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At what point during a big cube solve is parity determined? I'm guessing you're locked in to a particular parity once you finish the Reduction Phase, but I don't know enough about cubing theory to back that up.
 
Parity is determined once you finish two adjacent centers, assuming you never make those centers more than one move from solved.

Surely that depends? On even layered cubes PLL parity is influenced by how you pair up edges, unless I'm much mistaken. [d, R F' U R' F] [u', F' R U' F R'], where d and u are inner layer turns only, creates PLL parity without 'unsolving' the centers. Though I guess if you want to get technical, PLL 'parity' isn't actually a parity since it is 2 2-cycles.
 
Has anybody done much development of a method for the Vulcan?

This is the method I've worked out for it:
1. Solve one side of the puzzle, minus the tips.
6jfrl1.jpg
2. Solve the next layer.
j6pjdg.jpg
3. Solve the tips.
b96eeh.jpg
4. Pair the edges up with Ro U' Ro' U' Ro U' Ro' and Ro U Ro' U Ro U Ro'.
s5ejom.jpg
5. Parity fix with ???.
6. Solve last layer like in normal pyraminx

- For parity I just mess around with the edges a bit, and hope it solves, I haven't worked out an actual algorithm.
- "o" means outer layer. So Ro means the Right outer layer. It's weird, sorry. Is there any better alternative?
- If you want to get advanced you can place edges while pairing them for the Last Layer, rather than pairing then solving.

EDIT: Found a parity fix: Ro U Ro' U Ro U Ro U Ro U Ro' U Ro

It's in my weird notation, sorry. If you don't understand, tell me and I'll elaborate. Even better, present me with better notation methods.
 
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Maybe just a lowercase letter instead of "o" for outer layers? Besides that, I can't really think of any other notation methods, but I think your's works just fine.
 
For Fridrich, ZZ or Roux, does anyone solve corner permutation before finishing F2L/2nd block? Still trying to find a good system for recognition..
 
For Fridrich, ZZ or Roux, does anyone solve corner permutation before finishing F2L/2nd block? Still trying to find a good system for recognition..
If you're referring to ZZ-d (or otherwise reducing the LL to 2GLL), this can actually be done somewhat easily;
Let's say you were doing a ZZ solve, and had done the left block (so you have your 2x2x3, basically, with oriented edges.
From there, get one corner (probably DBR) in place. This should be a maximum of 2 moves, because you don't care about orientation.
From there, there are only 6 cases. :)

But then there's recognition, which is the same as my CPLS recognition "method."

Basically:
you can used a shortened/optimal CPLS if you'd like.
 
Sounds like a little too much overhead but I guess I'm getting you wrong. What I basically do is putting the D corners into D (not essentially permuted) and apply one of two algs (L U' (R'/R2) U L') to solve CP. Continuing with a <R,U,Rw,M> right block, the LL will be a 2GLL case.
 
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Average of 5: 1.86
1. 1.79 R' F R U' F' R2 U R' U' R' U'
y R U R U' R' U' y' R2 U R2' U

2. 1.54 F' R2 F2 R' F2 R' U R' U' R' U'
x y' F' U' R U R' U' R2 U' R2' U

3. (2.75) R F2 U' F' R U F' R U' R' U'
z' R' U2 R B2 R2' F R U R' U' F' U2

4. (1.16) R F2 R' F' U R2 U' F' U' R' U'
F2 R U R' U' F'

5. 2.25 F' R F U' F U' F R2 F' R' U'
z2 R U' R' U R U2' L U2 R' U R' F'

I lold
 
Remember when Yumu got his 10.83 WR average not THAT long ago and most people were like "SUB 11?!!"?


Andrew Kang and Paolo Moriello are tied at 50th right now with 10.87 averages.
 
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F R' U2 D' B' R L' D R' F2 R D' L B' L B' F2 u D2 f' D' F2 L2 f D2 R' D U' B' D2 R B D2 F U2 u2 r2 L R2 B

4x4 scramble from qqtimer with only 5 double layer turns :P

I got 50s OLL parity on this, good for me
 
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