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 Post subject: LL Methods
PostPosted: Thu Jul 14, 2005 6:58 am 
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At the moment, I do a 4-look LL which takes on average around 13secs, and am looking to reduce this.

Are there any other more uncommon methods that people use apart from the normal fridrich-type ones? Im interested in knowing about other unconventional methods.

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 Post subject: Last Level
PostPosted: Thu Jul 14, 2005 11:20 am 
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Hi Nicky,

Generally I do Corners then Edges or CLL ELL.
The 4 look version is CPLL COLL EOLL EPLL.

There are only 2 Corner positions, and 6 corner orientations which need to be changed. Permutation alone takes either 6 or 7 moves. Orientation leaving permutation unchanged takes 14, 11, 9 or 7 moves. Although doing permutation and orientation separately seemingly has a maximum of 21 moves for the corners, they can be done in such a way that there is never more than 14 moves. In addition, it is easy when combining both to cut the total number of moves.

There are only 3 edge orientations, and 3 edge positions which need to be changed. Done separately the maximum to change orientation is 12 moves, and the maximum to change position is 7 moves, 19 total. Combined has a maximum of 16, minimum of 7.

I'm slowly headed toward having all of the Corners done in one algorithm and also all of the Edges done in one algorithm.

Regards,

David J


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 Post subject: Re: Last Level
PostPosted: Thu Jul 14, 2005 12:23 pm 
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David J wrote:
The 4 look version is CPLL COLL EOLL EPLL.

David J


David,

Doesn't COLL combine corner permute and orient into one alg? At least according to Lars Vandenberg, that's what he calls it.
http://www.cubezone.be/coll.html

COLL almost seems like a waste to use on a 3x3x3, at least to me (it works nicely on a 2x2x2 though). 40 algorithims when you can just learn 7 corner orient algs (sune or OLL for just corners) and 21 PLL's.

I don't know if you are talking about Lars version of COLL or just corner orient when you mention COLL.

Sure that way you only need 4 edge permute algs but you need 40 COLL algs = 44 algs for everything except LL edge orient.

with PLL you only need 21 LL permute algs and 7 corner orient algs = 28 algs for everything except LL edge orient.

I've really been trying to figure out the best way to do LL lately and I thought I was onto something until my friend Clancy pointed out I was talking about ZBLL which had already been invented. It seems like I'm always a step behind the curve. Oh well.

I'm interested in learning more efficient LL methods with as few algs as possible, but it seems like everything has already been considered.

-mike

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 Post subject:
PostPosted: Fri Jul 15, 2005 1:20 pm 
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skeneegee,

The most common meaning behind COLL is the 40 algorithms that solve the corners and orient the edges, but I believe David has the correct meaning here: orienting the corners of the last layer (Corner Orient Last Layer). I'm guessing the more common usage got this acronym as a combo between CLL(Corner Last Layer) and EOLL(Edge Orient Last Layer) because it finishes both of those steps.

I hope that all made sense. :P :wink:

-Bill

By the way, here's a link explaining several different methods for solving the LL: http://yem.ai.univ-paris8.fr/~bh/cube/

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I've gone off the idea of progress. It's overrated. - Arthur Dent


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 Post subject: Re: Last Level
PostPosted: Fri Jul 15, 2005 2:26 pm 
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David J wrote:
The 4 look version is CPLL COLL EOLL EPLL.

David J


Hi Mike,

skeneegee wrote:
David,

Doesn't COLL combine corner permute and orient into one alg? At least according to Lars Vandenberg, that's what he calls it.
http://www.cubezone.be/coll.html


No. COLL stands for Corner Orientation Last Level.

skeneegee wrote:
COLL almost seems like a waste to use on a 3x3x3, at least to me (it works nicely on a 2x2x2 though). 40 algorithims when you can just learn 7 corner orient algs (sune or OLL for just corners) and 21 PLL's.


Many sites have OLL at 57 algorithms then 21 PLL.

skeneegee wrote:
I don't know if you are talking about Lars version of COLL or just corner orient when you mention COLL.


I have no idea what Lars has on his site.

skeneegee wrote:
Sure that way you only need 4 edge permute algs but you need 40 COLL algs = 44 algs for everything except LL edge orient.


I don't know where you're getting your numbers but there are only 3 edge permutations on the LL that need solving. And I don't know how you get 40 COLL when CLL, which is COLL plus CPLL, is only 31.

skeneegee wrote:
with PLL you only need 21 LL permute algs and 7 corner orient algs = 28 algs for everything except LL edge orient.


There are only six corner orientations and three edge orientations which need solving.

If you want a minimum set four algs will suffice, though they may need to be done more than once:

Permute corners R U' L' U R' U' L (U)
Orient corners R U R' U R U2 R' (U2)
Orient edges r U r' U2 r U r'
Permute edges F2 U r U2 r' U F2

I'm not sure but you might be able to solve the LL corners with R U' L' U R' U' L (U) because it does both, and the LL edges with r U r' U2 r U r' because it does both, too.
But It might be possible to only use R' U' F' U F R because it does all four, but that would be hellishly complex to solve all possible positions using it.

skeneegee wrote:
I've really been trying to figure out the best way to do LL lately and I thought I was onto something until my friend Clancy pointed out I was talking about ZBLL which had already been invented. It seems like I'm always a step behind the curve. Oh well.


