\documentclass[12pt]{article} %\documentclass[10pt,runningheads]{book} \usepackage{amsfonts} \usepackage{color} \definecolor{DarkOlive}{rgb}{0.1047,0.2412,0.0064} \definecolor{FireBrick}{rgb}{0.5812,0.0074,0.0083} \definecolor{RoyalBlue}{rgb}{0.0236,0.0894,0.6179} \definecolor{RoyalGreen}{rgb}{0.0236,0.6179,0.0894} \definecolor{RoyalRed}{rgb}{0.6179,0.0236,0.0894} \definecolor{LightBlue}{rgb}{0.8544,0.9511,1.0000} \definecolor{Black}{rgb}{0.0,0.0,0.0} \definecolor{FuncColor}{rgb}{1.0,0.0,0.0} \usepackage{hyperref} \usepackage{fancyheadings} \usepackage{amsbsy} \usepackage{amsthm} \usepackage{amsgen} \usepackage{amsfonts} %\usepackage[german,english]{babel} \usepackage{array, rotating} \usepackage{graphicx, amsmath} \usepackage{texdraw} \usepackage{amssymb} \usepackage{makeidx} % allows for index generation %\input amsfonts.sty \usepackage[all,arc,dvips]{xy} %\usepackage{chess} %\pagestyle{fancyplain} %\lhead[\fancyplain{}{\rightmark}]{\fancyplain{}} %\rhead[\fancyplain{}]{\fancyplain{}{\rightmark}} \setlength{\parindent}{8pt} %\pagenumbering{roman} %\pagestyle{empty} %\setcounter{page}{4} \newtheorem{theorem}{Theorem}[section] \newtheorem{corollary}{Corollary}[section] \newtheorem{conjecture}{Conjecture}[section] \newtheorem{lemma}{Lemma}[section] \newtheorem{proposition}{Proposition}[section] \newtheorem{definition}{Definition}[section] \newtheorem{example}{Example}[section] \newtheorem{axiom}{Axiom}[section] \newtheorem{remark}{Remark}[section] \newtheorem{exercise}{Exercise}[section] \newtheorem{ponderable}{Ponderable}[section] \def\aaa{\mathbb{A}} \def\nnn{\mathbb{N}} \def\fff{\mathbb{F}} \def\qqq{\mathbb{Q}} \def\rrr{\mathbb{R}} \def\ccc{\mathbb{C}} \def\zzz{\mathbb{Z}} \def\ppp{\mathbb{P}} \def\hhh{{\cal H}} \def\pf{{\bf Proof: \ }} \def\qed{$\Box$} \def\beqn{\begin{equation}} \def\eeqn{\end{equation}} \makeindex \begin{document} \pagestyle{fancyplain} \lhead[\fancyplain{}{\rightmark}]{\fancyplain{}} \rhead[\fancyplain{}]{\fancyplain{}{\rightmark}} \author{David Joyner} \title{Adventures in Group Theory: Rubik's Cube, Merlin's Machine, and Other Mathematical Toys\thanks{New puzzles}} \date{5-17-2007} \maketitle \tableofcontents \vskip .1in \section{Whip-it/Puzzle Tower} This sliding piece puzzle consists of several discs stacked up into a tower, and they can rotate about a central axle. There are usually 3 such discs (called the Whip-it puzzle, or Puzzle 6) but 6 discs is also quite common in cheaper knock-offs. Along the side of the tower are six columns of sliding pieces, in six colours. The pieces can slide up or down from one disc to another because one of the white pieces is missing leaving a gap. By rotating the discs, the pieces and the gap are moved around to the other columns. There is a version called Varikon, which has 5 colours with 5 balls of each, placed in 5 columns and 5 discs. There is also a gap but this is in the sixth disc, so one column is in effect a bit longer than the others. The same puzzle also exists in versions with 4 columns of 4 balls, and this one has been used as a promotional item for Smarties. This puzzle is the simplest in a family of related puzzles which includes the Babylon Tower, the Nintendo Barrel, and Missing Link. \subsection{The number of positions} On the 6 layer version there are 36 pieces if we include the space as a piece, which gives a maximum of 36! positions. This limit is not reached because: \begin{itemize} \item the pieces of each colour are indistinguishable (6!5ˇ5!) \item the colours are equivalent (6!) \end{itemize} This last factor comes about because it does not matter which color goes in which column. This leaves $36!/(6!^6 5!) = 22,251,481,138,642,910,394,240\approx 2.2\cdot 10^{22}$ positions. The 3 layer version has $18!/(3!^52!6!) = 571,771,200$ positions. The 5 column Varikon has $26!/5!^6 = 135,061,494,343,776$ positions, and the 4 column version has $17!/4!^5=44,669,625$ positions. \subsection{Solution} The tower can be solved layers, disc by disc from the bottom up, but it is just as easy to solve column by column. The latter is much easier to describe, so that is the method used here. \begin{enumerate} \item Decide which colours the columns will have. To speed things up, you could rotate some discs to partly solve some columns already. \item Look at the column containing the gap, and determine which colour the column has. \item Find a piece of the required colour in a different column. If there are none (i.e. the gap is in the completed white column, and the puzzle is not yet solved) then find any piece on the puzzle that is incorrect. \item Slide pieces in the column up or down until the gap is in a layer next to the layer that contains the piece. \item Rotate the layer with the gap until it lies above or below the piece. \item Slide the piece up or down into the gap. \item Rotate the layer back to its original position, which brings the piece into the column as required. \item Repeat steps 2-7 until the puzzle is solved. \end{enumerate} Note that in practice it is quicker to move not just the one layer, but all the layers above/below it as well. \section{Rashkey} This puzzle consists of two overlapping circular disks. The distance between the centres is exactly one radius length. The disks are made up of many pieces with curved sides, which allow them to be rotated any number of quarter turns. Each disk has 4 squares, 4 diamonds, and 12 triangles. As some pieces are shared between the two disks, there are all together 6 squares, 7 diamonds, and 20 triangles. The puzzle comes in three possible colour schemes: \begin{enumerate} \item The central 4 triangles of one disk are red, those of the other disk are blue, and all the rest is yellow. \item One disk is completely yellow, and the remaining parts of the second disk are red. \item One disk is red and the other is blue, except that the pieces which are shared between them which are yellow. \end{enumerate} These are in order of difficulty - 1 is easiest and 3 is hardest. Rashkey was invented by Oleg Raschkov, and patented on 2 September 1999, DE 29,904,348 U. \subsection{The number of positions} There are two ways of counting moves - a turn of a disc by any amount is a single move ({\it face turn metric}), or each quarter turn of a disc is counted as a move ({\it quarter turn metric}). The triangles lie in two sets of 10 which cannot intermingle. The 33 pieces fall into sets (orbits) of 10, 10, 7, and 6 pieces which can each be mixed. There are some parity restrictions, but since all colour schemes have many identical pieces, these restrictions have no discernible effect. There is a restriction on the set of squares (similar to Turn'Push), but only the colouring used in Rashkey 3 is detailed enough for it to be noticeable. %\begin{tabular}{|l|l|l|l|l|l|} \hline %Puzzle & Triangles 1 & Triangles 2 & Diamonds & Squares & Positions\\ \hline %Rashkey 1 & 4, 6 & 4, 6 & 7 & 6 & $(10!/4!6!)^2$ \ ,\ 44,100\\ \hline %Rashkey 2 & 2, 8 & 4, 6 & 3,4 & 2,4 & $10!/2!8!$ $10!/4!6!$ $7!/3!4!$ $6!/2!4!$ \ ,\ 4,961,250\\ \hline %Rashkey 3 & 2, 2, 6 & 2, 2, 6 & 1, 3, 3 & 2, 2, 2 & $(10!/2!2!6!)^2$\ $7!/1!3!3!$\ 60 \ ,\ 13,335,840,000\\ \hline %\end{tabular} \begin{tabular}{|l|l|} \hline Puzzle & Positions\\ \hline Rashkey 1 & $(10!/4!6!)^2= 44,100$\\ \hline Rashkey 2 & $(10!/2!8!)(10!/4!6!)(7!/3!4!)(6!/2!4! = 4,961,250$\\ \hline Rashkey 3 & $(10!/2!2!6!)^2(7!/1!3!3!)