# Slider Puzzles

The goal in this puzzle is to order the numbers sequentially in their natural order starting with 0 in the left upper corner and proceeding right and down. One should do this by sliding either whole columns or whole rows. But there is a caveat. In the original position rows and columns rotate not just slide. This means, for example, that the first element becomes the last when a row is rotated left. This is very much akin to the Rubik's Cube but played on the flat surface of a torus. There are two additional surfaces on which the game can be played - Klein bottle and Projective plane. The difference is in how points on the sides of the square are identified with each other. Two small squares on the left and right from the game board indicate the selected surface.

What if applet does not run? |

One reason I decided to implement this puzzle is that upon logging into my account I discovered e-mail from Microsoft with a free trial offer of their VisualJ++. The file is about 13M. So downloading it took a while which I fruitfully used to write this applet.

Edward de Bono once wrote a book (Six Action Shoes) on a flight from London, England, to Auckland, New Zealand on a Psion MC400 mobile computer. To create this puzzle, I, acting on his advice and donning pragmatic Brown Brogues, used a Pentium 100 mongrel and reusable components to accomplish my job more effectively.

#### NOTE:

Superficial similarity notwithstanding, this puzzle is much different from the Sam Loyd's Fifteen. For example, swapping the counters 14 and 15 does not result in an unsolvable arrangement. Furthermore, I am going to prove the following

### Proposition

On a torus, every 4x4 configuration is solvable.

The proof is based on the assertion that for any two adjacent counters there exists a sequence of moves that leaves all counters but the given two untouched. The given two counters, as the result of this sequence of moves, swap their locations.

The Graph Theory sets this puzzle in a general framework from which we can glean, if not complete answers, at least some information on solvability of this puzzle. An essential instrument here is the notion of transposition. The basic moves in the Sliders puzzle are cycles. According to a known identity one basic Sliders move in the nxn puzzle is an even permutation iff n is odd. Assume, n is odd. Then Sliders moves will not change the parity of the starting configuration. Applying a few results of the Group Theory to the table of moves it's possible to prove that puz(Sliders) consists of just two connected components. On the other hand, if n is even, then with every move the parity of the labelings changes. Again based on the table of moves it's possible to show that puz(Sliders) is connected.

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