4x4 Linear Fewest Moves Challenge

[cuBerBruce]

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.

 

[irontwig]

wtf? No more takers?

 

[Kirjava]

I'll have a go tonight.

 

[Cubenovice]

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?

 

[Kirjava]

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

 

[MaeLSTRoM]

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

 

[Christopher Mowla]

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.

 

[DavidWoner]

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

 

[cuBerBruce]

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)

 

[cubernya]

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

 

[aronpm]

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

 

[cubernya]

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

 

[cuBerBruce]

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.

 

[reThinking the Cube]

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?

 

[cuBerBruce]

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.

 

[reThinking the Cube]

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>?

 

[uberCuber]

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.

 

[cuBerBruce]

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).

 

[reThinking the Cube]

@CuberBruce

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

Do you know God's number for that?

 

[Cubenovice]

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?