I remember when Rubik's cube first arrived on the scene (yes, I'm that old). My brother got one from somewhere, played with it and got it hopelessly scrambled, left it lying around, I picked it up..Please Register or Log in to view the hidden image! It took me a while (over the next 3 weeks), picking it up and putting it down (mostly the latter), then I got hold of a treatise of sorts, about solving them, about how you can discover all these rotational symmetry groups ('slice groups'), etc. I figured out how to solve it, by using an algorithm that the article describes as "the driveway method", you solve two of the layers using a general algorithm (with tweaks), then solve the final layer using specific sets of transforms, and their inverses. You go through a whole heap of math, by solving a 3x3 cube even if you don't think about it that way. The last cube I owned was a "Professor's cube", a 5x5 brain teaser, that's really just a 3x3 with extra 'edge cubies', as the individual 'cubelets' along each edge between the "corner cubies", is called in the literature. The 5x5 I didn't play with as much, not being keen to scramble it too far. But eventually I did and got the thing 'locked' into some kind of bistable solution where it would oscillate between two "half-solutions", implying I had to scramble it some more, undoing the 'stuck' state it was in. I then made my usual mistake of pulling it apart, with the always good intention of reassembling the thing at the 'start' state. Which of course never happened. Nowadays, there are also 2x2, 4x4, 7x7 cubes, there's a sphere with 20 triangular sections like those earlier number-puzzle squares, 4 colors of 5 triangles each, each numbered 1-5. You play with the orientations and positions of the numbers and the colors. It's like a kind of sudoku sphere with 4 'layers' or sides (like 4 puzzles at once), and only 5 numbers (10/2). Pentagonal color-shapes mean clockwise/anticlockwise number solutions, etc. It's also something like a quasi-periodic tiling, except the tiles are an exact fit over the sphere so you need to generalize the 5 triangles x 4 colors to something else - Penrose tilings use different triangles, not isosceles for starters. But you can get other periods in the numbering (say you make 3 pentagons, there's a pattern distributed over the numbers in the 3x5-tile pentagonal shapes, which is 'programmed in'. The 4th 'shape' can't be a pentagon, either - it has to 'diverge' when you have a maximum of 3 pentagons, which I call a 'hadron', because of the three equal colors, and the fact the single red 'orientation piece' (the triangle you have to remove from the puzzle, to move any other pieces) isn't then joined to another red piece. Anyways, who else is into Rubik's cubes? Do you think you can model things like quark-antiquark pairs, like the article says you can (two opposite corners of a 2x2 or higher cube, each with a 1/3 twist)? Have you worked out a code?
It's been 25 years. (I'm older than practically everybody here; I remember hula hoops although I never had one.) I remember developing my own notational system so I could put the thing down and pick up where I left off the next time. I doubt that I discovered the quickest path to a solution, but I did develop a general set of moves that could be used to reach a solution. I only did it once.
The original Rubik's cube can be seen as having subsets of 2x2(x2), or you can consider the 2-cube as having only 8 corner pieces (each with 3 equal sides). The real 2x2x2 cube has no edges or center cubelets. You can model a complex space with one, because you generally 'process' each 3x3 layer (a 'main colour', and 4 'side colours' = 6 -1 colours), then a 3x1 row (which has 2 'side colours' , and 2 'end colors' = 6 - 2). You flip the corners, and the edges come along for the ride when you move 2 corners as a row. If you imagine a center square, on any face, is 00, (0,0), or {0,0}, you can see the square above it as 01, and so on, the square to the right of 00 as 10, so the upper right corner is 11. Then the 2x2 squares are like a general kind of "map" of the rest of the cube, just add 1 of 6 colours. If you use a notation like (0 0), and (0 1), etc, you can model a probability density matrix (like in QM). You need to add a rule, that a clockwise rotation, means the rows become columns, and you have row and column vectors. If you add in that x and y are complex probability amplitudes, and allow diagonal 'products' between vectors, it models a 2x2 tensor.
