Riding Polyominoes

From an idea of Paolo Licheri

Wrote by Gabriele Carelli

I apologise for my bed English, please let me know if you find some mistakes.

On  february 6 2004 Paolo Licheri wrote the following mail in the italian newsgroup it.hobby.enigmi.

[The traslation is of my own, any mistake are attributable only to myself]

I define (if nobody thought about them) *riding* polyominoes those in which we can touch all the cases passing from one to another one with the movement of the knight of the chess. By now I found a riding heptominoe:

and this riding octominoes (it also tiles the plane)

One first osservation: in the riding polyominoes we always move from one white case to one black, or viceversa: maybe there are some interesting parity problems.

Paolo

1,0,0,0,0,0,2,10,57,191,571,1546,....

Here are avaible the *.txt  files with all the solutions found by Livio ( 7-8-9-10, 11, 12)

Here are, as examples, the first not trivial riding polyominoes:

Paolo Licheri proposed an improvement challenge with the following mail:

[The traslation is of my own, any mistake are attributable only to myself]

The Y pentomynoes is not a riding one. But if we join two of them in an opportune way we obtain a riding decaminoe:

therefore we say that the Y pentominoes is riding of order 2.

If we join four dominoes we obtain a riding octominoes:
 

therefore we say that the dominoes is riding of order 4.

I bet that someone will like to amuse himself finding some others.

Paolo
 

I and Livio analyzed all the polyominoes until N=6, let's look our solutions:

Since do not exist riding hexominoes is not possible to have riding trominoes of order 2 and therefore we can be sure  that 3 is the minimal order of riding.

From the directory of the riding octominoes it is easy to see that the L tetrominoes is the only one riding with order 2 [ in 3 different ways ]. For parity reasons it is not difficult to prove that the T cannot have order 3 and therefore 4 is the smallest one.

With parity reasons it is easy to prove that the X cannot be riding of order 3 and therefore order 4 is the smallest one [ order 2 is easy analyziable by hand in exaustive way ].

Let's look the hexominoes: 29 has order 2, 3 has order 3 and only one has order 4 [but I'm not sure that it can be found better solution for the last one].

Other possible extension of the "game" is to join a pairs of different pentominoes to found a riding decominoes. Once again parity reasons assure us that the X pentominoes cannot be joined to an other one, for this reason the X is not present in the following table.

I would like to fill the table above, I'm sure that some pairs do not have solution but perhaps, of that still empty, someone it can be filled up. So do not hesitated to send me new solutions.

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Thi page is entrusted to the inexpert hands  of Gabriele Carelli

First edition  27/03/2004 , last edition: 15/04/2004

If you have comments, questions or corrections feel free to write me