Don't get discouraged. No one knows all of the algs of ZBLL. And very few use their own algs anyway. I've found math errors all over the place. People copy things and they don't understand them and the don't step through them. A few days ago someone posted "25 corner positions" on the Yahoo speedcubing list when there are 32 total.

skeneegee wrote:
I'm interested in learning more efficient LL methods with as few algs as possible, but it seems like everything has already been considered.

-mike


It doesn't matter what has already been done. ZBLL is only a threoretical claim. Sure people have found 1211 sets of algs but no one has them all memorised, never mind found them all by themselves. As far as I know I'm one of the only speedcuber using entirely his own algorithms, and I'm very slowly approaching that full set of algs. I've been working on it since 1980, but someone else got to name it. So, don't you feel cheated.

Why not look for what *you* think is simple and logical?

Regards,

David J


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 Post subject: Re: Last Level
PostPosted: Fri Jul 15, 2005 5:28 pm 
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David J wrote:
I'm not sure but you might be able to solve the LL corners with R U' L' U R' U' L (U) because it does both


That's correct, just try ((R U' L' U R' U' L) U')*3.

Quote:
and the LL edges with r U r' U2 r U r' because it does both, too.


No, that one doesn't work (which is fairly easy to prove ;-)).


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 Post subject: Re: Last Level
PostPosted: Fri Jul 15, 2005 7:08 pm 
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Hi Stefan,

StefanPochmann wrote:
David J wrote:
I'm not sure but you might be able to solve the LL corners with R U' L' U R' U' L (U) because it does both


That's correct, just try ((R U' L' U R' U' L) U')*3.


That's not quite what I meant! :P

Quote:
and the LL edges with r U r' U2 r U r' because it does both, too.


StefanPochmann wrote:
No, that one doesn't work (which is fairly easy to prove ;-)).


Actually, wise guy, (r U r' U2 r U r') *3 works the same as your first example, that is, staring with a solved cube, it returns to the solved state.

But that's no leap since all algs do that. In fact, that is part of the *definition* of an algorithm!

Anyway, regarding what I meant, I said I wasn't sure, but I didn't feel like taking the time to check it out. :)

David J


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 Post subject: Re: Last Level
PostPosted: Fri Jul 15, 2005 9:51 pm 
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David J wrote:

No. COLL stands for Corner Orientation Last Level.


That makes sense to me I think Lars Vandenberg should call it COP

David J wrote:
Many sites have OLL at 57 algorithms then 21 PLL.


I guess we were talking about 2 different things.

David J wrote:
I have no idea what Lars has on his site.


It works but I think it's too many algs for what it does.
http://www.cubezone.be/coll.html

David J wrote:
I don't know where you're getting your numbers but there are only 3 edge permutations on the LL that need solving. And I don't know how you get 40 COLL when CLL, which is COLL plus CPLL, is only 31.


Well there's "U", "U'", "Z", and "H", right?

yeah about the 40 I posted, I was too lazy to count them all since he said there were 40 at the top of the page, I just took his word for it. I guess there's only 38 there. Have you posted your 31 cases anywhere that I could see them?

David J wrote:
There are only six corner orientations....

Code:
   █→   ←█   █→   ██   ██   ←→   ←→
   ←↓    ↓→   ↓█     ↓↓   ←→  ←→    ↓↓


David J wrote:
If you want a minimum set four algs will suffice, though they may need to be done more than once


I mean I know PLL pretty good but I was wondering if theres better way than OLL, PLL. OLL seems almost overwhelming to learn, but I thought the same about PLL before I learned them all. Maybe I should just give OLL a try.

David J wrote:
I've been working on it since 1980, but someone else got to name it. So, don't you feel cheated.


25 years! You must be dripping with ambition, I'm impressed at your dedication.

David J wrote:
Why not look for what *you* think is simple and logical?


I'm trying, there's alot to this little toy and I love it!

Thanks for your time David

-mike

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"It's like an alarm clock, WOO WOO" -Bubb Rubb


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 Post subject: Re: Last Level
PostPosted: Sat Jul 16, 2005 6:28 am 
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David J wrote:
StefanPochmann wrote:
That's correct, just try ((R U' L' U R' U' L) U')*3.


That's not quite what I meant! :P


I wish you had a little more faith in me. Are you sure you noticed the extra U' in there? It does *not* return to a solved state.

In the meantime I also proved that (R' U' F' U F R) can't solve the complete LL. Was actually even easier than the proof for the edges algorithm.

Here's an algorithm that can do ELL completely: [R U R' U', r]. Call it #, then (# U2 # U # U) permutes three edges preserving orientation.

Using your corners algorithm and mine for edges we can solve the LL completely using these two algorithms (plus U turns). Is it possible using just one algorithm (plus U turns)?

A similar question I've asked myself is how many Sune "variants" must be applied to do OLL. For example, to flip four edges you can do with 3 "Sunes":
((Rr) U2 R' U' R U' (Rr)') U' ((Ll)' U' L U' L' U2 (Ll)) (R' U' R U' R' U2 R)
I know 4 is an upper bound since edges can be oriented with 2 or less, and then corners with 2 or less. But maybe it's always possible with 3? If not, which OLLs need 4?