If you take a Rubik's cube apart, what you find is a spindle which forms the 3d frame, six arms with squares at each end, that can rotate - the centre or 00 squares for each face. Mr Rubik illustrates, literally, the rotational symmetries of (a) 3-dimensional space. The other pieces lock together in a way that makes the movable pieces - the corners and edges - a mechanical solution, to finding a path over a spherical surface for each corner, with each accompanying edge piece. The movable pieces form a shell, with their interlocking bases. The base of each corner piece is triangular so it has 3 possible directions around the sphere, along paths constrained by the cubical geometry of the 'outer' shape. The original 'R-cube' is a finite set of colour combinations, where each square on each face can have 1 of 6 colours, the possible combinations are constrained by the 'start' state where each face is the same colour, all 9 squares per face. Starting from a scrambled state the 2x2x2 sub-cube is a part-solution, and a 2x2 sub-face of 4 squares is a part solution. If you get 4 squares the same colour on one face, you can use the 'unsolved' squares by rotating the puzzle along their 'cubical' or layer-of-pieces geometry, leaving the 2x2 area unchanged. A 2x2 square (in a corner, as in the above model), if you view it as a face, must have only 2 other colours for each edge piece, and for the corner piece, which you have to turn the puzzle to see. But if you don't, then there are 2 'imaginary' colours, which must match, i.e. you know you have a 2x2x1 solution. You need to get to a 2x2x2, then a 2x2x3, and so on. Sometimes the remaining 'outer' layer(s) will resolve into a solution as you do this, sometimes you have to do specific transforms, like changing the 2x2x2 sub-cube's color - you 'annihilate' one part solution and construct a different coloured one, in another corner. It's also a colour-mapping problem, from the sense (the basis) of 2x2 squares, 2x2x2 etc sub-cubes. The 6 colours restrict the range of discrete combinations, and so the number of discrete rotations needed to solve a scrambled cube. Because the corner pieces have a triangular base, this links the puzzle to the Rubik's sphere, where all the pieces are triangular 'bases', and the pieces find spherical paths, by rotating individually from one apex, along 1 of 2 sides. Each piece has pi/3 ways to rotate, is 1 of 4 colours and 1 of 5 numbers. Different degrees of freedom, different numbers of discrete states, combinations of rotations, or exchanges between degrees of freedom.
The rotational symmetries are, starting from a solved puzzle (R-cube), that i) each corner square on each face can occupy another corner (after a certain transform or sequence of 'quarter-turns'), then ii) you can transpose the edges - (01) and (10) - and restore the corner square so its 2 edge-colours are now 90 deg. out of phase. The (11) square, can occupy the (-1-1) position, or the (1-1) or (-11) positions, on a face, while you swap the edge pieces. The irrotational centre square (00), is 1/2 of a 2x2 colour sub-space. It's distributed over the (01) and (10) squares, the next 1/2 of the centre - i.e. (00) + (01) = 1 = (10) + (00). The (11) square is irrotational if it's the same colour as (00). Squares with a 1, have an extra colour (01) and (10); (11) has 2 extra colours. If the corner (11) cube is a different color, that represents a twist (if 1 of its other colours is the same as the centre, the corner has either a 1/3 or a -1/3 twist). 8 of the squares can be a different colour than the centre (00) square, but there are 6 colours. A face with 7 or 8 colours is an impossible state. However, you could build a computer model of a 3x3 or higher dimension puzzle - a 3d matrix with a spherical 'hidden' geometry. A computer model could have more than 6 colours - or you could just use your own coloured sticky labels, for the 'extra' squares on a face. You could give a puzzle 2 extra colours on each face, after preparing a 6-colour state per face, and solving the puzzle to get a transform. Because there are 6 faces, that's up to 12 extra colours.
Does the model fit? Does a 2x2 subspace of a 3x3 (6-)colour matrix model a tensor, or a field? To get to a complex space, you have to generalize it. The simplest 'useful' matrix, has 2x2 elements; a (mxn) matrix with 1 row and 1 column can only be a row or a column (it can rotate around an axis, but not translate - it has to look along the same circle of directions, which depend on how many elements are in the matrix 'around' it). Say you represent the start of a row with '<', and the start of a column with '|' then since the elements of a row are the start of a column, a 1x1 matrix is its own row and column. The end of a column is an element of the last row. Then (01) is <1| and (10) is <0|. (01)(10) = <00>, and (01)+(10) = <11>. Add imaginary colours, you have a complex space in which a|0> + b|1> is some point corresponding to an angle between the left, or vertical edge of a square matrix (or equivalently the bottom edge), and a line from the centre of rotation to the point, which if projected to an outer edge intersects with a circular arc between the points (01) and (10), or <1| and <0| (|1> and |0>). If you consider the line from the lower left, to the upper right bisects the matrix, then any line with an angle greater than pi/4 will not intersect with the outer righmost edge, only the upper edge, any any angle lower than pi/4 will only intersect with the right edge. Now you just need the idea of a conjugate transpose, which is I suppose the idea of inverting the (11), or <11> to get (-1-1), etc. You have to change its colour, maybe. Inversion and rotation are connected in that a rotation or a transpose are trivially reversible. A quarter turn of a square, or a layer of a Rubik's cube is easily reversed. Over a sphere, in 3 dimensions you get an extra rotation, which is not. The "Magic Cube" has a few mathematically remarkable properties. You could say it has 3 general dynamic phases - apart from its geometric/mechanical structure: a solid phase, when it isn't being 'rotated'; a liquid phase when it is; and a 'gas' phase when you separate the edge and corner pieces from a stable matrix and try to fit them together two or more at a time, to see how the puzzle ticks, as it were.
