# 4x4 Linear Fewest Moves Challenge



## StachuK1992 (Aug 24, 2011)

I'm not actually interested in FMC for big cubes; I am making this thread with the hope that it will encourage users to create new, good big cube methods.

I simply don't think we have near-perfect big cube methods.

Rules:
-Don't take more than 20 minutes. There's no need for this to take longer, because it should be done linearly.
-I don't care what (Sign/WCA) notation you use; they're both fairly universal as far as I'm concerned.
If you don't specify, I'm assuming that r=slice, R=outer, Rw=fat/wide turn.
-200 moves max. Slice+wide turn metric. 
(r2 = 1 move, as does R and Rw)

Have fun. Be creative ;D

statue


(Yes, I know that FMC tactics might not be the best for speedsolving methods. That's why I'm making you do it linear. Hehe)


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## cubernya (Aug 24, 2011)

I was actually just thinking about big cubes FM yesterday, but not linear!


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## caseyd (Aug 24, 2011)

linear?


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## StachuK1992 (Aug 24, 2011)

It means you can't rescramble, and if you 'undo' a move, both the moves you did and the moves you did to undo those moves count against you.
Your first solve is your only solve.


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## caseyd (Aug 24, 2011)

oh
ok


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## Mike Hughey (Aug 24, 2011)

Restricting to under 20 minutes might affect how practical it is to figure out a parity skip. (I know Chris Hardwick had some good ideas of how this might be achieved with a linear solve.) Give us an hour instead of 20 minutes, and I suspect it's very likely someone can come up with a consistent method for parity skip, but at 20 minutes, the time might be too tight to do all the counting that might be necessary.


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## StachuK1992 (Aug 24, 2011)

Good! I don't want you to develop a way to detect parity skip! 
I want the methods to be able to eventually used for regular solving. Too much time gives you too many impossible options that you couldn't take in a real solve.


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## irontwig (Aug 24, 2011)

Parity, but otherwise ok I guess, but a bit boring: 
http://tinyurl.com/algB2Uw-B-UwC14
94 moves, 6 pairs, Petrus


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## StachuK1992 (Aug 24, 2011)

If you could include a name of the method with the post, that would be neat.
For later quick comparison.


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## Kirjava (Aug 24, 2011)

Mike Hughey said:


> Give us an hour instead of 20 minutes, and I suspect it's very likely someone can come up with a consistent method for parity skip, but at 20 minutes, the time might be too tight to do all the counting that might be necessary.


 
Are you kidding? It's possible to completely avoid parity in normal speedsolves.


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## Lucas Garron (Aug 24, 2011)

Kirjava said:


> Are you kidding? It's possible to completely avoid parity in normal speedsolves.



Hmm. Do you get to see the scramble, Stachu?

Also, there was a while when we did this for the weekly competition (I think), but I'm not sure where that went.


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## StachuK1992 (Aug 24, 2011)

Why would you need to see it? Sure, you can see it; just don't inverse it or do something similar. 

One scramble, one solve.


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## Julian (Aug 24, 2011)

StachuK1992 said:


> Why would you need to see it?


Because, I assume, they can count quarter turns (or something similar) in the scramble to figure out whether or not there will be parity. (Similar to counting quarter turns in a BLD scramble to determine if there will be parity.)


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## Kirjava (Aug 24, 2011)

No need to look at the scramble - edge cycles can be traced in ~10 seconds.


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## cmhardw (Aug 25, 2011)

Kirjava said:


> No need to look at the scramble - edge cycles can be traced in ~10 seconds.


 
I tried very hard to do this a few years back and could occasionally detect parity in under 15 seconds, but I usually averaged near 25-30 seconds. I did not trace the cycles of all 24 wings. I traced out the arrangement of always 12 cycles (with or without parity basically) it would take to reduce to a scrambled 3x3x3 with oriented edges.