Stefan

P.S. Thanks for noticing that I'm a wise guy ;-).


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 Post subject: Re: Last Level
PostPosted: Sun Jul 17, 2005 5:56 am 
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David J wrote:
Actually, wise guy, (r U r' U2 r U r') *3 works the same as your first example, that is, staring with a solved cube, it returns to the solved state.

But that's no leap since all algs do that. In fact, that is part of the *definition* of an algorithm!


Hmm, that can't be quite correct, even if you mean using repetitions. I have a two move *algorithm* for my Square-1 that when I apply it repeatedly starting from solved state, will *not* return it to solved state. Actually I can't even repeat it because after the first time it blocks the first move of the algorithm.

Cheers!
Stefan


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 Post subject: Re: Last Level
PostPosted: Mon Jul 18, 2005 1:46 pm 
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David J wrote:
A few days ago someone posted "25 corner positions" on the Yahoo speedcubing list when there are 32 total.

I believe the posting on Yahoo was correct, and may have been quoting the results of Bernard Helmstetter's work (see the link in insanity_cubed's post above). Note that cases that are related by symmetry or inversion (reversal) are counted only once by BH, and that the solved case is included among the 25.

If you reckon that there are missing cases, please quote a sequence for one of them! I'd be interested, as I thought I'd checked the figure of 25 some months ago.

[In practice, of course, you need to know more than 24 sequences, unless you can reflect and invert them at high speed.]


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 Post subject: Re: Last Level
PostPosted: Mon Jul 18, 2005 2:07 pm 
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skeneegee wrote:
David J wrote:
I have no idea what Lars has on his site.


It works but I think it's too many algs for what it does.
http://www.cubezone.be/coll.html


Yeah, he repeats 9 with mirror images. It's a useful way to list them, and not a problem.

skeneegee wrote:
David J wrote:
I don't know where you're getting your numbers but there are only 3 edge permutations on the LL that need solving. And I don't know how you get 40 COLL when CLL, which is COLL plus CPLL, is only 31.


Well there's "U", "U'", "Z", and "H", right?


I don't know what those letters stand for.

skeneegee wrote:
yeah about the 40 I posted, I was too lazy to count them all since he said there were 40 at the top of the page, I just took his word for it. I guess there's only 38 there.


You were right the forst time - there are 40 listed on Lars page. Nine of them are reflections.

skeneegee wrote:
Have you posted your 31 cases anywhere that I could see them?


I don't have them posted anywhere. I wish I had software which allowed me to have text and images next to each other, but I don't.

David J wrote:
There are only six corner orientations....

Code:
   █→   ←█   █→   ██   ██   ←→   ←→
   ←↓    ↓→   ↓█     ↓↓   ←→  ←→    ↓↓


The first two are reflections.

David J wrote:
If you want a minimum set four algs will suffice, though they may need to be done more than once


skeneegee wrote:
I mean I know PLL pretty good but I was wondering if theres better way than OLL, PLL. OLL seems almost overwhelming to learn, but I thought the same about PLL before I learned them all. Maybe I should just give OLL a try.


Whatever floats your boat. Don't forget that there are the odd ones - ones that no one I know of are working on, like ELL then CLL, CPEOLL then COEPLL, and COEPLL then CPEOLL.

David J wrote:
Why not look for what *you* think is simple and logical?


skeneegee wrote:
I'm trying, there's alot to this little toy and I love it!

Thanks for your time David
-mike


Sure anytime, Mike.

David J


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 Post subject: Re: Last Level
PostPosted: Mon Jul 18, 2005 2:20 pm 
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Mike G wrote:
David J wrote:
A few days ago someone posted "25 corner positions" on the Yahoo speedcubing list when there are 32 total.

I believe the posting on Yahoo was correct, and may have been quoting the results of Bernard Helmstetter's work (see the link in insanity_cubed's post above). Note that cases that are related by symmetry or inversion (reversal) are counted only once by BH, and that the solved case is included among the 25.

If you reckon that there are missing cases, please quote a sequence for one of them! I'd be interested, as I thought I'd checked the figure of 25 some months ago.

[In practice, of course, you need to know more than 24 sequences, unless you can reflect and invert them at high speed.]


I did a quick check yesterday. He's missing [0123] [0021] for example.

David J


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 Post subject: Re: Last Level
PostPosted: Mon Jul 18, 2005 2:25 pm 
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David J wrote:
I did a quick check yesterday. He's missing [0123] [0021] for example.

That's the inverse of [0123][0012], which is Helmstetter's case 2. He doesn't count inverses as separate cases.


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 Post subject: Re: Last Level
PostPosted: Mon Jul 18, 2005 3:11 pm 
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HI Stefan,

StefanPochmann wrote:
David J wrote:
StefanPochmann wrote:
That's correct, just try ((R U' L' U R' U' L) U')*3.


That's not quite what I meant! :P


I wish you had a little more faith in me. Are you sure you noticed the extra U' in there? It does *not* return to a solved state.


Sorry. I assumed you got an idea right but just had a typo...

StefanPochmann wrote:
In the meantime I also proved that (R' U' F' U F R) can't solve the complete LL. Was actually even easier than the proof for the edges algorithm.

