Message #271
From: Sebastian Dumitrescu <portoseb37@hotmail.com>
Subject: 2^5 cube solved!!
Date: Thu, 08 Jun 2006 20:31:59 -0000
Hello all!
I haven’t posted any messages on this forum except my introductory
message last summer, when I solved the 3^4 cube. This time I solved
the 2^5 cube. I was quite lucky for my quick try because one of my
classmates gave me the link, otherwise I wouldn’t have known about the
existence of a 5D cube program.
It took me on the whole three days to solve the 2^5 after I downloaded
the program. My decision of trying the 2^5 came quickly after I played
with the 3^5 for a while, because it would have been too time-
consuming to try that one. And anyway, I had solved the 2^4 before the
3^4. Another argument for doing this one first was that it only has
one kind of piece, namely the 5-face piece which I call a "corner" by
analogy to the 2^3 cube. Obviously it’s easier to solve it when there
is only one kind of piece.
The solution resembles my 2^4 solution. First I built blocks on one
side, adding the pieces one by one until I had a block of 8 pieces
(2x2x2). Then I added two pairs of blocks (1x2) and then finally a 2x2
block to complete one face. Then I had one face left to solve. I first
changed the orientation of all the pieces so that their green colour
(the home colour of that face) was placed in the green face (so the
piece was not necessarily oriented correctly, it was "half-oriented"
). Then I permutated the pieces using a commutator I found myself,
and that went quite smoothly. The way i permutated them was I first
separated the pieces into "up" and "down" subsets, and for each of the
two I further seprated them into "in" and "out" so that I had 4
subsets of 4 pieces, the correct position of all the pieces being a
sum of some permutations of the four subsets, i.e. one piece only
needed to be moved inside one of those subsets so that it would be in
its correct position. After I placed the pieces correctly I did the
other half of the orientation process using the same subsets as
before, again with a nice commutator I found myself (with some
difficulty though).
I did use macros, otherwise I would have finished the cube sometime
tomorrow or the day after tomorrow. But I only used 3, two of them
found by myself and the third one being a combination of two 3^3 algs.
I’m so glad!!! I’m so glad I won’t try the 3^5 because I will only
feel exhausted after solving it, if I will manage to solve it. :)
Sebastian