## free rubiks cubes

Submitted by whatsupdog1 on Thu, 05/06/2010 - 21:17.## All the Syllables, Corners Group, Quarter Turn Metric

Submitted by Jerry Bryan on Thu, 04/22/2010 - 22:03.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

Submitted by Jerry Bryan on Sun, 04/11/2010 - 11:19.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

Submitted by nooneimportant on Fri, 04/02/2010 - 11:13.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 ?

thanks

## Second largest coset solved yet of actual Rubik's positions

Submitted by mdlazreg on Fri, 04/02/2010 - 05:43.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*)].

Enjoy:

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

## basic algorithms and schreier sims ?

Submitted by nooneimportant on Sat, 03/27/2010 - 17:30.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

Submitted by rokicki on Sat, 03/20/2010 - 13:43.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

Submitted by jaap on Fri, 03/12/2010 - 03:32.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

Submitted by rokicki on Sun, 03/07/2010 - 19:43.12h 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

Submitted by mdlazreg on Sun, 02/28/2010 - 23:32.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