\cdot 60 = 13,335,840,000$\\ \hline \end{tabular} The number of positions for each number of moves from the start of all positions of Rashkey types 1 and 2 have been computed. The results are shown on the WWW site \cite{Jwww} and tell us that Rashkey 1 needs at most 12 face turns (or 15 quarter turns) and Rashkey 2 at most 19 face turns (23 quarter turns). Rashkey 3 has too many positions to calculate fully. \subsection{Solution methods} Notation used in solutions: Let a clockwise quarter turn of the left disk be denoted by $L$. Rotations of 180, or 270 degrees are then denoted by $L^2$, and $L^3$. Turns of the right disk are denoted in the same way as $R$, $R^2$ or $R^3$. \subsubsection{Rashkey 1} I will use the 'the left center' to mean the four central triangles of the left disk, and similarly 'right centre' are those of the right disk. I will assume that the red pieces belong in the left centre when it is solved, and the blue pieces in the right centre. Thus in the mixed puzzle the red pieces will be in the left centre and the rim of the right disk, whereas the blue lie in the right centre and the rim of the left disk. You may have to turn the puzzle upside down to achieve this. It is actually remarkably tricky to find a short straightforward solution to this puzzle. Much of the solution below is intuitive once you see what you are trying to do, but looks complicated when written out in such detail. \vskip .1in {\bf Phase 1}: Place all the red pieces in two pairs at the top and right of the right disk. \begin{enumerate} \item If there is not yet a pair of red pieces next to each other, then make such a pair. Now bring that pair to the top of the right disk. This is easy so I will not elaborate on it. \item There are several possibilities for the other two red pieces. If they are together at the right hand side of the right disk, then continue with phase 2. \item If they are together at the bottom of the right disk, then do $R L^2 R L^2 R^2$ to put them on the right, and then continue with phase 2. \item If there is no red piece at the left centre, then do $R L^2 R^3$ to bring (at least) one such piece there. \item Bring the other loose red piece to the left centre as well, by turning the right disk. \item If the two red pieces in the left centre are not adjacent, then do the following steps to make put them next to each other: \begin{enumerate} \item Turn the left disk so that the two red pieces are at the bottom left and top right of the left centre. \item Turn the right disk so that there are no red pieces that lie in both disks. \item Do L. \item Turn the right disk to bring the two loose pieces together. \end{enumerate} \item Turn the left disk so that its adjacent red pair lies on the left. \item Turn the right disk to bring its pair to the bottom. \item Do L2 R2, and the red pieces will be at the top and right of the rim of the right disk. \end{enumerate} \vskip .1in {\bf Phase 2}: Make the two blue piece pairs on the rim of the left disk. The red piece are out of the way, so we can move the blue pieces without disturbing the reds. \begin{enumerate} \item If there is a blue piece at the top right of the right centre, then you need to remove it as follows: \begin{enumerate} \item Make sure there is a yellow piece immediately to the left of the blue piece, by turning the left disk. \item Do $R^3 L^2 R^2 L^2 R^2$. \end{enumerate} This swaps the order of the red pairs, dislodging the blue piece in the process. \item Turn the left disk through 360 degrees, and as you do so check out how many of the loose blue pieces visit the top left position of the right centre (suppose there are x of these) and also how many visit the bottom left position of the right centre (suppose there are y of these). \item If x is at least 2, then you can make a blue pair as follows: \begin{enumerate} \item Turn the left disk to bring one of the loose blue pieces to the top left of the right centre. \item Do $R$. \item Turn the left disk to bring another loose blue to the top left of the right centre. \item Do $R^3$. Now go back to step b. \item If y is at least 2, then you can bring together two of those pieces to make a blue pair as follows: \begin{enumerate} \item Turn the left disk so that the two loose blue pieces do not lie at the right center, and nor does any previously solved blue pair. \item Do $R$. \item Turn the left disk to bring one of the loose blue pieces to the bottom left of the right centre. \item Do $R^3$. \item Turn the left disk to bring another loose blue to the bottom left of the right center. \item Do $R$. \item Turn the left disk so that the right centre is completely yellow. \item Do R3 to bring the red pairs to the top and right as before. \end{enumerate} Now go back to step b. \item If x and y are both 1, then the two loose pieces cannot be paired up immediately. One of them has to be shifted as follows: \begin{enumerate} \item Turn the left disk to bring one of the loose blue pieces to the bottom left of the right centre. \item Do R. \item Turn the left disk so that the right centre is completely yellow. \item Do R3. \end{enumerate} Now go back to step c to pair them up. \end{enumerate} \end{enumerate} {\it Phase 3}: Bring the pairs together. \begin{enumerate} \item Turn the left ring to bring one blue pair to the right centre, and the other at the left or top of the left disk. The red pairs will still be at the right and top of the right disk. \item If the second blue pair is on the left side of the left disk, then do the move sequence: $R^2 L^2 R$. \item If the second blue pair is at the top of the left disk, then do the move sequence: $R L^3 R^3 L R L^3 R L^2 R$. \end{enumerate} Pretty Patterns for Rashkey 1 \begin{enumerate} \item Wheels 1: $R^2 L R^3 L^3 R^2 L^2 R L R L^2 R^3 L^3$. \item Wheels 2: $R^2 L R L^3 R^2 L^2 R^3 L R^3 L^2 R L^3$. \end{enumerate} \vskip .2in \subsubsection{Rashkey 2} I will assume that the red pieces belong in the left disk when it is solved. Thus in the mixed puzzle there should be more than two red triangles on the rim of the left disk. If this is not so, you have to turn the puzzle upside down. {\it Phase 1}: Solve the 6 red triangles on the outside of the left disk. Note that you can equally well think of this phase as solving the 4 yellow triangles in the center of the right disk. \begin{enumerate} \item Turn the left disk until there are (at least) two adjacent yellow triangles in the center of the right disk, and then turn the right disk so that the two yellow triangles lie on the right of the centre. \item Turn the left disk so that there are at least three yellow pieces at the right center. If the right centre is completely yellow, continue with phase 2. \item Turn the right disk so that its single red centre triangle is at the top right. \item Turn the left disk so that the right centre has two red triangles. \item If the two red triangles in the right centre are not adjacent, then \begin{enumerate} \item Turn the left disk so that there is only one red triangle in the right centre. \item Turn the right disk a quarter turn clockwise (so the centre red triangle is at the bottom right) \item Turn the left disk back, so the right centre has two adjacent red triangles. \end{enumerate} \item Turn the right disk so that its red centre triangles are on the left. \item Turn the left disk so that there are no red pieces in the right centre. \end{enumerate} \vskip .1in {\it Phase 2}: Solve the 2 red triangles in the left half of the center of the left disk. \begin{enumerate} \item If the two remaining red triangles are already correct, i.e. on