tag:blogger.com,1999:blog-264226589944705290.post8584687149637226147..comments2022-08-07T01:05:01.413-07:00Comments on God Plays Dice: An upper bound for the order of an element of the Rubik's cube groupMichael Lugohttp://www.blogger.com/profile/15671307315028242949noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-264226589944705290.post-74623328133289489482008-10-15T01:23:00.000-07:002008-10-15T01:23:00.000-07:00According to Magma (and assuming the generators in...According to Magma (and assuming the generators in S48 of the group R that I found <A HREF="http://web.ew.usna.edu/~wdj/book/node198.html" REL="nofollow">here</A> are correct), the maximal order is 1260 (and there are 8 conjugacy classes of order 1260, out of 81120).Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-17494210602164651542008-10-14T18:02:00.000-07:002008-10-14T18:02:00.000-07:00In my modified cycle notation (outlined here) the ...In my modified cycle notation (outlined <A HREF="http://unapologetic.wordpress.com/2007/03/21/rubiks-group-and-the-solution-to-the-cube/" REL="nofollow">here</A>) the move I mean has cycle structure:<BR/><BR/>(+ufl ulb ubr rdf) (+urf) (+dlf dbl drb) (+uf ul ub ur rf) (+df dl db dr lf bl rb) (f l b r)<BR/><BR/>Okay, here's another one that *does* satisfy your requirement of not making cube moves. It doesn't even use middle-layer moves:<BR/><BR/>RU^2D^{-1}BD^{-1}<BR/><BR/>with cycle structure:<BR/><BR/>(-ufl lbu rfu) (+ubr fdl dfr rbd ldb) (+uf lb dr fr ul ur bu) (+dl rb) (df db)<BR/><BR/>Orders 9, 15, 14, 4, 2. Least common multiple is 1260.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-18881056001979319002008-10-14T17:48:00.000-07:002008-10-14T17:48:00.000-07:00John,I'm not sure that counts as a counterexample....John,<BR/><BR/>I'm not sure that counts as a counterexample. Implicit in my framing of the problem is the idea that the cube stays in a fixed position, so you're really talking about the move RBLF in the usual notation, which would have order 1260/4 = 315. (That is, (RBLF)^315 is the identity and no smaller power is -- assuming your claim of 1260 is correct.)Michael Lugohttps://www.blogger.com/profile/15671307315028242949noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-11091496003212849312008-10-14T17:43:00.000-07:002008-10-14T17:43:00.000-07:00I should, of course, show a counterexample as well...I should, of course, show a counterexample as well: Twisting the right face of the cube by 90 degrees and then turning the whole cube around the vertical axis by 90 degrees has order 1260. And that <I>is</I> maximal.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-70973660393016557472008-10-14T17:42:00.000-07:002008-10-14T17:42:00.000-07:00John,I had a feeling I was oversimplifying things....John,<BR/><BR/>I had a feeling I was oversimplifying things. Thanks.Michael Lugohttps://www.blogger.com/profile/15671307315028242949noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-47982093985867297432008-10-14T17:37:00.000-07:002008-10-14T17:37:00.000-07:00The operation isn't exactly componentwise. There ...The operation isn't exactly componentwise. There are two wreath products going on, so permuting the edges (or corners) doesn't commute with a collection of edge-twists (or corner-flips).Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-50876810882744946032008-10-14T16:31:00.000-07:002008-10-14T16:31:00.000-07:00Once, I tried to solve a Rubik's cube without havi...Once, I tried to solve a Rubik's cube without having any prior knowledge of algorithms for solving the cube, or any sequences of moves accomplishing desired tasks, etc. (I succeeded after 10 hours; note that I also wasn't using paper. By the way, it seemed to me that the hardest thing by far was having a solved cube except for two adjacent corner pieces.)<BR/><BR/>At some point, I ran out of ideas for how to accomplish a certain sub-task. I decided to get more ideas by choosing "random" elements of the Rubik's cube group with low order. I reasoned that a good way of generating those would be to pick a random element <I>g</I>, figure out its order <I>n</I> "the dumb way", and then take <I>g</I> to the power of <I>n/k</I> for some small divisor <I>k</I> of <I>n</I>. This would produce a "random" element of order <I>k</I>.<BR/><BR/>Even though I succeeded in finding good maneuvers, let's just say that I also became acquainted with some elements of large order ...Unknownhttps://www.blogger.com/profile/06229416831400925212noreply@blogger.com