# 3 color 3x3x3 algorithm thread



## supercube (Jun 8, 2015)

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 (Jun 8, 2015)

supercube said:


> 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.
> 
> ...


That J Pll has B moves :/ How would you execute it?


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## 2180161 (Jun 8, 2015)

rotations?
rotate x'


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## supercube (Jun 8, 2015)

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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## qqwref (Jun 8, 2015)

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)


Spoiler



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)


Spoiler



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)


Spoiler



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)


Spoiler



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)


Spoiler



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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## supercube (Jun 9, 2015)

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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## qqwref (Jun 9, 2015)

Yeah, that should be it (those 5 cases + mirrors). For probabilities I think you can just add up the probabilities for the normal PLL cases.


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## cuBerBruce (Jun 9, 2015)

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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## qqwref (Jun 9, 2015)

That's a good point, Bruce. Some cases may have faster algs with one AUF than another - for instance, the E perms I got (starting from edges solved) were all 7 htm.


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## supercube (Jun 9, 2015)

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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## cuBerBruce (Jun 9, 2015)

supercube said:


> 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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## supercube (Jun 10, 2015)

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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## cuBerBruce (Jun 10, 2015)

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.

```
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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## supercube (Jul 4, 2015)

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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## TDM (Jul 4, 2015)

supercube said:


> 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.


Do you mean ZZ? Roux doesn't have EOLines


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## supercube (Jul 4, 2015)

sorry yeah I think I was trying to say ZZ


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## supercube (Jul 4, 2015)

20 move superflip
U R2 F B R B2 R U2 L B2 R U’ D’ R2 F R’ L B2 U2 F2

16 move superflip
U R2 F B R B2 R L B2 R U’ D’ R2 F R’ L


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## qqwref (Jul 5, 2015)

I found a 14 move superflip: U R L F L R B F D' U' F L' R' U


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## cuBerBruce (Jul 5, 2015)

supercube said:


> 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.




qqwref said:


> 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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## supercube (Jul 6, 2015)

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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## ch_ts (Jul 6, 2015)

fyi there's an online solver here:
http://www.irbsystems.com/test2/rc/index.html


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## JustinTimeCuber (Jul 6, 2015)

how about a 2-color cube? Three red sides, three blue sides. Red opposite blue for all of them. How many PLLs would there be?


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## cuBerBruce (Jul 6, 2015)

supercube said:


> 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.



Yes, I guess you can think of dealing with AUF case and angle case at the same time rather than independently.

Taking "F2 U' L2 U F2 U" as our standard algorithm, we have the two versions:

"a" = F2 U' L2 U F2 U
"b" = R2 U' F2 U R2 U (rotated alg)

We can then solve any combination of the 4 angle cases and 2 pre-AUf cases according to the following table.


```
("Good" AUF case)      ("Bad" AUF case)
  Solved       Edges mis-aligned      Edges aligned
corners on:        with F2L              with F2L

    F              "b"                    U "a"
    B              U2 "b"                 U' "a"
    L              "a"                    U' "b"
    R              U2 "a"                 U "b"
```

A bit more complicated than simply treating the angle case and AUF case independently, if you ask me. (And the cases using a U2 are not FTM-optimal, but cube rotations are totally avoided using this scheme.)



supercube said:


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


Yes.

Mathematicians in the early days generally assumed <U,D,F,B,L,R> (face turns only) solving, so they simply referred to the metrics involving face turns as QTM (quarter-turn metric) and HTM (half-turn metric). But in the speedsolving community, we don't generally assume solving via face turns only, so we typically say "FTM" (face turn metric) to indicate anything that is a face turn counts as a single move, rather than saying HTM which doesn't describe "what type" of half-turns are allowed.


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## supercube (Jul 6, 2015)

I think it is very exciting to know that there are easy and advanced methods for the 3 color cube that can both be used in speedsolving. avoiding cube rotations and U moves is definitely something to work for long term without getting bored.


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## supercube (Aug 5, 2015)