the left half of the left centre, then go directly to the next phase. \item Turn the right disk to bring one of the unsolved red triangles to the left centre. \item If that loose red is at the top right of the left centre, then do $L R^2 L^3 R^2 L$, but if it lies at the bottom right of the left center then do $L^3 R^2 L R^2 L^3$. These sequences will move the loose red into the left half of the left center. \item Turn the right disk to bring the last unsolved red triangle to the left centre. \item There are 4 cases for the last two reds, depending on where they lie in the left center. \newline Bottom left and top right: $L R^2 L^3 R^2 L$. \newline Bottom left and bottom right: $R^2 L^3 R^2 L^3 R^2 L^2 R^2 L^3$. \newline Top left and bottom right: $L^3 R^2 L R^2 L^3$. \newline Top left and top right: $R^2 L R^2 L R^2 L^2 R^2 L$. \end{enumerate} \vskip .1in {\it Phase 3}: Solve the 2 red squares \begin{enumerate} \item If the top left square is not red, then turn the right disc to bring a red square to the bottom centre of the puzzle, and do $L R^3 L^3 R L R^3 L^3 R L$. \item If the bottom left square is not red, then turn the right disc to bring a red square to the top centre of the puzzle, and do $L^3 R L R^3 L^3 R L R^3 L^3$. \item Repeat these steps until both red squares are correct. \end{enumerate} \vskip .1in {\it Phase 4}: Solve the 3 red diamonds \begin{enumerate} \item If the diamond on the far left of the puzzle is not red, then turn the right disc to bring a red diamond to the far right, and do $L R^3$, followed by $L R^3 L^3 R$ ten times, and then $R L^3$. \item If the diamond on the top left of the puzzle is not red, then turn the right disc to bring a red diamond to the top right, and do $L R^3 L^3 R$ ten times. \item If the diamond on the bottom left of the puzzle is not red, then turn the right disc to bring a red diamond to the bottom right, and do $L^3 R L R^3$ ten times. \end{enumerate} {\vskip .2in \subsubsection{Rashkey 3} I will assume that the red pieces belong in the left disk when it is solved. Thus in the mixed puzzle there should be more than two red triangles on the rim of the left disk. If this is not so, you have to turn the puzzle upside down. \vskip .1in {\it Phase 1}: Solve all the red triangles This is exactly the same as phases 1 and 2 of Rashkey 2 (but where yellow is mentioned, it should say ``non-red'', i.e. ``yellow or blue''). \vskip .1in {\it Phase 2}: Solve the 6 blue triangles on the rim of the right disc. Note that you can equally well think of this phase as solving the 2 yellow triangles in the center of the left disk. \begin{enumerate} \item Turn the right disc until there is at least one yellow triangle in the left centre. \item If the yellow triangle is at the top right of the left centre, then bring the other yellow triangle in the left center as well by doing the appropriate sequence below: \newline Top left: $L^3 R^3 L^3 R L^3 R^2 L^2 R L^2 R L^3$. \newline Top right: $L^2 R^2 L^3 R L^2 R^3 L R^2 L^2$. \newline Right top: $L^3 R^2 L^3 R^2 L R^2 L$. \newline Right bottom: $L^3 R^2 L^2 R L R L^3 R^2 L^3 R^2 L^3 R^2$. \newline Bottom right: $L^3 R L^3 R^3 L^3 R^2 L^2 R^3 L^2 R^3 L^3$. \newline Bottom left: $L^2 R^2 L^3 R^3 L^2 R L R^2 L^2$. \item If the yellow triangle is at the bottom right of the left centre, then bring the other yellow triangle in the left centre as well by doing the appropriate sequence below: \newline Top left: $L^2 R^2 L R L^2 R^3 L^3 R^2 L^2$. \newline Top right: $L R^3 L R L R^2 L^2 R L^2 R L$. \newline Right top: $L R^2 L^2 R^3 L^3 R^3 L R^2 L R^2 L R^2$. \newline Right bottom: $L R^2 L R^2 L^3 R^2 L^3$. \newline Bottom right: $L^2 R^2 L R^3 L^2 R L^3 R^2 L^2$. \newline Bottom left: $L R L R^3 L R^2 L^2 R^3 L^2 R^3 L$. \end{enumerate} \vskip .1in {\it Phase 3}: Solve the triangles in the right centre \begin{itemize} \item Do one of the following sequences, depending on the position of the two blue triangles in the right center: \newline Right: Do nothing as the triangles are solved already. \newline Bottom: $R L R L^3 R^3 L R L^3 R^3 L R^3$. \newline Left: $R L^2 R^2 L^2 R^2 L^2 R^3$. \newline Top: $R^3 L^3 R^3 L R L^3 R^3 L R L^3 R$. \newline Top left and bottom right: $R^2 L^3 R^3 L^3 R L R L R^3 L^3 R^3 L R$. \newline Top right and bottom left: $R^2 L R L R^3 L^3 R^3 L^3 R L R L^3 R$. \end{itemize} \vskip .1in {\it Phase 4}: Solve the squares \begin{enumerate} \item If the top left square is not red, then turn the right disc to bring a red square to the top right, do $L R^3 L^3 R$ five times, and turn the right disc back to its original position. \item If the bottom left square is not red, then turn the right disc to bring a red square to the bottom right, do $L^3 R L R^3$ five times, and turn the right disc back to its original position. \item Depending on the positions of the two blue squares, do one of the following move sequences: \newline Top right, bottom right: Do nothing as they are solved already. \newline Bottom centre, bottom right: Do $L^2$, then $L R^3 L^3 R$ five times, and $L^2$. \newline Top centre, top right: Do $L^2$, then $L^3 R L R^3$ five times, and $L^2$. \newline Bottom centre, top centre: Do $L^3 R^3$, then $L R^3 L^3 R$ five times, and $R L$. \end{enumerate} Note that phase 4 uses the sequence $(L R^3 L^3 R)^5$, which swaps the center squares as well as the top left and top right squares. It possible to solve the squares more quickly than the above method by using different conjugates of that sequence to solve more pieces at the same time. In fact you need to apply that sequence at most twice and often only once is enough. \vskip .1in {\it Phase 5}: Solve the diamonds \begin{enumerate} \item If the left diamond (of the left disc) is not red, then turn the right disc until a red diamond is at the top, and do L, repeat $L R^3 L^3 R$ ten times, do $L^3$, and turn the right disc back to its original position. \item If the top left diamond is not red, then turn the right disc until a red diamond is at the top, repeat $L R^3 L^3 R$ ten times, and turn the right disc back to its original position. \item If the bottom left diamond is not red, then turn the right disc until a red diamond is at the bottom, repeat $L^3 R L R^3$ ten times, and turn the right disc back to its original position. \item Depending on where the yellow diamond is, do one of the following sequences: \newline Centre: Do nothing as the puzzle is solved already. \newline Top right: Do $L^3 R^3$, then $R L^3 R^3 L$ ten times, and $R L$. \newline Far right: Do $L^3 R^3$, then $L R^3 L^3 R$ ten times, and $R L$. \newline Bottom right: Do $L R$, then $R^3 L R L^3$ ten times, and $R^3 L^3$. \end{enumerate} Note that phase 5 uses the sequence $(L R^3 L^3 R)^{10}$, which cycles the centre, top left and top right diamonds. It possible to solve the diamonds more quickly than the above method by using different conjugates of that sequence to solve more pieces at the same time. \vskip .2in {\it Pretty Patterns for Rashkey 3} \begin{itemize} \item Batman: $R^2 LRLR L R^3 L^2 R L^3 R^3 L^2 R L^2 R^3 LRL R^3 L^3 R^3 L^3$. \item Butterfly: $L^3 R L^2 R^3 L R L R^2 L^3 R^2L^2R^2L^2 R^2 L R^3 L R^2 L R^3 L R^2 L R^2$. \end{itemize} \section{K\'ep Korong} Also sometimes called ``Rubik's Cheese''. The K\'ep Korong is a predecessor of the Hockey Puck, and has the same mechanism but with fewer pieces. It has the shape of a thick disk. In the centre are two semicircular parts, and around these are 6 segment pieces. The centre can