We'll use the notation:
U up face (top)
D down face (bottom)
L left face
R right face
F front face
B back face.
Then an edge cube is denoted by two faces, e.g. UL. And a corner cube by three, e.g. RFU.
To rotate the left face 90 degrees CW, we'll denote it L+, and CCW, L-, or 180 degrees (either way) L2.
The problem:
Imagine a complete cube but with DF, DR, DB, DL oriented incorrectly (see attached image).
What is the quickest way to complete the cube from here?
So far I've managed to do it in 64 moves*. I tried an online solver, but the best it could do was 84 moves.
Points are available for: i) a quicker solution, ii) proof/evidence that a quicker solution even exists, iii) proof/evidence that this is the quickest solution, iv) a number for the absolute minimum possible number of sequences.
* The winning sequence so far (the spacing reveals a slight repetitive pattern):
R+ L- F+ L+ R- D+ R+ L- F+ L+ R- D+ R+ L- F2 L+ R- D+ R+ L- F+ L+ R- D+ R+ L- F+ L+ R- D2
B+ F- R+ F+ B- D+ B+ F- R+ F+ B- D+ B+ F- R2 F+ B- D+ B+ F- R+ F+ B- D+ B+ F- R+ F+ B- D2
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This article claims 29 moves or less for any position
http://www.bambooweb.com/a
They also have a page on Optimal Solutions:
http://www.bambooweb.com/a
Note that optimal solvers tend to take non-obvious, non-sensical paths.
Richard Korf and his research...
http://www.seas.ucla.edu/h
I guess you could do it in 48 moves using the Dedmore H pattern described on this site :
http://www.chessandpoker.c
>All you need are 4.3 x10^19 buttons and 3.9 x10^20 pieces of string
I was actually using a database and a route finding algorithm, but whatever floats your boat I suppose. My intention was to find the longest way home - or what is the most mixed up a cube can be. Managed to map all the corners and which perm they changed to when making all the individual face turns, but I was momentarily stuck on a way to get it to recognise identical permutations but on a rotated cube - if I can do this then I think I have enough working space to map the edge pieces as well.
My hands-on solution for the 4 inverted edges runs to many more than 64 turns. It's 13 turns initially, but as these mess up the relative positions of those edges and all the corners on that face it takes many moves to sort those out. When solving I fix the edges before the corners so the extra work isn't apparent.
Here's my 48, subtly shorter than the sequence in the question although similar. It does turn the edges on the top face instead of the bottom. If you're not allowed to invert the cube then all that is necessary is to rename the faces. Each line inverts one edge and turns the top. Alternate lines mix and unmix the lower two layers.
L- R+ F+ R- L+ D- L- R+ F2 R- L+ U+
L- R+ F2 R- L+ D+ L- R+ F- R- L+ U+
L- R+ F+ R- L+ D- L- R+ F2 R- L+ U+
L- R+ F2 R- L+ D+ L- R+ F- R- L+ U+
It should shorten as it seems a bit wasteful doing all those turns and only inverting a single piece at a time.
( I think I got it in either 18 or 24 moves while trying this, one of those where you forget where you are and try and back up. The result was that the bottom edges had all inverted and remained in position - not the top as I was trying to work on. Can't reproduce that at the moment unfortunately.)
Here's my "human" version based on the OLL & PLL techniques I use. (I would actually rotate the cube for parts of it, but for simplicity's sake I'm listing it all from a constant point of view)
First, a move that flips around two of the edge pieces (+ side effects)
R' U' F' U F R
Then flip over the other edges (+ side effects)
F' L' U' L U F
Straighten out the corners with a Sune
B U B' U B U2 B'
Lucky me, the corners are in the right places (a 1 in 24 chance).
Then a move that applies a clockwise permutation to the three edge pieces that are out of place. (Bob Burton's PLL #07)
R2 U R U R' U' R' U' R' U R'
That's 6+6+7+11 = 30 moves
Here's another approach that happens to work in 30 moves too.
R U D' F2 U2 D2 B U B' D2 U2 F2 D U' R'
That's a commutator that flips two of the edges and leaves the top turned clockwise.
This does the exact same thing, but flips the other two and reverses the top's turn.
L U D' B2 U2 D2 F U' F' D2 U2 B2 D U' L'
mmm, looks different this morning. I think I got my cw and acw confused because your Sune is working for me now. I can by way of recompense save you one move from your Bob Burtons #7, although I add a couple of x2 moves - depends how you count them.
Bob Burton's PLL #07) R2 U R U R' U' R' U' R' U R'
can be replaced with: R2 D' U2 R' L F2 R L' D R2
Now we have 6+6+7+10 = 29 moves
> I have a slight issue with the D2 turns claiming to be one
This is actually the subject of a Very Old debate on cubing, namely what the turn metric should be.
The Face Turn Metric people think that a turn of a face counts as one move. F, F2, and F' are each one move.
The Quarter Turn Metric people think that a quarter turn of a face counts as one move. F and F' are one move, but F2 is two.
In the end, the Face Turn Metric more or less won out, kind of like the way Blu-Ray beat out HD-DVD. Quarter turn is mathematically simpler, but Face Turn is closer to the physicality of manipulating a cube. Cube Explorer uses the Face Turn metric. The famous unproven proposition that the cube can be solved from any position in 20 moves or less is based on the Face Turn metric.
An optimal solver on the QT metric would probably do better than 22 moves.
> Bob Burton's PLL #07) R2 U R U R' U' R' U' R' U R'
> can be replaced with: R2 D' U2 R' L F2 R L' D R2
Burton's sequences are optimized for physical execution time, not for move count. But that's what I had memorized, so that's what I used. Notice that all the moves are R and U, which means the left hand stays static while the right hand can go very fast. The last R' U R' can be impressively done with a single snap, but it's not so impressive when an edge piece goes flying off.
>not so impressive when an edge piece goes flying off
I used to get the same trouble with my R2 D'... sequence. Just as your audience were coming round to the idea that you were actually solving their randomised mantlepiece ornament it could collapse into a heap of small bricks in your hands. Like you it was the sequence I learned many years ago so it's the one I use. I mentioned it only because it is one step shorter that your sequence and a shorter sequence is what we are looking for.
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by: InteractiveMindPosted on 2009-06-09 at 08:53:25ID: 24582387
Hmm, I did edit the question and attach an image, but hasn't shown. Here it is (hopefully)...
Here, the orange face is the bottom (D).