If you have ideas on how to do the tracing in ~10 seconds I would be extremely interested in talking more with you! I've only managed to do this in under 15 seconds less than 10 times.

One anticipatory comment, if you do trace all 24 wings, then how do you handle cases such as where you have accounted for 20 pieces and are left with 4 remaining pieces (you know that none are solved)? It can sometimes take approximately 4-5 seconds just to locate _one_ of those 4 pieces, let alone the cyclic structure of all 4.

It's not that I don't believe you, it's more that I'm completely at a loss for what your method would be! I would be very interested in taking this conversation to PM as well, since it is completely irrelevant to this thread (according to Statue's request).


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## riffz (Aug 25, 2011)

I think he underestimated how long it takes to trace them, but his point was that given 20 minutes you can determine parity from the start relatively quickly without looking at the scramble.

Amiright?


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## Kirjava (Aug 25, 2011)

Not at all. I know of at least one person who can do it in under 15 seconds on average. 

I'll PM you my ideas when I get home. I'll also do an FMC solve - I have some ideas for lowering movecounts.


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## Meep (Aug 25, 2011)

I might try this when I get home. Though I feel you should lower the max move count; I just counted the moves in a regular 5x5 speedsolve (with redux) and got 190.


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## RNewms27 (Aug 25, 2011)

Meep said:


> I might try this when I get home. Though I feel you should lower the max move count; I just counted the moves in a regular *5x5* speedsolve (with redux) and got 190.


 
This is 4x4.

Ah I see it now. Yeah limit should be ~150.


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## Meep (Aug 25, 2011)

RNewms27 said:


> This is 4x4.


 
Exactly. I was noting that if a regular 5x5 _non-FMC_ speedsolve was sub-200 moves, anything coming close to 200 in 4x4 wouldn't seem like an effective speed method.


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## cuBerBruce (Aug 25, 2011)

riffz said:


> I think he underestimated how long it takes to trace them, but his point was that given 20 minutes you can determine parity from the start relatively quickly without looking at the scramble.
> 
> Amiright?


 
Well, I agree that 20 minutes allows plenty of time to allow persons to determine parity from inspection of the cube. But I also think most people can determine parity from inspecting the scramble alg faster than they can determine parity from inspecting the cube. If one way is faster than the other, why wouldn't you use the faster method (assuming the faster method is not considered cheating)?

In my opinion, ideally in linear FMC, the competitor should not see the scramble alg at all. He should just get the cube already scrambled, and then have to solve it with every move being counted. (Perhaps minor exceptions such as when two turns of the same layer are done consecutively.) Of course in an online competition, I think people will expect to be allowed to perform the scramble by themselves, and thus, this gives rise to the issue of people making use of the scramble alg for purposes other than just getting the cube to the correct initial state.

In regular FMC, it's standard to provide the competitor with the scramble alg, and the competitor must do the scrambling himself/herself. If regular 4x4x4 FMC (not linear 4x4x4 FMC) were an official event with basically the same format as the existing event, there would be no point in forbidding the competitor from counting the number of wide quarter turns in the scramble, since there would basically be no way a judge could figure out whether or not the competitor determined the parity that way.

Another issue: should a competitor be allowed to use a 5x5x5 cube for 4x4x4 FMC? Since the 4x4x4 cube does not have fixed center pieces like the 3x3x3, a 5x5x5 cube is useful for allowing the competitor not to have to keep track of which way the cube is oriented, just like the fixed centers on the 3x3x3 allow a person not to have to keep track of how the cube is oriented in normal 3x3x3 FMC. For that reason, I would generally prefer to use a 5x5x5 cube for 4x4x4 FMC.


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## irontwig (Aug 30, 2011)

wtf? No more takers?


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## Kirjava (Aug 30, 2011)

I'll have a go tonight.


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## Cubenovice (Aug 30, 2011)

Haven't solved a 4x4x4 in over a year, don't know any parity algs anymore so this seems like a fun thing to try!