rotate with respect to the segments, and one of its halves can be given a 180 degree turn to change the order of the segments. Each side has a picture of a cartoon character. A neat feature is that in its solved state the centre must be out of alignment to complete the pictures. Rubik's Cheese is a predecessor of the Rubik's UFO, and has the same internal mechanism except that it has only one layer instead of two. It is extremely rare and hard to find. It is essentially the same puzzle as the Kép Korong, except that it has no rotating centre, so any three adjacent segments can turn at all times. Each segment has different colours on top and bottom, and in the solved position each side of each piece shares its colour with a piece next to it. The solved position therefore shows three colours on each side of the puzzle. The picture above shows the version where there are three identical pairs of pieces, but it is more common for the pieces to have six different colour pairings so that in the solved position the three coloured regions of one layer do not coincide with the three regions of the other. The US patent for Rubik's Cheese was filed on 9 November 1981 and granted 18 October 1983, US 4,410,179, but there is an earlier Hungarian patent, 9 November 1980, HU 2679. \subsection{The number of positions} The pieces come in two sets of three which cannot intermingle. In fact, the Rubik's Cheese has a mechanism where three pieces are fixed on axes connected to a centre core, while the other three pieces are held between them. Each piece has two possible orientations, so at first sight there are at most 3!ˇ26 = 384 positions. This limit is not reached because the number of flipped pieces of each set of pieces has the same parity as the permutation parity of the other set. This leaves only 3!ˇ24 = 96 positions for the Rubik's Cheese. \begin{center} \begin{tabular}{cc} Moves & Positions \\ \hline 0& 1\\ 1& 3\\ 2& 6\\ 3& 12\\ 4& 18\\ 5& 24\\ 6& 23\\ 7& 9\\ \hline Total& 96 \\ \hline \end{tabular} \end{center} \subsection{Solution} {\bf Notation}: Mentally label the segments/pieces of the disk clockwise from A to F. Any twist of three segments can then simply be denoted by the letter of the middle segment of those three. \vskip .1in {\it Phase 1}: Orient the pieces \begin{enumerate} \item If there are any three adjacent pieces of which two or three are flipped, then twist that half. \item Repeat step a as often as possible. \item If there are still pieces flipped, for example piece A, then do CDE and turn over the puzzle. If necessary, repeat until no flipped pieces are left. \end{enumerate} \vskip .1in {\it Phase 2}: Arrange the segments 1. Consider pieces A, C, and E correct, and compare pieces B, D, and F to them. To cycle around B->D->F, do the moves CDECDE. To go in the opposite direction B->F->D do EDCEDC. \vskip .1in {\it Phase 3}: Fix the centre (Kép Korong only) \begin{enumerate} \item During the last move done in the previous phases, you can always ensure that at least one half of the centre facing the correct way. Due to parity, the whole centre should then be facing the correct way. If you have to turn over the centre anyway: \newline Do any move (i.e. turn over one half) \newline Rotate the centre 180 degrees \newline Turn over one half again. \item Rotate the centre until the picture on the front is correct. \end{enumerate} \section{Rubik's Rings / Hungarian Rings} This puzzle consists of two intersecting rings made up of a number of coloured balls. The rings of balls intersect at two places, so they share two of the balls. Each ring of balls can be turned, so the balls can be mixed. The Rubik's Rings version has 34 balls of three colours. The intersections divide each ring into sections; there are 5 balls on the inner sections between the two intersection points, and an outer section with 11 balls. In its solved state the 11 blue, 11 red, 12 yellow balls are arranged so that the outer sections are red or blue, and the inner sections and intersections are yellow. The Hungarian rings puzzle has 38 balls of four colours, intersections which lie 5 apart (i.e. 4 balls in between them). Two colours have 9 balls (yellow and blue) and two colours have 10 balls (black, red). When solved, the balls of each colour must form a continuous row. Rubik's Rings has the two rings intersecting at an angle, making a 3-dimensional shape. This allows for a neat ratchet system to stop the balls from moving unintentionally. It might be interesting to quote from the afterword of the Rubik's Cubic Compendium [p212] here. It has a picture of the Hungarian rings and the following text by David Singmaster: \begin{quotation} Closer to Rubik's Magic Cube are 'interlocking cycle' puzzles where several rings of pieces cross each other. Endre Pap, a Hungarian engineer, invented a flat version with two rings which was marketed as the Hungarian Rings. The idea was not entirely new, as there is an 1893 patent for it. \end{quotation} That patent is US 507,215 by William Churchill, filed on May 28 1891, granted on October 24, 1893. \subsection{The number of positions of Rubik's Rings} There are 34 balls, which can be arranged in at most $34!$ ways. This limit is not reached because: \begin{itemize} \item The yellow balls are indistinguishable (12!) \item The red balls are indistinguishable (11!) \item The blue balls are indistinguishable (11!) \item The red and blue balls are equivalent (2) \end{itemize} The last point is because the puzzle can be solved both with red or with blue on the left hand side. The total number of positions is therefore $34!/(2\cdot 11! 11! 12!) = 193,413,243,572,640$. \subsection{The number of positions of Hungarian Rings} There are 38 balls, which can be arranged in at most $38!$ ways. This limit is not reached because: \begin{itemize} \item The yellow balls are indistinguishable (9!) \item The blue balls are indistinguishable (9!) \item The red balls are indistinguishable (10!) \item The black balls are indistinguishable (10!) \item The yellow and blue balls are equivalent (2) \item The red and black balls are equivalent (2) \end{itemize} The total number of positions is therefore $38!/(2\cdot 9! 10!)^2 = 75,406,424,215,922,599,800$. Note however that (even when taking into account the color equivalences) there are still 8 possible solutions. \subsection{Solution to Rubik's Rings} This solution is different to the one in the booklet supplied with the puzzle. I think this solution is quicker. It is also more intuitive and therefore easier to remember. \vskip .1in {\it Phase 1}: Solve the yellow balls (the inner area) In this phase all the yellow balls will be placed in position, in the inner sections of the rings including the intersection points. \begin{enumerate} \item First construct a row of 5 yellow balls in the left ring. This is quite easy to do as follows: \begin{enumerate} \item Find a yellow ball in the ring on the right that does not lie at an intersection. \item Rotate the left ring to bring whatever row of yellow balls you already have into the outer section of the ring, but adjacent to the intersection point nearest your chosen yellow ball. \item Turn the right ring to bring the yellow ball to the intersection, joining it up with the row of balls in the left ring. \end{enumerate} Repeat this process until you have a row of 5. \item Turn the puzzle upside down so that the row of 5 balls is in the right ring. \item Now we will put the 7 remaining yellow balls in a row