December 8, 2007:Rowe Hessler broke the UWR for solving cube BLD, 37.54 seconds.

[cmhardw]

Hmm. Freestyle 3-cycles that are not commutators? That's possible for edges of course. Please tell me an easy corner 3-cycle that is not a commmutator or a conjugated/shifted commie

-Per

First I thought of algs like R2 D' L2 D R2 D' L' R' F2 B2 R' L U R' L F2 B2 R' L or something equally ridiculous.

But, since the inverse of D' L2 D (which I will call algorithm "A") is it's own inverse, and D' L' R' F2 B2 R' L U R' L F2 B2 R' L is equivalent to algorithm A then D' L' R' F2 B2 R' L U R' L F2 B2 R' L is equivalent A'.

Thus even this is still written in the form BAB'A' where:
B=R2
A=D' L2 D

only A' is replaced with the equivalent algorithm D' L' R' F2 B2 R' L U R' L F2 B2 R' L.

So yeah I can't think of an easy corner 3 cycle that is not a commutator. Though I guess that depends on your definition of the word easy, because I can sure think of corner 3 cycles that are not commutators ;-)

Chris

 

[cmhardw]

I would argue that this 3 cycle is easy to execute, and it is not a commutator:

R' U R' U' R' U' R' U R U u R' U R' U' R u' R2 F' U F

That leads to an interesting question, what is the shortest HTM (or STM) algorithm that cycles 3 corners only and is not a commutator?

Chris

 

[cmhardw]

Two concatenated commutators can make a non-commutator 3 cycle.

R U' R' D R U R' D' U' L D L' U L D' L'

A B A' B' C D C' D'
I wonder if it's possible to have a three cycle where permutation B' is the inverse of permutation C causing a cancellation. You would have to require that D is not the inverse of A' though.

 

[ThePizzaGuy92]

this has also spawed Kai to post his sub-minute record, so major props to him for being the 3rd human to ever do a BLD solve under a minute!

Huh? The 3rd ever? I'm not sure.

Matyas/Dannyang/Marcus -> Chi Chu -> Rafal -> Suraimu -> Kai.
I'm pretty sure that is the order so far. There is also probably people who havn't posted times.

Edit:
Didn't realise Marcus had aswell.

why did i not knw so many people have achieved that! thats insane. i'm having trouble with success, i'm not even worried about time yet. haha

 

[Lucas Garron]

10 QTM: R U' F' U R' U' R F R' U
I think it's essentially a commutator, but it's not quite in pure form. Does it qualify here?

EDIT 1: Removed redundant algs.
EDIT 2: How about U' R' U2 F' U2 F R U F U2 R U2 R' F'
and U R' F' R F R U' R2 U R F' R' F U' R ?

 

[watermelon]

How about R U R' F' r U R' U' r' F R2 U' R' ?

 

[Lucas Garron]

How about R U R' F' r U R' U' r' F R2 U' R' ?

I mentioned someowhere else that I was going to post it here, but then found out it was a conjugated Niklas..
Where'd you get that alg? I haven't seen anyone else who knew it...

 

[mrCage]

10 QTM: R U' F' U R' U' R F R' U
I think it's essentially a commutator, but it's not quite in pure form. Does it qualify here?

EDIT 1: Removed redundant algs.
EDIT 2: How about U' R' U2 F' U2 F R U F U2 R U2 R' F'
and U R' F' R F R U' R2 U R F' R' F U' R ?

Hi

R U' F' U R' U' R F R' U is a shifted version of R' U R U' F' U R' U' R F
which again is [R' U R U',F'] = [[R',U],F'] nested commutator. So, yes it's disqualified

U' R' U2 F' U2 F R U F U2 R U2 R' F' is interesting. It can be split in 2
U' R' U2 F' U2 F R + U F U2 R U2 R' F'. The second part is the mirror of the first part. I cannot see how this one is a commutator derivative. It may still be however ;-)

U R' F' R F R U' R2 U R F' R' F U' R is same as U R' F' R F R U' R + R U R F' R' F U' R. I don't see any commutator stuff in this one either.

However i wouldn't really say that either of these 2 latest ones are easy. But that's just my subjective opinion of course

- Per

 

[mrCage]

How about R U R' F' r U R' U' r' F R2 U' R' ?

R U R' F' r U R' U' r' F R2 U' R'=R U R' F' L F R' F' L' F R2 U' R' in purely uppercase notation. R U R' F' L F R' F' L' F R2 U' R' is the same as R U R' + F' L F R' F' L' F R + R U' R'. Then it is recognised as a conjugated commutator :-s

-Per

 

[mrCage]

Hi

As a quick remark i will just inform that the beloved sune is also a double commutator.

R' U' R U' R' U2 R = [R' U'][U2,R]

-Per

 

[mrCage]

Hi

As a quick remark i will just inform that the beloved sune is also a double commutator.

R' U' R U' R' U2 R = [R' U'][U2,R]

-Per

This conjugate is similar to sune and orients edges
R B [B,L'] B' R'

- Per

 

[AvGalen]

First: No, you are not talking to yourself, someone else is reading this
Second: A useful button exists with the text "Edit".

I like the sune!

 

[watermelon]

How about R U R' F' r U R' U' r' F R2 U' R' ?

I mentioned someowhere else that I was going to post it here, but then found out it was a conjugated Niklas..
Where'd you get that alg? I haven't seen anyone else who knew it...

I found that a while ago while experimenting with an alternative J-perm (my main alg is the LUR version) and r turns. I always thought of the RFU J-perm as a conjugated R U R' U' R' F R F' OLL...

 

[mrCage]

First: No, you are not talking to yourself, someone else is reading this
Second: A useful button exists with the text "Edit".

I like the sune!

Haha, i know about the edit button. I might have appended it to the first sune post, but i chose not to. Sune can be very useful for fewest moves actually

-Per