I have run all 8020 symmetrically distinct "last layer" positions of Rubik's Cube with the optimal solver of the Cube Explorer program. All these positions could be solved in 16 face turns or less. I also used the number of positions associated with each of these 8020 symmetry class representatives to determine the precise distribution of distances of all 62208 "last layer" (abbreviated LL) positions. Note that this analysis considers solving the last layer with respect to the already solved first two layers.

I note that Helmstetter (see here) has done a similar analysis previously, but his analysis basically only considers solving the last layer pieces relative to themselves, and does not consider what cases may need an additional move to align the last layer properly with the first two layers. My analysis includes all moves needed to solve the last layer with respect to the first two layers. So Helmstetter considered only 1212 cases (15552 "relative" positions reduced by symmetry and antisymmetry), while I considered 8020 cases (62208 "absolute" positions reduced by symmetry).

(Reconstructed from the Drupal archives.  Much thanks to Mark for all the work he does in supporting this site.)

This message addresses both the Non-Trivial Identities thread and the Generalizing Dan Hoey's Syllables thread.  I thought that I had a good handle on the relationship between Non-Trivial Identities and Duplicate Positions, but I find a confusing discrepancy.

Consider the following four positions.

     w = F  R' F' R  U  F'    w' = F  U' R' F  R  F'
     x = U  F' L' U  L  U'    x' = U  L' U' L  F  U'
     y = U' R  U  R' F' U     y' = U' F  R  U' R' U
     z = F' U  L  F' L' F     z' = F' L  F  L' U' F

It is the case that w=x=y=z, what I call a duplicate position.  This duplicate position is obviously related to the list of 1440 non-trivial identities from the Non-Trivial Identity Thread.  Therefore, I was thinking that (for example) we would find wx', wy', and wz' in the list of 1440 non-trivial identities.  But we don't, or at least not exactly.  We find wx' and wy', but not wz'.  Why not?

Folks, I'm in the middle of fighting with the drupal database. While trying to add the ability to contact the admin I managed to mess up the database in some way. As a result I'm reverting the database back to 5 am this morning and anything added after that won't show up.

I'm going to add in the ability for anonymous users to read all the posts and comments but they won't be able to post. A new drupal is in the cards but the problem is transferring all the info from the old drupal to the new is rather difficult.

I'm going to try to add in the lost posts later tonight. Sorry for the trouble everyone, but I'm still learning the idiosyncrasies of drupal. Turning on the menu module was the cause of the trouble and I won't be touching that particular part again :-/

As of last Tuesday, with 1200 QTM cosets solved, all at a distance of 25 except two at 24, I've been able to show that 31 moves suffices in the QTM to solve every cube position.

Surprised by the dearth of 26s (none found out of more than 20 trillion positions solved), I decided to take a cue from Radu (and others) and explore the symmetrical positions. In the past few days I've run about 500 cosets with 16-way, 8-way, and 4-way symmetry and have not found any distance-26 positions other than the one that is known.

This is in contrast to the face turn metric, where distance-20 positions pop up much more frequently.

Using my brandnew Core i7 920 CPU machine (Vista 64 bit with 6 GB of RAM) I solved 100,000 random cubes optimally with a rate of about 4-5 cubes / minute. So the computation took less than two weeks. I got the following results:

14f*:     18
15f*:    197
16f*:   2710
17f*:  26673
18f*:  67099
19f*:   3303

No 20f* cube was encountered. According to Toms results this indeed would be very unlikely and for 100000 cubes the probability should be less than 10^-6. The 95% confidence interval for the probability of 18f* is about 0.671 ±0.003, for 17f* 0.267 ±0.003 for 19f* 0.033 ±0.001 and for 16f* 0.027 ±0.001.

I have used Cube Explorer to solve all "tripod finish" cube configurations. All positions were solved in no more than 15 face turns. I calculated an average distance of about 12.746 face turns per position.

A "tripod finish" configuration has all cubies solved except four corners and three edges that make a configuration resembling a tripod. For example, corners URF, UFL, UBR, and DFR along with edges UF, UR, and FR make up one representative tripod configuration. For the above tripod configuration (or any equivalent one), there are a total of 7776 possible legal arrangements of the cubies. These 7776 arrangements can be reduced by symmetry to 1317 cases.

I have modified my coset solver to work in the quarter turn metric, and
with 396 cosets solved, I can announce that every cube position can be
solved in 32 or fewer quarter turns.

I am running phase one to a depth of 19 and letting phase two complete
the coset; each run takes about 12 minutes and approximately 63% of
the runs yield an upper bound of 25; the other 37% yield an upper
bound of 26.

No coset I have run yet has required more than 26 moves to solve, and
the possible distance-26 positions that I have run through an optimal
solver have all yielded distances less than 26, so I do not have a

The thread on this forum entitled "Non Trivial Identities" has had the most responses of any thread to date (39 responses so far).  The thread contains a great deal of useful information, including a detailed analysis of non-trivial identities of length 12q.  The thread may be found at http://cubezzz.homelinux.org/drupal/?q=node/view/114 (you have to login to see all the responses).

I tend to view the problem of analyzing identities more in terms of duplicate positions than in terms of identities, but you can of course easily transition from one view of the problem to the other.  In any case, the problem of finding a formula for the number of moves at each distance from Start is closely tied to the problem of finding duplicate positions and/or non-trivial identities.

Inspired by Jerry Bryan's recent Cubelovers article, I decided to run a few large randomly-selected cosets such that optimal solutions were found for *all* positions (rather than just proving all positions were within 20, as I usually do). From this, I hoped to gain some insight into how long it would take to enumerate the entire cube space and find counts for every distance. I also hoped to get an estimate for the total count of 20s in the entire cube space.

I first wrote a program to select a random coset, and then set those cosets to running, specifying a maximum search ply of 18. This way

Some observations on the Rubik's Cube Group, its sub-groups and solution algorithms.

When approaching the problem of auto-solving Rubik's cube in my Virtual Rubik program I chose to proceed in a three step process by first solving a 2 x 2 x 2 block using turns of all six faces. Then using turns of the three faces not intersecting the solved block, the block is extended to a 3 x 2 x 2 block, leaving two faces unsolved. And finally the last two faces are solved using just turns of those two faces.