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Date: Sun, 26 Jul 1998 21:24:44 +0200
Reply-To: Rainer.adS.BERA_GmbH@t-online.de
Organization: BERA Softwaretechnik GmbH
To: Cube Mailing List
Subject: 4*4*4 patterns
From: Rainer.adS.BERA_GmbH@t-online.de (Rainer aus dem Spring)
Dear cube lovers,
as promised the other day here comes my collection of 4*4*4 patterns.
My favorites are the single twisted rings. I still find it surprising
to see that there is no second ring on the "other" side.
The maneuvers use all sorts of slice moves which are probably not
accepted as moves by most cubeologists. I am too lazy to rewrite them.
Does anybody have any idea which format I should post for people
without a TeX system?
Plain ASCII is not really what one needs to display cube maneuvers.
I can offer Mathematica notebook, postscript, WINWORD (arrrrgh)and
(perhaps?) pdfd.
Does anybody know of a 4*4*4 emulator or even solver? Anything like
a (sub)optimal solver is probably beyond the current PC powers.
Rainer
PS
I am NOT a LaTeX expert :) hints are welcome !
[Moderator's note: The notation is fairly straightforward, but may
be new to cube-lovers. Subscripts (encoded with underscores) show
which 1x4x4 slabs relative to the given face are turned; omitted
subscripts apparently mean the outer face, as "_1" would mean if it
were used. Exponents have their usual meanings as repetition. The
only advantage to running this through TeX seems to be that you get
true superscripts and subscripts and somewhat nicer fonts.
I've added a few commands that seem required by LaTeX. I've also
replaced a number of narrow spaces (coded as backslash-comma) with
ordinary spaces, so that this is more readable as text and so that
some of the worst long-line problems are reduced. Perhaps some of
these processes would be more readable if printed on multiple lines,
or punctuated somehow. I wonder if there could be some
simplification with the [X,Y] = X Y X^{-1} Y^{-1} commutator notation
or the X^Y = Y^{-1} X Y conjugate notation, or if this would make the
processes too hard to follow. --Dan]
\documentstyle{article}
\begin{document}
\section{Patterns}
\subsection{Dot Patterns}
\subsubsection*{2 Dots (u,d)}
$(R^2_2 F_{23}^2)^2$
Since this pattern exists it is obviously possible to create all dot
patterns, i.e.\ all $6! = 720$ elements of the dot permutation group.
\subsubsection*{2 Dots (f,r)}
$D_2 R_{23}^2 D_2^{-1} L^2 B^2 U_2^{-1} R_{23}^2 U_2
R_{23}^2 B^2 L_{123}^2$
\subsubsection*{3 Dots (f,b,r)}
$L_{123}^2 B^2 U_2^2 R_{23}^2 U_2 R_{23}^2 U_2 B^2 L^2 D_2 R_{23}^2
D_2^{-1}$
\subsubsection*{3 Dots (f,u,r)}
$F_2^{-1} U^2 F_2 D_{23} F_2^{-1} U^2 F_2 D_{23}^{-1}
B_2 U^2 B_2^{-1} D_{23} B_2 U^2 B_2^{-1} D_{23}^{-1}$
\subsubsection*{4 Dots (f,r)(l,b)}
$D_2 R_{23}^2 D_2^{-1} U_2^{-1} R_{23}^2 U_2$
\subsubsection*{4 Dots (f,b)(r,l)}
$R_{23}^2 D_{23} R_{23}^2 D_{23}^{-1}$
\subsubsection*{4 Dots (f,u)(r,l)}
$R_2^{-1} U^2 R_2^2 B^2 R_2 R_{23}^2 D_{23} R_{23}^2 D_{23}^{-1}
R_2^{-1} B^2 R_2^2 U^2 R_2$
\subsubsection*{6 Dots (f,b)(r,l)(u,d)}
$D_{23} F_{23}^2 D_{23}^{-1} R_2^2 F_{23}^2 R_2^2$
\subsection{Brick Patterns}
\subsubsection*{Exchanged 1*1*1 Cubes}
$B^{-1} U^{-1} B L^2 F^{-1} D R_2^2 B_2^2 R_{12}^2 B_2^2 R^2 B_2^2
F^{-1} D^{-1} F^2 L^2$
\subsubsection*{Exchanged 1*1*2 Bricks}
$R^2 U^2 R_{123}^{-1} D_{12}^{-1} R_{123} U^2 R_{123}^{-1} D_{12}
R_{123} U^2 F_{12} U^2 F_{12} U^2 F_{12}^{-1} U^2 R_2^2 F_{12}^2
R_2^2 F_{12}^2 R_{12}^2$
\subsubsection*{Exchanged 1*1*3 Bricks}
$B D^{-1} U^2 B^{-1} R^{-1} B U^2 F^{-1} L F^{-1} L^{-1} F^2 D B^{-1}$
Of course, this is a 3*3*3 maneuver.