in the left ring. Nearly the same method can be used as in step a, as long as we make sure that the row of 5 remains in the outer section if the right ring and so takes no part in the action. It is however possible that while building your row of 7, all the remaining loose yellow balls also lie in the left ring so that you get stuck at step a1 above. In this case move the left ring so that the yellow row is still in the outer section, but one of the loose yellow balls lies at an intersection point. You can now move the right ring a little and continue with step a2. \item You now have a row of 5 and a row of 7 yellow balls. First make sure both rows are away from the intersection points, next turn the right ring to put the row of 5 in place between the intersection points, and finally turn the left ring to put the row of 7 in place. \end{enumerate} \vskip .1in {\it Phase 2}: Separate the red/blue balls. In this phase the red and blue colours are separated, thus solving the puzzle. Before you start this phase however, you have to decide which colours the outer sections should be. It is usually best simply to find out which colour dominates a section, red or blue, and then consider that to be the colour it is to be when solved. Balls of the wrong colour in each ring will be called simply 'wrong balls', and balls of the correct colour in each ring 'correct balls'. This phase will attempt to swap a wrong ball on the left with a wrong ball on the right, and thus make them correct. Instead of clockwise or counter-clockwise turns, I will use the terms 'inwards' and 'outwards' turns. An inwards turn of a ring brings some balls from the outer section of the ring inwards towards the top intersection point. In other words, an inwards turn of the left ring will be clockwise but an inwards turn of the right ring will be counter-clockwise. Outwards turns are the opposite of inwards turns. \begin{enumerate} \item If there is a wrong ball in the top half of each ring, then the following sequence will correct them: \begin{enumerate} \item Turn the right ring inwards to bring a correct ball to the top intersection. \item Turn the left ring inwards to bring the left wrong ball to the top intersection. \item Turn the right ring to bring the right wrong ball to the top intersection. \item Turn the left ring back (outwards) into position. \item Turn the right ring back (outwards) into position. \end{enumerate} Note how at the bottom intersection there will always be a yellow ball, so no changes occur there. \item Repeat step a as often as you can. Note that as it stands, the wrong ball in the right hand ring can even be in the exact middle of the outer section (6 balls away from the intersection) and the sequence will still work, but the left wrong ball must be above the middle. If the left ball is in the middle and the right hand one is not, then you can use the same sequence but with left and right interchanged in each of the steps a1 to a5. \item If there are wrong balls in the lower halves of both rings then turn the puzzle upside down and go back to step a to correct them. \item If there are wrong balls in the top half of one ring, and in the bottom half of another, then this can be solved by bringing one wrong ball to the middle, and going back to step a to correct it. To bring a wrong ball in the top half to the middle, do the following sequence: \begin{enumerate} \item Turn the ring with the wrong ball 6 steps inwards, bringing the ball that was in the middle to the top intersection. \item Turn the other ring inwards till a correct ball lies at the top intersection. \item Turn the first ring outwards till the wrong ball lies at the top intersection. \item Turn the other ring back (outwards) into position. \item Turn the first ring back (outwards) into position. \end{enumerate} \item The only case that can not be handled by the previous steps is when there are only two wrong balls, both in the middle. This can be solved by the following sequence: \begin{enumerate} \item Turn one ring inwards 6 steps. \item Turn the other ring inwards 6 steps. \item Turn the first ring outwards 6 steps. \item Turn the other ring outwards 6 steps. \end{enumerate} \end{enumerate} Note: Very often you can run various instances of the sequence in phase 2 step a together. The first turn brings a correct ball to the top intersection, and from then on each turn brings a wrong ball to the intersection. You must take care not to turn too far outwards, as the bottom intersection must always have a yellow ball. When you have no wrong balls within reach, turn the rings back to their original positions. This will speed up the process considerably. \section{Dino Cube / Rainbow Cube / Brain Twist} The Dino Cube is a cube shaped puzzle, and like the Skewb, it has eight axes of rotation centred around the corners. Its cutting planes go diagonally through the square faces, cutting off triangular pyramidal corners. There are twelve moving pieces, one on each edge of the cube. It is called a dino cube because it originally had pictures of dinosaurs on the sides. Versions with 6 colours (pictured above) were also made, as well as with only 4 or 2 colours. The second puzzle pictured above is the Rainbow Cube. It has the shape of a cuboctahedron. It is very much like a Dino Cube in which the corners have been cut off, giving it 8 triangular faces as well as 6 square ones. The puzzle still has only 12 moving pieces, but now there are also stationary centers in the triangular faces. There are two colour schemes. One has 14 colors, the other has only 7 colours with opposite faces the same colour. As the puzzle with the 7 colour scheme does not have any identical pieces, the two colour schemes give puzzles of the same difficulty. The Brain Twist is a new puzzle by Hoberman. The third picture above shows what it looks like when unfolded in its star shape. To make a move, the points of the star are pushed together causing it to fold into a tetrahedron shape, after which the four corners of the tetrahedron can be twisted. You can unfold it and then fold it in the other direction to get a different tetrahedron with four new corners to twist. The equivalence with the dino cube is most easily seen when the puzzle is in its star shape, and imagining the effect of a twist of three pieces around a corner. There are two solution patterns - one with each tetrahedron face a single colour, and one with each tetrahedron corner a single colour. The second solution does not look as nice when in star shape however. It was patented by Charles Hoberman and Matthew Davis on 12 May 2005, US 2005/098947. Jackpot, also known as Platypus, is a new puzzle from Meffert's, and the fourth picture above shows a prototype version. Unlike the other puzzles on this page, on the Jackpot the eight axes are not identical. It is not cube shaped but is based on the tetrahedron, with the corners extended in a triangular cylindrical shape. There are four large hexagonal faces, marked with card suits, as well as four small triangular faces, marked with J, Q, K, and A card values. The pieces are all similarly marked on their sides. For an extra challenge, on some versions of the Jackpot the hexagonal faces also have coloured marks along their edges, so that their orientations matter. It was patented by Yusuf Seyhan on 16 January 2003, WO 03/004117. There is also another puzzle that is equivalent to the Dino Cube, but I do not know much about it. It is in the shape of a Stella Octangula, of which the corners can rotate. There are small pieces in between them along the edges of the internal octahedron, and the three pieces around each tip form a circle on the surfaces of the adjacent tips. Each circle should be of one colour. \subsection{The number of positions} There are 12 moving pieces, which seemingly have 2 possible orientations giving at most $212\cdot 12!