Here's an algorithm that can do ELL completely: [R U R' U', r]. Call it #, then (# U2 # U # U) permutes three edges preserving orientation.

Using your corners algorithm and mine for edges we can solve the LL completely using these two algorithms (plus U turns). Is it possible using just one algorithm (plus U turns)?


Yeah, that's the main question.

StefanPochmann wrote:
A similar question I've asked myself is how many Sune "variants" must be applied to do OLL. For example, to flip four edges you can do with 3 "Sunes":
((Rr) U2 R' U' R U' (Rr)') U' ((Ll)' U' L U' L' U2 (Ll)) (R' U' R U' R' U2 R)


Fun! I do that kind of stuff all the time.

PS thanks for using my notation.

StefanPochmann wrote:
I know 4 is an upper bound since edges can be oriented with 2 or less, and then corners with 2 or less. But maybe it's always possible with 3? If not, which OLLs need 4?

Stefan

P.S. Thanks for noticing that I'm a wise guy ;-).


You're welcome ;)

I'm home with a bad head cold. My brain is half fried (and it's not even Friday!)Maybe I'll give this stuff a look in a few days.

Thanks,

David J


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 Post subject: Re: Last Level
PostPosted: Tue Jul 19, 2005 1:22 am 
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David J wrote:
there are the odd ones - ones that no one I know of are working on, like ELL then CLL, CPEOLL then COEPLL, and COEPLL then CPEOLL.


David,

Wow! thanks for the info, I see now what CLL is good for, I had no idea ELL existed. Not only that but that combo is fewer algs than OLL/PLL! Exactly what I was trying to find.

thanks again -mike

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 Post subject: Re: Last Level
PostPosted: Tue Jul 19, 2005 4:30 am 
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skeneegee wrote:
Exactly what I was trying to find.

If you are seriously contemplating "CLL then ELL", you should have a look at Masayuki Akimoto's pages on it. He has a well-developed recognition system, though he explains it only for the blue-opposite-white colour scheme. There is also useful information at speedcubing.com (under "Algorithms").


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 Post subject: Re: Last Level
PostPosted: Tue Jul 19, 2005 11:13 am 
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Mike G wrote:
David J wrote:
I did a quick check yesterday. He's missing [0123] [0021] for example.

That's the inverse of [0123][0012], which is Helmstetter's case 2. He doesn't count inverses as separate cases.


That one isn't an inverse. That's the trouble with theory sometimes - it doesn't match practice.

Two corners with the Up color facing away from each other is different from two corners with the Up color facing the same direction. We're talking about Last Level positions, not last level and another side.

I realize that many speedcubers use OLL/PLL because they follow other people's methods and that one is the most well known. (OLL makes a distinction between those two orientations.) Anyway, as such I can believe that few have really looked hard at the other cases, but there are 7 cases missing from the 32 corner positions, not just one, do I have to point them all out?

David J


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 Post subject: Re: Last Level
PostPosted: Tue Jul 19, 2005 11:28 am 
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StefanPochmann wrote:
David J wrote:
Actually, wise guy, (r U r' U2 r U r') *3 works the same as your first example, that is, staring with a solved cube, it returns to the solved state.

But that's no leap since all algs do that. In fact, that is part of the *definition* of an algorithm!


Hmm, that can't be quite correct, even if you mean using repetitions. I have a two move *algorithm* for my Square-1 that when I apply it repeatedly starting from solved state, will *not* return it to solved state. Actually I can't even repeat it because after the first time it blocks the first move of the algorithm.

Cheers!
Stefan


First off I'm not talking about Square-1.

Second while a recipe may be a type of algorithm, and even one's "method" is a type of algorithm, the definition of an alogithm in mathematics includes the recursive aspect.

On Rubik's cube you can make a series of movements: not only will that series have the same effect upon the cube when repeated, but repetions of that sequence will return the cube to the initial state eventually. There are no exceptions.

Regards,

David J


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 Post subject:
PostPosted: Tue Jul 19, 2005 12:02 pm 
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David,

I have a couple of questions about your LL 2 look.

How many total algs do you use in the LL? What would you do if the LL was oriented (luckily) as soon as you finish F2L? Would you use PLL in that case?

Does that happen 71/1211?

I find this stuff facinating, I'm learning lots in this thread.

-mike

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 Post subject:
PostPosted: Tue Jul 19, 2005 4:48 pm 
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Hi Mike,

skeneegee wrote:
David,

I have a couple of questions about your LL 2 look.

How many total algs do you use in the LL?


I'm in the middle of an inventory - I use probably between 300 and 400. I only need fifty, but I'm learning to better allow for interesting occurrances. For example most of the edge flipping algs are 12 moves but I found two 11 move ones which change corners, too. I've a lot which solve the corners and flip two edges over. So I'm commonly left with a 7 move EPLL.

skeneegee wrote:
What would you do if the LL was oriented (luckily) as soon as you finish F2L? Would you use PLL in that case?


I've been working on that, too. Almost there, I need two more algs.

skeneegee wrote:
Does that happen 71/1211?


I'm beginning to question that often quoted 1211. The way I solve the F2L that full orientation happens more often than that.

skeneegee wrote:
I find this stuff facinating, I'm learning lots in this thread.

-mike


Cool.

