]>
Domain of the Cube Forum - Cube lovers returns
http://cubezzz.dyndns.org/drupal
To promote mathematical discussions about the Rubik's Cube and related puzzles.enThe 2018 Computer Fewest Moves Challenge is Underway
http://cubezzz.dyndns.org/drupal/?q=node/view/567
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.<br />
<br />
For those who are interested, visit:<br />
<br />
http://cubesolvingprograms.freeforums.net/Thu, 18 Oct 2018 07:53:28 -0400Lower bound for the diameter of the 3-gen subgroup <U,F,R>
http://cubezzz.dyndns.org/drupal/?q=node/view/566
What are the known bounds for the diameter of this subgroup? <a href='http://cubezzz.duckdns.org/drupal/?q=node/view/190'>This thread</a> 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:
<pre>U R F2 R U2 F2 U' F' R U F U R' F R2 U F R U2 R F'</pre>
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:
<pre>
Depth PositionsSun, 02 Sep 2018 23:00:35 -0400Mad Octahedron
http://cubezzz.dyndns.org/drupal/?q=node/view/565
<p align="justify">I have written a computer simulation of the octahedral twisty puzzle. It is available as freeware on the Apple App Store:</p>
<a href="https://itunes.apple.com/hr/app/mad-octahedron/id1400458838?mt=12" target="_parent">Mad Octahedron</a>Thu, 21 Jun 2018 10:47:47 -0400Future URL recommendation for the forum
http://cubezzz.dyndns.org/drupal/?q=node/view/564
Hello everybody :)<br />
<br />
In the future I think it would be a good idea if we all used URLs of the form:<br><br />
http://forum.cubeman.org/?q=node/view/563#comment<br><br />
rather than<br><br />
http://cubezzz.dyndns.org/?q=node/view/563#comment<br><br />
dyndns.org has raised their prices every year and I'm considering <br />
dropping their service. Unfortunately if we do that a lot of old URLs will stop <br />
working, so I'm open to any clever ideas on what is the best way to deal with <br />
this problem. I'm committed to keeping the maxhost.org and <br />
cubeman.org domain names working for the long term, but<br />
I'm very unhappy with dyndns.Tue, 08 May 2018 17:47:42 -0400Three Million Positions in Four Metrics
http://cubezzz.dyndns.org/drupal/?q=node/view/563
I optimally solved three million positions in four distinct metrics.<br />
These positions are distinct from the three million positions I ran<br />
some years back. Random numbers were generated with the Mersenne<br />
Twister algorithm. The four metrics I ran were quarter-turn metric,<br />
half-turn metric, slice-turn metric, and axial-turn metric (equivalent<br />
to the robot-turn or simultaneous-turn metric on the 3x3 cube).<br />
<br />
The generators for each metric are strict super- or sub-sets of the<br />
generators for the other metrics. The quarter-turn metric has 12<br />
generators, the half-turn metric has 18 generators, the slice-turnMon, 23 Apr 2018 11:52:52 -0400Presentation for the Mathieu Group M24 from dedge superflip
http://cubezzz.dyndns.org/drupal/?q=node/view/562
Following on from a highly symmetric presentation I supplied for the miniature Rubiks cube group
<a href="http://cubezzz.homelinux.org/drupal/?q=node/view/177">Presentation for the 2x2x2 Rubiks cube group</a>
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:
<i>
<p style="font-size:17px;color:grey"
<pre><b>< a,b,c | a<sup>2</sup> = b<sup>2</sup> = c<sup>2</sup> = 1,
<p style="font-size:17px;color:magenta"
<b> (ab)<sup>6</sup> = [(bc)<sup>6</sup>] = [(ca)<sup>6</sup>] = 1,
<p style="font-size:17px;color:red"
<b> bacabacacabacababacabac = 1,
<p style="font-size:17px;color:green"
<b> (ababacbc)<sup>3</sup> = 1,
<p style="font-size:17px;color:blue"
<b> bababcbcbcbabab = cacabacacabacac</b> >
</i>
</pre>Sat, 21 Apr 2018 11:25:04 -0400Gear cube extreme can be solved in 25 moves
http://cubezzz.dyndns.org/drupal/?q=node/view/561
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.<br><br>
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.<br><br>
Phase 1 distribution:
<pre>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</pre>
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.Thu, 15 Feb 2018 23:07:51 -0500Kilominx can be solved in 34 moves
http://cubezzz.dyndns.org/drupal/?q=node/view/560
Last night, I found <a href='https://www.speedsolving.com/forum/threads/kilominx-gods-number-bounds.67278/'>this thread</a> on the speedsolving forums which proves an upper bound of 46 moves. First, the puzzle is separated into two halves, which takes at most 6 moves. Each half is then solved in at most 20 moves (= 7 moves for orientation + 13 moves for permutation, after orientation is solved), for a total of 6+2*(7+13) = 46. xyzzy writes
<pre>The ⟨U,R,F⟩ subgroup, while much smaller than G_0, is still pretty large, having 36 billion states. It's small enough that a
full breadth-first search can be done if symmetry+antisymmetry reduction is used, but I will leave this for another time.</pre>Sun, 11 Feb 2018 12:58:51 -05005x5 sliding puzzle can be solved in 205 moves
http://cubezzz.dyndns.org/drupal/?q=node/view/559
Consider a 5x5 sliding puzzle with the solved state
<pre> 1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 </pre>
We can solve the puzzle in three steps. First solve 1,2,3,4,5,6,7, then solve 8,9,10,11,12,16,17,21,22, and finally solve the 8 puzzle in the bottom right corner.