Me, obviously. I'm more into plain-old speed-solving rather than the actual mathematical theory behind it although I did start writing a computer program to solve a cube from any position. I was trying to devise my own algorithm, based on the way I solve the cube. I didn't necessarily want to find the optimal solution, because the only solutions I could think of at the time were WAY too inefficient (we're talking like O(n^n)). That was several years ago, before I started college. It's still on my todo list as far as programming projects, but at this point in my schooling I have basically no time. I haven't even had time to practice my speed-solve in a long time.
Dang, I didn't explain my row, column notation that well. The start of a row is an element of a 'first' column, and the end of a column is an element of a 'last' row. So the '|' symbol really means a 'pivot operation', at the end of a row or the start of a column it 'pivots' on the last row or column element, respectively. Compare how a 2x2 square matrix with a 6-colour map resembles a Hermitian 2x2 'N-colour' matrix: You get \( |\psi\rangle\, =\, \alpha|0\rangle \,+\, \beta|1\rangle \). For a 'closed' quantum system of N bits: \( |\Psi(t)\rangle \,=\, e^{-iHt} |\Psi(0)\rangle \), where H is the Hermitian \( 2^N\, \mathtt{x} \, 2^N \) matrix. In that sense, a 2x2x2 Rubik's cube is a classical representation of a \( 2^6\, \mathtt{x} \, 2^6 \) matrix, where the colours represent different bits in a classical space, then each element is in some 'fixed' phase with each element of the same colour (same 'bit'), over the R-cube. If the 6 fixed colours could 'oscillate' between two or more extra colours, you get more possible states; in Hilbert spaces you get at least two states, they have a complex phase relation which is non-classical. You can see that a geometric square made of 4 'sub-squares', is bisected by the line from lower left to upper right apex, x=y (or a=b), and by the line from the upper left to lower right apex (from the first element of the first row/column, to the last element of the last row/column), which is x+y = 1 (or a+b = 1). These diagonal lines are inverses of each other, or where one has an 'up' slope, the other has a 'down' or negative slope, I guess depending on if you go from 00 to 11, or the inverse way. Or from 01 to 10 or the inverse.
If you haven't seen it yet, the link between a mathematical/mechanical puzzle, that looks like a 3-d cube that you can 'rotate' different layers of - because a layer is somehow a part of the overall cube shape, and QM is: cyclic permutations. The simplest cube-puzzle is a 2x2x2. Assume a puzzle with only 2 colours - then each face can be all one colour, or all the other colour, or some combination. Call the colours black & white, or perhaps colour(0) and colour(1). Then the possible combinations are all the cyclic permutations of pairs of: (00), (01), (10), (11), as 2x2 matrices, 2 rows and 2 columns. Rows and columns can switch, by rotating the face a quarter turn clockwise or anticlockwise. In fact each permutation corresponds to a set of rotations (q-turns), between 'pure' states like all 0, or all 1.
OK, now assume a puzzle with 2x2 colours, and triangles instead of squares, what's the smallest number that can fit over a spherical surface? How do you transform rotations through \( \pi/4 \) with a 'square' puzzle, to ones through \( \pi/3 \) for a 'triangular' puzzle? [ed:] what I meant with \( \pi/4 \), is that each face on a R-cube can be rotated a quarter-turn, but the R-sphere can have each of 3 triangles change its colour, move wise - the fact there's a missing piece makes the puzzle a different mechanical solution, but since the piece can be replaced as part of each move, it's trivial. It's a different kind of map-colouring problem. Piece-wise, the 2x2x2 cube is moved 4 pieces at a time, through \( \pi/2 \) physical radians, then an outer row/column (as half of 4 pieces) another 4 ways, through another \( \pi/2 \), that's \( \pi \); each location on the sphere "changes colour", by exchanging places with the missing red piece, after a single \( \pi/3 \) rotation, in which the red location exchanges with one of 3 pieces around it, then goes back in the 'hole'; the algorithm for colour-exchange on the R-sphere. It takes 3 of these to get \( \pi \), and the red piece can exchange in 3x3 ways - 9 paths, the cube has 2x4x2 paths, for each of 2 (or 3 etc) pieces as rows/columns, there has to be a connection in there somewhere. On the R-cube, when you move a 2-piece row/column, it has 4/6 colours, (which is the maximum) and no label, the three exchange pieces for the hole on the R-sphere (the red piece), have labels (oriented symbols that point to 1/3 vertices) and 1/4 colours. Maybe that's the minimum number that makes sense to the algorithm it uses. The spaces are different - the R-cube is a closed surface colour-wise; the R-sphere is like a 4-colour box with sides made out of 5 triangles - with two colour 'ends' missing which are non-adjacent. an open cube. The affine logic as linear transforms in each space is different.
