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3 color 3x3x3 algorithm thread

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the 3 color 3x3x3 that has yellow opposite yellow, red opposite red, blue opposite blue can be solved using a different set of algorithms. I did not find a thread for this so I made one. there are 4 of each edge. there are 3 unique edge color combinations. there are 4 of each corner piece. there are 2 unique color patterns on the corners. the corners are mirrors of each other. the puzzle is solved in many alternate positions so long as each face contains 9 stickers of the same color. there are two of each center so be careful when you calculate the number of possible states of the cube.

so far I have only one algorithm. I noticed that I can use Gc Ga or Rb perm to finish the cube if the case is Gc Ga or Rb. it does not matter which one you use so you can just use the shortest one every time. from this we can also make a hypothesis that average number of moves to PLL will be reduced compared to 6 color 3x3x3. we can make another hypothesis that the number of PLL cases will be reduced from 19 to some smaller number. we already eliminated 4 cases if we drop all 4 G perms. H perm is a PLL skip also so we can drop that.

I want to know if there are shorter PLL's that swap pieces from the bottom layer with pieces from the top layer.

J PLL B' R' F R2 B' R F' R' 8 moves compared to R' U L' U2 R U' R' U2 R L U' 11 moves
T PLL R B2' R' U2 R2 U2 R B2' R' 9 moves compared to R U R' U' R' F R2 U' R' U' R U R' F' 14 moves
A PLL R' U R' D2 R U' R' 7 moves compared to R' U R' D2 R U' R' D2 R2 9 moves. this makes the T PLL obsolete.
Z PLL M2 U M2 U M2 5 moves replaces M2 U M2 U M' U2 M2 U2 M' U2 10 moves
 
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Berd

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Bertie Longden
#2
the 3 color 3x3x3 that has yellow opposite yellow, red opposite red, blue opposite blue can be solved using a different set of algorithms. I did not find a thread for this so I made one. there are 4 of each edge. there are 3 unique edge color combinations. there are 4 of each corner piece. there are 2 unique color patterns on the corners. the corners are mirrors of each other. the puzzle is solved in many alternate positions so long as each face contains 9 stickers of the same color. there are two of each center so be careful when you calculate the number of possible states of the cube.

so far I have only one algorithm. I noticed that I can use Gc Ga or Rb perm to finish the cube if the case is Gc Ga or Rb. it does not matter which one you use so you can just use the shortest one every time. from this we can also make a hypothesis that average number of moves to PLL will be reduced compared to 6 color 3x3x3. we can make another hypothesis that the number of PLL cases will be reduced from 19 to some smaller number. we already eliminated 4 cases if we drop all 4 G perms. H perm is a PLL skip also so we can drop that.

I want to know if there are shorter PLL's that swap pieces from the bottom layer with pieces from the top layer.

here is a faster J PLL. U2 B' R' F R2 B' R F' R' B2' 10 moves compared to R' U L' U2 R U' R' U2 R L U' 11 moves
That J Pll has B moves :/ How would you execute it?
 
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F,B become L,R. then R becomes F. it is a rotation yes.

edited J PLL U2 B' R' F R2 B' R F' R' B2' in original post to remove U2 at the beginning
edited again B2 redundant. B2 removed.
 
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#5
Here are some HTM-optimal PLLs - the numbers in parentheses are the counts without AUF. I put mirrors together to make it more useful.

Solved, H, Na, Nb: (0)

Ua, Ub, V, Y: (5)
R2 U F2 R2 F2 U
R2 U F2 R2 F2 U'
R2 U F2 L2 B2 D
R2 U F2 L2 B2 D'
R2 U B2 R2 B2 U
R2 U B2 R2 B2 U'
R2 U B2 L2 F2 D
R2 U B2 L2 F2 D'
R2 D R2 F2 L2 U
R2 D R2 F2 L2 U'
R2 D R2 B2 R2 D
R2 D R2 B2 R2 D'
R2 D L2 F2 R2 U
R2 D L2 F2 R2 U'
R2 D L2 B2 L2 D
R2 D L2 B2 L2 D'
B2 U' R2 F2 L2 D
B2 U' R2 F2 L2 D'
B2 U' R2 B2 R2 U
B2 U' R2 B2 R2 U'
B2 U' L2 F2 R2 D
B2 U' L2 F2 R2 D'
B2 U' L2 B2 L2 U
B2 U' L2 B2 L2 U'
B2 D' F2 R2 F2 D
B2 D' F2 R2 F2 D'
B2 D' F2 L2 B2 U
B2 D' F2 L2 B2 U'
B2 D' B2 R2 B2 D
B2 D' B2 R2 B2 D'
B2 D' B2 L2 F2 U
B2 D' B2 L2 F2 U'