Wouldn't it be easier to compare methods if we all use the same scramble?


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## Kirjava (Aug 30, 2011)

F' B' r f R2 u2 B F' u B U2 u r2 B2 D r B2 f' r U' R2 L2 F r U2 L2 B' u2 U r' L B2 L r2 f' D2 L' D F2 r'

LFw'F'l'F'r2 x'y Rw'U2F2Rw

z F2Rw2S z'R'Ul2URwU'Rw'UFRwD'RF'x2

U'RwURw'U2Rw'(3Rw')U2RwURw'l2U'lRwUR'U2RU2R2U' z'

Uw'L2dR2d'L2DR2D y 

dF'UFd'F'U'F

R2'U'RFR'UR2U'R'F'RU

y'F2R'Fl2F'RFl2F

l2UrD2r'U'rD2r'l2

L2F'uFU2F'u'FU2L2

106 + parity

this is hard


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## MaeLSTRoM (Aug 30, 2011)

I'll try this tomorrow, I think I have a way to detect parity. Not sure if it works all the time though >_>


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## Christopher Mowla (Aug 31, 2011)

I was trying to use irontwig's scramble, but I accidentally solved the inverse of that scramble instead.

Not that this is a fully developed method, but it's basically "optimized reduction." It is basically {outer layer turns, Rw U2 Rw', outer layer turns, Rw U2 Rw', ...} (with cube rotations too, of course), with some exceptions here and there. Basically, Rw U2 Rw' is usually used for center row placement (with adjacent faces), but it also affects the wings (obviously). By mere observance of how it affects the wings, one can manage to pair wings and solve centers simultaneously. I do admit it took about an hour and a half to get the hang of it, but when I tried the scramble below a second time, I got this solution within 10 minutes! The solution reduces the 4x4x4 completely to a 3x3, but I didn't try to avoid either reduction parity (so if we choose my approach, we should incorporate what we know about avoiding parity into this method).



Spoiler



(Uw L' Rw' R2 F' R' Uw' F' L D U R2 Rw B2 D' B' Fw D F2 B Rw D2 Fw Uw Rw B U' L2 D' L' B2 Uw' F Fw' R2 L2 U' Fw2 D F')'

D B2
Rw U' Rw 
R2 L U' Fw
Fw R Fw'
L D L' D 
Lw B2 Lw'
D'
Fw' R2 Fw 
L2 F 
Lw' F2 Lw 
L' F' L'
Rw R' F Rw'
M2 D
Rw' U2 Rw
R2 D U'
Fw' U2 Fw
(Link here)

The solution cancels to a 41 btm reduction, which means we need no more than 61 moves if we solve it optimally with a 3x3x3 solver.
(Uw L' Rw' R2 F' R' Uw' F' L D U R2 Rw B2 D' B' Fw D F2 B Rw D2 Fw Uw Rw B U' L2 D' L' B2 Uw' F Fw' R2 L2 U' Fw2 D F')'

D B2
Rw U' M' x' r
U' Fw2
R Fw'
L D L' D 
Lw B2 Lw'
D'
Fw' R2 Fw 
L2 F 
Lw' F2 l
F' L'
r F Rw'
M2 D
Rw' U2 R' r
E' y'
Fw' U2 Fw
(Link Here)


I don't have much time right now to really see why this solution worked out so well (see the first version, not the canceled one!), but I thought I would share my current investment of time.


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## DavidWoner (Aug 31, 2011)

101 with reduction. 3-3-3-3 pairing. CFOP for 3x3 phase, no parities. I was too lazy to write it down.


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## cuBerBruce (Sep 1, 2011)

First, I note that irontwig miscounted his moves. The 3rd line contains 8 moves, not 7. However, he can make up for this by using l2 x' y U2 in place of Lw2 x' y B2 U2, and still claim 94 moves total.

