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4x4 Linear Fewest Moves Challenge


Oct 27, 2010
Belluno, Italy
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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.


Oct 8, 2006
Malden, MA, USA
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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.

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?

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.
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.

Christopher Mowla

Premium Member
Sep 17, 2009
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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.
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


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'
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.

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.

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).