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



## pjk (Dec 8, 2007)

December 8, 2007Rowe Hessler broke the Unofficial World Record for solving Rubik's Cube blindfolded, with a new record of 37.54 seconds.

Posted on speedcubing. Rowe, do you have the scramble by any chance? Also, if you posted your solution, that would be awesome. Thanks


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## joey (Dec 8, 2007)

Good to see him getting good times again.


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## masterofthebass (Dec 8, 2007)

I e-mailed him, like most of us, and I offered to put up his method on here. This looks extremely awesome, and I can't wait to see the corner method.


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## Lucas Garron (Dec 8, 2007)

Tried creating the scramcle, but the info's incomplete.
(Rowe: Were the cycles really all independent orientation-wise?)

Anyhow, I think he's using freestyle, in a way; it's about the fastest way you can cycle, and that scramble is great for it.

By the way: I got him to sign up on this forum a while ago, but I think he couldn't post...


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## philkt731 (Dec 8, 2007)

If its three cycle, its gotta be something like Pedro's. Maybe its two cycle but doing edges and corners at the same time...


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## watermelon (Dec 8, 2007)

His method is basically 3-cycle freestyle with commutators thrown in. It's extremely flexible (setting up corners on faces other than U and D for example), and seems like it would indeed be one of the fastest methods with enough practice.


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## Hiram (Dec 8, 2007)

This scramble looks extremaly easily and would be really nice for any method. Still, this time is incredible...


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## alexc (Dec 9, 2007)

Wow!!!!!!!!!!!!!!!!!!!!


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## AlexandertheGreat (Dec 9, 2007)

that's insane......hopefully he gets something like that in competition!


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## ThePizzaGuy92 (Dec 9, 2007)

thats incredible, sub-20 memo is amazing enough, but if someone could pull off a sub-20 execution... of a BLD solve?! wow


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## Derrick Eide17 (Dec 13, 2007)

yeah basically waht everyone has mentioned i though too. it is just simply an AMAZING record... but of course we all know lmao.... secretly... .someplace... SOMEWHERE... Matyas has gotten a better time and just hasnt posted it yet and prolly never well LMAO. oh yeah GREAT JOB ROWE!


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## ThePizzaGuy92 (Dec 13, 2007)

Derrick Eide17 said:


> yeah basically waht everyone has mentioned i though too. it is just simply an AMAZING record... but of course we all know lmao.... secretly... .someplace... SOMEWHERE... Matyas has gotten a better time and just hasnt posted it yet and prolly never well LMAO. oh yeah GREAT JOB ROWE!



haha XD...


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## ThePizzaGuy92 (Dec 13, 2007)

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!


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## joey (Dec 13, 2007)

ThePizzaGuy92 said:


> 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.


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## MarcusStuhr (Dec 13, 2007)

I had a few sub-minute times around when Danyang and Matyas did -- not sure who did what in what order though.


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## mrCage (Dec 13, 2007)

watermelon said:


> His method is basically 3-cycle freestyle with commutators thrown in. It's extremely flexible (setting up corners on faces other than U and D for example), and seems like it would indeed be one of the fastest methods with enough practice.



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


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## AvGalen (Dec 13, 2007)

mrCage said:


> watermelon said:
> 
> 
> > His method is basically 3-cycle freestyle with commutators thrown in. It's extremely flexible (setting up corners on faces other than U and D for example), and seems like it would indeed be one of the fastest methods with enough practice.
> ...


R' F R' B2 R F' R' B2 R2 is not really a commutator


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## joey (Dec 13, 2007)

AvGalen said:


> R' F R' B2 R F' R' B2 R2 is not really a commutator



It is a move cancelling conjugated commutator!
R2 [R F R',B2] R2


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## philkt731 (Dec 13, 2007)

joey said:


> ThePizzaGuy92 said:
> 
> 
> > 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!
> ...


And whre does Rowe fit in here?


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## joey (Dec 13, 2007)

I think he comes after Chi Chu, and before Rafal. As I said it's not a fully complete list


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## cmhardw (Dec 14, 2007)

mrCage said:


> 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


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## cmhardw (Dec 14, 2007)

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


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## cmhardw (Dec 14, 2007)

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.


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## ThePizzaGuy92 (Dec 14, 2007)

joey said:


> ThePizzaGuy92 said:
> 
> 
> > 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!
> ...



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


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## Lucas Garron (Dec 15, 2007)

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 ?


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## watermelon (Dec 18, 2007)

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


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## Lucas Garron (Dec 18, 2007)

watermelon said:


> 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...


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## mrCage (Dec 19, 2007)

Lucas Garron said:


> 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.
> ...



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


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## mrCage (Dec 19, 2007)

watermelon said:


> 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


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## mrCage (Dec 19, 2007)

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


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## mrCage (Dec 19, 2007)

mrCage said:


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



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

- Per


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## AvGalen (Dec 19, 2007)

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!


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## watermelon (Dec 20, 2007)

Lucas Garron said:


> watermelon said:
> 
> 
> > How about R U R' F' r U R' U' r' F R2 U' R' ?
> ...


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...


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## mrCage (Dec 20, 2007)

AvGalen said:


> 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


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