As calculated by TDM, 576 (taking his word for it). Is this still acceptable?
The New Method / Substep / Concept Idea Thread
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As calculated by TDM, 576 (taking his word for it). Is this still acceptable?
Logiqx
Veteran
Here's another discussion CF discussion for you - http://www.speedsolving.com/forum/showthread.php?37600-Columns-First-underestimated-method
Renslay
Legend
I did not say it's a revolutionary method. It is a method. Or a variation, if you prefer; but I think saying Hoya and Redux is nearly the same is like saying CFOP and Petrus is nearly the same.
Moreover, there are many things on the wiki which is not a revolutionary method, or not even a method (like the Winter Variation).
supercavitation
Veteran
Given that it's 576 cases, I think we would need to figure out what all of the cases are before I commit to being willing to generate cases. We can figure this out by PM without clogging up this thread for others.
TDM
Legend
I'm beginning to think I may have made an error, but I'm struggling to see where. I've included AUF for both the EO and the PLL you'd get by flipping edges; this should mean there are actually 4 times less cases. 576/4 equals 144. But if I recalculate, I can either say that for each of 22 PLLs, there are 8 possible edge flips leaving with me with 176 cases, but if I say that for each of 4 edge flips there are 72 PLLs I have 288 cases... can someone else please help me out here?
btw, all three of these numbers seem very low to me. 576 looks closer to what it should be, but as that is 72*8 I know that I must have repeated some cases... :confused:
(this post was originally supposed to be me explaining what all the cases are, but that didn't happen xD)
JKNK
Regular
Anyone like my hoyau method idea
TDM
Legend
Wouldn't the U layer centre be much harder to do? Also, as (if I understand correctly) this method is designed to make the three cross edges easier to pair... but isn't it easier to pair them as you usually do with Yau?
Time to put my pathetic high school mathematical skills to the test!
I think you need to include AUF for only edges or only corners but not them both, as one can change cases while the other stays the same, to create a different case. There are also cases where the algs would have already been found, such as ELL with all corners permuted, that we wouldn't have to calculate. What I mean:
6 CPLL cases: One adjacent swap, one diagonal swap, Aa, Ab, E, X/solved. (X would technically be an ELL case, but as ELL would be a substep of this substep (substepception?!), we'll include it.)
6*(ELL cases including parities = 4!*4 = 96) = 576, what TDM came up with.
Doing it the other way around (with edges fixed instead) obviously just yields the same result.
However, out of these 576 cases, we can subtract a few that don't need to be calculated, namely PLL and ELL cases.
576 - 25(ELL) - 21 (PLL) = 530
So there are 530 cases that need an algorithm to be found. Again, this doesn't seem right... maybe someone more advanced will lend their assistance.
Edits in red.
I think you need to include AUF for only edges or only corners but not them both, as one can change cases while the other stays the same, to create a different case. There are also cases where the algs would have already been found, such as ELL with all corners permuted, that we wouldn't have to calculate. What I mean:
6 CPLL cases: One adjacent swap, one diagonal swap, Aa, Ab, E, X/solved. (X would technically be an ELL case, but as ELL would be a substep of this substep (substepception?!), we'll include it.)
6*(ELL cases including parities = 4!*4 = 96) = 576, what TDM came up with.
Doing it the other way around (with edges fixed instead) obviously just yields the same result.
However, out of these 576 cases, we can subtract a few that don't need to be calculated, namely PLL and ELL cases.
576 - 25(ELL) - 21 (PLL) = 530
So there are 530 cases that need an algorithm to be found. Again, this doesn't seem right... maybe someone more advanced will lend their assistance.
Edits in red.
