Random Cubing Discussion

[Phlippieskezer]

'Method' I've been fiddling around with on 4x4+...
I wouldn't be surprised if it's been thought up before, but (and you can probably guess its inspiration):

1. Solve two opposite centres.

2. Solve "Rouxblock" (i.e. 4x3x1 for 4x4x4 cube).

3. Solve opposite "Rouxblock."

4. CMLL.

5. Solve centres on M.

6. E(M)LL.

 

[JonnyWhoopes]

Pretty sure that's Stadler.

 

[Phlippieskezer]

Pretty sure that's Stadler.

(Assuming you're talking to me...)
Neat. Thanks.

Though, the wiki isn't giving me any results, but Google did. So, you're right.

Guess that's my new big cube method. :3
(The only difference in what I was doing before is that I was directly solving the edges with commutators, whilst it looks like he [Stadler] sort of pairs them up and then solves them).

 

[qqwref]

Is this well-known?

(r' F r) (F R U2 R' F') (and the inverse)

 

[Stefan]

Is this well-known?

(r' F r) (F R U2 R' F') (and the inverse)

Darn, I'm kinda using the inverse ((F R U2 R') (F' L' U' L)) but never thought about inverting it.

 

[teller]

Is this well-known?

(r' F r) (F R U2 R' F') (and the inverse)

I published (F R U2 R' F') and all of it's mirrors in my first rotationless F2L vid, and the reaction I got to the general case was that no, it's not well known. I do solve your case that way, but I think about it as targeting the general setup rather than a standalone alg.

 

[Athefre]

It's possible to get Kinect working on PC. I wonder how well it would work as a control method for computer cubes.

It probably wouldn't be able to track above a certain TPS or as accurately as a solver would like, but it seems like it would be fun.

 

[bobthegiraffemonkey]

Might be useful to some people, and it does end with a random lol. Wasn't really sure where to post this tbh, I figured this thread would do.

 

[Godmil]

It's possible to get Kinect working on PC. I wonder how well it would work as a control method for computer cubes.

AFAIK The drivers have been for people to develop for it no the PC... however it would be kinda pointless cause you don't need depth perception to recognise the colours on a cube face. Also the non ir cam part isn't that great. The Playstation Eye would be better as it's a pretty good webcam (and cheap too).

 

[macky]

Is this well-known?

(r' F r) (F R U2 R' F') (and the inverse)

My site lists L'ULFRU2'R'F with the note "reduction by L'UL." Because you know, FRU2'R'F should be well known.

 

[Athefre]

AFAIK The drivers have been for people to develop for it no the PC... however it would be kinda pointless cause you don't need depth perception to recognise the colours on a cube face. Also the non ir cam part isn't that great. The Playstation Eye would be better as it's a pretty good webcam (and cheap too).

I've also considered Playstation Eye before - to be used as a way to time a physical cube solve without a physical timer. But, what I meant with my previous post, and what I thought would be fun, is for a solver to be able to control a computer cube with Kinect tracking hand movements. Not having to hold a physical cube. It wouldn't be as useful as my first idea, but I think it would be fun in a silly way to feel like you have telekinetic powers.

 

[JonnyWhoopes]

Could somebody explain to me why [L', U2] [L, F2] [R, U2] works?

 

[qqwref]

The Kinect idea might be fun, but it definitely won't be as fast as a hi-games type cube unless you add in some kind of fingertrick macros (which I don't recommend in general).

Jonny: I don't have an explanation for why it works as written (other than that the three edge cycles cancel each other out), but it's equivalent to L' U2 L U2 R U2 L' U2 L U2 R' U2 = [L': [U2, L U2 R U2 L']] which is a pretty intuitive 3-cycle commutator.

 

[Godmil]

I've also considered Playstation Eye before - to be used as a way to time a physical cube solve without a physical timer. But, what I meant with my previous post, and what I thought would be fun, is for a solver to be able to control a computer cube with Kinect tracking hand movements. Not having to hold a physical cube. It wouldn't be as useful as my first idea, but I think it would be fun in a silly way to feel like you have telekinetic powers.

Oh yeah, that's a really cool idea precision may be a slight issue with the kinect, but it would be cool.

 

[DavidWoner]

I often see oriented edges defined as those solvable within <R, L, U, D, F2, B2> Isn't this redundant? Why don't we say <R, L, U, D> since F2 and B2 already fall under that moveset?

There is probably something more elegant but for now this proof suffices: F2 = R L' U2 R' L U D R U R2 U' R' U' R' U2 R D' alg.garron.us

 

[Kirjava]

I think that's a bit silly.

 

[qqwref]

It's a good point mathematically, but I think the point of showing all 6 moves on the generator is to easily see the differences between cube groups. It's obvious that the difference between <U,L,R,D,F2,B2> and <U,L,R,D,F,B> is that you can't do quarter turns on F or B; it's less obvious when you write them as <U,L,R,D> and <U,R,D,F,B>.

Actually, I think it's been a long trend in cubing to use mathematical formulations that make more intuitive sense, whether or not they are actually formally correct.

 

[cuBerBruce]

I often see oriented edges defined as those solvable within <R, L, U, D, F2, B2> Isn't this redundant? Why don't we say <R, L, U, D> since F2 and B2 already fall under that moveset?

There is probably something more elegant but for now this proof suffices: F2 = R L' U2 R' L U D R U R2 U' R' U' R' U2 R D' alg.garron.us

It's a good point mathematically, but I think the point of showing all 6 moves on the generator is to easily see the differences between cube groups. It's obvious that the difference between <U,L,R,D,F2,B2> and <U,L,R,D,F,B> is that you can't do quarter turns on F or B; it's less obvious when you write them as <U,L,R,D> and <U,R,D,F,B>.

Actually, I think it's been a long trend in cubing to use mathematical formulations that make more intuitive sense, whether or not they are actually formally correct.

Well, there is nothing mathematically incorrect with including generators that are redundant in this way when using group generator notation. I think <R, L, U, D, F2, B2> is generally used simply to indicate you can turn the F and B layers without misorienting edges (but only half-turns). I suppose this convention goes back to the Thistlethwaite algorithm where you certainly want to make use all of the basic moves at each step, and including the F2 and B2 in the group generator notation emphasizes you can still use those basic turns in that particular step. (Yeah, I know, Thistlethwaite actually used <L, R, F, B, U2, D2> instead, but basically nobody uses his notion of orientation these days.)

 

[MTGjumper]

U2

Is it a half turn or a double turn? (And I've realised as I've been writing this that someone could say "no, it's a band")

 

[5BLD]

Half turn. Double (layer) turn is like Uw.
That's what I think at least.