OK, I am giving a solve, except it is a computer-generated solve, so it certainly doesn't count as a human FMC solve. The solve is based upon a solution from my 5-step solver program. The solver only looks for one solution for each step, so it is arguably a linear solve. Since the OP seemed to be interested in unconventional methods, and I think this certainly reflects an unconventional method, I thought it may be fitting to post this as an illustration of the "potential" of an unconventional method. Clearly for human FMC or speedsolving, the steps need to be broken down further into substeps to make it practical for a human to master, much like Heise's "Human Thistlethwaite" method breaks down some of the four Thistlethwaite steps into substeps to make it simple for a human to learn. "Humanizing" this five-step method will increase the typical move count, obviously, just as "Human Thistlethwaite" increases the typical move count over a standard computer-optimized Thistlethwaite-style solution.

The solver uses these five steps:
Step 1: Make the cube <U,u,d,D,L2,l2,r2,R2,F2,f,b,B2>-solvable.
Step 2: Make the cube <U,u2,d2,D,L2,l2,r2,R2,F2,f,b,B2>-solvable.
Step 3: Make the cube <U,u2,d2,D,L2,l2,r2,R2,F2,f2,b2,B2>-solvable.
Step 4: Make the cube <U2,u2,d2,D2,L2,l2,r2,R2,F2,f2,b2,B2>-solvable.
Step 5: Finish solving the cube.

The scramble is from the weekly forum contest 2011-34, first speedsolve scramble:
L' Fw' L' Fw D' R2 D' Uw' L2 B' F2 L Rw2 F' Rw R2 B Fw Uw U2
L R' D R Uw2 B2 Fw F2 D' L Fw' Rw R2 B2 D2 F' D' L' D2 B'

The solution is:
Step 1: U2 Lw' D' d Lw' B2 U x' (7/7)
Step 2: L2 d Fw2 d R2 Dw' b Lw2 d f y (10/17)
Step 3: D f2 D U2 b' Lw2 d2 f2 D f (10/27)
Step 4: D R2 D F2 D B2 Lw2 D2 R2 D Bw2 U (12/39)
Step 5: Lw2 F2 D2 f2 u2 l2 b2 U2 L2 Bw2 Dw2 (11/50)


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## cubernya (Sep 1, 2011)

By the way (if it's ever been calculated) what's god's number for a 4x4?


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## aronpm (Sep 1, 2011)

theZcuber said:


> By the way (if it's ever been calculated) what's god's number for a 4x4?


 it wasn't long ago that God's number for 3x3 was proven... 4x4 is monumentally more difficult to prove


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## cubernya (Sep 1, 2011)

That's why I'm saying if it's been done. I know 2x2 is 11 and 3x3 is 20, but don't know any higher


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## cuBerBruce (Sep 1, 2011)

theZcuber said:


> By the way (if it's ever been calculated) what's god's number for a 4x4?


 For block turns, somewhere in the range of 29 to 67. Mostly likely somewhere in the low to middle thirties.


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## reThinking the Cube (Sep 1, 2011)

cuBerBruce said:


> Step 4: Make the cube <U2,u2,d2,D2,L2,l2,r2,R2,F2,f2,b2,B2>-solvable.
> Step 5: Finish solving the cube.



Do you know God's number for Step#5?


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## cuBerBruce (Sep 1, 2011)

reThinking the Cube said:


> Do you know God's number for Step#5?


 
Yes.
16 block turns, or
19 single-layer turns, or
20 twists ("face turns").

That is, if you only use 180-degree turns.


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## reThinking the Cube (Sep 1, 2011)

cuBerBruce said:


> 16 block turns, or
> 19 single-layer turns, or
> 20 twists ("face turns").



Very interesting.

How does this compare to a 3x3x3 that can be solved using only <U2,D2,R2,L2,F2,B2>?


