The One-Answer Question Thread [closed]

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Can I say I am sub 15 now? :)

stats: (hide)
number of times: 50/50
best time: 10.518
worst time: 23.823

current mo3: 15.634 (σ = 1.28)
best mo3: 12.662 (σ = 1.89)

current avg5: 14.630 (σ = 0.49)
best avg5: 13.356 (σ = 1.11)

current avg12: 14.811 (σ = 1.28)
best avg12: 13.512 (σ = 1.22)

current avg50: 14.815 (σ = 1.96)
best avg50: 14.815 (σ = 1.96)

session avg: 14.815 (σ = 1.96)
session mean: 14.969
 
arvind1999 said:
Can I say I am sub 15 now? :)

stats: (hide)
number of times: 50/50
best time: 10.518
worst time: 23.823

current mo3: 15.634 (σ = 1.28)
best mo3: 12.662 (σ = 1.89)

current avg5: 14.630 (σ = 0.49)
best avg5: 13.356 (σ = 1.11)

current avg12: 14.811 (σ = 1.28)
best avg12: 13.512 (σ = 1.22)

current avg50: 14.815 (σ = 1.96)
best avg50: 14.815 (σ = 1.96)

session avg: 14.815 (σ = 1.96)
session mean: 14.969
Click to expand...

It actually isn't really important, but yeah, definitely.
 
Smiles said:
... and if your solution to get to 396 is so complicated, maybe that's why your answer is so far off. my solution was like 1 line and maybe that's all you need? try simplifying it and just doing the basics. what i did at first was EP x CLL for an approximation of 480
Click to expand...

I know that my 396 is not correct. For some reason I added L3C cases onto it even though they are included in L4C.

I figured this by using OCELL (orients last layer corners and permutes last layer egdes) algs and matched them with the amount of cases for corner permutation. The way I figured OCELL, there are 14 cases that include swapping 2 opposite edges and 22 that include swapping 2 adjacent edges. I came out with that plus the 7 algs with edges in place to orient the corners and get 43. This 43 I know is correct. For each of the cases that have 2 edges needing swapped, the corners have 8 possible permutation because of lack of the ability to AUF. Cases with edges correct are all included in L4C with 84 algorithms. L4C actually has 96 algorithms , but some are the same by doing a cube rotation. An example of a few of these would be the CPLL cases. Because there are 36 (14+22) cases with 8 possible corner orientations, this gives you 36x8 or 288 cases that need 2 edges swapped. 288+84+24=396 I should not have included the 24 cases for L3C. This would bring it down to 372 cases or 384 if you count L4C as 96. The L4C can be calculated by multiplying the 8 possible corner orientations achievable with cube rotations, and multiplying this by 12 which is the number of possibilities for corner permutation without AUFs or cube rotations calculated by 4x3x2x1/2. 12x8=96. The only thing I can think of that could have possibly made my answer off is if I got the wrong number for possible corner permutations that have 2 edges wapped (I got 8 possibilities). I got this by counting them out, so I'm pretty sure it is correct.

EDIT: I was wrong in the amount of possible corner permutations. I left out the two simplest cases. Rotate all 4 corner clockwise or counterclockwise. This bring me to 10 cases making this 344 or 456 cases. That is more accurate, but it still isn't the 493 or 501.
 
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sneaklyfox said:
Average of 5 is really 3 of 5, not counting best time and worst time.
Average of 12 is really 10 of 12, not counting best time and worst time.
Click to expand...

The site has separate listings. One side says Avg. 5 and Avg. 12. The other has 3 of 5 and 10 of 12. And the times are all different, so I'm still confused. Thanks!
 
happymeal98 said:
The site has separate listings. One side says Avg. 5 and Avg. 12. The other has 3 of 5 and 10 of 12. And the times are all different, so I'm still confused. Thanks!
Click to expand...

I just looked. It says Avg. 5 and Avg. 10. I believe the left side is just average of all/last 5/last 10. Right side is best.

Edit: Ok, I'm not really sure. I think the right side might be best 3 of 5 (but of the last 5 solves).

Edit 2: After playing around a bit more... yeah, the 3 of 5 and 10 of 12 excludes the best and worst times, but only for the last solves. So it doesn't actually show you your best "Average of 5" or "Average of 12".
 
