============================================================================== Citation: Scientific American, Oct 1987 v257 p170(4) ------------------------------------------------------------------------------ Title: Now there is Rubik's Magic, a new puzzle that provides a study in permutation operators. (the Amateur Scientist) Authors: Walker, Jearl ------------------------------------------------------------------------------ Subjects: Rubik's Magic_analysis Mathematical recreations_analysis Puzzles_analysis Reference #: A5218307 ============================================================================== Full Text COPYRIGHT Scientific American Inc. 1987 Now there is Rubik's Magic, a new puzzle that provides a study in permutation operators Ern o Rubik, famous as the inventor of Rubik's Cube, recently produced a new and equally enchanting puzzle called Rubik's Magic. It consists of eight plastic squares aligned in a flat two-by-four array. The squares are connected by a nylon thread that runs along diagonal grooves in the plastic. The squares can be rotated about their edges. One side of the puzzle displays three separated rings. On the back side the rings are in disarray. The challenge is to realign the squares and thereby form the rings on the back side. If you succeed, the rings interlock. The puzzle is then an incomplete three-by-three array missing one corner square. One can go about solving the puzzle by random manipulation of the squares, flipping them in various ways as allowed by the thread hinges. A systematic study offers a better approach and displays several subtle mathematical properties of the puzzle. Recently Wolfgang Glebe presented such a study in Spektrum der Wissenschaft, the German edition of Scientific American. Here I repeat his analysis and give a solution to the puzzle that was found by one of his readers. It may be the shortest solution in the sense that it requires the fewest manipulations of the squares. Turn the disarrayed side face up, with the long edge horizontal and the copyright symbol in the upper row. Attach a small lable to each square. Beginning with the upper left square and moving to the right, number the labels. This arrangement of the squares and numbers is said to be state 0, symbolized as Z0. In the state the squares are in a certain location with respect to one another and the numbers are in a certain orientation. How many other unique states can the puzzle have when it is in a two-by-four array? You can change the locations of the squares while also reorienting the number on each square. Arrangements that amount to a simple rotation of the entire puzzle should not be considered unique. Mirror-image arrangements should also be excluded. Nevertheless, a calculation suggests there should be about 1.3 billion unique states. Glebe found that the calculation is misleading: there are only 32 unique states. The reduction in number results from the way the squares are attached by the thread hinges. A square is always attached to the same two neighboring ones, although their relative orientations can change. For example, square 1 is always attached to squares 2 and 5 and square 2 is always attached to squares 1 and 3. The states can be categorized as either symmetrical or asymmetrical. In a symmetrical state, of which there are 16, the numbers 1 through 4 are lined up in a horizontal row or are grouped in a square on the(ìdft or the right side of the two-by-four array. Any other arrangement is said to be an asymmetrical state. Begin with the puzzle in Z0. How can you manipulate it into the other 31 possible states? One transformation entails what Glebe calls operation R. Fold the puzzle away from you and around the horizontal center line. Make the upper row end up as the top layer in the resulting double layer. Now "roll' the double layer by moving the top layer to the right and the bottom layer to the left, each by one square. Next unfold the puzzle by rotating the bottom layer about the back edge of the top layer. Notice that it unfolds in only one direction. Operation R can be employed with any state of the puzzle. The success of the roll depends, however, on which way the puzzle is initially folded. Sometimes you must fold it toward you for the layers to roll. The direction in which you finally unfold the puzzle also varies according to the initial state. The direction of the roll can also be varied by moving the top layer to the left and the bottom layer to the right. Operation R transforms the puzzle from the symmetrical state Z0 to an asymmetrical state with 6, 7, 8 and 4 (the 4 is inverted) on the upper row and 5 (inverted), 1, 2 and 3 on the lower row. Note that the companion squares are all still attached. From this state perform operation R again. The puzzle returns to Z0. The rearrangement of the squares through an operation such as R is called a permutation. When an operation done twice returns the puzzle to its former state, it is said to be involutory. Symbolized, RR is equal to i, where i represents the identical state. The next task is to find ways of transforming Z0 into the other 15 symmetrical states. Glebe designates one method as operation D. Here again the first step is to fold the rows around the horizontal center line, either backward, which is appropriate when the initial state is Z0, or forward. The success of the second step depends on the choice of folding direction. If the puzzle cannot be made to take the second step, return to the initial arrangement and do the folding the opposite way. In the second step you expand the puzzle into a loop and then push in the left and right sides, allowing the top and bottom pieces to collapse onto them. The next step requires the stack to be opened. Depending on the state of the puzzle when you begin the operation, you may need to turn the stack over at this point. You cannot go