[Help Thread] FMC Discussion

[irontwig]

Good old Petrus:
U2 F' D' R' D R' B R L' D2 L [S1+2+3+4]
B D' F' D B' D' F [S5]
F D F' D F D2 F' [S6+7] (24)

 

[Cubenovice]

Good old Sune

U2 F' D' R' D R' B R # L' D2 L F2L + EO
R D R' D R D2 R' Sune
D leaves 3 corners
at # insert B R B' L' B R' B' L to cancel 2 moves

U2 F' D' R' D R' B R B R B' L' B R' B' D2 L R D R' D R D2 R' D 25 HTM

 

[okayama]

Scramble: L2 R2 U' B F' R2 B D2 L' B' R2 B2 F U' R D' R2 D B2 L2 D' F2 B2 R2 F2
Solution: U2 F' R' D B' L B2 L2 R D' L D R' L' D' L D L' D' L D (21 HTM)

2x2x3 block: U2 F' R'
All but 5 corners: D B' L B2 L'
All but 3 corners: L' * D' L D L' D' L D L' D' L D

Insert at *: R D' L D R' D' L' D

I found the skeleton in 5 min or so, and wrote down the full solution within 15 min.

 

[PalashD]

Scramble: L2 R2 U' B F' R2 B D2 L' B' R2 B2 F U' R D' R2 D B2 L2 D' F2 B2 R2 F2
Solution: U2 F' R' D B' L B2 L2 R D' L D R' L' D' L D L' D' L D (21 HTM)

2x2x3 block: U2 F' R'
All but 5 corners: D B' L B2 L'
All but 3 corners: L' * D' L D L' D' L D L' D' L D

Insert at *: R D' L D R' D' L' D

I found the skeleton in 5 min or so, and wrote down the full solution within 15 min.

I also found this skeleton but I suck at insertions!!

 

[ilikecubing]

What are skeletons?

 

[nathanajah]

Java FMC Day 2010

First Round:
Scramble: R2 U R2 U L2 D' R2 B2 U2 F' L D2 R2 U L' D R' U2 B' F2
Solution: B' R L D' R2 D L' D' R2 L' B2 D F' U2 F2 B2 D R' B' R D' L' D' L D L' D2 L D' R D' R2 U' R D2 R' U R F (39 moves)
Explanation:
Premove: F
2x2x3: B' R * D' L' B2 D F' U2 F2 (9/9)
Orient edges: B2 D R' B' R (5/14)
F2L-1: D' L' D' L D L' D2 L (8/22)
Solve edge: D' R D' R' D2 (6/28)
Leave 3 corners: D2 R' U' R D2 R' U R (6/32) cancel 2 moves
Undo Premove: F (1/33)
Insert at *: L D' R2 D L' D' R2 D (6/39) cancel 2 moves

Final Round:
Scramble: B2 L2 D' B2 U2 R2 U L' D' B2 R F' L D B' F2 L2 D' F'
Solution: R L' U' R2 U L U' D' R U R' D R D F R F' R' D' F' R2 F' L D' L B' L' R U B L' U2 (32 moves)
Explanation:
NISS solve
Inverse scramble:
1x2x3: U2 L B' U'
2x2x3: R' L B L' D L'
F2L-1: F R2
Normal Scramble:
Premove: R2 F' L D' L B' L' R U B L' U2
F2L: R U' * R' U ** (4/4)
Leave 4 corners: D F R F' R' D F' (7/11)
Undo Premove: R2 F' L D' L B' L' R U B L' U2 (12/23)
Insert at *: U L' U' R2 U L U' R2 (5/28) cancel 3 moves
Insert at **: U' R' D' R U R' D R (4/32) cancel 4 moves

 

[irontwig]

What are skeletons?

A skeleton solves most, but not all pieces. The most common skeletons are "corner-skeletons" that leave 3-5 corners to be solved with inserted commutators.

 

[kinch2002]

New PB Just got 23 moves on the Christmas Competition Scramble#3. I'll post the solution when the competition is over.

 

[PalashD]

http://fmc.mustcube.net/results.php

I beat Guss and Okayama!!
Round 325 Classic!

 

[Attila]

http://fmc.mustcube.net/results.php

I beat Guss and Okayama!!
Round 325 Classic!

Congratulations, much more succesful solutions

 

[ssb150388]

http://fmc.mustcube.net/results.php

I beat Guss and Okayama!!
Round 325 Classic!

Congrats dude...
Hope to see you soon!!

 

[Kenneth]

Mabye this is not the tread for this but how about some 4x4 3-flip FMC?

M' U2 M U' S U L2 U' S' U Lw2 U2 B2 l' B2 l B2 l2 U2 r' U2 r B2 l' (24)

Shortest I found after 27, 26 and 25 using 2 algs, that is using 3.

