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Cube Lagoon

All the Syllables, Corners Group, Quarter Turn Metric

My recent posting on |EndsWith| values for the corners of the 3x3x3 in the quarter turn metric was really secondary to another project I was working on.  Namely, I was working on determining all the syllables for the corners of the 3x3x3 in the quarter turn metric.  I discovered many more syllables than I was expecting.  As a part of determining why there were so many syllables, I discovered that the |EndsWith| values were much higher than I was expecting.  To a certain extent, one can think of the higher |EndsWith| values as being the "cause" of there being a larger number of syllables than expected.

EndsWith Values, Corners Group, Quarter Turn Metric

My calculations of God's algorithm have generally included an analysis of the distribution of EndsWith values.  My best God's algorithm results for the quarter turn metric are out to 13q. Tom Rokicki has since calculated out to 15q.  Out to 13q, it is the case that for the vast majority of positions x we have |EndsWith(x)| = 1.  Tom may be able to speak to the 14q and 15q cases, but I cannot.

Given the predominance of |EndsWith(x)| = 1 out to 13q, I have assumed that for the vast majority all of cube space it's probably the case that |EndsWith(x)| = 1.  I no longer believe that assumption is correct.  Which is to say, I now have an EndsWith distribution for the entirety of the corners group in the quarter turn metric.  Beyond a certain distance from Start in the corners group, there are no instances of |EndsWith(x)| = 1 whatsoever, and overall the |EndsWith(x)| = 1 cases are very much in the minority.  It now seems to me that the same is probably true for the complete cube group including corners and edges.  I suspect the reason we have not seen this effect for the complete cube group is simply that we have not yet been able to calculate out far enough from Start.

algorithm for generating permutations for the rubik cube


let's say I want to distribute permutation checking over say 10 computers.
so if there are a total of n permutations the first machine will check n/10.
the second will check from n/10 to 2n/10
third will check from 2n/10 to 3n/10.
and so forth.
the algorithm that generates permtuations needs to generate the ith permutation in O(1)
so that I can efficiently start each machine's work.
what algos for generating permtuations do you know that can do this ?


Second largest coset solved yet of actual Rubik's positions

After solving the trivial coset [All positions that have the same orientation as the START position] published here Largest coset solved yet of actual Rubik's positions
I went ahead and calculated the full distribution of the flipped coset [All positions that have the same orientation as Reid's position (The only known 26q*)].

Depth         Trivial coset           Flipped coset
0                         1                       0
1                         4                       0
2                        10                       0

basic algorithms and schreier sims ?

hi, where can I find a comprehensive exposition with explanations and examples, the kind of stuff that you can really learn from ?
I'm trying to implement this for solving the cube, it's a hobby project of mine.

I've noticed that another approach(if I use some basic algos like http://www.ryanheise.com/cube/beginner.html ) would be to pattern match the cube and "hardcode" all the cases of ryan heise and this would also yield a solution.

I'm a begginer with the cube, I almost solved it in reality and would like to write code to solve it.

(I have already set up an opengl simulation and

UF and RBL generate the whole cube group

The two simple generators UF and RBL (without any rotations) generate the whole cube group.

It's surprising to me that these two very simple generators suffice.

It's easy to see no shorter set of generators (expressed in face turns) suffice because you need at least five faces.

Presentation for Rubik's cube

I just found a recent post by "secondmouse" on sci.math that deserves a wider audience. I'll quote it here in full.
I found the following short presentation for the
miniature 2x2x2 Rubik's cube of order 3674160:

    < a,b,c | a^4 = b^4 = c^4 = 1,
                 ababa = babab,
                 bcbcb = cbcbc,
                 abcba = bcbac,
                 bcacb = cacba,
                 cabac = abacb,
                 (ac)^2 (ab)^3 (cb)^2 = 1 >

See the following link for more info as to why

1,000,000 cubes optimally solved in both QTM and FTM

I have solved all 1,000,000 random cube positions from the earlier article now in both QTM and HTM. Here is the resulting grid:
    12h 13h 14h  15h   16h    17h    18h   19h     sum
15q   1   1   3    2     -      -      -     -       7
16q   -   2  18   48    35      -      -     -     103
17q   -   3  23  143   347    354      -     -     870
18q   -   5  40  305  1713   4520   2034     -    8617
19q   -   1  40  505  5190  29711  33363   474   69284
20q   -   2  39  674  9932 100164 212466  7213  330490
21q   -   -   9  345  7697 104052 301668 16371  430142
22q   -   -   -   41  1533  28173 120449  9720  159916

Largest coset solved yet of actual Rubik's positions

Using Tom Rokicki's coset solver as well as his optimal solver I managed to do a full analysis of the corner and edge permutations of all the 3x3x3 cube positions that have the orientation in the solved state.

Here is the distribution table:
0              1
1              4
2             10
3             36
4            123
5            368
6           1336
7           4928
8          16839
9          63920
10        257888
11       1019992
12       4317941
13      20240924
14     102343680
15     568081384
16    3458261494
17   22676234692
18  153062896516