Z, E: (5)
L2 R2 U B2 F2 D
L2 R2 U B2 F2 D'
L2 R2 U' B2 F2 D
L2 R2 U' B2 F2 D'
L2 R2 D L2 R2 U
L2 R2 D L2 R2 U'
L2 R2 D' L2 R2 U
L2 R2 D' L2 R2 U'
B2 F2 U L2 R2 D
B2 F2 U L2 R2 D'
B2 F2 U' L2 R2 D
B2 F2 U' L2 R2 D'
B2 F2 D B2 F2 U
B2 F2 D B2 F2 U'
B2 F2 D' B2 F2 U
B2 F2 D' B2 F2 U'

Aa, Ab, F, T: (5)
U F2 U' L2 U F2 U
U F2 U' L2 U F2 U'
U F2 U' L2 D R2 D
U F2 U' L2 D R2 D'
U F2 D' B2 U L2 D
U F2 D' B2 U L2 D'
U F2 D' B2 D F2 U
U F2 D' B2 D F2 U'
U B2 U L2 U' B2 U
U B2 U L2 U' B2 U'
U B2 U L2 D' R2 D
U B2 U L2 D' R2 D'
U B2 D F2 U' L2 D
U B2 D F2 U' L2 D'
U B2 D F2 D' B2 U
U B2 D F2 D' B2 U'
U' F2 U R2 U' F2 U
U' F2 U R2 U' F2 U'
U' F2 U R2 D' L2 D
U' F2 U R2 D' L2 D'
U' F2 D B2 U' R2 D
U' F2 D B2 U' R2 D'
U' F2 D B2 D' F2 U
U' F2 D B2 D' F2 U'
U' B2 U' R2 U B2 U
U' B2 U' R2 U B2 U'
U' B2 U' R2 D L2 D
U' B2 U' R2 D L2 D'
U' B2 D' F2 U R2 D
U' B2 D' F2 U R2 D'
U' B2 D' F2 D B2 U
U' B2 D' F2 D B2 U'
D R2 U' F2 U R2 D
D R2 U' F2 U R2 D'
D R2 U' F2 D B2 U
D R2 U' F2 D B2 U'
D R2 D' L2 U F2 U
D R2 D' L2 U F2 U'
D R2 D' L2 D R2 D
D R2 D' L2 D R2 D'
D L2 U F2 U' L2 D
D L2 U F2 U' L2 D'
D L2 U F2 D' B2 U
D L2 U F2 D' B2 U'
D L2 D R2 U' F2 U
D L2 D R2 U' F2 U'
D L2 D R2 D' L2 D
D L2 D R2 D' L2 D'
D' R2 U' F2 U R2 D
D' R2 U' F2 U R2 D'
D' R2 U' F2 D B2 U
D' R2 U' F2 D B2 U'
D' R2 D' L2 U F2 U
D' R2 D' L2 U F2 U'
D' R2 D' L2 D R2 D
D' R2 D' L2 D R2 D'
D' L2 U F2 U' L2 D
D' L2 U F2 U' L2 D'
D' L2 U F2 D' B2 U
D' L2 U F2 D' B2 U'
D' L2 D R2 U' F2 U
D' L2 D R2 U' F2 U'
D' L2 D R2 D' L2 D
D' L2 D R2 D' L2 D'

Ga, Gc, Ra, Rb: (6)
R2 U B2 L2 U L2 U
R2 U B2 L2 U L2 U'
R2 U B2 L2 D F2 D
R2 U B2 L2 D F2 D'
R2 D L2 F2 U F2 D
R2 D L2 F2 U F2 D'
R2 D L2 F2 D R2 U
R2 D L2 F2 D R2 U'
L2 U F2 U' F2 L2 U
L2 U F2 U' F2 L2 U'
L2 U F2 U' B2 R2 D
L2 U F2 U' B2 R2 D'
L2 U F2 D' R2 F2 U
L2 U F2 D' R2 F2 U'
L2 U F2 D' L2 B2 D
L2 U F2 D' L2 B2 D'
L2 D R2 U' R2 F2 D
L2 D R2 U' R2 F2 D'
L2 D R2 U' L2 B2 U
L2 D R2 U' L2 B2 U'
L2 D R2 D' F2 L2 U
L2 D R2 D' F2 L2 U'
L2 D R2 D' B2 R2 D
L2 D R2 D' B2 R2 D'
B2 L2 U F2 U' R2 D
B2 L2 U F2 U' R2 D'
B2 L2 U F2 D' F2 U
B2 L2 U F2 D' F2 U'
B2 L2 U' R2 U F2 D
B2 L2 U' R2 U F2 D'
B2 L2 U' R2 D R2 U
B2 L2 U' R2 D R2 U'
B2 L2 D R2 U' B2 U
B2 L2 D R2 U' B2 U'
B2 L2 D R2 D' R2 D
B2 L2 D R2 D' R2 D'
B2 L2 D' F2 U L2 U
B2 L2 D' F2 U L2 U'
B2 L2 D' F2 D F2 D
B2 L2 D' F2 D F2 D'

Gb, Gd, Ja, Jb: (5)
B2 U R2 U' R2 U
B2 U R2 U' R2 U'
B2 U R2 D' F2 D
B2 U R2 D' F2 D'
B2 U' B2 U B2 U
B2 U' B2 U B2 U'
B2 U' B2 D L2 D
B2 U' B2 D L2 D'
B2 D B2 U' B2 D
B2 D B2 U' B2 D'
B2 D B2 D' R2 U
B2 D B2 D' R2 U'
B2 D' R2 U R2 D
B2 D' R2 U R2 D'
B2 D' R2 D B2 U
B2 D' R2 D B2 U'
 
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Thread starter #6
first I would like to thank you for your time. :tu would it be correct to say that these represent the 5 cases for PLL? 6 cases total if you include PLL skip? are the probabilities known?
 
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#8
I would argue that qqwref's optimal counts are not correct, though.

I believe there is no such thing as a 5-move J-perm, for example. Likewise, I believe there is no such thing as a 5-move A-perm, F-perm, or T-perm. Regardless of the AUF case.

Since the maneuvers may move last layer (U) pieces to the D-layer, the alignment of the last layer with the bottom 2 layers at the start matters, unlike what is the case for the normal cube.
 
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you can rotate the algorithm to eliminate AUF but you will inevitably require a D move %50 of the time. it is possible that there might be another set of algorithms that consolidates the D move for some or all of the cases. we will not know until we prove it (if it appears) or disprove it by exhaustive testing of the entire order 7 group. 7 is Ga, Gc, Ra, Rb: (6) + AUF (or ADF alternatively) to prove that if it does exist, that it would have no additional benefit over existing algorithms. likewise we would look for order 6 in the other groups with moves (5) + AUF.

I think I might need to draw up some pictures that include the lower layer and the top layer for each algorithm.

next challenge, if we only have 2 + 7 OLL then 5 PLL for a 3 color cube, perhaps that ZBLL for the 3 color cube is easy to learn with 7*6-1 ZBLL's? this could be very fun to speed solve I think. something easier than 3x3 but you can still ZB it!
 
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#11
you can rotate the algorithm to eliminate AUF but you will inevitably require a D move %50 of the time.
Nope.

If you try to avoid the AUF, it will still end up costing you a move somewhere, hence no reduction in move count. You would need an alg that keeps all pieces in their same layers, but then you will need to use a suboptimal alg.

You can call R-perm 5 moves + AUF, but in that case, the AUF (or possibly an ADF) is 100% guaranteeed to be required. You can not eliminate the AUF without adding another move back somewhere.

For the record, this is what I get for the cases:

U perm: The canonical AUF case (corners solved) requires 6 moves. The other AUF case (the 4 corners not solved wrt the first 2 layers) requires 7 moves.

Z perm: The canonical Z perm case (corners solved) requires 6 moves. The AUF case corresponding to the canonical E-perm (edges solved) requires 7 moves.

A perm: The canonical A perm AUF case (edges solved) requires 7 moves. The other AUF case (the 4 edges not solved wrt the first 2 layers) requires 6 moves.

R perm: All AUF cases are essentially equivalent. 7 moves are required. You can call it 6 moves + AUF, but it is still guaranteed 7 moves total.