```
X INDICATES NOT SOLVED

   B
  0X0
L 00X R
  000
   F

R2 U F2 R2 F2 U

   B
  0X0
L X0X R
  0X0
   F

L2 R2 D L2 R2 U

   B
  XXX
L X0X R
  0X0
   F

R2 D L2 D' R2 U

   B
  X0X
L 00X R
  0X0
   F

L2 U F2 R2 U R2 U

   B
  XXX
L X00 R
  000
   F

R2 U' R2 U R2 U
```


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## supercube (Jul 13, 2017)

ch_ts said:


> fyi there's an online solver here:
> http://www.irbsystems.com/test2/rc/index.html



I started using this for the first time today. I should have checked it out a long time ago. the methods contained here are fantastic. instead of layer by layer OLL + PLL solving you can solve using petrus or something else. the methods used in this link are 3 move or 5 move solutions to finish the 3 color cube. I don't think I will switch to a %100 corners first solve style but I think if I can influence corner rotation intuitively during a LBL or petrus solve then I have a good chance at getting a case with finishing move length 3. also interesting is that corners first solvers can use roux for edge orientation and permutation with great success due to the limited edge identity (increased lucky edge permutation). I regret not using a speed cube but the diansheng is working reasonably well.


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## supercube (Jul 13, 2017)

*I am now %100 convinced that the corners first method is better for fewest moves, less algorithms to memorize, shorter algorithms, easier recognition, and most likely faster times.*

after all, think about this, there are only two types of unique corner pieces. like binary, each corner is correct or it is not correct. the maximum number of moves to permute (but not care to orient) one corner piece from a scrambled cube is 1 move in QTM (with one exception). there is however a very un-lucky rare case of having all 8 corners in a specific configuration that makes this statement untrue slightly. in %1.42857142857 of scrambles (4/8*3/7*2/6*1/5) you will have for example corner type A in LFU RFU RFB RFD also with corner type B in LBU LFD LBD RBD. given that in this example that RFU should contain type B as the correct corner piece, for this special case you can not move the correct corner piece into the RFU position with 1 move in QTM because LFU RFB RFD all contain type A pieces. a type B corner piece must be moved from one of the LBU LFD LBD RBD positions to the RFU position using 2 moves in QTM. also 3 moves QTM if the corner piece is taken from LBD to be placed in RFU.

on average there will be 4 corners not permuted. I do not know if the distribution is flat or gaussian. I do know that if you plan your moves carefully that you can position 1, 2, 3, or 4 corners pieces simultaneously using 1 move in QTM. the lower bound for move length in QTM to place but not orient all 8 corners from any given scramble would likely be based on the lower bound to generate the worst case scenario given in the example above from a solved cube. I can manually try the upper bound by hand. we know it is less than or equal to the 14 move superflip provided in this thread. I do not have proof but I think given the 1 move QTM solutions to place any given corner it would certainly be around 8 plus or minus a few.

in conclusion I think that a corners first solver can orient and permute all the edges after permuting and orienting all the corners with a combination of slice moves and 180 face turns (L2 R2 U2 D2 F2 B2) that will preserve the appearance of corner sticker colors. there are indeed even more algorithms that do not conform to this convention but do preserve the appearance of corner sticker colors. it is likely that none of these algorithms are greater than 7 moves unless we have a system of solving all 12 edges at once. a 32 move speed solve is reasonably easy to achieve using this method. that number will be reduced by combining edge orientation and edge permutation into a one look. we can also use edge influence when solving corners intuitively since most corners first solves of a 3 color cube have many equally efficient alternate mirrors. that would bring a speed solve down to 24 moves.


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## supercube (Jul 13, 2017)

the most important alg you will need for solving with permuting the corners as step 1. I use this to swap two bad corners. if you have 4 bad corners on the same face you can always turn that face a quarter turn in either direction. having only 2 bad corners looking at the entire cube is 5 moves HTM. 1 move QTM vs 5 move HTM. the difference is huge.

```
B
  XXX
L X00 R
  000
   F

R2 U' R2 U R2 U
```

8 bad corners can be solved by


```
U D
U D'
U' D
U' D'
L R
L R'
L' R
L' R'
B F
B F'
B' F
B' F'
```

the optimal solution for 6 bad corners is unknown to me at this time but you can use D if 4 of the bad corners are in the D layer. then you will have 2 bad corners in the U layer. see above.

also very interesting is that all of these moves preserve corner orientation so you could in theory orient corners as step 1 before permuting them. or you can permute them as step 1. there is absolutely no difference here. right now I am permuting corners first and using sune after cube rotations with color neutral LL. I do this because I am very good at corner permutation recognition but not very good at corner orientation when picking up a scrambled cube. they all look the same almost.


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## supercube (Jul 14, 2017)

switched back to layer by layer but almost never start with a cross or cross hunting exactly. I have found that color neutral solving is of great importance. I also found that one can not be strictly petrus or strictly beginner/CFOP/fridrich. the 3 color cube has very little in the way of %100 unique pieces (cubies?). it does have many alternate pieces with the same identity or interchangeable identity. the flexibility of the solver in combination with the flexibility of the cube itself will exponentially yield more efficient solves. I use edge influence with the corner slot pair. I do OLL for the corners only. then I use the modified PLL in this thread. I keep my eyes on it while executing the OLL slow so I can execute the PLL in a fluid motion. memorizing all reflections, rotations and inverse of the modified PLL has proved to be very valuable with very little difficulty. ladies and gentlemen this is in my opinion the fastest way to solve this puzzle.


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## supercube (Jul 18, 2017)

I have a new question. Starting from a solved 3 color cube, what are the possible 2 move, 3 move, 4 move scrambles (HTM) with the 48 redundant symmetries removed.

for example, in 2 moves, there is only


```
U R
U R2
U D
```

Notice there is a lot less than 18*15 unique 2 move HTM scrambles before we remove redundant symmetries on a 6 color cube. A 3 color cube will not ever start a scramble with a 180 face turn. Nor will the second term be a 180 face turn on an opposite face from the first term. Therefor, before symetries are removed, a 3 color cube can have only 12*14 scrambles 2 moves long. And only 3 possible states with unique properties. In 1 move HTM there is only one scrambled state of a 90 face turn.

So far we have Fibonacci 1,1,3 starting from solved to scrambled. Any good programmers want to cover some new territory here? If God's number is 14 then maybe it would be similar to the 15th term of the Fibonacci sequence? Is the rate of growth perfectly logarithmic?


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