David J


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 Post subject: Re: Last Level
PostPosted: Thu Jul 21, 2005 3:44 am 
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David J wrote:
That one isn't an inverse. That's the trouble with theory sometimes - it doesn't match practice.
Two corners with the Up color facing away from each other is different from two corners with the Up color facing the same direction. We're talking about Last Level positions, not last level and another side.

Sorry, but you are mistaken here, assuming that you are talking about CLL as a preliminary to ELL. To be explicit about this case, the optimal sequence
R B' R B2 U' B' U' B U B' R2 (U)
solves the case of 2 up colours facing in the same direction, and
(U') R2 B U' B' U B U B2 R' B R'
does the case where they are facing away from one another. I've included the leading/trailing move to make it quite obvious that these solutions are inverse to one another.
Quote:
(OLL makes a distinction between those two orientations.)

That's because the OLL problem doesn't define where the cubies are going to relative to one another, so it is quite meaningless to speak of an inverse in most cases, including this one. For the CLL problems, on the other hand, the corner permutation is defined (up to rotation of the whole layer), so it is legitimate to speak of inverse problems there.
Quote:
Anyway, as such I can believe that few have really looked hard at the other cases, but there are 7 cases missing from the 32 corner positions, not just one, do I have to point them all out?

Given your objection to the case we've discussed, I'd expect your 7 "missing" cases to be the inverses of the 7 cases in Helmstetter's list that are not marked with an asterisk: these are the cases where the inverse is not just a (geometrical) symmetry variant of the position listed. I don't mind being proved wrong, btw, though I (and some others) might not agree that "few have looked really hard at the other cases". Surely we are not alone. :wink:


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 Post subject: Re: Last Level
PostPosted: Thu Jul 21, 2005 1:03 pm 
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Hi Mike G,

Mike G wrote:
David J wrote:
That one isn't an inverse. That's the trouble with theory sometimes - it doesn't match practice.
Two corners with the Up color facing away from each other is different from two corners with the Up color facing the same direction. We're talking about Last Level positions, not last level and another side.

Sorry, but you are mistaken here, assuming that you are talking about CLL as a preliminary to ELL. To be explicit about this case, the optimal sequence
R B' R B2 U' B' U' B U B' R2 (U)
solves the case of 2 up colours facing in the same direction, and
(U') R2 B U' B' U B U B2 R' B R'does the case where they are facing away from one another. I've included the leading/trailing move to make it quite obvious that these solutions are inverse sequences.


Let me see: you rotate two corners once, then twice, and the third time they are all standing up straight. By your definition all corners upright is an inverse of these two. Good example of the recursive aspect of algorithms. You give and an interesting example of slight of hand considering that at
http://yem.ai.univ-paris8.fr/~bh/cube/
it reads, > permutation and orientation of corners (25 configurations, average : 9.18), <
They are presented as "permutation," "orientation" and "configurations," *not* sequences.

Quote:
(OLL makes a distinction between those two orientations.)

Mike G wrote:
[ That's because the OLL problem doesn't define where the cubies are going to relative to one another, so it is quite meaningless to speak of an inverse in most cases, including this one. For the CLL problems, the corner permutation is defined (up to rotation of the whole layer), so it is legitimate to speak of inverse problems here.


You simply aren't paying enough attention.
Examples 10 and 12 are inverses of one another:
10 (QU2) is solved by B' R B L' B' R' B L a sequence which produces 12 (QU).
So that page *has* both of your "inverse sequences."

Quote:
Anyway, as such I can believe that few have really looked hard at the other cases, but there are 7 cases missing from the 32 corner positions, not just one, do I have to point them all out?

Mike G wrote:
Given your objection to the case we've discussed, I'd expect your 7 "missing" cases to be the inverses of the 7 cases in Helmstetter's list that are not marked with an asterisk: these are the cases where the inverse is not related by purely geometrical symmetry to the position listed.


Nope, not even close.
Examples 10 and 12 are not marked with an asterisk, and aren't among the missing.

Mike G wrote:
I don't mind being proved wrong, btw, though I (and some others) might not agree that "few have looked really hard at the other cases". Surely we are not alone. :wink:


C'mon, you haven't even noticed that one of the main "two corners out" positions for many, often in the 1980s the first alogorithm found by people studying the cube, R U' L' U R' U' L (U) is missing. [0 1 3 2 ] [10 1 1 ] And, it is it's own inverse.

I recommend that you, and those who agree with you, look again.

David J


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 Post subject: Re: Last Level
PostPosted: Thu Jul 21, 2005 2:23 pm 
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Hi David,

I'll have to keep this short, but...
David J wrote:
Let me see: you rotate two corners once, then twice, and the third time they are all standing up straight. By your definition all corners upright is an inverse of these two.

I said no such thing, and made no "definition" here, but no matter... I'd rather not repeat myself.
Quote:
Examples 10 and 12 are inverses of one another:
10 (QU2) is solved by B' R B L' B' R' B L a sequence which produces 12 (QU).

Your sequence does indeed produce 12, but it does not solve 10. Look again.
Quote:
Mike G wrote:
Given your objection to the case we've discussed, I'd expect your 7 "missing" cases to be the inverses of the 7 cases in Helmstetter's list that are not marked with an asterisk

Nope, not even close.