J perm: All AUF cases are essentially equivalent. 6 moves are required. You can call it 5 moves + AUF, but it is still guaranteed 6 moves total.
 
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Thread starter #12
I have studied what you wrote and it does make more sense now. the R perm case is fascinating as it will always start or end in AUF (or ADF) and it will never vary in move count. it will always be 7 moves. the other cases average 6.5 moves and 5.5 moves respectively. I arrived at the 0.5 moves (%50 probability) by converting all AUF to ADF then counting all 4 rotations of D and eliminating the two redundant rotations (180 mirrors).
 
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#13
The U-, Z-, and A-perm (if you consider both AUF cases) average 6.5 moves because if you get the "bad" AUF case you can first apply a pre-AUF (U or U') and then do the 6-move alg for the "good" AUF case. Since you can expect the "good" and the "bad" AUF cases an equal number of times, the average is 6.5.

The J-perm, like the R-perm, has both AUF cases being symmetrically equivalent, so it always takes 6 moves no matter what. You can't solve the cube from any J-perm case in 5 moves, so you can't get average of 5.5 for J-perms.

EDIT:
Here is what I get for ZBLL and LL case counts. Warning: These number include the solved case as a case.
Code:
        Raw Count   Cases  Cases with mirrors considered the same
ZBLL       972       130     77
LL        7776      1004    550
EDIT2:

The expected number of moves on average for PLL (using FTM-optimal algs) is:

(1/36)*0 + (1/36)*1 + (1/36)*6 + (1/36)*7 + (1/9)*6 + (1/9)*7 + (1/9)*6 + (1/9)*7 + (2/9)*6 + (2/9)*7 = 37/6 = 6.1666...

This assumes:
* the first two layers are completely solved (including that the D layer is color-aligned with the E layer) at the start of the PLL step
* the whole cube is solved at the end of the PLL step (all 3 layers color-aligned)
* no deliberate influencing of the PLL case is done while solving the first two layers and OLL
 
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so I got my stickers on my diansheng cube. the T perm may require both a cube rotation (%50 chance) and a AUF (%75 chance). you can always reduce this to a %50 AUF with no cube rotations but it is imperative that you know 3 rotations of the same algorithm. you only need 3 (not 4) algorithms because 2 of the 4 U rotations can AUF U or U'. if you know how to do it you can choose the correct U or U' move to force all pre AUF cases to use the same algorithm. perhaps it is better to redraw the recognition picture and omit the first U move for the non-AUF case? none of these algorithms have recognition pictures that you would need to use the algorithms but I think I can post that sometime soon. I want to play with it more first.

on a side note it may be a good idea to use roux method on this cube. EO line as first step is very very low move count.
 
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#19
so I got my stickers on my diansheng cube. the T perm may require both a cube rotation (%50 chance) and a AUF (%75 chance).
Well, if you're claiming a 75% chance of needing an AUF, then you must be using the AUF to rotate the U layer in the align the U layer for executing a particular algorithm. Then, you will still need to verify whether or not the U layer is in the desired alignment with the bottom two layers. There will be a 50% chance that you will need to rotate the bottom two layers a quarter turn to have the proper relationship between the top layer and the bottom two layers. You can not simply use a cube rotation because that does not change the relative alignment between the U layer and the bottom two layers.

Alternatively, you could AUF (50% chance needed) to get the proper relative alignment between the layers first, and then use a cube rotation (75% chance) to get the the U layer rotated for the desired algorithm. With this, your probabilities for AUF and cube rotation are reversed. Well, I note you could substitute U2 for a y2 rotation in the case that the U layer is off 180 degrees from desired.

For FTM-optimality purposes, of course, we would always use a cube rotation or rotated algorithm. That leaves just a 50% probability of whether or not a pre-AUF is needed before applying one of the the basic 6-move T/A/F-perm algorithm.


I found a 14 move superflip: U R L F L R B F D' U' F L' R' U
Yes, 14 moves is optimal in FTM and QTM for superflip. Rokicki determined God's number is 15 in FTM, 17 for QTM.

By the way, I've been meaning to mention this related thread on the Domain of the Cube forum.
 
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if you include U, U', U2 as a setup %75 of the time, you can always reduce it to only 2 rotations of the algorithm that need to be memorized. I think that might be similar to what you were saying with the probability being reversed.

in that other thread, is HTM another way to say FTM?
 
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