Tiling problem

[jovial patio]

Consider the following polyomino of size 4:

x
xxx

Which rectangles can it tile (allowing translations, rotations and reflections) ? Well, we can prove that a necessary condition for a rectangle to be tilable by it is that its area must be divisible by 8. It can be proved using a coloring argument : https://www.cut-the-knot.org/blue/LtilingOfRectangle.shtml#solution

Now consider a similar one, of size 5:

x
xxxx

I made some tests with a program and it seems that if this one tiles a rectangle, then the area of the rectangle is divisible by 10.
Generally, consider the polyomino of size n:

x
xxx ... x

Then it seems that it's necessary for the area of the rectangle to be a multiple of 2n.
However, I haven't managed to find a suitable coloring to show it (like in the case n = 4). I tried with the same one along with some variations.
It may be useful to look for more advanced tools, and I came across the use of gaussian integers in the following article : https://www.sciencedirect.com/science/article/pii/S0304397508001527
Each case is represented by a complex number x + iy where x and y are integers and we try to find an invariant f.
The article suggests to take f(x + iy) = (x + iy)^m (mod v) where m is a positive integer and v is a gaussian integer such that the image of a polyomino by f is always 0 no matter where it is placed not its orientation. Yet I cannot find a suitable v and m for the case n = 5.

[daring jewel]

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[devout halo]

Can we try recreate the original example using the article's suggestion?

[devout halo]

I suppose m=1 and v=2,i

[jovial patio]

Do you want to try f(x, y) = (x + iy) (mod 2+i) ?
Then it is indeed a translation invariant since 2+i divides 5, but one must also check whether it's a rotation/reflection invariant (D-coloring) or not

Gaussian integers of the polyomino (noted τ) : 0, 1, 2, 3, i
So f(τ) = 0+1+2+3+i = 6+i (mod 2+i)

For this rotation :

x
x
x
xx

We have :
0, 1, i, 2i, 3i

and : f(τ) = 1 + 6i

Both 6 + i and 1 + 6i are congruent to 4 modulo 2+i, which is a good thing

But

xxxx
x

gives f(τ) = 0+1+2+3-i = 6-i = 3 (mod 2+i)

[jovial patio]

My conjecture was wrong actually...

[devout halo]

Well that explains

[jovial patio]

+close

[daring jewel]

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