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JKNK
Regular
It's a little easier with hoyau because you can use any center expect the white for edge pairing the 3 cross edges. The only down side is a little harder yellow center which you can do just as fast as the regular one if one cross edge doesn't block it. Maybe ill do an example solve (im barely sub 2 so ill suck)
TDM
Legend
The bit in blue isn't right. You have to take away 72 (one for each PLL that you've counted in the 576) and then add 21 because there are 21 PLLs. Same thing with ELL: subtract 48 (because 12 EP cases*4 EO cases=48... again, I don't know if this is right) and then add 29 for the number of ELL cases. In addition, you can do -72+22 for each of the 4-flip cases. I may try to calculate this by counting the cases rather than trying to simplify it to a few calculations.
Sorry, I was wrong with 3 cross edge pairing: they're the same... at least I think so. Please could you list the substeps out on separate lines to make it clearer what your method is? I can't quite understand your explanation.
Edit:
Here's what I got:
PLL: 22
PLLEF: 22
Opposite flip: 72
Adjacent flip: 72
Total = 188
... that also doesn't seem right, but now I really can't see where I've gone wrong.
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JKNK
Regular
Step 1: Make a center
Step 2: Put in 3 cross edges (you can use the u layer and if you know hoya this will be really easy and you can use the left and right centers so its easier than hoya or yau)
Step 3: Solve the opposite center (this well be easy, the only con is that you'll have to move the d layer but only once)
Step 4: Solve the remaining centers without messing up the cross edges
Step 5: put in the reaming edge and finish with 3-2-3 edge paring.
Step 6 Finish the Cube like a 3x3
If anyone is still confused just say so and i'll make an example solve
TDM
Legend
It won't be easier than Hoya or Yau? I find Yau edge pairing is more efficient than Hoya, so having the option to do either doesn't help. I can't see the advantage. There's no disadvantage though.
You won't have to move it only once? What if the four U layer centre pieces are on F, L, B and R?
From what I've understood, this is just Yau but with the U layer centre done in a different place (that makes it harder to do).
JKNK
Regular
It is actually easier scince you can use the all centers expect for the d layers and in most cases you only have to move the d layer once
Is there anyway someone could generate algorithms for a 1- look last layer?
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TDM
Legend
What is? You don't specify what you're talking about here: cross edges or U layer centre? Please could you give some example solves?
Cube Explorer.
JKNK
Regular
TDM
Legend
The last bit of my post was the brute force method. I went through every PLL case and determined how many times each one was counted for each EO case using rotational symmetry. I thought that it'd be more accurate than using a formula and having the chance of missing something out, but of course it gave me my lowest result so far.
After thinking some more, there are fewer EO cases than CO cases, so maybe there should be fewer cases than ZBLL. Maybe this is actually right?
Ty. Unfortunately, I don't know a fast way of making algs for every single case: Cube Explorer requires each case to be input one at a time (unless there's something to make this possible I don't know of). However, there are some links at the bottom of the wiki page to lists that people have already created, so you shouldn't need to generate your own.
Out of interest, why do you want these?
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JKNK
Regular
I just thought it would be a good idea for a 1 look ll. I mean it will reduced time down since the last layer takes about 4 seconds or less for world class solvers.
TDM
Legend
... It's not a good idea. Even ZBLL is not a good idea. There are too many cases, and even if you do learn all of them, you'll have trouble recognising which case is which and remembering the alg to use in every solve; because there are so many you will almost never see the same case twice, so you won't be able to practise what you know. You'll then forget everything and will have difficult ever doing the LL in one look, even if you know all the algs, except for the cases that are very easy to recognise (e.g. r U R' U' r' F R F').
Yeah sorry, that was stupid of me. But I counted 40 for opposite flip:
Ua, Ub: 2+2=4
H, Z: 1+2=3
Aa, Ab: 2+2=4
E, F: 2+2=4
Ga, Gb, Gc, Gd: 2+2+2+2=8
J, L: 2+2=4
Na, Nb: 1+1=2
Ra, Rb: 2+2=4
T, V, Y: 2+2+2=6
4+3+4+4+8+4+2+4+6=39, plus one for solved permutation = 40.
If I'm right, that leaves 156 cases, or 155 that aren't a LL skip. A lot less than the 576 I originally calculated.
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