$ positions. This limit is not reached because \begin{itemize} \item The pieces cannot actually be flipped (212) \item The pieces must have an even permutation (2) \item On the Dino Cube, Brain Twist, and Stella, the orientation of the puzzle is immaterial (12). \end{itemize} This leaves $11!/2 = 19,958,400$ positions on the Dino Cube, Brain Twist, and Stella, and $12!/2 = 239,500,800$ positions on the Rainbow Cube and standard Jackpot. Some versions of the Jackpot puzzle have colours on the sides of the hexagonal faces, making their orientation visible. As these faces can be twisted independently, there are 34 times as normal, viz. $34\cdot 12!/2 = 19,399,564,800$ positions. The Dino Cube actually has two solutions, which are mirror images of each other. This is similar to the two solutions of the Brain Twist. The minimal number of moves to achieve each possible position of the Dino Cube/Brain twist needs to solve was computed by the second author (J.S.). There are 1004 positions of the puzzle at maximum distance from start. For the Rainbow cube, the number of moves in each position was calculated by Claude Crépeau and Thanh Vinh Nguyen. There are 366 positions of the puzzle at maximum distance from start. In each case, the tabulations are shown in the table in \cite{Jwww}. \subsection{Solution} These are very easy puzzles to solve. There is really only one easy sequence of moves that you need to be able to do. Consider two adjacent axes on the puzzle, with the five piece positions they contain numbered as shown here: \begin{center} \begin{verbatim} 1 2 3 4 5 \end{verbatim} \end{center} You are trying to put the correct pieces at positions 1 and 2, and you haven't yet solved 3, 4 and 5 yet (though it is quite possible that one or more of them happen to be correct, ignore them). Solving the first piece, at position 1 is trivial, since you can just put it there without disturbing previously solved pieces. Next, you want to put the correct pieces at position 2, without disturbing the piece at 1. \begin{enumerate} \item First bring the piece belonging at 2 to position 4 or 5, without disturbing any of the pieces you have previously solved. \item Then turn the top face/corner clockwise, putting the piece that currently is at position 2 to the intersection at position 3. \item Turn the bottom face/corner to bring the piece belonging at 2 to position 3, replacing the old piece. \item Turn the top face/corner anti-clockwise, back to its previous position, which puts the correct pieces at both 1 and 2. \end{enumerate} Once you understand this simple technique, the puzzle can be easily solved by solving the pieces in the following order: \begin{enumerate} \item Solve three adjacent pieces that share an axis (a triangular face in the Rainbow Cube, three pieces at one corner in the Dino Cube, etc). \item Solve the six pieces that are adjacent to the first three you solved. You need not move any of the first three solved pieces at all. \item Solve the last three pieces (which are the ones directly opposite the first three). This is trivial, as you only need at most a single move to do this (though see note below). \end{enumerate} Sometimes however it is hard to recognise which piece belongs where. On the Rainbow Cube each colour is used on opposite faces of the puzzle, which means that each piece also has a mirror image twin. If you just keep track only of the colours of the triangular faces, then you cannot go wrong as the mirror piece will have its colours swapped if you try to use it instead. On the Dino Cube adjacent pieces have only one colour in common, so it seems that you could put the wrong pieces next to each other. This is not the case however - if you try to place the wrong piece, it will seem to be flipped. On the Brain Twist it is easy to get confused because of the constant need to flip the puzzle inside out, but this puzzle has another neat twist in store. It may happen that you solve the puzzle completely but for two pieces that need to be swapped. This is unfortunately not possible (it is an odd permutation and every move is an even permutation). What has happened is that you have built the colour pattern in mirror image. You have to practically re-solve the whole puzzle - choose any face you had solved, swap two of its pieces, and continue from there. If you have a puzzle on which the face orientations are visible, then the techniques outlined above are not quite sufficient. With a little care you can ensure the face orientations are correct of all but the last face (step c above). If you are solving a Jackpot where the hexagonal faces are marked, then you can simply leave an unmarked triangular face till last instead. Pretty Patterns and other moves: {\bf Spot Patterns}: The rainbow cube has two types of spot patterns, patterns in which only the triangular centres have a different colour. One has 8 spots, the other only 6 spots. \begin{itemize} \item The 8 spot is made like this: Hold the puzzle with a square on top. Make a slice move, i.e. turn any triangular face clockwise, and its opposite face anti-clockwise (so the two faces have actually moved as a unit). Turn the whole puzzle a quarter turn around the vertical axis. Repeat this a number of times. After 6 or 10 times (depending on the direction you turn the puzzle) you will get the 8 spot. \item The 6 spot is done much the same way, except that a triangular face is on top, the slice move is done on any of the other faces, and the puzzle is rotated through 120 degrees each time. After doing this 8 or 10 times, depending on the direction, the 6 spot will appear. \end{itemize} Swapping between solutions The Dino Cube has two solutions which are mirror images of each other. To change a solved Dino Cube to show the other solution, perform the following move sequence: \newline \verb+urb urf urb' ulf urf drb' urb' ulb dlb' dlf+. There is a longer method which is much easier to remember: newline \verb+(urb ulf drf dlb) (urf' ulb' drb' dlf') (urb ulf drf dlb)+. This first turns one tetrad of corners clockwise, then the other tetrad counter-clockwise, and then the first tetrad clockwise again. The same method works on the Brain Twist to swap from the faces-solution to the corners-solution, and vice versa. Simply twist all corners clockwise, flip it, twist all corners counter-clockwise, flip again, and turn all corners clockwise. \section{Swissmad} Swissmad is a puzzle with 12 square tiles arranged the shape of a cross. The tiles are contained in a flat frame which has transparent front and back plates, so that both sides of the tiles are visible. The cross has a 2 by 2 square in the centre and arms 2 tiles wide and one tile long. The two columns of the vertical arms can slide up or down one tile, independently of each other, by pushing the two parts of the transparent front plate of the puzzle up or down. Similarly, the two rows of the horizontal arms of the cross can individually slide left or right by pushing the transparent back plate sections left or right. Each tile is red or white on the front and red or white on the back. This gives