From: roger.broadie@iclweb.com (Roger Broadie)
To: "Cube-Lovers"
Cc: "Jonathan Tuliani"
Subject: Re: MEGAMINX
Date: Sat, 17 Jan 1998 22:54:40 -0000
In my opinion Jonathan is quite right in his analysis of the problem and
its solution. When cubes or similar puzzles are coloured ambiguously, it
is always possible that the puzzle will be in an apparently impossible
configuration which must be cured by a move which changes identically
coloured pieces in an invisible way. I am lucky enough to have a
dodecahedron puzzle from the 80s called the Supernova, which was made in
Hungary and sold in the UK by Pentangle. That used twelve different
colours, and the problem did not arise. Perhaps the instructions for the
Megaminx were originally written for this form.
Other puzzles can show similar effects - the variant of the cube with the
vertical edges bevelled so that it is octagonal in horizontal cross-section
has edge-pieces in the middle horizontal slice that have only one colour,
so their orientation is ambiguous and the top and bottom edge-pieces can
appear to have impossible flip states, such a single edge-piece flipped.
Jonathan remarks that every sequence of moves seems to rotate 3 edges
cyclically. The reason (as usual) is to be found in parity considerations.
A single turn of a face of the dodecahedron yields two 5-cycles, one of
the edge-pieces and one of the corners. Both these are even permutations,
and it is therefore impossible to create a permutation of odd parity by any
combination of turns. Hence a 3-cycle is the minimum possible, and a
single 2-cycle is impossible.
Obviously any 3-cycle of edges can in principle be conjugated into a form
to solve Jonathan's problem, but I thought I'd look for a process for a
3-cycle of edge-pieces which would take one piece from the bottom into the
top and preserve orientation (in the sense that the faces in the top and
bottom remain so). I don't have the instruction booklet that goes with the
Megaminx, so I don't know what notation it used, or indeed what anyone else
may have used for this puzzle, so with the usual apologies if I'm ignoring
a standard notation:
Position the puzzle on a table, so there is a horizontal top plane and
bottom plane, and turn it so that one of the faces in the top band directly
faces you. Call that face F, the top T, the two faces on either side of F
respectively Ru on the right and Lu on the left, and the two faces in the
lower band that join in the centre Rl on the right and Ll on the left (i.e.
u=upper and l=lower). The arrangement of faces you see is thus (nb use a
non-proportionally spaced typeface);
T
Lu F Ru
Ll Rl
D
Then
Rl Ll' T Ru Lu' F^2 Ru' Lu T' Ru Lu' F^-2 Ru' Lu Ll Rl'
does (TRu, TF, Drl).
Yes, you can swap pairs of edges - this is an even permutation which can be
composed out of 3-cycles, and in general any even number of swaps is
possible.
Roger Broadie
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Date: Sun, 18 Jan 1998 02:28:59 +0100 (MET)