I want to tile a $12 \times 12$ grid using L-shaped triominoes. There must be no overlaps or missing spaces, and I know that it is possible to do so.
Now, I want to know about a new condition: each row and column intersects the same number of triominoes. Intuitively it doesn't seem to work, is there a rigorous proof? I have been unable to make one.
I've previously proven each row/column intersects 8 triominoes, and determined the bottom row must look something like this:
A A a A A a A A a A A a A a a A a a A a a A a a
No other way to fit. So I think that if each two rows have to be like that, we can only vary the 3x2 blocks, which still result in the same intersection. Trivially, every column 2 differs from column 1, even though rows are invariant
Here, I am wondering if there is any irregular way of arrangement that can overcome the issues. I have been unable to come up with any as of so far, is there any way to rigorously prove the non-existence?
It's not possible.
As you mentioned, we can consider an edge of the square. It needs to have $8$ trominoes, each of which will intersect it in either one or two cells, so there must be four trominoes taking up one cell on the bottom row and four taking up two cells. Call these Type A and Type B trominoes.
Consider one of the Type A trominoes:
There is only one way to fill the cell marked with
*:This means that we can asssociate each type A tromino with a unique type-B tromino by looking at the cell in its "elbow", and vice versa to locate the A tromino from the B tromino. Since there are four of each, they'll need to be paired up one-to-one.
But now consider the corner of the grid! We have two possibilities up to symmetry:
In the first case, both the western and southern edges demand contradictory placements of A-type trominoes:
And in the second case, there is no way to pair up the tromino with a type-A tromino on the southern edge in a way consistent with our requirements.