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Всемирный Мега Царь

Post subject: 3×3×3 — algorithm of maximum order 2520 Posted: Tue Jun 14, 2011 9:20 pm 

Joined: Fri May 27, 2011 11:26 am

corner elements — 3 orientations, 3cycle, 5cycle +edge elements — 2 orientations, 4cycle, 7cycle рmax = 2³ × 3² × 5 × 7 = 2520(x R' F² R² L' U R B' R² L F' L' U² L U)²⁵²⁰ = 0
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alaskajoe

Post subject: Re: 3×3×3 — algorithm of maximum order 2520 Posted: Wed Jun 15, 2011 8:16 am 

Joined: Mon Feb 06, 2006 12:52 am

What does it mean? Is this the longest algo you can do that brings you back to the solved state without performing any movesequence (s) more than once? I don't really understand because scrambling a million years and than solving could be called an algorithm couldn't it? Or are you SUPPOSED to perform a certain amount of moves 252 times to get back to solved state? EDIT: I just klicked on the blue stuff and ended up on your site. Ok so the 252^0 in the exponent is supposed to be a 2520 and you are supposed to repeat those handfull of moves that often to get backto the solved state. Yes, that is impressive. I will not try it out.
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schuma

Post subject: Re: 3×3×3 — algorithm of maximum order 2520 Posted: Wed Jun 15, 2011 3:49 pm 

Joined: Thu Jul 23, 2009 5:06 pm Location: Berkeley, CA, USA

I think when people say the max order is 1260, they assume the moves like x, y, z are not allowed. In other words, the centers are fixed.


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NType3

Post subject: Re: 3×3×3 — algorithm of maximum order 2520 Posted: Wed Jun 15, 2011 5:41 pm 

Joined: Mon Oct 18, 2010 10:48 am

I'm fairly certain what he's saying is this is the algorithm which need be repeated the most times (2520) to bring the cube back to a solved state.
_________________ Noah
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Allagem

Post subject: Re: 3×3×3 — algorithm of maximum order 2520 Posted: Wed Jun 15, 2011 7:33 pm 

Joined: Sun Oct 08, 2006 1:47 pm Location: Houston/San Antonio, Texas

I believe schuma is right, only face moves are allowed in traditional analyses. By reorienting the centers like that, you have introduced a way to change the parity on the edges without altering the corners. So typically if you look at the permutations/orientations of the state after a single iteration of any algorithm, and write it down like this: corner elements — A orientations, B1cycle, B2cycle, B3cycle, etc.... +edge elements — C orientations, D1cycle, D2cycle, D3cycle, etc... The standard restrictions are: A=0 or 3 C=0 or 2 B1+B2+...<=8 (there are only 8 corners) D1+D2+...<=12 (there are only 12 edges) (B11)+(B21)+...+(D11)+(D21)+... must be EVEN! (parity error) [an easier way of thinking about this last restriction is there must be an even number of B's and D's that are even. Read that sentence carefully... ]Then the order of the algorithm would be lcm(A*B1,A*B2,A*B3,...,C*D1,C*D2,C*D3,....) Strictly speaking, the A and C coefficients are only distributed to cycles that they apply to. Or better yet, maybe would should say there are A1, A2, A3, etc and C1, C2, C3, etc and each A can be either 0 or 3 (or what?... 3 where the sign doesn't count in the lcm calculation?) and each C can be either 0 or 2, but there must be an even number of nonzero C's, and (how can I say this mathematically?) the sum of the A's must be 0 modulo 9? Yeah, it's just easier to use the above  it's less acurrate in terms of what each element represents, but it's mathematically equivalent since we are only looking for the maximum order...Under the above restrictions, the maximum achievable order is 1260 with A=3 B's=3,5 C=2 D's=2,2,7 order=lcm(9,15,4,4,14)=1260 However, assuming center orientations ARE allowed, then you introduce a third category: the centers, and a second type of move: slice turns (whole cube reorientation is acheived by composition). A face turn toggles the parities of corners and edges simultaneously, but a slice turn toggles the parities of edges and faces simultaneously, allowing the corners and edges to have different parities from one another, thereby removing that last constraint. All center permutations will have order of either 2,3, or 4, so they don't add any new factors that weren't already acheivable. So considering all of this, Всемирный Мега Царь (had to copy and paste that one! ) is right. If you allow face reorientations, the maximum order for any algorithm is 2520, and the above algorithm is indeed an example of that. Cool stuff actually. Has anyone calculated the maximum algorithm orders of other puzzles? Peace, Matt Galla


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GuiltyBystander

Post subject: Re: 3×3×3 — algorithm of maximum order 2520 Posted: Wed Jun 15, 2011 9:42 pm 

Joined: Wed May 13, 2009 4:58 pm Location: Vancouver, Washington

Reminds me of Jaap's Devil's Algorithm for the floppy cube. http://www.jaapsch.net/puzzles/floppy.htmCode: P = L FRFRFRFRFRF L FRFRFRFR L RFRFRFRFRFR L RFRFRFRFRFR L R. Devil = PFPB PFPB It's an impressive algorithm that visits every possible position once, but as far as "algorithm order," it only ranks 2.
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contrabass

Post subject: Re: 3×3×3 — algorithm of maximum order 2520 Posted: Thu Jun 16, 2011 4:53 pm 

Joined: Sun Oct 28, 2007 5:23 pm

There are a couple of things wrong with Allagem's analysis, but nothing major (I agree with the answers). A should be 1 or 3 and B should be 1 or 2 (silly little thing). Also, for those multipliers to take effect, say A, there must be a higher exponent of 3 in the corner permutations that the edges, and vice versa for B and the factor of 2. This is because the two orientations of the edges can only be the multiplier if it is on the permutation with the highest exponent of 2 or else it will be absorbed by the factor in another permutation. Sorry for the unclear explanation. I also did some analysis for the megaminx. Without changing the orientation, the largest order is 720,720 (A=3; B=3, 5, 11; C=2; D=2;7;8;13). With changing the orientation, the largest order is 1,113,840 (A=3; B=3, 17; C=2; D=2, 7, 8, 13; a 5fold rotation of the core around a face). One more thing is that for a NxNxN puzzle, there is never more than 24 of the same piece, so the largest order is 23*19*17*16*13*9*7*5=486,748,080, with no orientation adjustment for the reason above.
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