Sorry, I shouldn't have presumed to guess.
Quote:
C'mon, you haven't even noticed that one of the main "two corners out" positions for many, often in the 1980s the first alogorithm found by people studying the cube, R U' L' U R' U' L (U) is missing. [0 1 3 2 ] [10 1 1 ]

I didn't notice that because, in fact, that case is not missing: your sequence generates Helmstetter's case 14.

It was, indeed, one of the first sequences I found when studying the Cube in the 1980s.
Quote:
I recommend that you, and those who agree with you, look again.

Please would you do the same?

All the best
Mike


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PostPosted: Thu Jul 21, 2005 3:24 pm 
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hey, I know you guy's are in the middle of figuring something out but could one of you please explain your numbering system to me? The [0123] you guys are using. If it's a long explaination, then could you point me in the right direction?


thanks -mike grimsley

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skeneegee wrote:
could one of you please explain your numbering system to me? The [0123] you guys are using. If it's a long explaination, then could you point me in the right direction?

It's not complicated, but it's probably quickest for you to look at Bernard Helmstetter's own description at http://www.ai.univ-paris8.fr/~bh/cube/

Just follow the first link in that page.

PS: I am taking a two-week break from this thread. ;)


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PostPosted: Fri Jul 22, 2005 7:21 pm 
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skeneegee wrote:
hey, I know you guy's are in the middle of figuring something out but could one of you please explain your numbering system to me? The [0123] you guys are using. If it's a long explaination, then could you point me in the right direction?


thanks -mike grimsley


Hi Mike,

You already know these are for the Last Level.

The first set of numbers within brackets is the position and the second within brackets is the orientation.

[ 0 1 2 3 ] means that every corner is in the right place, noted clockwise starting at the lower left: 0 is Left Front corner, 1 is Left Back corner, 2 is Right Back corner and 3 is the Right Front corner. The numbers pertain to the cubies themselves, the position of the numbers pertain to the place on the cube.

To reverse them the convention chose was the switch 1 and 3 like [ 0 3 2 1 ] and to swap two adjacent corners they chose the right side [ 0 1 3 2 ].

The second set of brackets are for orientation. [ 0 1 1 1 ] would mean that the Left Front corner was upright, and the other three were rotated clockwise. A 2 means that the corner in question was rotated counterclockwise.

OK?

I have nicknames for each of the orientations. :)

David J


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 Post subject:
PostPosted: Fri Jul 22, 2005 9:50 pm 
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Mike and David,

Thanks, I get it now!

-the other mike g

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 Post subject:
PostPosted: Mon Jul 25, 2005 1:30 pm 
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Hi Mike G,

I did look again.

I apologise for making mistakes in my argument, thus wasting your time and clouding the issue.

The labels to the page specify orientations, permutations, and configurations, yet it is missing seven. If they were labeled "sequences" exclusively, that would be different.

Mike G wrote:
Given your objection to the case we've discussed, I'd expect your 7 "missing" cases to be the inverses of the 7 cases in Helmstetter's list that are not marked with an asterisk

David J wrote:
Nope, not even close.

Mike G wrote:
Sorry, I shouldn't have presumed to guess.


You are right that the seven missing configurations *are* the inverses of the unasterisked ones.

Each of the different orientations has its own numbers of occurrances where two adjacent corners are out of place. All of the orientations have one occurrance where all corners are in place and one occurrance where the corners are reversed.

Here are my nicknames for the orientations, and the number of occurrance, called "Two Out," where two adjacent corners are out of place:
All Up, 1
Three Down, 4
Headlights, 3
Sidelights, 3
Oddlights, 2
Wheels, 2
Barrow, 3.

The page in question
http://yem.ai.univ-paris8.fr/~bh/cube/solutions_c1.html
has all the permutation occurances of the orientations called All Up, [ 0 0 0 0 ] and Sidelights [ 0 0 1 2 ].
The other orientations' missing occurrances are listed here with a new case number. For convenience, I've labeled them as 2a, 10a, 11a, 12a, 16a, 17a, and 18a.

The missing seven configurations are:
Three Down, Two Out # 11a [ 0 1 3 2 ] [ 0 2 2 2 ]
Headlights, All In #2a [ 0 1 2 3 ] [ 0 0 2 1]
Headlights, Two Out # 10a [ 0 1 3 2 ] [ 0 2 1 0 ]
Headlights, Two Out # 16a [ 0 1 3 2 ] [ 2 1 0 0 ]
Oddlights, Two out # 12a [ 0 1 3 2 ] [ 0 2 0 1 ]
Wheels, Two Out # 18a [ 0 1 3 2 ] [ 2 1 2 1 ]
Barrow, Two Out # 17a [ 0 1 3 2 ] [ 2 1 1 2 ]

I don't know why Sidelights was given the place of honor. If I made a page like that one, leaving out inverse sequences, I'd give precendence to "Headlights," not "Sidelights."
The fewest number of moves making for the greatest amount of change on the Last Level is R' U' F' U F R and its inverses and mirrors. The algorithm reverses the order of the corners, the corner orientation is changed to "Headlights," two edges are flipped over, and the edge position is changed, and all of this in only 6 moves.