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arvind1999 said:
Can I say I am sub 15 now? :)

stats: (hide)
number of times: 50/50
best time: 10.518
worst time: 23.823

current mo3: 15.634 (σ = 1.28)
best mo3: 12.662 (σ = 1.89)

current avg5: 14.630 (σ = 0.49)
best avg5: 13.356 (σ = 1.11)

current avg12: 14.811 (σ = 1.28)
best avg12: 13.512 (σ = 1.22)

current avg50: 14.815 (σ = 1.96)
best avg50: 14.815 (σ = 1.96)

session avg: 14.815 (σ = 1.96)
session mean: 14.969
Click to expand...

i think sub-15 means you average under 15.
so you were sub 15 during that session, but are you always? O:

i have been sub-13 before for like 200 solves, but my normal session averages are like 13.4 so i wouldn't say i'm sub-13. getting there, though :P

elrog said:
I know that my 396 is not correct. For some reason I added L3C cases onto it even though they are included in L4C.

I figured this by using OCELL (orients last layer corners and permutes last layer egdes) algs and matched them with the amount of cases for corner permutation. The way I figured OCELL, there are 14 cases that include swapping 2 opposite edges and 22 that include swapping 2 adjacent edges. I came out with that plus the 7 algs with edges in place to orient the corners and get 43. This 43 I know is correct. For each of the cases that have 2 edges needing swapped, the corners have 8 possible permutation because of lack of the ability to AUF. Cases with edges correct are all included in L4C with 84 algorithms. L4C actually has 96 algorithms , but some are the same by doing a cube rotation. An example of a few of these would be the CPLL cases. Because there are 36 (14+22) cases with 8 possible corner orientations, this gives you 36x8 or 288 cases that need 2 edges swapped. 288+84+24=396 I should not have included the 24 cases for L3C. This would bring it down to 372 cases or 384 if you count L4C as 96. The L4C can be calculated by multiplying the 8 possible corner orientations achievable with cube rotations, and multiplying this by 12 which is the number of possibilities for corner permutation without AUFs or cube rotations calculated by 4x3x2x1/2. 12x8=96. The only thing I can think of that could have possibly made my answer off is if I got the wrong number for possible corner permutations that have 2 edges wapped (I got 8 possibilities). I got this by counting them out, so I'm pretty sure it is correct.

EDIT: I was wrong in the amount of possible corner permutations. I left out the two simplest cases. Rotate all 4 corner clockwise or counterclockwise. This bring me to 10 cases making this 344 or 456 cases. That is more accurate, but it still isn't the 493 or 501.
Click to expand...

When you say 2 edges swapped, do you mean adjacent swap? Or opposite swap? or both? I didn't really check too much of your math but the clarification would be needed.
And as for corner permutations that you mentioned in your edit, what about X perms? (H perm + U2)

I find that it's easier to say that the CLL is fixed and the edges move around it. Edges are just easier to visualize.
But if you think about corners moving around the edges and start including impossible CPLL cases like clockwise 4-cycles, you have to start including edge parity, which just becomes a different case when you AUF.

you're either putting too much in some parts, or leaving things out in some part, but thinking about the whole thing doing with:
CLL + EPLL, and PLL (corners oriented) as a separate category
 
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When two edges are swapped, which two are swapped doesn't matter when refering to corner permutation. Only the fact that 2 are swapped matters. Edge parity wouldn't matter because you cannot AUF or you would change the case as you said. You AUF to get atleast 2 edges in the correct position. You then can have 2 adjacent edges needing swapped, or two opposite, or all 4 solved. You will always position the cube so that your opposite swaps are UF and UB and the diagnal ones would be UF and UR (you could do it with different edges, but the important thing is that you hold it the same every time). With opposite edge swaps, you can have 14 possible corner orientations. With diagnal swaps you can have 22 possible corner orientations. The permutation of corners will be 10 possible cases with having 2 edges swapped (it doesn't matter which ones, just that they are). Thus, with having 22 possible orientations for corners with cases that have an adjacent edge swap, and 10 possible permutations for corners, you know that there is 220 cases with an adjacent edge swap. With an opposite edge swap, there are 14 possible corner orientations, and 10 possible permutations for corners giving you 140 cases with an opposite edge swap. 220 + 140 + 84 = 444 The 84 is for cases that have all edges correct (L4C).
 
your organizational system for edge swaps is a little unorthodox, since you're disallowing UL/UR swaps and AUFing instead, among other things.
try having them fixed, whatever case is whatever you have. F2L doesn't matter, you can do Dw turns all you want and the same edges are in the same place, solved or not only matters relative to the other Uedges.