First is M' U2 M U2 3-cycle, followed by another 3-cycle com to move pices back but flipped, where I use the last L2->Lw2 to start the pure 1-flip, The last U2 of the 1-flip cancles the last U2 of the first two algs.

Shorter anyone?

(I tried the wide move 1-flip and then a solver to find an alg for what it left but that was 15f* making a total of 30)

 

[Christopher Mowla]

Mabye this is not the tread for this but how about some 4x4 3-flip FMC?

M' U2 M' U' S U L2 U' S' U Lw2 U2 B2 l' B2 l B2 l2 U2 r' U2 r B2 l' (24)

Shortest I found after 27, 26 and 25 using 2 algs, that is using 3.

First is M' U2 M U2 3-cycle, followed by another 3-cycle com to move pices back but flipped, where I use the last L2->Lw2 to start the pure 1-flip, The last U2 of the 1-flip cancles the last U2 of the first two algs.

Shorter anyone?

(I tried the wide move 1-flip and then a solver to find an alg for what it left but that was 15f* making a total of 30)

I think you have typed the algorithm wrong because...
M' U2 M' U' S U L2 U' S' U Lw2 U2 B2 l' B2 l B2 l2 U2 r' U2 r B2 l

But, assuming that you did mean to construct a pure 3 2-cycle algorithm (not double-parity), yes, there is a shorter one. This could be optimal:

M U l' x' U2 r B2 r U2 l2 U2 x U2 r D2 r' U2 r2 B2 U' M' (19 btm)


Derivation

Spoiler

[1] Taking the standard edge flip algorithm:
r2 B2 U2 (l) U2 r' U2 r U2 F2 r F2 l' B2 r2

[2] Shifting it,
F2 r F2 l' B2 r2
r2 B2 U2 (l) U2 r' U2 r U2

[3] Reversing the parenthesized move, (usually done for the purposes of development of a symmetrical doubleparity algorithm)
F2 r F2 l' B2 r2
r2 B2 U2 (l') U2 r' U2 r U2

[4] Add a set-up move:
r2
F2 r F2 l' B2 r2
r2 B2 U2 (l') U2 r' U2 r U2
r2

[5] Restore the centers on the left side adding on a piece:
r2
F2 r F2 l' B2 r2
r2 B2 U2 (l') U2 r' U2 r U2
r2
(y2 r U2 r2 U2 r2 U2 r' U2 r2 U2 r2 U2 y2)

[6] Adding the same outer-set-up moves as Lucasparity,
r U2 r'
r2
F2 r F2 l' B2 r2
r2 B2 U2 (l') U2 r' U2 r U2
r2
(y2 r U2 r2 U2 r2 U2 r' U2 r2 U2 r2 U2 y2)
r U2 r'

[7]
a) Putting this into cube explorer:
R U2 R' R2 F2 R F2 L' B2 R2 R2 B2 U2 L' U2 R' U2 R U2 R2 y2 U2 U2 R U2 R2 U2 R2 U2 R' U2 R2 U2 R2 U2 y2 R U2 R'

b) Choosing the 17f solution:
R2 F2 R D2 R F2 B2 U2 R' D2 B2 L' B2 R D2 L2 B2

c) Translating the result to the 4x4x4,
r2 F2 r D2 r F2 B2 U2 r' D2 B2 l' B2 r D2 l2 B2

We do have a 3 2-cycle of wings, but we also have a 2 2-cycle of composite edges in S. So,

[8] If we take this 17f solution and add on the following piece:
R2 F2 R D2 R F2 B2 U2 R' D2 B2 L' B2 R D2 L2 B2 + S2 U2 S2 U2
running the program again for this (applying this to a clean cube), we get a 15f solution for the pure case!
r U2 l' B2 l' U2 r2 U2 F2 l' B2 l F2 l2 U2

[9] If we wish to put all 3 2-cycles into the last layer, we can add set-up moves again:
x'
M
F
r U2 l' B2 l' U2 r2 U2 F2 l' B2 l F2 l2 U2
F'
M'
x
=x' M F r U2 l' B2 l' U2 r2 U2 F2 l' B2 l F2 l2 U2 F' M' x

This can be rewritten as:
M U l' x' U2 r B2 r U2 l2 U2 x U2 r D2 r' U2 r2 B2 U' M' (19 btm)

EDIT:

Perhaps this could be faster:
M'2 U l x U2 r' F2 r' U2 l2 U2 x' U2 r' D2 r U2 r2 F2 U' M'2 (19 btm)

 

[TMOY]

Assuming you meant to construct a pure triple edge-flip, you didn't succeed, sorry. Your algs affect centers, more precisely it turns the B center by 180°. This can be fixed the following way:

M U l' x' U2 r B2 r U2 l2 U2 x U2 r D2 r' U2 r2 B2 U' x L2 U2 R' L' U R L U2 R' L' U
(30 moves, probably not optimal but I can't think of a better one right now)

 

[Kenneth]

Of course I assume centres are solved, it is not possible else, some centres must be swapped.