******

David J wrote:
Examples 10 and 12 are inverses of one another:
10 (QU2) is solved by B' R B L' B' R' B L a sequence which produces 12 (QU).

Mike G wrote:
Your sequence does indeed produce 12, but it does not solve 10. Look again.


I could have stated this better.

The inverse of 12, retaining the corner position, is among the missing.
Example 10 is the correct orientation for the inverse of 12.
The missing one of that set is Oddlights, Two out # 12a [ 0 1 3 2 ] [ 0 2 0 1 ].

My point here is that one orientation's "inverses" can produce more than one other orientation.

For example, case 2 has the same orientation as case 12 - Sidelights. The inverse of case 2 produces the "Headlights" orientation, and the inverse of case 12 produces the "Oddlights" orientation. This can cause confusion.

******

David J wrote:
C'mon, you haven't even noticed that one of the main "two corners out" positions for many, often in the 1980s the first alogorithm found by people studying the cube, R U' L' U R' U' L (U) is missing. [0 1 3 2 ] [10 1 1 ]

Mike G wrote:
I didn't notice that because, in fact, that case is not missing: your sequence generates Helmstetter's case 14.
It was, indeed, one of the first sequences I found when studying the Cube in the 1980s.


David J wrote:
I recommend that you, and those who agree with you, look again.


Mike G wrote:
Please would you do the same?


My error; I gave the right case, but the wrong algorithm. It should have been
# 11a [ 0 1 3 2 ] [ 1 0 1 1 ] Three Down Two Out L R' U' R U L' U2 R' U2 R .

Mike G wrote:
All the best
Mike


Thanks! You, too.

David J

On 10/18/2005 I made this correction:Three Down, Two Out # 11a [ 0 1 3 2 ] [ 0 2 2 2 ] *it was: Three Down, Two Out # 11a [ 0 1 3 2 ] [ 1 0 1 1 ]


Last edited by David J on Tue Oct 18, 2005 8:18 pm, edited 2 times in total.

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 Post subject:
PostPosted: Mon Aug 15, 2005 4:16 am 
[Gone. I guess I should look into that "troll" button thing! - Sandy]


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PostPosted: Mon Aug 15, 2005 8:19 am 
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euphemist wrote:
valtrex -

Erk. Is that just a helpful suggestion for us tired old puzzlists, or should this spammer be ejected?


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PostPosted: Mon Aug 15, 2005 4:08 pm 
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Mike G wrote:
euphemist wrote:
valtrex -

or should this spammer be ejected?


I hope Sandy comes back very soon. I hope to God this doesn't get out of hand. Yes I think they all should be :evil: ejected :evil: that post that stuff that has nothing to do with cubing at all. What do you think cubing/puzzles friends?


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PostPosted: Mon Aug 15, 2005 6:38 pm 
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At the Daily WTF (http://www.thedailywtf.com/) the forums have a troll button.

If enough logged in uers click the troll button, the post goes away...maybe more ;)


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 Post subject: missing position
PostPosted: Tue Oct 18, 2005 3:40 pm 
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Hi Mike G.,

Would you please tell me where on this page
http://yem.ai.univ-paris8.fr/~bh/cube/solutions_c1.html I might find this position, its inverse or its mirror: [ 0 1 3 2 ] [ 1 0 1 1 ] ?

David J


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 Post subject: Re: missing position
PostPosted: Wed Oct 19, 2005 9:40 am 
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Hi David

Sorry, I didn't spot your post till now. And *oops* I've only just seen that you brought this up before!
David J wrote:
Would you please tell me where on this page http://yem.ai.univ-paris8.fr/~bh/cube/solutions_c1.html I might find this position, its inverse or its mirror: [ 0 1 3 2 ] [ 1 0 1 1 ]?

It is the inverse of the mirror of case 11: front/back reflection changes [0132][1011] to [0132][0222], and taking the inverse of that then gives [0132][0111], the version listed by Helmstetter.

To be more explicit about it, a sequence to set-up your case [0132][1011] in the correct orientation is
W = U R' U2 R U2 L U' R' U R L'
I simply read that off from your own solution, of course.

Helmstetter's solution for his case 11 is
X = (U) R B2 L' B2 R' B L B' (U2)
with the (implied) trailing U-move included. Now, F/B reflection changes sequence X into
Y = (U') R' F2 L F2 R F' L' F (U2)
and the inverse of Y is Y' = (U2) F' L F R' F2 L' F2 R (U)

So Helmstetter's solution to your case [0132][1011] would be (U2) F' L F R' F2 L' F2 R (U); or, what amounts to the same thing here, B' R B L' B2 R' B2 L (U').

I've just tried that last sequence on your case (set up using "your" W), and it quite definitely works.

Best regards
Mike


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 Post subject:
PostPosted: Sun Dec 18, 2005 3:16 pm 
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Just found something quite interesting. When doing the T-alg [(R U R' U') (R' F) (R2 U') (R' U' R U) (R' F')], if you use Rr instead of R, then it does a 3corner cycle.

Are there any other interesting alg adaptions like this?