\subsubsection*{Exchanged 1*2*2 Bricks}
$R_{12} B R_{12}^{-1} F_{12}^2 R_{12} B^{-1} R_{12} D R_{12}^2
F_{12}^2$
\subsubsection*{Exchanged 1*2*3 Bricks}
$D^{-1} B_{12}^{-1} L^2 U^2 F_{12}^{-1} R_{123}^{-1} D_{12}^{-1}
R_{123} U^2 R_{123}^{-1} D_{12} R_{123} U^2 F_{12} U^2 F_{12} L^2
B_{12} D$
\subsubsection*{Exchanged 1*3*3 Bricks}
$F_2^2 R_{23}^2 F_2^2 B^{-1} U^{-1} B L^2 F^{-1} D L_2^2 B_2^2
L_{12}^2 B_2^2 L^2 B_2^2 F^{-1} D^{-1} F^2 L_{123}^2$
\subsubsection*{Exchanged 2*2*2 Cubes}
$B_{12}^{-1} U_{12}^{-1} B_{12} L_{12}^2 F_{12}^{-1} D_{12}
F_{12}^{-1} D_{12}^{-1} F_{12}^2 L_{12}^2$
Of course, this is a 2*2 maneuver.
\subsubsection*{Exchanged 2*2*3 Bricks}
$U_2 L_{12}^2 U_2^{-1} D_2^{-1} L_{12}^2 D_2 R_{12} B R_{12}^{-1}
F_{12}^2 R_{12} B^{-1} R_{12} D R_{12}^2 F_{12}^2$
\subsubsection*{Exchanged 2*3*3 Bricks}
???
\subsubsection*{Exchanged 3*3*3 Cubes}
$F^2 L^2 D F^{-1} B_{12}^2 R^2 B_2^2 R_{12}^2 B_2^2 R_2^2 U^{-1} R B^2
R^{-1} D_{23} R F_{23}^2 R^{-1} D R $
\subsubsection*{4 Chess Boards}
$U^2 D_2^2 R^2 L_2^2 F^2 B_2^2 R^2 L_2^2$
\subsubsection*{6 Bars}
$F^2 R^2 F_{23}^2 L^2 F_2^2 D_{23}^2 F_{12}^2 D_{23}^2$
\subsubsection*{2 Twisted Rings}
$L_{12}^2 F_{12}^{-1} L_{12}^{-1} F_{12} L_{12}^{-1} U_{12}
B_{12}^{-1} U_{12} B_{12} U_{12}^2 (B^{-1} U^2 B R^{-1} U^2 R)^2$
Certainly not the shortest maneuver.
\subsubsection*{1 Small Twisted Ring}
$F^{-1} L_2^2 F R^2 B_2 U^{-1} B_2^{-1} D_{12}^{-1} B_2 U B_2^{-1}
D_{12} R^2 F^{-1} L_2^2 F$
\subsubsection*{1 Large Twisted Ring}
$F_{12}^{-1} R_{12} D_2^2 R_{12}^{-1} U_{12}^{-1} R_{12} D_2^2 U L_2
D_2^{-1} L_2^{-1} U^{-1} L_2 D_2 L_2^{-1} R_{12}^{-1} U_{12} F_{12}$
\subsubsection*{4 Diagonals}
$U(R^2 F R^2 D_{23}^2)^2 U_{12}^{-1} F^2 R_{12}^2 D_{23} R_2^2
D_{23}^{-1} R^2 F^2 U_2$
\subsubsection*{2 Small Twisted Peaks}
$B_2^2 D_2^2 L_{12}^2 U F^2 L^2 D^{-1} L^{-1} D L^{-1} F U^{-1} F
L_{12}^2 D_2^2 B_2^2$
\subsubsection*{2 Large Twisted Peaks}
$D_{12}^2 R_{12}^{-1} D_{12}^{-1} R_{12} D_{12}^{-1} F_{12}
L_{12}^{-1} F_{12} L_{12} F^2 U_2^2 R_2^2 F^{-1} R D_{23}^2 R^{-1}
U^{-1} R D_{23}^2 R^{-1} U F R_2^2 U_2^2 F_2^2$
\subsection{Snakes}
\subsubsection*{Snake 1}
$L_{12}^{-1} U_2^2 L_{12} F_{12}^2 L_2 F_{12}^2 R_2^{-1} D_{12}^2 R_2
U_{12}^2 R_2^{-1} D_{12}^2 R_2 U_{12}^2$
\subsubsection*{Snake 2}
$F^2 B^2 D^2 L_2 D^2 L_2^{-1} D^2 R_2 D^2 R_2^{-1} B^2 L_2 F^2 R_{12}
U_2^2 R_{12}^{-1}$
\subsubsection*{Snake 3}
$D_{12} R_{12}^2 F_{12} R_{12}^{-1} B_2 R_{12} F_{12}^{-1} R_{12}^2
D_{12}^{-1} R_{12}^{-1} B_{12}^2 D_{12}^{-1} B_{12} L_2^{-1}
B_{12}^{-1} D_{12} B_{12}^2 R_{12}$
\end{document}