Second M'->M but it is not only that, the applet can't handle all notation, I think it is the Lw that messes things...

19 is nice but I guess the optimal solution is possibly shorter if you allow any face turn to move around paired deges. Problem is that you get thousands of algs if you use the solver and you can't test them all. It is much easier to sort out the ones you already know will work.

 

[TMOY]

Of course I assume centres are solved, it is not possible else, some centres must be swapped.

Sprry but you're just wrong. On the 4^3 you can perform any permutation of wings without affecting centers at all.
(It is not true anymore on cubes bigger than 4^3, though.)

 

[Christopher Mowla]

Assuming you meant to construct a pure triple edge-flip, you didn't succeed, sorry. Your algs affect centers, more precisely it turns the B center by 180°. This can be fixed the following way:

M U l' x' U2 r B2 r U2 l2 U2 x U2 r D2 r' U2 r2 B2 U' x L2 U2 R' L' U R L U2 R' L' U
(30 moves, probably not optimal but I can't think of a better one right now)

You honestly thought I believed that I found a center-preserving algorithm? If I found an algorithm to do an odd permutation and preserve the centers, I would have mentioned "(center-preserving)". This algorithm is not only brief for what it does, but it can be applied to all size cubes, not just the 4x4x4's "special case". When I heard the term "4x4x4", I immediately thought "big cubes" because most common algorithms, which work on the 4x4x4, are transferable to all big cube sizes, or at least to higher order even cubes (speaking of individual orbits as well).

Even if I was able to find a center-preserving algorithm for the 4x4x4, it would in no way be fewer moves than swapping two 1x2 center blocks within the same face or even swapping two X-center pieces on two different faces (where two are swapped on each face).

Sorry but you're just wrong. On the 4^3 you can perform any permutation of wings without affecting centers at all.
(It is not true anymore on cubes bigger than 4^3, though.)

I believe Kenneth knows this as we all do.

For a cube size n=4 or greater with r orbits with inner-layer odd parity, there are exactly \( 2r\left( n-2 \right)-4r^{2} \) T-Center pieces and/or obliques which cannot be solved back due to the odd permutation.

Kenneth probably didn't even think twice about what you were thinking because, in his mind, he was interested in fewest moves to flip 3 edges, not "fewest moves to preserve the centers".


Both Kenneth and I have made some algorithms which are pretty impressive, so I don't think you should so quickly assume that we don't know some elementary facts such as these. People use different terminologies for describing different things. In addition, when someone has a goal, and something else obviously works against that goal, that particular item is subconsciously blotted out of his or her mind.

EDIT:

19 is nice but I guess the optimal solution is possibly shorter if you allow any face turn to move around paired deges.

Well maybe one day we will know the answer.

 

[TMOY]

You honestly thought I believed that I found a center-preserving algorithm?

No. I'm only fighting against a bad terminology.

Even if I was able to find a center-preserving algorithm for the 4x4x4

Does it imply that you aren't ?

I believe Kenneth knows this as we all do.

Just lol. Of course everybody knows it, that's certainly why I have already had to teach it to a lot of people.

For a cube size n=4 or greater with r orbits with inner-layer odd parity, there are exactly \( 2r\left( n-2 \right)-4r^{2} \) T-Center pieces and/or obliques which cannot be solved back due to the odd permutation.

From the point of view of the mathematician I am, your statement just makes no sense. Orbits of what ? T-centers ? Oblique centers ? Other centers ? Wings ? Alots ? I'll give you a less poorly worded one:

On a cube of size n=4 or greater on which exactly r orbits of wings are oddly permuted and the corners are evenly permuted, there are exactly \( r\left( n-2 \right)-2r^{2} \) orbots of centers which are oddly permuted, and these centers are all either T-centers or oblique centers.

Both Kenneth and I has made some algorithms that are pretty impressive, so I don't think you should so quickly assume that we don't know some elementary facts such as these

I don't assume anything. I'm just stating the obvious fact that ha has made a mistake in his post concerning that "elementary" fact.

People use different terminologies for describing different things

It doesn't imply that all terminologies are good. For me, calling "pure" something which does a lot of hidden things on the cube is definitely a poor choice of word.

 

[Christopher Mowla]

Does it imply that you aren't ?

Well, I did get one move less using MrCage's commutator for a 2 2-cycle of X-center pieces:
M'2 U' r F2 (l' r) r U r l d2 l' r' U' r l Dw2 l' F2 l F2 D2 l F2 l' U2 r U' M'2 (28 btm)

I don't think I can approach something like this any other way. Either I do what you did, or I do what I did. My approach just happened to be one move less. It is slightly more applicable to other size cubes besides the 4x4x4 than yours, but I guess for our purposes, that does not matter anyway.

 

[4.]

What is FMC?