Random Cubing Discussion

[Kirjava]

"One of the goals of this method is to solve the last layer in the fewest moves possible, or approximately the fewest moves. Currently I do this around 20% of the time, and I'm convinced that I can push that to 100% or near 100% within a few months."

The claim about 100% near optimal LL solutions (like many of his) will likely never come to fruition, and the fact that the claim of currently 20% of LL cases being 1 look for him quite frankly absurd. Being able to cover 1/5 of LL cases that have at least two edges oriented would require nearly 800 algs.

There used to be a definition of "lucky" that said that if something happened less than 20% of the time, it was lucky.

 

[Jostle]

anyway, D' with left ring. Yeah, I could use it, I may even change to this for my E perm.

 

[uberCuber]

http://www.youtube.com/watch?v=BzkSSu4KjS0

"This 37 turn solve is a good example of Snyder Method 2"

How is skipping a step demonstrating a good example?

I love how he spent more than 15 seconds scrambling the cube and then had 2 opposite white edge pieces already solved.
Also, lol at how his talking is just cut off at the end of the video

 

[ben1996123]

http://www.youtube.com/watch?v=plQbnfxG2sA

2:52.88 square 1 "bld"

 

[uberCuber]

http://www.youtube.com/watch?v=plQbnfxG2sA

2:52.88 square 1 "bld"

Clearly a legitimate bld solve with nubvandenbergh method

 

[Escher]

http://www.youtube.com/watch?v=BzkSSu4KjS0

"This 37 turn solve is a good example of Snyder Method 2"

How is skipping a step demonstrating a good example?

Wowwwwwwww he is terrible at cubing.

 

[oll+phase+sync]

the fact that the claim of currently 20% of LL cases being 1 look for him quite frankly absurd.

I always belived this 20% claim comes from "enforcing more" while enforcing 2 oriented edged like "orient 4 edges" or an "2x2x1 Block". That sounded reasonable to me, if some knows a lot of LL cases.

But I can't see a path leading to 100%


P.S. Personally - an event happening 20% of all time is not lucky but already very regular.

 

[riffz]

D' with left ring finger

y/n?

I use it all the time for algs like R U2 R D R' U2 R D' R2 in BLD.

 

[Kirjava]

lolwut

 

[Cubenovice]

These guys make me ashamed to be Dutch...

 

[Kirjava]

Are they making fun of his english? I thought they were but I don't see why >_>

 

[Cubenovice]

Are they making fun of his english? I thought they were but I don't see why >_>

Yep and off his youtube videos and his puzzle "fetish"...
They say that there is no better entertaiment that watching his videos, "old guy with child like enthousiasm about puzzles in less than phantastic English with bad pronounciation".

Good thing that their own English in the prank phone call was much worse.

I especially disliked their laughter with the comment about "living life to the fullest" in the audio fargment where Oscar mentioned he was dedicating one of the puzzles to his father who had passed away.

Oh well,Ooscar mentioned himself that he has a big fat accent and apart from that I think his English is pretty fine.

 

[Kirjava]

Reminds me of this

 

[irontwig]

Good thing that their own English in the prank phone call was much worse.

Yes, very obviously Dutch, not that it's anything wrong with having an accent as long as you're understandable.

 

[Kirjava]

http://web.archive.org/web/20091026193545/http://geocities.com/abcmcfarren/math/math.htm

 

[qqwref]

Oo, nice find. It's like a proto-Jaapsch.net

The cube solutions don't look very efficient, but they're interesting approaches nonetheless.

 

[mr. giggums]

<U,D,R,r> solving for 4x4

I just tried it and it took a little while but I got it with only comms. There seems to be no parities (or I got lucky).

Here is my first approach...

Spoiler

1. centers
2. paired cross edges (cross on left)
3. Left Layer
4. Middle Layer edges
5. CLL
6. ELL

This is like k4 on left.

^That solution isn't very efficent so I might try a different one and see how it goes.

 

[qqwref]

There are definitely parities for that. I think I'd do centers -> pair edges -> orient edges -> UDR 3x3 with parities (so, similar to yours).

 

[mr. giggums]

I'm not counting PLL parity (it isn't actually a parity only a 2x2 cycle) I'm still not sure about OLL parity but I did another solve and didn't get it. I'll try your method.

 

[mr. giggums]

I just tried with your method and found that OLL parity is possible. So does any one know of a <U,D,R,r> OLL parity alg.