## God's Algorithm out to 14q*

Submitted by rokicki on Wed, 06/24/2009 - 09:48.First, positions at exactly the given distance:

d mod M + inv mod M positions -- ------------ ------------- -------------- 0 1 1 1 1 1 1 12 2 5 5 114 3 17 25 1068 4 130 219 10011 5 1031 1978 93840 6 9393 18395 878880 7 86183 171529 8221632 8 802788 1601725 76843595

## God's Algorithm out to 12f*

Submitted by rokicki on Tue, 06/23/2009 - 10:22.The first table is the count of positions with exactly the given depth.

d mod M + inv mod M positions -- ------------ ------------ -------------- 0 1 1 1 1 2 2 18 2 8 9 243 3 48 75 3240 4 509 934 43239 5 6198 12077 574908 6 80178 159131 7618438 7 1053077 2101575 100803036

## Twenty-Nine QTM Moves Suffice

Submitted by rokicki on Mon, 06/15/2009 - 20:35.we have shown that there are no positions that require 30 or more

quarter turns to solve. All these sets were run on my personal

machines, mostly on a new single i7 920 box.

These sets cover more than 4e16 of the total 4e19 cube positions,

when inverses and symmetries are taken into account, and no new

distance-26 position was found. This indicates that distance-26

positions are extremely rare; I conjecture the known one is the

only distance-26 position.

In order to take the step to a proof of 28, I would need a couple

## Inappropriate links

Submitted by cubex on Sun, 05/10/2009 - 21:00.For newbies or younger readers:

If you find some of the posts are too difficult to understand please go ahead and ask questions! The people here are willing to help explain things.

Mark

## Interesting Problem/Puzzle/Game

Submitted by dukerox7593 on Sun, 05/03/2009 - 22:32.your opponent has a secret number that is 4 digits long. the digits are 0-9 and no digit can be repeated in the number (in other words all 4 numbers are different)

examples: 1234, 1948, 4950

you have to guess the number

when you guess a number: your opponent gives you back 2 numbers (x,y)

number x is the amount of numbers in the right place

number y is the amount of numbers right but in the wrong place

anyone have an algorithm to solve this problem using the 2 numbers given back?

## subgroup enumeration

Submitted by B MacKenzie on Sun, 04/12/2009 - 15:07.I've been playing around with the Rubik's cube subgroup generated by the turns: (U U' D D' L2 R2 F2 B2) which I refer to as the D4h cube subgroup after the symmetry invariance of the generator set. This, I believe, is the subgroup employed by Kociemba in his two phase algorithm. Anyway, I have performed a partial enumeration of the subgroup and its coset space. I was wondering if anyone might be able to confirm these numbers as a check on my methodology.

Six Face/D4h Coset Enumeration (q turns) Depth Cosets Total 0 1 1 1 4 5

## Twenty-Five Random Cosets in the Quarter Turn Metric

Submitted by rokicki on Sun, 03/01/2009 - 13:42.I decided to run the same 25 cosets I ran for the half turn metric. The results are summarized in the table below.

Sequence 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

## Thirty QTM Moves Suffice

Submitted by rokicki on Thu, 02/19/2009 - 00:17.that 30 or fewer quarter turns suffice for every Rubik's cube

position. Every coset was shown to have a bound of 25 or less,

except the single coset containing the known distance-26 position.

I also solved every coset exhibiting 4-way, 8-way, and 16-way

symmetry, and each of these also were found to have a bound of

25 or less. Thus, if there is an additional distance-26 or

greater position, it must have symmetry of only 2, 3, or 6, or

no symmetry at all. I believe, based on this, that it is likely

that on other distance-26 positions exist.

## Last Layer Optimal Solving

Submitted by Bruce Norskog on Tue, 02/03/2009 - 01:07.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).

## Relationship of Duplicate Positions and Non-Trivial Identities

Submitted by Jerry Bryan on Thu, 01/29/2009 - 08:36.(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?