Step 1 requires 91 moves:
<pre>depth new total
0 18 18
1 6 24
2 13 37
3 27 64
4 54 118
5 117 235
6 231 466
7 443 909Fri, 26 Jan 2018 17:46:05 -0500God's algorithm for the <2R, U> subset of the 4x4 cube
http://cubezzz.dyndns.org/drupal/?q=node/view/558
Here I'm using sign notation, so 2R is the inner slice only. There are 10 edges, 10 centres in sets of 2, 2, 2 and 4, and 4 permutations of the corner pieces for a total of 4*10!*10!/(2!2!2!4!) = 274,337,280,000 positions. From July 4th 2017 to July 6th 2017, I ran a complete breadth first search of this puzzle in around 60 hours. God's number is 28.
<pre>Depth New Total
0 1 1
1 6 7
2 18 25
3 54 79
4 162 241
5 486 727
6 1457 2184
7 4360 6544Wed, 24 Jan 2018 22:00:30 -0500Do we have a 3x3x3 optimal solver for stm metric?
http://cubezzz.dyndns.org/drupal/?q=node/view/557
I thought it might be interesting to run an optimal solver using the slice turn metric (including face turns) on some pretty patterns. I don't remember anyone releasing an optimal solver that uses stm but maybe there is one by now? <br />
<br />
Also is it true we don't know if using slice turns plus face turns could reduce God's Number from 20 to less than 20?Thu, 10 Aug 2017 07:52:23 -0400More details about my new program
http://cubezzz.dyndns.org/drupal/?q=node/view/556
<p>
<strong>Introduction</strong>
</p>
<p>
On 02/23/2016, I posted a message about a new program I had developed that had succeeded in
enumerating the complete search space for the edges only group. It was not a new result because Tom Rokicki
had solved the same problem back in 2004, but it was important to me because the problem served as a testbed
for some new ideas I was developing to attack the problem of the full cube.
I am now in the process of adapting the new program to include both edges and corners.
In this message, I will include some additional detail about my new program that was not included in the first message.
</p>Thu, 08 Jun 2017 15:55:54 -0400Pattern databases for the 5x5 sliding puzzle
http://cubezzz.dyndns.org/drupal/?q=node/view/555
<p>In 2002, Korf and Felner [1] used pattern databases to solve optimally 50 random instances of the 5x5 sliding puzzle. They used a static <nobr>6+6+6+6</nobr> partition of tiles (described below), along with its reflection in the main diagonal. In a 2004 paper by Felner, Korf and Hanan [2], the authors describe in a footnote the way they handled the empty tile as 'not trivial'; the empty tile was taken into account when precomputing the database, but then the tables were compressed by discarding the information about the empty location. The authors do not provide the distribution of values from the pattern databases, but do provide maximum values and the number of nodes generated when solving random instances of the 5x5 sliding puzzle.</p>Sun, 02 Apr 2017 12:37:51 -0400A cubic graph with cubic diameter
http://cubezzz.dyndns.org/drupal/?q=node/view/554
<p><a href="https://en.wikipedia.org/wiki/15_puzzle" target="_blank">The Fifteen puzzle</a> is sometimes generalized to a sliding puzzle on an arbitrary simple connected graph <i>G</i> with <i>n</i> vertices in the following way. <nobr><i>n</i> − 1</nobr> movable pieces numbered 1, 2, ... <nobr>(<i>n</i> − 1)</nobr> are placed on vertices of the graph <i>G</i>. At most one piece is placed on each vertex. One vertex of <i>G</i> is left unoccupied. A move consists in choosing a vertex <i>v</i> adjacent to the currently unoccupied vertex <i>v</i><sub>0</sub> and 'sliding' the piece at <i>v</i> along the edge (<i>v</i>; <i>v</i><sub>0</sub>). The aim is to restore the order so that piece numbered <i>i</i> occupies vertex numbered <i>i</i>, for <nobr><i>i</i> = 1 .. <i>n</i> − 1</nobr>. In the case of the Fifteen Puzzle, the underlying graph <i>G</i> is the <nobr>4 × 4 grid graph</nobr>, and a degree-2 vertex is left unoccupied in the goal configuration.</p>Mon, 06 Mar 2017 03:53:47 -0500Is there a way to evenly distribute face turns for 12 flip?
http://cubezzz.dyndns.org/drupal/?q=node/view/553
Back in Jan 1995 Mike Reid found this process for the 12 flip:<br />
<br />
R3 U2 B1 L3 F1 U3 B1 D1 F1 U1 D3 L1 D2 F3 R1 B3 D1 F3 U3 B3 U1 D3 24q<br />
<br />
This process has 24 q turns, so I'm wondering could there be a 24 q turn process that evenly distributes the turns so that each side turns 4 q? The idea just seemed elegant to me, 6 faces each turning 4 q turns. <br />
<br />
MarkMon, 14 Nov 2016 22:17:19 -0500