Can odd parity be detected in a big cube, prior to being solved?

[KwS Pall]

If you know the quater inner moves done in scramble you can solve odd parity while solving 3rd center!
I did sub 1 while counting moves and I do sub 55 regulary without counting all at home or school

 

[LNZ]

I did some internet research into God's algorithm for the 4x4x4 cube and it has been shown that the maximum half turn metric turns to optimally solve a 4x4x4 cube is between 30 and 68. As the 4x4x4 cube has about 7x10^45 possible states, progress in this area will be painfully slow.

For a real 4x4, I can't detect parity from a scrambled cube. But I can see it after solving the first three layers through.

 

[Themancube]

Anyway give me your average move of fewest move 2x2x2 only ?


http://www.youtube.com/watch?v=fhXnBI7JSV0

 

[hawkmp4]

If you know the quater inner moves done in scramble you can solve odd parity while solving 3rd center!
I did sub 1 while counting moves and I do sub 55 regulary without counting all at home or school

Well of course you can. That's trivial and frankly, useless.

 

[Stefan]

I did some internet research into God's algorithm for the 4x4x4 cube and it has been shown that the maximum half turn metric turns to optimally solve a 4x4x4 cube is between 30 and 68.

How was that lower bound determined?

 

[Derrick Eide17]

I did some internet research into God's algorithm for the 4x4x4 cube and it has been shown that the maximum half turn metric turns to optimally solve a 4x4x4 cube is between 30 and 68.

How was that lower bound determined?

sighkick powurz ob vee is lee

 

[Lucas Garron]

I did some internet research into God's algorithm for the 4x4x4 cube and it has been shown that the maximum half turn metric turns to optimally solve a 4x4x4 cube is between 30 and 68.

How was that lower bound determined?

So I wonder, too. I don't remember what we got when I helped qq, but if I'm not being silly, a lower bound of 32 on 4x4x4 is easy (fix DBL, 9 generators with 3 turn possibilities each).

 

[cuBerBruce]

I did some internet research into God's algorithm for the 4x4x4 cube and it has been shown that the maximum half turn metric turns to optimally solve a 4x4x4 cube is between 30 and 68.

How was that lower bound determined?

So I wonder, too. I don't remember what we got when I helped qq, but if I'm not being silly, a lower bound of 32 on 4x4x4 is easy (fix DBL, 9 generators with 3 turn possibilities each).

LNZ should have been more specific on what he/she was allowing for single moves. It appears to me that LNZ is not using the same "half turn metric" as defined in the WCA regulations.

For lower bounds I concur that 32 is an easy lower bound if you are only considering single-layer turns or only "twists" (twisting the cube along only one cut-plane at a time).

Using <U, u, d', Uud', Uu, ud', R, r, l', Rrl', Rr, rl', F, f, b', Ffb', Ff, fb'>, I came up with 29 as a lower bound (allowing half-turns as single moves) if I set up my recurrence relations correctly. This set of moves is what I've called "block turns."

The upper bound of 68 I'm assuming is from my analysis showing 68 block turns suffice, but I later improved that to 67 in this thread.