That path thing isn't very well done - it turns out you can go through 3x3x3 paths, as you swap triangles. but you can also have situations where all 3 edges that can rotate into the space are the same colour (which is where the numbers and their orientation comes into it, i.e. the symbols have a value and a direction, so mathematically are vectors in a space). Or where there are two the same, so the dimensions of a colour-swap, in respect of a single exchange can be 1 2 or 3 colours. The vectors in or on the R-(3)cube are unnumbered, but oriented in respect of each centre square, or 'home' face. The corners orient in 3 colours, the edges in 2. A corner's colour orientation is a vector with 3 possible states, in any corner it occupies, in respect of 3 'home' squares; an edge's is a vector with 2 possible 'phase angles' in respect of 2 'homes'. It meets 2 centres, the corners meet 3. If you transform the 'meet' sense of a piece's orientation to the R-sphere, it's the corner the number points to. There are 3 meets that can rotate to join with a side, around the 'hole', tracking the hole is equivalent to tracking a path for the removable 'red' piece, like a kind of 'north pole'. Any construction based on the idea of a logical machine needs 2 things to describe it - the states of individual components and their relations or connectedness, how they 'go together' - i.e how to generate the states or transform between them. The idea of an operation or 'recipe' that builds the machine is also implied, and once realised it will then 'perform' the business of producing the states as operations are performed, i.e. the states are calculations of the system. The states of a triangular piece on the R-sphere are its colour, and its orientation (in respect of the pole), and its position, in terms of how many moves away it is from the 'pole piece'. The operations are: rotate a piece around 1 of 2 axes through 1/3 of pi radians (the 3rd axis, or apex is 'stuck', and has to be pulled along 1 of 2 sides), into the 'hole' (swap places with the pole piece); this will rotate the 'number' in 1 of 2 ways wrt the pole. Then any piece has a 'path' as successive 'swaps' with the pole, around the sphere, any calculation is a list of swaps, which change the selected piece according to which solution is being applied to the start state of the 'puzzle'. Note that, with an even-numbered R-cube, the fixed centre piece disappears. A (4)-cube has a 2x2 subcube at the centre of each face.
K, so this puzzle thing hangs on the symmetries, and how they 'appear' in the two kinds of puzzle. Obviously, Rubik's games aren't the only mechanical-solution puzzles around, as a quick check of wikipedia shows. But I think the R-cube and the R-sphere illustrate something fundamental about spaces, or spherical surfaces at least and how to 'carve' them. The other kinds of N-dimensional puzzle, include triangular and tesseract type constructions, and thanks to PCs there are puzzles with 4 spatial dimensions - hyperpuzzles I suppose. Start with a R-cube; a simple one is a 2x2x2. You have 3 'views' of any 3d cube, 1 face, 2 faces, or 3 faces. 3 faces is the maximum, it hides 3 'opposite' faces - each hidden face is the colour corresponding to the 6 edges you can see, because there's a vertex between each adjacent pair, and is 90 deg 'out of phase' with. IOW, each R-cube has 3+3 colour dimensions, although it could be fewer. An R-cube can be modeled, then, as 3 'real' and 3 'imaginary' colour dimensions, in which any rotations that 'send' a colour to a hidden face will mean an 'imaginary' colour will be seen on a visible face. This means the entire 2x2x2 cube is 4+4+4 coloured squares, with 4 corners, one corner is the 'top', and its hidden opposite is the 'bottom'. Like 2 poles (1 real, 1 hidden). To give the 'north pole' a twist, you have to give the 'south pole' an inverse twist. This generalises to any 'swap' or cycle around the cube. A 'cycle', is the 4-cycle described by Mr Hofstader in the article. A quarter-turn 'cycles' the edges 4 ways - each edge is rotated to face another colour, and each 4-cycle can be forwards or backwards (clockwise or counter-clockwise). Because of the necessarily missing piece, and the triangular shapes, the R-sphere has a different colour-symmetry. The 'key' or pole-position is always occupied by the 'key-piece' which is 1 of 4 colours. Any view around this space is always 1/2 of a sphere. The pole can have, or 'see' 1, 2, or 3 colours. The swap-colours for the pole are the same symmetry as the sides of the R-cube. A cycle is different too, but there are 2 ways to swap a piece with the pole (as there are 2 ways to give 1 of 3 faces of the R-cube a q-turn) which rotates the number on it. It takes 3 swaps, 1 of which can be a return to a previous position, to rotate a number by 180 deg. 3 'layer swaps' on the R-cube will rotate a corner by 180 deg so it has the orientation it had at the start of the swap-sequence, but may stiil be 'out of phase' with its home-position (I of 3 'home colours' that are visible). The home-position on the 4-colour R-sphere is the pole's position. It looks like a box. with a missing lid and a bottom 'surface' that appears when you move the coloured triangles' spatial orientations with each other - it has a translational d.o.f. the R-cube doesn't.
