The Eliac is a complex deep-cut 2-gen circle puzzle:



The left circle rotates in increments of 90 degrees and the right circle rotates only by 180 degrees. There is a simulator of the puzzle here.

Using ksolve++ I made an optimal solver modified it slightly to turn it into a coset solver. The subgroup I used for the coset solver is the subgroup of positions where the 18 small triangles, 10 diamonds, and 2 squares are solved. There are 1600300800 arrangements of those 30 pieces and each coset has 3024000 solvable positions. Unfortunately since the puzzle is 2-gen, there isn't a good way to select a subgroup generated by a subset of the generators of the whole puzzle, which (as far as I can tell) is what is required in order to make the "pre-pass" trick work for sub-optimally solving cosets very quickly. So each coset needs to be solved optimally using a pure DFS, which takes quite a long time (about 1.5 hours on my laptop). Notice that the puzzle has a horizontal reflection symmetry so we only need to solve one coset in each symmetry class.

Hi folks,

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Hello, I am new to this forum and this is my first post. I know that maybe its content will seem trivial to some of you, but I am afraid that I am not so well versed in mathematics and programming, so I try to do at least something within my limited capabilities (I am 15.34 at the moment).


Based on Dan Hoey’s calculations, I was able to calculate the number of essentially different positions up to symmetry of the four-dimensional analogue of the Rubik’s cube. However, it is quite probable that I have made some mistakes, as I have done it only by hand. Nevertheless, from the patterns observed I was able to make some interesting generalizations, presented later in this post.

For the first time, the FMC will have solvers in the "NxNxN" category for up to 9x9x9 competing to solve 10 random scrambles in as few moves as possible. Also of note: There are now four solvers in the 5x5x5 section.

For those who are interested, visit:

http://cubesolvingprograms.freeforums.net/

What are the known bounds for the diameter of this subgroup? This thread gives a lower bound of 26 QTM but doesn't discuss HTM. I seem to recall a lower bound of 20 HTM from somewhere, but no upper bound. However, I recently found a position that requires 21 HTM:

U R F2 R U2 F2 U' F' R U F U R' F R2 U F R U2 R F'

Is this the first one known? Out of all 186624 positions with all pieces correctly permuted, this position and the inverse are the only ones requiring 21. Here's the full distribution:

Depth   Positions

I have written a computer simulation of the octahedral twisty puzzle. It is available as freeware on the Apple App Store:

Mad Octahedron

Hello everybody :)

In the future I think it would be a good idea if we all used URLs of the form:

http://forum.cubeman.org/?q=node/view/563#comment

rather than

http://cubezzz.dyndns.org/?q=node/view/563#comment

dyndns.org has raised their prices every year and I'm considering
dropping their service. Unfortunately if we do that a lot of old URLs will stop
working, so I'm open to any clever ideas on what is the best way to deal with
this problem. I'm committed to keeping the maxhost.org and
cubeman.org domain names working for the long term, but
I'm very unhappy with dyndns.

I optimally solved three million positions in four distinct metrics.
These positions are distinct from the three million positions I ran
some years back. Random numbers were generated with the Mersenne
Twister algorithm. The four metrics I ran were quarter-turn metric,
half-turn metric, slice-turn metric, and axial-turn metric (equivalent
to the robot-turn or simultaneous-turn metric on the 3x3 cube).

The generators for each metric are strict super- or sub-sets of the
generators for the other metrics. The quarter-turn metric has 12
generators, the half-turn metric has 18 generators, the slice-turn

Following on from a highly symmetric presentation I supplied for the miniature Rubiks cube group Presentation for the 2x2x2 Rubiks cube group I investigated whether the Mathieu Group M24 could be similarily presented taking full advantage of the plentiful symmetry inherent in the Rubiks Revenge cube (4x4x4). The answer was indeed yes - here is the presentation I found - again on three involutions:

< a,b,c | a² = b² = c² = 1,

   (ab)⁶ = [(bc)⁶] = [(ca)⁶] = 1,

   bacabacacabacababacabac = 1,

   (ababacbc)³ = 1,

   bababcbcbcbabab = cacabacacabacac >

Write the puzzle as the group <R,F,U,D> where R and F are 180 degree moves. We use a two-phase algorithm to first reduce the state of the puzzle to the subgroup <R3,F3,U,D>, and then finish the solve in the second phase. The subgroup <R3,F3,U,D> is the group of all positions where all of the gears are oriented, because R3 is the same as R' except the gear orientation remains unchanged.

The first phase is easy to compute. There are only 3^8 = 6561 positions because each gear has only 3 different orientations, despite having 6 teeth.

Phase 1 distribution:

Depth	New	Total
0	1	1
1	4	5
2	8	13
3	78	91
4	102	193
5	1064	1257
6	920	2177
7	3576	5753
8	592	6345
9	216	6561
10	0	6561

The second phase is harder. The number of positions is 24*8!^2/2 = 19,508,428,800, since it turns out that the permutation of the 3 unfixed edges on the E slice is completely determined by the permutation of the centres. This phase was solved with a BFS and took around 7 and a half hours to complete.