One-Answer 3x3x3 Question Thread

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If AUF is correct but the orientation is 180° wrong to execute your PLL Alg, do you do U2 + Alg + U2 or do you do a cube rotation?

In the moment I do cube rotation but probably I should change.
But I still have problems with double flicks, so my U2 are not that fast.
 
OldSwiss said:
If AUF is correct but the orientation is 180° wrong to execute your PLL Alg, do you do U2 + Alg + U2 or do you do a cube rotation?

In the moment I do cube rotation but probably I should change.
But I still have problems with double flicks, so my U2 are not that fast.
Click to expand...
U2 alg U2.
 
Weird question but how many unique commutators are there for corners and edges respectively on 3x3? I was trying to figure it out for corners and used the following rationale ( Note: I'm counting inverses as unique therefore UFR UBR UBL =/= UFR UBL UBR ) =>

(1) There are 378 unique cases for each buffer with 24 possible buffer locations
=> Total: 9072
(2) We need only to use 8 buffer locations (one for each corner permutation ) since every com covers 3 orientations of other coms ( UFR UBR UBL = RUF BUR LUB = FUR RUB BUL )
=> Total: 3024
(3) We can reduce the the number of buffers down to 6 since the final 2 buffers are covered by the preceding 6.
=> Total: 2268
(4) For each new buffer location we subtract any cases containing the previous buffer piece. Example: If we start with UFR as our first buffer and then move onto UBR as our second, we can remove the 36 cases in the UBR set since 36 cases contain UFR. As the number of sets increase so does the number of cases we subtract ( in multiples of 36, one for each preceding buffer piece ).
=> Total: 1728

Lmk if this is correct.
 
Silky said:
Weird question but how many unique commutators are there for corners and edges respectively on 3x3? I was trying to figure it out for corners and used the following rationale ( Note: I'm counting inverses as unique therefore UFR UBR UBL =/= UFR UBL UBR ) =>

(1) There are 378 unique cases for each buffer with 24 possible buffer locations
=> Total: 9072
(2) We need only to use 8 buffer locations (one for each corner permutation ) since every com covers 3 orientations of other coms ( UFR UBR UBL = RUF BUR LUB = FUR RUB BUL )
=> Total: 3024
(3) We can reduce the the number of buffers down to 6 since the final 2 buffers are covered by the preceding 6.
=> Total: 2268
(4) For each new buffer location we subtract any cases containing the previous buffer piece. Example: If we start with UFR as our first buffer and then move onto UBR as our second, we can remove the 36 cases in the UBR set since 36 cases contain UFR. As the number of sets increase so does the number of cases we subtract ( in multiples of 36, one for each preceding buffer piece ).
=> Total: 1728

Lmk if this is correct.
Click to expand...
Is your question about commutators or about 3-cycle cases? Every 3-cycle can be solved with a commutator, but there are also many things that are not 3-cycles which are nevertheless solvable with commutators.

The total number of 3-cycle cases is:
\( \frac{24\times21\times18}{3\times3}=1008 \)

24×21×18 counts each distinct 3-cycle 9 times: 3 times because the orientation doesn't matter (this is your "(2)"), and 3 times again because A-B-C, B-C-A and C-A-B are all the same 3-cycle. I don't follow your reasoning for "(4)".
 
xyzzy said:
Is your question about commutators or about 3-cycle cases? Every 3-cycle can be solved with a commutator, but there are also many things that are not 3-cycles which are nevertheless solvable with commutators.

The total number of 3-cycle cases is:
\( \frac{24\times21\times18}{3\times3}=1008 \)

24×21×18 counts each distinct 3-cycle 9 times: 3 times because the orientation doesn't matter (this is your "(2)"), and 3 times again because A-B-C, B-C-A and C-A-B are all the same 3-cycle. I don't follow your reasoning for "(4)".
Click to expand...

Yes, the number of 3-cycles. "(4)" Was about after counting all of the UFR cycles, when we go to count the UBR cycles we want to ignore cycles in the UBR set that contain UFR. So Aa and Ab are both contained in the UFR, UBR, and UBL sets. We don't want to count this 3 times.. only once, so when we are going through the UBR set everytime we see UFR pop up we don't count it since that 3-cycle was already counted in the UFR set. Don't know if that better explains it.. or if it's correct T-T
Obviously I was trying to brute force which is the problem ( I suck at probability/combinations etc )
 
Is it bad to do R’ U2 (J perm) U R when you get a J perm from a U2 away? It adds a move at the start and end but it flows better for me than U2 J perm U. My guess is that it’s because I’m faster at doing a j perm from thumb down grip.
 
Is it advised to use home grip for scrambling? I have, without realising it, found myself with right thumb on bottom, fingers on top, normal grip with the left, because it makes B2 and F2 moves much faster. I would rarely do those moves in a solve, but some scrambles have a lot of B2 and F2 moves, which I suck at from home grip. Still, learning to do awkward moves faster from home grip is probably more useful, so should I try to stick to that during scramble?
 
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