Message #749

From: Chris Locke <>
Subject: Re: [MC4D] Chronicles of a Rubik junkie’s experience with the {5}x{5}
Date: Sat, 31 Oct 2009 09:24:31 +0900

Hello Roice, and congrats on solving the {5}x{5} 3, and for sharing your
story with us!

I took inspiration from the fact that last night, when you uploaded your
solution to {5}x{5} 3, that there was still puzzles other than the {5}x{4}
which I failed to finish first, and as such, was able to start a second
puzzle. I was quite fascinated by how the uniform duoprisms worked,
especially how whereas the {5}x{4} has multiple kinds of pieces based on
which faces they are touching (like there are 3c pieces that touch two 5 and
one 4 block, and 3c pieces that touch one 5 and two 4 blocks, each requiring
different sequences), the uniform duoprisms only have one kind of piece
since the two torii that make up the duoprism are formed by the same kinds
of blocks (fun trying to ‘visualize’ it as two interlocking torii :D). This
meant that I would only need algorithms for 2c(within torus), 2c(between
torus), 3c, 4c.

So I decided to try to solve {6}x{6}, it being the next largest uniform
duoprism. From my experience with the {5}x{4}, I was able to rather quickly
solve all 2c pieces without macros, and for 3c and 4c, was able to rather
quickly find new algorithms. Basically, all I ended up needing was a
3-cycle for 3c pieces, and a 3-cycle for 4c pieces. Also, along the way you
notice that the colors of each torus, stay on it’s respective torus. So if
you have a white face on one torus, you won’t find a white sticker on the
other torus. Helps to keep this in mind when you are working away. You are
able to solve all kinds of parity situations with just these moves by
careful usage of conjugation. For instance, if on one face you have all 3c
in place except you need to swap two, you can bring down one from an
unsolved layer, 3 cycle it into the face you’re working on (my 3-cycle macro
only does swaps in a plane, but conjugation can make it do almost anything),
then put that piece back up into the face it came from but in an adjacent
position, then take the piece you want to bring back down, and pull it
down. The result being you do two pair swaps, one in the face you’re
working on, one in the other face you don’t care about. For orientation,
you can do a similar thing, but by commuting with a twist of one of the
surrounding torii’s faces (it temporarily messes up a couple 2c(within
torus) pieces, but the commutation fixes that right up). It really helps to
also have scrap paper to use and carefully keep track of where you move
pieces and whatnot when you are trying to find the proper conjugations
needed, but after a while you can start to see the bigger picture and do
these fixes on the go.

Interestingly, the exact same methodology applied for the {5}x{4} puzzle
too, only I needed separate algorithms for the two kinds of 3c pieces each.
And upon solving the {5}x{4} 3 and {6}x{6} 3, I have the feeling that the
same algorithms can be adapted to solve {n}x{n} 3 for any size duoprism of
length 3, it would just obviously take more time. So to answer your
question Roice, I think that while these are definately parity cases we run
into, since they can be solved by 3-cycles and conjugation for both the
uniform and non-uniform duoprisms I solved, they are probably a prevalent
feature of all duoprisms. But unlike {4}x{4} 3, I never ran into the case
where a single 4c piece needs to be flipped. It might be that I was lucky,
but the whole puzzle just felt like complicated parity cases like that are
non-existent. Again, I could be wrong, but that’s just what I felt (hard to
draw accurate results from a single trial though :D). As for 4 length
puzzles… the parity possibilities with that scare me!

Another note, Roice, I two typically find my double swap 4c macro first
also, but it’s fairly simple to take that and make it into a straight-up
3-cycle by putting it into a commutator I discovered.

In conclusion, most of these length 3 puzzles I feel can be solved fully
once you isolate a 3-cycle for each of the corresponding pieces. I had more
algorithms for when I did {5}x{4} 3, but if I did it again I’d probably do
it similar to how I solved the {6}x{6} 3 and cut down on my algorithms
greatly (I had one realllly crazy 174 move algorithm just to do a 3-cycle of
one kind of 3c pieces, but it can probably be greatly shortened if I start
from scratch :D).

Anyway, hopefully my train of thought put into text makes sense ^^