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## uberCuber (Sep 1, 2011)

Scramble: f2 u D' r u U2 D2 f R F' u' r' L D' F D' L2 R2 u F2 u2 B' U2 B D2 u' L U' D R' r D2 f L2 U' B' f D' r2 f'

y' L u' y L' F R u F' L2 u R2 u2

y R' u' R' u2 F2 y' L' U2 L2 U2 L' u'

x2 y' l' R' U' R U l x F' U r U' R' U r' x2 U' R' x' u' R F' U R' F 2U z x' u' R F' U R' F u

z' x' F' R2 x2 U' F' D2 y R' U' F' R' U F' U' F'

y R U R' F L' U2 L y' U R U2 R2 U' R' U R U2 R' U R U' R'

y2 3r U R' U' 3r' F R F' U R' U' R y R2 3u R' U R U' R 3u' R2

121 moves. SiGN Notation. Clicky

Redux with terrible use of Petrus for 3x3 stage.


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## cuBerBruce (Sep 1, 2011)

reThinking the Cube said:


> How does this compare to a 3x3x3 that can be solved using only <U2,D2,R2,L2,F2,B2>?


 
15 turns (although only 13 face turns if quarter-turns are permitted).


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## reThinking the Cube (Sep 1, 2011)

@CuberBruce

How many distinct positions for a 3-color 4x4x4 (red=orange, white=yellow, blue=green)?

Do you know God's number for that?


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## Cubenovice (Sep 1, 2011)

cuBerBruce said:


> Since the OP seemed to be interested in unconventional methods, and I think this certainly reflects an unconventional method, I thought it may be fitting to post this as an illustration of the "potential" of an unconventional method. Clearly for human FMC or speedsolving, the steps need to be broken down further into substeps to make it practical for a human to master, much like Heise's "Human Thistlethwaite" method breaks down some of the four Thistlethwaite steps into substeps to make it simple for a human to learn. "Humanizing" this five-step method will increase the typical move count, obviously, just as "Human Thistlethwaite" increases the typical move count over a standard computer-optimized Thistlethwaite-style solution.


 
I tried this yesterday (as I'm on a Thistlethwaite binge lately...) and in searching if it had been done before I came across your solver.
Nice job!

I first had a go at the 1st weekly-35 scramble and already got stuck after EO (not too bad) and solving corners...
Will look in more detail into your programmed solves as I noticed some funny stuff:
step 1: corner oreintation
step 2: EO? but missing two edges?


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## porkynator (Sep 1, 2011)

Scramble: Rw' Fw2 B2 Rw' D B2 D2 L U' L Fw2 B' D2 F2 B2 Uw Fw' U L U Uw' Fw2 D2 L2 R Fw2 B' D F' R2 Rw2 F2 U Fw' Uw D2 U2 B' U2 R'

FIrst center:
R Fw

2x3x3 block:
U2 F2 U2 F2
B U2 B' Uw' R' U R' Uw2 B L' B U' L' U L F U2 F'

Expand to 2x3x4:
R' U R Uw' U2 r B2 r' B' Uw R2
B2 U' B2 R U2 R'

Expand to 3x3x4:
B' Uw B' Uw2
R' U' R B' Uw B2 Uw'
Rw B Rw' B Uw' B Uw B2 U' B U B Uw' B2 Uw

Last 2 centers:
Rw B' Rw' B Rw B2 Rw'

Finish edge pairing:
R' U R Uw
U R' U' R Uw'
B' (put a paired edge on D to avoid bad cases and improve look-ahed when speedsolving)
R' U R Uw'
L U2 L' Uw
B' U B Uw'
L U2 L' Uw

Finish 3x3 Petrus-style
R B' R'
U' B U2 B' U2 B2 U B U' B' U
x' y'
R U2 R' U' R U' R' L' U2 L U L' U L
U' r2 U2 r2 Uw2 r2 u2
L2 U L U L' U' L' U' L' U L'

138 moves

This is a method I invented (maybe it has already been invented before, I don't know) a few weeks ago. I'm using it for speedsolving too, but with a less block-building approach. I thought it was more efficent, by the way.