four possible tile types, and there are exactly three of each type of tile. In the standard solved position the back of the cross is half white and half red, while the front of the cross is in quarters of alternating colour. Embossed on the frame are eight patterns that you can try to make. The names are reproduced below with the names they have been given in the leaflet provided with the puzzle. \begin{tabular}{cccc} Time & Interaction& Excellence & Balance \\ Diversity & Audacity & Mutual help & Security\\ \end{tabular} The leaflet quite amusingly tries to justify the names of these patterns. The leaflet itself is also in the shape of a cross, normally with the arms folded over to make it square. The arms of this leaflet show the patterns 'Time', 'Excellence', 'Balance' and 'Diversity', so by changing the order in which you fold these arms, you can have any of these patterns on the front. By some clever folding, you can however also get it to show the 'Mutual help' pattern instead. Swissmad was invented by Olivier Pahud, and patented on 16 June 2005, WO 2005/053809. \subsection{The number of positions} The packaging states ``Two faces, two colours, one million combinations''. As you will see, this is not quite accurate. Lets assume the rows and columns are centred. There are 12 pieces, in four sets of identical triplets. These can be arranged in $12!/3!^4 = 369,600$ positions. If you don't assume the rows and columns are centred, then this number must be multiplied by 34 because each row/column has 3 states. Then there are $12!/2^4 = 29,937,600$ positions. \subsection{Solution} There is a very simple sequence of four moves that cycles three pieces around. Slide a column down, a row to the right, the column back up, and finally the row back to the left. The three pieces that are moved are the tile at the intersection of the row and column, as well as the tile above and the tile to the left. Basic move sequence Any three adjacent tiles that do not lie in a straight line but in a kind of L formation, any such three tiles can be cycled around. Simply shift the row and column alternately to push the two outside tiles towards the middle one, and then move the row and column back. Using this sequence it is quite easy to solve any position. To solve a position, the best general strategy is to first solve the arms of the cross one by one. With the move sequence you can move any tile from an arm into the central area, then move it around the central area to the side you want, and finally move it out into an arm of the cross. Once all the arms are correct it is usually an easy matter to solve the central area. Always remember to check both sides of the tile are correct for the spot you are moving it to. One minor complication that can arise with this method is that you may end up needing to swap two pieces without moving anything else. This usually happens when all the arms have been solved, and the central area contains all four types of tiles. In this case replace any tile on an arm by the identical one from the central area. After this, the puzzle can be solved easily. When solving the standard pattern (time/diversity), I usually do things differently. I solve the 'time' side first, putting all the reds in one half, all the whites in the other. When doing this I don't need to worry about the other side of the puzzle. Then I turn it over and separate each of those halves again, into red and white quarters. By never letting any tile from the other half enter the half you are solving, the 'time' pattern on the back will not be disturbed. The advantage of this method is that you don't need to keep track of the reverse colours of the tiles at all. \section{Diamond 8-Ball Puzzle} This puzzle consists of 8 numbered balls which can roll around a $3\times 3$ square. There is a restriction that differentiates it from a simple $3\times 3$ version of the fifteen puzzle. There are walls on the left and right side of the centre position, so that a ball can enter or leave the centre in a vertical direction only. In the solved state the centre spot is empty, and balls are in clockwise numerical order starting from 1 at the top left corner. This puzzle was invented by Joshua Frankel, and manufactured by Binary Arts, now called ThinkFun. \subsection{The number of positions} If we count only those positions with the gap in the centre, then there are 8 pieces which can be any order, giving a maximum of $8! = 40,320$ positions. As with the fifteen puzzle, only even permutations are achievable, so there are actually $8!/2 = 20,160$ positions with the centre empty, or $9!/2 = 181,440$ positions with the empty spot anywhere. Note however that there are really two solutions - you could solve it with the 1 ball at the bottom right, and then turn the puzzle upside down. The second author (J.S.) performed a computer search for this puzzle. The following table shows how many positions there are (with the center empty) for each number of moves from the solved position. A move consists of moving the space around the left or right side in either direction, shifting five pieces in the process. The first column shows the number of positions at each distance from only one of the two solutions, the second if either solution is allowed. \begin{center} \begin{tabular}{|l|l|l|} \hline Moves & One solution & Two solutions\\ \hline 0& 1& 2\\ 1& 4& 8\\ 2& 12& 24\\ 3& 32& 64\\ 4& 82& 156\\ 5& 204& 360\\ 6& 496& 792\\ 7& 1,153& 1640\\ 8& 2,431& 3202\\ 9& 4,325& 5132\\ 10& 5,829& 5556\\ 11& 4,408& 2916\\ 12& 1,133& 306\\ 13& 50& 2 \\ \hline Total& 20,160& 20,160 \\ \hline \end{tabular} \end{center} \subsection{Solution} {\bf Notation}: There are 2 loops in the puzzle; the left half and the right half. Denote these by L and R. A clockwise shift of all the five balls in a loop is indicated by the relevant letter (L or R), and an anti-clockwise shift by the letter followed by an apostrophe (i.e. by L' or R'). \vskip .1in {\it Phase 1}: Solve balls 3,4,5, and 6. \begin{enumerate} \item Do any moves to bring ball 5 to the top right corner, i.e. the position where 3 will be when solved. \item If ball 4 lies on the right (directly below the 5, or in the bottom right corner), then do RR LL R'R' so that 4 lies in the left loop and 5 is again top right. \item Shift the left loop to bring ball 4 to the top position, and then do R. You now should have ball 4 at the top right, and 5 directly below it. \item If ball 3 lies at the bottom right corner, then do R L R'. \item Shift the left loop to bring ball 3 to the top position, and then do R. You now should have balls 3, 4 and 5 in their correct positions. \item Shift the left loop to bring ball 6 to the bottom position, where it belongs. \end{enumerate} \vskip .1in {\it Phase 2}: Solve the rest. \begin{enumerate} \item Find ball 7. Depending on its position, do one of the following sequences: \newline top-left: $L'RL'R' LLRL'L'R'$ \newline top: $RLL R'LR LLR'$ \newline left: $R'LRLL R'L'L'RL$ \item Find ball 8. Depending on its position, do one of the following sequences: \newline top-left: $LR'L'RL'R'LLRL$ \newline top: $L'R'L'RL'R'LLRL'$ \end{enumerate} \vskip .1in {\it Nice sequences}: $LRLRLRL$: Constructs the second solution, with the 1 at the bottom right corner. \section{Cmetrick} This puzzle consists of a frame containing a $3\times 3$ array of coloured balls. The balls are identical, and have 6 colours, arranged like the 6 sides of a cube. If you rotate any ball to the left or