wrong. If the stack cannot be opened, turn it over and try again. Finally, open the puzzle fully by folding the side flaps outward. Operation D transforms Z0 to Z5, another symmetrical state in which the rows are interchanged with no reorientation of the numbers on the squares. This operation is also involutory. If you do operation D on Z5 to return to Z0, you will find that the folding and unfolding usually go in directions opposite to those you employed previously. The table in the top illustration on the next page indicates how operation D alters the other symmetrical states. For example, it transforms Z1 into Z3 and vice versa. A concise way of writing the transformation is D(Z1) = Z3. Operation E is similar to D. You again fold the rows backward onto themselves and then pull the layers apart to form a loop. This time you make the top and bottom of the loop concave and then push the left and right sides inward to collapse on the top and bottom pieces. The next step requires that the group of squares be unfolded. For some initial states the step is immediately successful. For others you must turn the puzzle about a vertical axis for the step to work. You cannot go wrong here because the puzzle cannot be made to unfold erroneously. Next reach behind the plane and unfold the top and bottom flaps. The final product is Z3, another symmetrical state. Thus E(Z0) is equal to Z3. If you transform Z3 back to Z0 with operation E, the first fold is opposite to the one shown in the second illustration from the top on the next page. The transformations of the other symmetrical states resulting from operation E are listed at the bottom of the illustration. Operation F is quite different from D and E [see third illustration from top on opposite page]. Starting from Z0, fold the rows forward. When you try to pull the double layer into a loop, the far right and left sides resist. Pull the right side over the left side to form a stack. Then flip the top right square over the top left square while also flipping the bottom left square under the bottom right square. Turn the stack so that you can unfold the top and bottom layers. If the pieces resist, turn the stack and try again. The top and bottom layers can be unfolded in only one direction. Insert fingers into the near and far sides to separate the two layers of squares. The puzzle may resist to some extent until the threads adjust themselves. Guide the separation with your fingers. When the layers have been separated and the puzzle is tubular, push in the left and right sides of the tube to create two vertical layers. Unfold the layers by opening the near or the far end as is appropriate. Again the strings may resist slightly; you may need to guide the separation with your fingers. F transforms Z0 into Z2, another symmetrical state. The transformation is involutory. If you transform Z2 back to Z0, the directions of folding and unfolding are the same as before. So far the operations have produced Z2, Z3 and Z5, each of which can be turned into asymmetrical states by R. You can produce the rest of the symmetrical states from Z0 by employing two or more of the operations in succession. Try the permutation combination DE. (The notation implies that E is done first.) The result is Z1. Since the permutation ED produces the same state, D and E are said to be commutative, that is, their order is unimportant. F is a more powerful operation than D and E because it is not commutative. For example, DF(Z0) yields Z8, whereas FD(Z0) yields Z9. The table on this page indicates how all the symmetrical states can be obtained with various combinations of D, E and F operating on Z0. Each product state can be turned into a corresponding asymmetrical state by R. Suppose a series of operations is performed on one of the symmetrical states other than Z0. Is the product a new state? It is not, as can be seen in one of Glebe's examples. Consider DFEF(Z14). The first operation, F, produces Z11, as determined from the table in the illustration of F: DFEF(Z14) is equal to DFE(Z11). Continue the procedure: DFE(Z11) = DF(Z7) = D(Z3) = Z1, which is a known state. Since no new states can be produced with any combination of operations D, E and F, the permutations listed in the table are said to be closed. Glebe points out that in some cases a series of operations can be abbreviated. The permutation DE offers an example. Start with Z0. Fold the rows back onto each other. Then pull at the center on the top and bottom sides of the double layer to separate the layers into a loop. Continue pulling until the left and right sides touch, forming a new double layer. When you unfold the double layer, you have Z1. All this work is a warm-up for solving Rubik's Magic. The trick is to find a means whereby one of the 32 two-by-four states can be changed into a three-by-three state with one corner square missing. Glebe discovered two ways, one of which requires 20 operations. A reader of his article found a simpler method, which is shown in part in the bottom illustration on the opposite page. Begin with operation R on Z0 and then follow the instructions. The interlocked three rings are finally revealed when the flap on the right is folded to the right. Many other techniques of ferreting out a solution to the puzzle can be found in the book by James G. Nourse listed in "Bibliography' [page 183]. The book also contains procedures for forming the puzzle into a variety of shapes. Photo: Operation R changes the symmetry of a state Photo: The symmetrical states other than Z0 Photo: Rubik's Magic in its initial (top) and solved states Photo: Operation D interchanges rows Photo: Operation E Photo: Operation F transforms symmetrical states into other symmetrical states Photo: A solution to Rubik's Magic Photo: The permutations ==============================================================================