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## cuBerBruce (Sep 1, 2011)

reThinking the Cube said:


> @CuberBruce
> 
> How many distinct positions for a 3-color 4x4x4 (red=orange, white=yellow, blue=green)?
> 
> Do you know God's number for that?


 For 3-color 4x4x4 cube, I get the following breakdown:

Corners: 70*2187 = 153,090
Edges: 3,246,670,537,110,000
Centers: 9,465,511,770
Orientation of cube doesn't matter: 1/24

Total positions: 196,027,906,155,517,884,227,905,125,000

And no, I don't know God's number for it.



Cubenovice said:


> I tried this yesterday (as I'm on a Thistlethwaite binge lately...) and in searching if it had been done before I came across your solver.
> Nice job!
> 
> I first had a go at the 1st weekly-35 scramble and already got stuck after EO (not too bad) and solving corners...
> ...


 
If you are trying to figure out how my five steps affect the pieces, here (in the spoiler) is my description of how the first four steps affect the pieces.


Spoiler



Step 1
Orient the corner cubies, and put the u- and d-layer edges into those two layers. (A d-layer edge may be in u layer, and a u-layer edge may be in the d layer.)

Step 2
Put front and back centers onto the front and back faces into one of the twelve configurations that can be solved using only half-turn moves. Arrange u- and d-layer edges within the u- and d-layers so that they will be in one of the 96 configurations that can be solved using only half-turn moves.

Step 3
Put centers for left and right faces into the left and right faces so that they are in one of the 12 configurations that can be solved using only half-turn moves. This leaves the centers for the U and D faces arbitrarily arranged in the U and D faces. Put top and and bottom layer edges into positions such that the U or D facelet is facing either up or down. Also, put these edges into an even permutation.

Step 4
Put corners into one of the 96 configurations that can be solved using only half-turn moves. Put U and D centers into one of the 12 configurations that can be solved using only half-turn moves. Put all U- and D-layer edges into a configuration that can be solved using only half-turn moves. This consists of 96 possible configurations for the l- and r-layer edges, and 96 for the f- and b-layer edges.


I note that my solver program does not output any cube rotations. So steps 2, 3, and 4 may use a different set of allowed turns than what those steps are supposed to use. It is simply translating the moves due to the cube being in a different orientation than it's supposed to be for those steps.


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## Kirjava (Sep 1, 2011)

porkynator said:


> This is a method I invented (maybe it has already been invented before, I don't know) a few weeks ago.


 
When in doubt, assume it has.


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## Christopher Mowla (Sep 8, 2011)

Here's 2 more examples (explained). Both solutions are for cuBerBruce's scramble. Again, these are just reductions, not complete solutions. However, I did make a small effort to avoid odd parity and luckily in these two solutions, PLL parity is also absent.


Spoiler



*Scramble*:
L' Fw' L' Fw D' R2 D' Uw' L2 B' F2 L Rw2 F' Rw R2 B Fw Uw U2 L R' D R Uw2 B2 Fw F2 D' L Fw' Rw R2 B2 D2 F' D' L' D2 B'

*Solution* (1)
D2 B2 Rw2 F' L' Uw' F2 L Bw L2 Bw' R D F' U' R' F Dw2 R2 Dw2 L' F' Lw F2 Lw' U R2 U R' Dw2 F2 Dw2 L U' L2 F' Uw' L2 Uw D' Bw' U2 Bw M2 Bw' U2 Bw (47)
Link (A link to the break down is also provided below)


1) Complete 2 1x2 center blocks in D and dedge in DB
D2 B2 Rw2

2) Complete a 1x2 center block in L and dedge in BR (also breaks the odd permutation)
F' L' Uw'

3) Complete more 1x2 center blocks in U and F and pair one dedge.
x' z L2 U Rw U2 Rw'

Currently there are two unpaired dedges in UF and UR that contain the same wings. There is also two unpaired dedges in FL and UB that contain the same wings.