right, all three balls in the same row are rotated in the same way. If you rotate a ball up or down, then the balls in the same column rotate with it. The aim is of course to get all the balls in the same orientation, so that they show the same colour on the front and sides. The Cmetrick Mini is just the same as the ordinary Cmetrick, except that it is a 2×2 square, so only has 4 balls. It is of course easier to solve, but also easier to handle because it is easier to keep the balls aligned enough to do the moves. This puzzle is related to the Rolling Cubes Puzzle in that it has a $3\times 3$ array containing cubes which are reoriented. Mechanically it is somewhat related to the Rubik's Clock, as the nine parts are connected with an internal set of cogs. With regards to difficulty this puzzle is more like the Rubik's Clock than the Rolling Cubes Puzzle. CMetrick was invented and patented by Dror Rom, 8 January 2004, WO 2004/002587. \subsection{The number of positions} There are 9 cubes, each with 24 possible orientations, so this gives an upper bound of 249 positions. These are not all possible however because of parity restrictions. If you move the second and third columns so that the top row balls all have the same orientation parity, and similar the second and third rows to equalise the left column's parities, then the four balls in the bottom right corner will automatically have the same parity as the rest. This therefore means that the real number of positions is $24^9/2^4 = 165,112,971,264$. Note that there are in fact 24 solutions, so if we consider identical any positions that differ only by recoloring then there are really only $24^8/2^4 = 6,879,707,136$ positions. The parity restrictions can be worked out by using linear algebra the same way as the Lights Out puzzles, as explained on the Lights Out Maths page. Suppose we ignore the ball orientations, and only keep track of the orientation parities. There are then only six ways that the parities change by a move - changing all the parities in a column or a row. These six ways are not independent, since the sum of all six have no effect, but any five of them are. This means that there are exactly 25 ways to affect the parities, instead of the 29 which you would have if there were no restrictions. This is where the factor 24 comes from in the previous paragraph. There is another way to look at the same thing. Consider four balls in a 2×2 square. Any move on the Cmetrick will change the parity of zero or two of these balls. The number of balls with odd parity in this square will therefore always remain even. From this it is easy to see that once the balls on the middle row and column of the Cmetrick are known, the parities of the corner balls can be determined. Again, this is the factor 24 above. The Cmetrick Mini has 4 cubes, each with 24 possible orientations, so this gives an upper bound of 244 positions. On this puzzle there is only one parity restriction, so the number of positions is $24^4/2 = 165,888$. Again there are in fact 24 solutions, so if we consider identical any positions that differ only by recolouring then there are really only $24^3/2 = 6,912$ positions. Stefan Pochmann has calculated the number of positions at each depth of the Cmetrick, and his results are shown below left. It shows that in the worst case only 15 quarter turns are needed to solve the puzzle. There are $761,436$ such worst-case positions. Below right are the results of the second author (J.S.) for the Cmetrick Mini, showing you need at most 9 quarter turns to solve it. There are $204$ such worst-case positions. \subsection{Solution} {\bf Notation}: A ball can be rolled in four directions, Up, Down, Left, and Right. Quarter turns in these directions will be denoted by the letters U, D, L and R respectively. \vskip .1in {\it Phase 1}: Solve the edge balls, forming a cross. The centre ball will be considered already solved. In this phase, the balls adjacent to the centre will be solved. \begin{enumerate} \item Consider the ball below the center. In the next few steps this ball will be rolled until it matches the orientation of the center ball. \item Look at the front color of the center ball, and see where that color is on the below centre ball. \item Depending on where that colour is, do one of the following: \newline Right side: L \newline Left side: R \newline Up side: D R U L \newline Down side: U R D L \newline Rear face: R R \newline This should match up the front colours. \item Look at the sides of the below centre ball, and compare it to the sides of the centre ball. \item Depending on which way the ball needs to be moved to match the center, do one of the following: \newline Clockwise: U L D \newline Half turn: U L L D \newline Anti-clockwise: U R D \newline The two balls should now match exactly. \item Turn the whole puzzle a quarter turn. \item Repeat steps 1-6 three more times, so that all the balls adjacent to the center match, forming a cross. \end{enumerate} \vskip .1in {\it Phase 2|: Solve the corner balls. \begin{enumerate} \item Consider the ball at the bottom right corner. In the next few steps this ball will be rolled until it matches the orientation of the center ball. \item Look at the front color of the center ball, and see where that color is on the bottom right corner ball. \item Depending on where that color is, do one of the following: \newline Right side: L U R D \newline Left side: R U L D \newline Up side: D R U L \newline Down side: U R D L \newline Rear face: R R U L L D \newline This should match up the front colors. \item If the sides of the bottom right corner ball do not match the sides of the centre, then it needs a half turn which can be done by \verb+L D R R U L+. The ball should have now been solved. \item Turn the whole puzzle a quarter turn. \item Repeat steps a-e three more times, so that all the corner balls match the centre, and so the puzzle should be solved. \end{enumerate} You can now optionally reorient all the balls by turning all rows or all columns in the same way. \section{Topspin} The Topspin puzzle is by Binary Arts (now called ThinkFun), but is also sold as No. Crunch under their XEX brand name. It consists of 20 numbered round pieces in one long looped track. You can slide all the pieces of the loop along. There is also a turntable in the loop which can rotate any four adjacent pieces so that they will be in reverse order. This in effect swaps two adjacent pieces and the two pieces on either side of them. The aim is of course to place the pieces in numerical order. It was invented by Ferdinand Lammertink, and patented on 3 Oct 1989, US 4,871,173. \subsection{The number of positions} There are 20 distinct pieces, which can therefore be put in at most $20!$ possible positions. All these are attainable, so there are $20!=2,432,902,008,176,640,000$ distinct positions. If rotations of the loop are considered to be the same, then there are only $19!=121,645,100,408,832,000$ positions. \subsection{Solution} {\it Phase 1}: Place pieces 1-16. This is very easy. Simply find the next piece to solve and do any move that brings it closer to the solved part of the loop, but which does not move any already solved pieces. Then bring it to a position such that there are three pieces between it and the solved part and then a single turn will append it. \vskip .1in {\it Phase I1}: Place pieces 17-20. First I will show a method with few moves but which one might find hard to memorise. \end{document}