Message #753

From: Chris Locke <>
Subject: Re: [MC4D] Chronicles of a Rubik junkie’s experience with the {5}x{5}
Date: Mon, 02 Nov 2009 15:01:41 +0900

It was pretty easy for me to notice the torus locking stickers for me
because of my custom color scheme. For instance, here is the longest list I
have for 14 colors:

black, white, red, green, blue, yellow, magenta, teal, grey, brown, pink,
turquoise, dark green, peach

So you can see, the first torus was basically all primary/secondary colors,
and the second torus was all the more complicated tertiary colors, so the
pattern REALLY jumped out at me. The reason for this, is that unlike the
3^4 case, where every facet can be turned by a quarter turn in any axis, the
duoprisms are limited by geometry to only allow quarter turns within the
respective torus. And yeah, it seems that 4^3 is unique in this regard.
What about naming conventions though? How do you decide what to use to
notate the standard hypercube, {4}x{4} or {4,3,3}? I’m not sure if there
are other degeneracies in naming conventions, so it’s possible this is the
only case like this.

I have a question too: how do we decide which puzzles will be recorded on
the wiki record page? Since it’s a wiki, are we going to allow any puzzle?
I only ask because I realized that there are an infinity of duoprisms
possible to solve by using the "invent my own" feature, limited only by the
solver’s patience (and possibly computer specs :P). Couple this with the
fact that as you go higher, I don’t think the puzzles get fundamentally more
difficult, just much longer. I suspect that the algorithms can be
generalized for the duoprisms of higher order, much like how if you can
solve 3^4, 4^4, and 5^4, you should be able to adapt the algorithms to solve
any n^4 hypercube with enough time. Nevertheless, it is still quite a feat
to be able to solve these higher order puzzles, and I would not be opposed
to allowing such higher order records to stand. Also, will the next release
allow all possible twists for length 2 puzzles? What will happen to the
records of these puzzles set with the more limited twists? And yes, I
realize both of these questions put 3 of my 4 firsts on thin ice, which was
actual part of my motivation for going for the dodecahedral prism this
weekend, since that should be a permanent one :D. I need a break for a bit
now from all that cubing over the weekend (^_^’)

And yeah, the {5}x{5} 5 is one monster. If I still had my notes from when I
solved 5^4, I might be able to adapt the algorithms to solve some of the
interior pieces, but upon inspection, there are also MANY kinds of interior
pieces, all which will probably need separate, and again probably unique,
algorithms to solve. Beastly indeed.


2009/11/2 Roice Nelson <>

> My thanks to Chris for the cool insights about the duoprisms! And thanks
> to Melinda for the picture, dramatically showing the way stickers are slaved
> to their respective tori. I hadn’t mentally put some of this together as
> cleanly as in these emails, even after solving the {5}x{5}-3. Great stuff
> :D
> One thing that was neat for me was to think about while reading was how the
> {4}x{4} fit into it all. Since it has only one type of 2C piece, there
> seems to be a succession of symmetry breaking that happens with the
> progression of puzzles - {4}x{4} is more symmetrical than general {n}x{n},
> which in turn is more symmetrical than general {n}x{m}.
> In some ways, it’s hard to say which puzzles are more difficult. The
> {4}x{4} might take less sequences, but it’s stickers can move amongst both
> tori, so the scrambling result is more complicated (it wouldn’t produce a
> neat picture of dual tori like the {6}x{6} did for Melinda). But even if
> there is somewhat of a conservation-of-difficulty effect with the tradeoff
> between number of piece types and scrambling capacity, I guess it still
> feels like difficulty rises more quickly as the duoprisms become more
> general. Chris could say better of course.
> As an aside: the {3}x{3} (which we aren’t officially supporting yet because
> we’re still considering how the twisting works on it) also only has 1 type
> of 2C piece, but it is degenerate when it comes to 2C pieces around the
> "rings". And so unlike the flexible {4}x{4}, it’s pieces are again slaved
> to their respective tori. I always knew the 4^3 was special :)
> Chris, congrats to you as well on all the firsts your snagging :) And much
> thanks for all the issue updates and feedback along the way!
> All the best,
> Roice
> P.S. I agree with Melinda about the {5}x{5}-5 being a big prize. Not sure
> I’ll ever tackle it, but that had been my favorite from very early on.
> On Sat, Oct 31, 2009 at 12:42 AM, Melinda Green <>wrote:
>> [Attachment(s) <#124b2361b88a61fa_124a91e52cd6e093_TopText> from Melinda
>> Green included below]
>> Chris Locke wrote:
>> > 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.
>> >
>> I hadn’t noticed it before but you’re totally right that there are two
>> different kinds of 3c pieces! That means that sometimes we may need to
>> be clear on which type of 3c piece we’re talking about. That’s very cool
>> and I bet you’re right that this will be why the non-uniform duoprisms
>> will be harder to solve than the uniform ones.
>> > 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.
>> >
>> Yes, I’d noticed that when I was reimplementing scrambling. I just
>> happened to end up with mostly yellow/red colors on one torus and
>> blue/green colors on the other. Then when I performed a full scramble, I
>> ended up with two beautifully speckled toruses, one in each color
>> scheme. [See attached screen shot.] At first I was sure that I simply
>> had a bug in my scrambling algorithm. It took me a while before I
>> understood what you just pointed out.
>> Here’s a 12-color specification in case anybody wants to reproduce this.
>> Just be sure to unwrap the formatting and put this all on one line in
>> your facecolors.txt file:
>> 255,0,204 153,0,153 255,51,0 255,102,102 255,153,0 255,255,0 102,153,0
>> 0,255,0 0,204,153 0,255,255 0,0,255
>> 102,102,255
>> > 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!
>> >
>> This suggests to me that one of the next big prizes up for grabs by
>> those patient cubists among you who like big puzzles will be the {5}x{5}
>> 5. It has the same sort of notational appeal as the 5^5, and the parity
>> issues it brings should make it suitably frightening in that dimension
>> as well. To the couple of masochist among you who have been asking us to
>> implement 6^4 and larger: even though you now finally have those cubes
>> available, I suggest that you focus on this plumb new prize first! ;-)
>> > 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 ^^
>> The thing that I really like is that even though I’m a better coder than
>> a cubist, I found that that actually did make sense to me! Either I’m
>> learning, or you’re a good writer. Probably both!
>> Thanks for the report and the instruction. Oh, and a huge
>> congratulations on snagging the first {6}x{6} 3 solution, Chris!!
>> -Melinda