What i mean is, for example, if you have an opposite edge swap UF/UB, do a Dw2. With the exact same ZBLL case from the same angle, you now have a UR/UL swap, in which case you would then AUF.
I'm suggesting that if you're gonna use fixed edges, follow EPLL rules of 12 cases. 4 Ua perms, 4 Ub perms, 1 skip, 1 H perm, and 2 Z perms.
Now since you can have both front and back U perms, among others, rotating your CLL case by U2 will end up giving you identical cases. Therefore, have your CLL in fixed positions. One CLL case must always be in the same angle.

I can't help you too much with your method, since it just takes a little bit too much thinking to get to the solution.

But if I remember correctly, your original question was why the wiki said 493 when you got 396, and i think the [CLL (43 - 3) x EPLL (12) + PLL (21) - identical H-ZBLL cases (4 + 4) = 40*12+21-8 = 493] solution should suffice as an answer.

im sure if you tried hard enough you could find the error in your solution, but doing it this way makes things much simpler.
 
If it would be possible to reduce the number of cases by eliminating multiple cases that can give you the same case with another recognition pattern, then no its not a sufficient answer. I've been looking for errors and I can't see why it doesn't work when it makes perfect sense. I know that my first sentence sort of doesn't make sense, but I don't see why you would 493 when I can identify all of the cases with less than that.
 
6 different possible edge swaps (4 adj, 1 opp, 1 solved)
40 different possible distinct recognizable CLL cases
each CLL case can have X permutation done it it and remain the same recognized case, as long as you get the CO correct afterwards. However, the ZBLL case changes. (or you could say that each edge swap case can have H perm done to it, same result)
therefore, # of total cases not including PLL is doubled (PLL already accounts for H/X perms within it).
PLL = 21
2 H-CLLs are completely mirrors, therefore the # of distinct edge cases for each of them loses 2/4 (half) of their adj swaps. 6-2 = 4 EP cases for each instead of 6.

2[6(40-2) + 4(2)] + 21 = 493

I just did it a different way. This time i used edge swaps instead of EPLL and everything still worked out fine. I didn't separate out L4C because it's just an extra edge swap case (solved). I always separate out PLL because counting PLL cases in with all the CO jumble would be a waste of time. so that's why i used 40 as the number of non-CO'd CLL cases.

edit: the point i am making is that no matter your recog system, the # of cases are the same. previously, my recog system was to look at CLL first. for argument's sake, let's say my recog system was to solve COLL, look at the exact EPLL + AUF, then undo my COLL alg and i'll know which ZBLL case i have. but if i have just PLL, then i'll solve it with PLL. that's pretty much just saying COLL*EPLL+PLL = 40*12+21=501 and of course subtract the identical H-CLL cases. there is no possibility of a repeated case there because i looked at COLL and the resulting EPLL, if anything were the same i'd end up having ZBLL algs that don't correspond to a case.

your solution is too complicated for me to try to find the error, plus i cant 100% understand all of it without completely sorting through it and it'll all just be too much. there is definitely an error in your solution.

the only way to completely check your solution against 493 is to:
1. draw out all of the cases you have created
2. look for repeated cases, if any exist.
3. find the matching one to each of the 493 in the ZBLL cases list.
4. prove that each of the cases in the ZBLL list that didn't pair up with yours, isn't actually unique.

of course you're not going to do this, but it's easier to say that my solution is correct than yours, since it's so much less complicated.
 
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for a beginner, anything better than a rubik's storebought one is good. and that basically means no dollar store.
i got to 23-second average before using a cube better than the rubik's one (not DIY) so it actually really doesn't matter at all. the price seems great so i think you should get it.
other choices like FII, guhong, and other DIYs would all work. for beginners, anything that won't go out of control (like a zhanchi) is perfect.
 
Username said:
What do you think is faster/more useful?

Full OLL and Full PLL or Orient edges, COLL, EPLL? Are there any other LL methods that go with CFOP and don't require lots of algorithms?
Click to expand...

Full OLL and PLL are better. Or you can do WV and PLL. COLL can be useful for bigcubes and OH but not really for 3x3, unless you know the corners are going to be opposite (Y, N perm) and you want to manipulate it or something. COLL is also a prereq for ZBLL.
 
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