# Tiling rectangle with trominoes - an invariant

There are two types of trominoes, straight shapes and L-shaped. Suppose a rectangle $$R$$ admits at least one tiling using trominoes, with an even number of L-trominoes.

EDIT: we do not admit ALL orientations, but only those which are constructed by starting in a box, and next box is placed to the left, or below previous box. Hence, only 2 out of the 4 possible orientations of the L-tromino is allowed. Hence, these are the four trominoes that we are allowed to use:

$$\begin{matrix} \blacksquare & \blacksquare \\ \blacksquare & \\ \end{matrix} ,\qquad \begin{matrix} \; & \blacksquare \\ \blacksquare & \blacksquare \\ \end{matrix} ,\qquad \begin{matrix} \blacksquare \\ \blacksquare \\ \blacksquare \end{matrix} ,\qquad \begin{matrix} \blacksquare & \blacksquare & \blacksquare \end{matrix}$$

Prove that every tiling of $$R$$ must use an even number of L-trominoes.

This smells a lot like a classical tiling problem, but unlike the classical cases (dominoes and missing corners), the situation is not to show that a tiling is impossible... Some nice invariant is perhaps what I am looking for.

I actually know how to prove the above statement, but it requires a lot more machinery than I would like. Moreover, I am interested in a more general question, (regarding representation theory and cylindrical Schur functions, and the proof I know does not generalize to this situation), but I hope that a good proof of the above problem generalizes.

For the interested: The general setting supposes that the rectangle is a torus, and that we use $$k$$-ribbons, where $$k$$ is odd. We then want to show that the number of $$k$$-ribbons which occupies an even number of rows, occur an even number of times. Hence, a proof not relying on the fact that $$R$$ has boundaries, or heavily uses the fact that each shape has three squares, is of extra interest.

• hm really? matematika-shkolnikam.ru/43.jpg – Fedor Petrov Jan 22 at 10:18
• Also, since $3$ is prime, the sizes are $m\times n$ where one is divisible by $3$, so you always have a tiling with no L's which is even. – Ville Salo Jan 22 at 10:34
• I guess the tag should be tiling, not tilting. – Dag Oskar Madsen Jan 22 at 10:38
• A more general theory is due to Igor Pak, ams.org/journals/tran/2000-352-12/S0002-9947-00-02666-0/…. – Richard Stanley Jan 22 at 15:06
• Have you looked at diagonals skewed to your pieces? (Two of the pieces occupy 2 squares of a certain diagonal direction, while the others occupy only one;alternatively,two pieces lie within two diagonals, whereas the others cross three diagonals.) Gerhard "Hoping For A New Slant" Paseman, 2020.01.22. – Gerhard Paseman Jan 22 at 20:54