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 Post subject: Re: Last Level
PostPosted: Wed Dec 21, 2005 3:35 pm 
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StefanPochmann wrote:
A similar question I've asked myself is how many Sune "variants" must be applied to do OLL. For example, to flip four edges you can do with 3 "Sunes":
((Rr) U2 R' U' R U' (Rr)') U' ((Ll)' U' L U' L' U2 (Ll)) (R' U' R U' R' U2 R)
I know 4 is an upper bound since edges can be oriented with 2 or less, and then corners with 2 or less. But maybe it's always possible with 3? If not, which OLLs need 4?


I've answered my question today in the speedcubing group:
http://games.groups.yahoo.com/group/spe ... sage/24097

For OLL, at most 3 sunes are needed, and average is 2.01724 sunes.


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 Post subject: Re: Last Level
PostPosted: Thu Dec 22, 2005 3:53 pm 
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Hi Stefan:

StefanPochmann wrote:
A similar question I've asked myself is how many Sune "variants" must be applied to do OLL. For example, to flip four edges you can do with 3 "Sunes":
((Rr) U2 R' U' R U' (Rr)') U' ((Ll)' U' L U' L' U2 (Ll)) (R' U' R U' R' U2 R)


To flip four edges try two "sunes" this way:
b R U R' U R U2 R' U b' QU' Rr U2 R' U' R U' R'r' U'

Cheers,

David J


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 Post subject: Re: Last Level
PostPosted: Thu Dec 22, 2005 10:08 pm 
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David J wrote:
To flip four edges try two "sunes" this way:
b R U R' U R U2 R' U b' QU' Rr U2 R' U' R U' R'r' U'


Bah, that's cheating ;-). But you made me try some more variants and I found this nice one (I prefer M over S/b):
M U R U R' U R U2 R' M'
(sorry I don't use your notation, I just like to be able to check easily with an applet)

It can also be used to flip four edges with two "sunes":
(M U R U R' U R U2 R' M') (M' R U2 R' U' R U' R' M)

Which simplifies to:
(M U R U R' U R U2 M2 U2 R' U' R U' r')


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 Post subject:
PostPosted: Fri Dec 23, 2005 6:26 pm 
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Juast to confirm, the 3 sunes includes positioning the edge?


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 Post subject:
PostPosted: Fri Dec 23, 2005 8:04 pm 
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AndrewSeven wrote:
Juast to confirm, the 3 sunes includes positioning the edge?


Sunes may be varied by adding a slice turn or two.

DJ


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 Post subject:
PostPosted: Sat Dec 24, 2005 9:43 am 
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Ya, I noticed the effect of a slice on the sune before and started playing with the extra slice again when I saw the recent posts.

David J:
What does ""M" do? [edit : doh! got it]
I'm often uncertain about your notation, esp, the little images, are they on a web page somewhere?


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 Post subject:
PostPosted: Sat Dec 24, 2005 3:41 pm 
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AndrewSeven wrote:
Ya, I noticed the effect of a slice on the sune before and started playing with the extra slice again when I saw the recent posts.

David J:
What does ""M" do? [edit : doh! got it]
I'm often uncertain about your notation, esp, the little images, are they on a web page somewhere?


Hi AndrewSeven,

Do you mean the ones from skeneegee's post dated July 15th?

My software cannot correctly copy these. The ¦ means a solid block for the color of that side being upright. The arrows point to indicate the direction the color of that side is rotated to face.

¦→
←↓
Three Down: Three corners are rotated counterclockwise.

←¦
↓→
Three Down: Three corners are rotated clockwise

¦→
↓¦
Oddlights: Two non-adjacent corners are upright, the other two are rotated. For example: Right Rear is rotated counterclockwise and Left Front is rotated clockwise.

¦¦
↓↓
Headlights: Two corners are upright, two are rotated to face the same direction, like headlights on a car.

¦¦
←→
Sidelights: Two corners are upright, two are rotated to face away from each other.

←→
←→
Wheels: All four corners are rotated, two pairs face away from each other, like wheels on a car.

←→
↓↓
Barrow: All four corners are rotated, two face away from each other like wheels on the front of a wheel barrow, and two face the same direction like handles on a wheel barrow.

My notation uses lower case letters to mean "slices only," and "Q" to mean cube rotations. I don't generally use MES or x y and z.

Cheers,

David J


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 Post subject:
PostPosted: Sat Dec 24, 2005 5:30 pm 
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AndrewSeven wrote:
Juast to confirm, the 3 sunes includes positioning the edge?


No, that's an open question. My statement was for OLL only.

I asked a similar question at the end of my post: If you have solved F2L and LL corner permutation, how many of my 8 sune variants are necessary (avg/max) to solve the cube?

A trivial upper bound is 7. That's 3 for OLL, and twice 2 for edge permutation with done with two 3-cycles like this: (R U R' U R U2 R') y (L' U' L U' L' U2 L). But I believe the average is 3-4 and the maximum is 5. That's because with avg=2 and max=3 you can do OLL and then you're left with 1 out of 12 edge permutations. Surely there will be overlaps, but adding two more sunes will potentiall add a factor of 49 (7*7). It's not 8 because one of the 8 variants is always the inverse of the previously executed, so it's no use executing it. And if you look at my OLL table, the first factor is 6, the second is indeed 7. And I don't see a good reason why it should drop rapidly unless you're running out of different cases (after exploring 1+6+42 OLLs it drops deep cause there aren't that many left).

Cheers!
Stefan


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