# Message #802

From: Chris Locke <project.eutopia@gmail.com>

Subject: Length 4 Pentagonal Duoprism

Date: Thu, 03 Dec 2009 01:45:26 +0900

Hello everybody! This is Chris again, and I thought some of you might like

to hear an update on how my length 4 solve of the {5}x{4} duoprism went.

I thought that with the length 5 down, I would take a bit of a break, but

the gap in the record table felt so tempting, and I had be thinking about

how I already have done most of the work needed to solve it by doing the

length 3 and 5 puzzles. Turns out that my algorithms for 3 cycling face and

edge pieces were perfectly usable in the length 4 case to build up all faces

and edges. Only thing that caught me a little off during this phase was the

reappearance of, what I guess could be called degenerate pieces. In the

length 2 and 4 puzzle, there are some pieces in the edge and face blocks

that don’t have the same number of colors as the other pieces. As in, there

are some 1c pieces that are equivalent to corresponding length 5 2c pieces,

where the color that is part of the square torus is gone due to the shape of

the cuts. Likewise, there are 2c edge pieces sandwiched between two 3c edge

pieces. Turns out this isn’t too much of a problem, and it actually kind of

makes it easier as you don’t have to worry about an extra color matching up,

and the algorithms are exactly the same as the length 5 case.

The real problem with the length 4 is the possibility of parity issues. A

parity issue, for those unaware, occurs in even length puzzles precisely

because you don’t have ‘central’ pieces to each block, so when you build

your centers, faces, and edges, you can’t be sure that you are putting them

together in the correct permutation to be solvable or not when you reduce it

to its equivalent length 3 puzzle. Oh yeah, if you want to solve the puzzle

for yourself, you might want to opt out of the following discussion on the

parity issues present. In that case, you can skip the next paragraph.

For faces in this puzzle, it turns out that with a 3 cycle algorithm for

each of the kinds of face pieces, you can actually solve any potential

parity issue with just those macros alone. The faces between pentagonal

facets (2c(5,5)) seem to have no issues. The 2c(4,4) faces are pretty easy

to fix if you end up with a case where you need to flip one or swap two

blocks. This is because while those two cases have odd parity when you

consider the reduced length 3 puzzle, in terms of the individual pieces they

have even parity so can be fixed by the even 3 cycles (remember that 3 cycle

is an ‘even’ permutation). If you consider the 2c(5,4) faces, there are

basically two kinds of pieces. Pieces on a given diagonal within the face

can be cycled by algorithms, but you can never place a face piece outside of

its diagonal. You can discover this by playing around with the various

kinds of possible twists. So this case is similar to the 2c(4,4), except

you have to do each diagonal separately, whereas 2c(4,4) has all the pieces

identical. The real problem comes with edges. From the previous logic, it

can be seen that a single swap of two edges isn’t too bad as that is

equivalent to swapping two pairs of pieces, which is even. So that can be

fixed with a 3 cycle macro of edge pieces. But if you end up with a a

single flipped 3c(5,4,4) edge, then bad luck for you: that is the only real

parity problem that can occur in the length 4 pentagonal duoprism that can’t

be fixed by the macros you already have from length 5! I played around a

bunch before I finally stepped back to think about the problem and realized

what I just stated, which is that it is impossible to fix with just my 3

cycle. Once I did that, I spent a couple days of thinking about the problem

in my free time without actually touching the puzzle, because I didn’t

really know what to do. When I thought that it might be necessary to break

everything up and rebuild, I had to then categorize the possible twists, to

determine exactly which twist I need to do to apply an odd permutation to

edge pieces so that I get the correct parity. Turns out there are only two

possibilities when you think about it: twisting a pentagonal facet on a

2c(5,4) face / 3c(5,4,4) edge (equivalent), or doing the same twist while

holding down ‘2’ on your keyboard. I’m not sure sure what that is called,

so I’ll just call it twist2. Now, the first twist can be thrown out because

it does something else that is odd: it also does an odd permutation (to be

precise, a single pair swap) of the 2c(5,5) face central piece. This

problem can only be fixed by the very same twist, which unfortunately means

that no matter how you twist this way, you will have either your edges in

odd parity, or your 2c(5,5) faces in odd parity. So the only possibility is

the twist2. Now that I had this knowledge in hand, I proceeded first to try

to find a better way to use this knowledge than just rebuilding from

scratch. I won’t give the algorithm here, but turns out if you consider

just the 1c pieces, you can use this twist2 along with some other carefully

chosen twists to make a move sequence that changes parity of the edge

pieces, and keeps your 1c pieces intact. You have to remember though that

whatever your algorithm is, it ‘must’ have an odd number of these twist2s in

order to put the cube in a solvable orientation. Turns out that the

algorithm I used kept all edge blocks together, and only broke up a handful

of the 2c(4,4) face blocks. This is not as good as an algorithm that just

flips the edge I want, but it’s much better than the alternative of

rebuilding everything! So I then proceeded to use my 2c(4,4) face 3 cycle

macros to rebuild the faces, then solved as a length 3 puzzle, and voila!

Victory!

Well, it was quite an adventure, but I’m glad I followed this pentagonal

duoprism road this far. There are still be length 6 or 7 puzzles left

undone, and I’m quite sure that my solves were pretty inefficient and can be

toppled in twist count too though :D. Not sure I have the time and effort

to take on anything above length 5 though, but they are awesome achievements

which I’m sure someone will one complete one day I’m quite sure.

Not sure what I’ll attempt next, but one thing that is sorely missing from

my hypercubing experience, is the 5D cube. I have avoided that monster for

a long time! I like the 4D puzzles because the 3D projection that we can

work with in MagicCube 4D makes it possible to use innate 3D visualization

skills. Of course this is also why I have problems with really small 4D

puzzles like the simplex, because they are so small that the 4D-ness is

really important, whereas other puzzles are large enough that you can work

in locally 3D regions effectively. The 5D cube is another beast entirely

though, as you can’t rely on 3D instincts to guide you at all it feels.

Nevertheless, one day I’ll will make a serious push to solve that one too I

hope! Until then, I will enjoy this closer to 3D MagicCube 4D software as

much as I can :)

Chris