4) We line them both pairs of dedges (a total of 4 dedges) in M, as well as the 2 complete 1x2 green center blocks so that we can pair all 4 dedges, and solve the front center in just 3 moves.
y' D L F' R' D' F
Those three moves are
Lw2 D2 Lw2

5) We complete 2 more 1x2 center blocks in U and F as well as pair up another dedge in M.
y z' x L' F' Lw F2 Lw' 

6) Similar to how we paired 4 dedges and solved one center at the same time, we can set up for the same case, only this time we can only solve 2 dedges and a center.
x y R F2 R F' Lw2 U2 Lw2

7) We pair the last two dedges and complete the remaining 1x2 center blocks
x2 F R' F2 D' Rw' F2 Rw

8) We finish solving the centers
z' x F' x' Lw' U2 Lw S2 Lw' U2 Lw

Link


*Solution* (2)
D' Fw' B' Uw' R2 F' Uw' R F Dw' R2 Dw D' L Rw'2 B Rw'2 U' R2 f' D Bw D F2 L' Bw' L2 Bw M R2 Bw' D2 Bw L Fw D2 Fw' R2 B Uw' B2 Uw F R L' Fw' R2 Fw B D' B' Dw B2 Dw' (54)
Link (A link to the break down is also provided below)

1) Pair dedge in UR
D' Fw' 

2) Complete 1x2 center block in F
B' Uw' 

3) Pair FR dedge, complete 1x2 center in R, and break the odd permutation.
R2 F' Uw'

4) Complete a center and pair one dedge
z x2 U B Lw' U2 Lw

5) Complete another center.
y' z' F' R Lw2 U Lw2

6) Complete another 1x2 center block in F and pair two dedges.
z B' U2 l' F Rw

7) Complete 2 1x2 center block in U and F and pair a dedge.
x U L2 F' Rw' F2 Rw

8) Pair another dedge
x' E U2 Rw' F2 Rw

9) Pair another dedge
D Lw F2 Lw'

10) Pair another dedge
x' z' F2 U Lw' U2 Lw

11) Complete 2 more (the last 2) 1x2 center blocks and pair another dedge
x y L U D' Lw' U2 Lw

12) Solve the last 2 centers and pair last 2 dedges at the same time.
x' z' U R' U' Rw U2 Rw'
Link


Analysis. Obviously the beginning moves you choose makes all the difference in the solution. In the first solution, I paired up multiple dedges at a time. In fact, I paired up so many dedges at the beginning of the solution that I ran out of dedges to pair at the end. Hence the last step was to just complete the centers. So if you do try to use my method, try to focus on solving a reasonable amount of dedges and center sections throughout the solve.

The second solution had a mixture of solving centers only, pairing dedges only, and doing both. Probably it didn't have as good of a start as the first solution.

*Notes*
As I work out these solutions, I just choose beginning moves and try to break the odd permutation (if it exists). Then I pair up any dedges I can and try to solve the centers (at least form 1x2 center blocks) in as little moves as I can simultaneously as I pair the dedges.

*Extension*
Although this extension might not work out, those who are familiar with 3x3 FMC may design solutions like the above to set up an easier 3x3x3 scramble. The downside is that, due to the fact that you can create multiple solutions for the reduction (as shown in this post), modifying each one to affect the outer layers so that the pseudo 3x3x3 can be solved in fewer moves will require even more time and trials before a possibly worth while solution is achieved.

As far as I can see at this point, using the sequences Rw U2 Rw', Rw2 U2 Rw2, Rw2 U Rw2, Rw U' Rw', etc., seem to promise reduction move counts no less than in the 30's range (most likely in the 40's). Of course, if the given scramble has some dedges already paired and a very favorable situation, it could be less (but still not in the 20's).


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