edges in the right place) of say 10 and 2 edges, then the first will give 9 edges (not 10) and then 3, not 2. Sum 12 edges. No parity.

If you have parity you must have an odd number of edges in your solving cycle and vice versa. If you have an even number AND parity you have your cycles wrong (or an odd number and no parity, again your cycles are wrong).

When you break in to a new cycle you must start and end with the same piece (firstly to bring it to the buffer and lastly to put it in the right place).


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## Sora (Sep 26, 2009)

I think i still don't understand what is MatsBergsten try to explain to me...Maybe i a bit of dumb

Try scamble it:
R2 D2 B2 U L2 R U R' U' B2 L U2

My target list:
BL-UF-UB-DR-DL-FR-UL-BR-FL-DF(2nd cycle)-DB-DF >12 targets(even number),2cycle
but happens parity=.=

i using 
T1 [shoot to UL] R U R' U' R' F R2 U' R' U' R U R' F' 
T2 [shoot to LU] x' R2 U' R' U x R' F' U' F R U R' U' 
J1 [shoot to UF] R U R' F' R U R' U' R' F R2 U' R' U' 
J2 [shoot to UB] R' U2 R U R' U2 L U' R U L' 


If according to badmephisto video,there is a formula here:
(#items)=(#wrong pieces)+(#cycles)-2

for the example above,12(wrong piece)+2(cycles)-2=12(even number)
it should not happens parity..can anyone explain more clearly?


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## MatsBergsten (Sep 26, 2009)

To start from the real beginning then, is your buffer (for edges) UR?


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## Sora (Sep 26, 2009)

yup


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## MatsBergsten (Sep 26, 2009)

Sora said:


> I think i still don't understand what is MatsBergsten try to explain to me...Maybe i a bit of dumb
> 
> Try scamble it:
> R2 D2 B2 U L2 R U R' U' B2 L U2
> ...



You have the edge cycles all right. 9+3 = 12. So even, NO PARITY!!

There are an even number of moves to get the corners right.
Have you LUB as buffer?

(start at LUB) = LDB - FUL - RDB - LDF - RDF - URB = 6 moves (one cycle of 7 corners, URF in right place (although twisted)).

So why do you think you have parity?


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## Sora (Sep 26, 2009)

you said there is no parity for the case...But my UFR and UBR has swapped after i solve all the edges.... the corner buffer piece for me is LBU.

i holding the cube by yellow on top,orange at front.

I think i had figure out where is the problem,any1 knows what is the S alogarithms


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## MatsBergsten (Sep 26, 2009)

You have your edge cycle, which consists of 12 moves.

Every one move is either a T1/T2 or a J1/J2. Every one of those
swaps URB & URF. So after 12 swaps/moves they shall be where
they started. Otherwise you have solved something wrong.

So your list of moves is to short. You must have 12 T or J-perms
to solve the edges (not four).


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## trying-to-speedcube... (Sep 26, 2009)

Could you describe exactly what you're doing? I solved the edges of your scramble using Old Pochmann, and there is no parity. Simple as that. If you do have parity, you have to post exactly what you're doing so we can see where your mistake is.



Spoiler



L R U R' U' R' F R2 U' R' U' R U R' F' L'
R U R' F' R U R' U' R' F R2 U' R' U'
R' U2 R U R' U2 L U' R U L'
D2 L2 R U R' U' R' F R2 U' R' U' R U R' F' L2 D2
L2 R U R' U' R' F R2 U' R' U' R U R' F' L2
d2 L R U R' U' R' F R2 U' R' U' R U R' F' L' d2
R U R' U' R' F R2 U' R' U' R U R' F'
d2 L' R U R' U' R' F R2 U' R' U' R U R' F' L d2
L' R U R' U' R' F R2 U' R' U' R U R' F' L
D' L2 R U R' U' R' F R2 U' R' U' R U R' F' L2 D
D L2 R U R' U' R' F R2 U' R' U' R U R' F' L2 D'
D' L2 R U R' U' R' F R2 U' R' U' R U R' F' L2 D

Corners are still the same as before.


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