Tiling $3k\times (3k-2)$ board with $L$- Trominoes

Question

Consider a $3k\times (3k-2)$ board. For which values of $k$ can we cover the board with $L$- trominoes?

For this, It was clear that for $k$ even, we are done, because then we can divide the board into $2\times 3$ rectangles. And we can cover the $2\times 3,$ as shown in the image.

I then tried for odd $k$, which wasn't working out for me. So I think it's not possible. Hence I am trying to prove that for odd $k$ it's not possible.

Consider the following colouring, which I did for $9\times 7,$ but can be done for any odd $3k\times 3k-2$ (Just make sure that the corners have same colour):

I was first trying to prove for $9\times 7$ board.

Prove that in a $9\times 7$ board, we can't cover it with $L-$ trominoes.

In this coloring, note that there are $5\times 4 + 4\times 3=32$ yellow cells and $63-32=31$ white cells. Since $L-$ trominoes are of $3$ cells area, we get that a total of $21$ $L-$ trominoes were used to cover the board.

Also note that $L-$ trominoes, have $2$ yellow and $1$ white cells or $1$ white and $2$ yellow cells. Let $k$ be the number of $L-$ trominoes which have $2$ yellows cells. So we get $2\cdot k+(21-k)\cdot 1=32.$

From here, we get $k=11$. So there are $11$ $L-$ trominoes which have $2 $ yellow cells and $10$ $L-$ trominoes which have $2$ white cells.

Now consider the $10$ white dominated $L-$ Trominoes. Labelling the rows, with $1,2\dot 7$ (as shown in image ), we get that these white dominated $L-$ trominoes, can be placed only in row $2,4,6$ rows only ( so that it's white dominated).

Hence by PHP, we that that there will exist one row with $4$ white dominated cells, centred at the yellow cells of the row. But note that there are only $4$ yellow cells.

This couldn't give me a contradiction, but this is the progress I had till now.

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This question was actually a lemma I wanted to prove for another question.

The actual question was,

Consider an $n\times n$ board with the four corners removed. For which values of $n$ can you cover the board with $L-$ trominoes.

And by some coloring pattern, I reduced to prove the starting question.

Any hints?

Answer 1

Not an answer, but a diagram showing a possible $9\times 7$ tiling.

Answer 2

Using the below theorem, your rectangle can be tiled for all $k\ge 2$.

Theorem (Chu-Johnsonbaugh) For any integers $m,n\ge 2$, an $m\times n$ rectangle can be tiled with L-trominoes, except when $mn$ is not a multiple of $3$, or if one of the dimensions is $3$ and the other is odd.

Proof: We need to show how to tile any rectangle of the form $(3k)\times \ell$, where $k\ge 1,\ell\ge 2$, and where either

• $\ell$ is even, or

• $\ell=3$, and $k$ is even, or

• $\ell \ge 5$, $\ell$ is odd, $k\ge 2$.

The first two cases case are easy to deal with, using copies of a $3\times 2$ rectangle. From now on, assume you want to tile a $(3k)\times \ell$ rectangle, with $k\ge 2$ and $\ell\ge 5$ is odd. To do this:

• Write the integer $k$ as a sum of copies of $2$ and $3$. For example, if $k=9$, you could write $k=2+2+2+3$. Use this representation of $k$ to divide the $(3k)$ side of the rectangle into sections of $6$ and $9$.

• Write the odd integer $\ell$ as a sum of copies of $2$ and $5$. Similarly, this allows you to break the $\ell$ side of the rectangle into sections of length $2$ and $5$.

• Use the markings from the last two steps to break the entire $(3k)\times \ell$ rectangle into several smaller rectangles, whose dimensions are either $6\times 2,6\times 5,9\times 2,$ or $9\times 5$. Each of these sub-rectangles can be tiled with L-trominoes. The $6\times 2$ and $9\times 2$ cases are easy, while the $6\times 5$ and $9\times 5$ tilings are illustrated below: