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Random Cubing Discussion

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Is there a name for an LL method that orients the corners and separates opposite edges (not necessarily placing the two edge bars correct relative to each other)? Then PLL for the second step. It looks like it would be 23 cases - 14 for Orient+Separate, 9 for PLL.
For corners there are 7 orientations.

For edges there are 3 possibilities: edge-3-cycle, other-3-cycle, no-3-cycle

For corner oriented case, just the two 3-cycles are necessary.

3*7 + 2 = 23 - this way of couting does not use any kind of L/R - color placement (I didn't understand your plan, or how just 7 more algs are enough to make a differece)

All this cases can be solved by 9 PLLs (the PLLs not doing an edge-3-cycle A T H E N F Z),

It's like doing phasing during OLL instead of during F2L.
 
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qqwref

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I don't think I like how the 9 remaining PLLs contain N, F, and E :|

The 3*7 + 2 explanation is wrong because it's true that there are 3 edge possibilities and 7 COs, but with the solved CO there are actually only 2 edge possibilities (solved and adjacent swap). So there are 3*7+2 cases if you include the skip case, but only 3*7+1 if you don't.
 
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Athefre

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Yeah, after I made the post I seriously thought about what PLLs it included and moved on other ideas.
 
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... but only 3*7+1 if you don't.
correct just 22 instead of 23 cases. Recognition is also very nice for this step. And for the Sune-Pattern the 3 algs are just Sune, (U')Antisune and Niklas, for Bruno und Doublesune mirroring just gives you the 2nd 3-Cycle.

I never got any deeper into this myself, because using phasing during F2L reduces this step back from 22 to just 7 cases. Adding alternativ CO algs enables me again to get rid of not nice PLLs most of the time. (thats what my username stands for :))

@Athefre: Did you calculate the move count of your original methode (is it in this thread too?). My estimation was that it was even lower then COLL+EPLL?
 

Kirjava

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Well, I don't think statuenotation will catch on, but I'd like to propose something meaningful;

[X|Y|Z] = XYX'ZY'Z'

Used to use this for generating intuitive 5 cycles.
 

cmhardw

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How about calling that [X;Z:Y]? (Thus it is reminiscent of the [X:Y] notation, which it is related to.)
Wasn't the semi-colon already defined in a commutator notation on this board? I could swear Michael that either you or Lucas (or Stefan, etc.) already introduced this.

--edit--
iirc [a;b:c,d] was:
a b c d c' d' b' a'
 
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qqwref

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I think the idea of using ; as a substitute for :, and having some weird precedence thing, was all Lucas's. I don't really like it because it adds a lot of confusion (for humans at least!) while saving only a few characters. But even if you are using his notation there is no reason to ever write [a;b:c], because it would mean [a:[b:c]] which could be written easier as simply [ab:c].
 

riffz

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Wasn't the semi-colon already defined in a commutator notation on this board? I could swear Michael that either your or Lucas (or Stefan, etc.) already introduced this.

--edit--
iirc [a;b:c,d] was:
a b c d c' d' b' a'
Yea, I think the semi colon and colon were proposed both as conjugate notation, one having precedence over the commutator comma.
 

Erzz

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Section 2.03 of this, replace StarCraft II / SCII with "cubing". I think it fits quite well.

Edit: The next section is also applicable, if you make a few more substitutions.
 
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qqwref

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This sequence (in SiGN) on the 5x5x5 needs to be executed 281,801,520 times to return to solved again:
(B' 2D R' 3D' R 2D' R' 3D R B) (R2 r d2 u2) (f' l' f2 l f l u' f' l' f2 l)

http://www.randelshofer.ch/cube/professor/?B-NDR-MD-RND-R-MDRBR2TRTD2TU2TF-TL-TF2TLTFTLTU-TF-TL-TF2TL

This number (16 * 9 * 5 * 7 * 11 * 13 * 17 * 23) is probably the highest possible order for the 5x5x5. A factor of 19 is also possible (on 6x6x6+ the highest possible order is 19 times this or 5,354,228,880) but there just aren't enough orbits to use all of the factors on the 5x5x5. In this sequence, the three 24-piece orbits are being used as (23), (13,11), and (17,7), the middle edge orbit is (8,4) with a flip on both cycles, and the corner orbit is (5,3) with a twist on both cycles.
 
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Athefre

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I've been wondering how I went five years without knowing that most cubers scramble with white on U and green on F. It took that recent origin topic for me to find out. Maybe because I haven't been to a competition.

Here I've been scrambling with white/yellow on L/R and getting confused when I see Thom provide example solves and I end up with orange/red blocks. I've been thinking all this time that he was seeing easy color neutral solves.

I guess it doesn't matter and I don't plan to change my habit. It's just depressing that I didn't know.
 

Rpotts

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Here I've been scrambling with white/yellow on L/R and getting confused when I see Thom provide example solves and I end up with orange/red blocks. I've been thinking all this time that he was seeing easy color neutral solves.
I thought Thom always solved with white/yellow on LR?? Like, 4-colour neutral.
 

whauk

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This sequence (in SiGN) on the 5x5x5 needs to be executed 281,801,520 times to return to solved again:
(B' 2D R' 3D' R 2D' R' 3D R B) (R2 r d2 u2) (f' l' f2 l f l u' f' l' f2 l)

http://www.randelshofer.ch/cube/professor/?B-NDR-MD-RND-R-MDRBR2TRTD2TU2TF-TL-TF2TLTFTLTU-TF-TL-TF2TL

This number (16 * 9 * 5 * 7 * 11 * 13 * 17 * 23) is probably the highest possible order for the 5x5x5. A factor of 19 is also possible (on 6x6x6+ the highest possible order is 19 times this or 5,354,228,880) but there just aren't enough orbits to use all of the factors on the 5x5x5. In this sequence, the three 24-piece orbits are being used as (23), (13,11), and (17,7), the middle edge orbit is (8,4) with a flip on both cycles, and the corner orbit is (5,3) with a twist on both cycles.
wow thats really interesting.
can you produce a higher order if you allow rotations?
what are the highest orders for the other cubes? (2x2, 3x3 and 4x4)
 

qqwref

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can you produce a higher order if you allow rotations?
No. Moving the fixed centers doesn't let us improve any of the available factors, and adding parity to the middle edge and corner orbits doesn't help either.

what are the highest orders for the other cubes? (2x2, 3x3 and 4x4)
2x2 - 45
3x3 - 1260 without rotations, 2520 with rotations
4x4 - I think it's 765765


EDIT: Oh, and the highest order on any N-minx is (2^5 * 3^3 * 5^2 * 7^2 * 11 * 13 * 17 * 19 * 23 * 29 * 31 * 37 * 41 * 43 * 47 * 53 * 59) = 9,690,712,164,777,231,700,912,800. I think this requires at least 9 60-piece orbits, and the Teraminx has 8, so as far as I can tell this is first possible on the Petaminx.
 
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Athefre

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I've realized there are a lot of things I've yet to do.

- BLD
- Never timed a big cube solve
- 2x2 average (nor learned a 2x2 specific method)
- Went to a competition

I've created a couple of unique methods (well, my first one is no longer unique and my newest one could be considered a ZZ version of the most hated method during the Twisty Puzzle days). Most of my time has been spent figuring out non-matching Roux, ZZ, and Petrus and creating methods. I would like to do other things (especially BLD), but I lose the motivation almost immediately after I start.
 
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