Select polyominoes for a set (currently 1 or 2), for which tilings should be shown.
Then click "Show" button.
You may also see list of all polyomino sets for which data is available here.
Prime rectangles: ≥ 23.
Smallest rectangle (5x6):
Smallest square (6x6):
Blue number - strongly prime rectangle (which cannot be divided into two or more number of rectangles tileable by this set).
Green number - weakly prime rectangle (which cannot be divided into two rectangles tileable by this set, but which can be divided into three or more rectangles).
Purple number - prime rectangle (unknown if weakly or strongly prime).
Red number - composite rectangle (which can be divided into two rectangles tileable by this set).
Gray number - it is unknown whether rectangle is prime or composite.
Question mark (?) - solution count is unknown.
Click on underlined numbers to view picture with one solution.
$N(w; h)$ - number of ways to tile $w\times h$ rectangle (including symmetric solutions)
$T(w; h) = \begin{cases} 1, & N(w; h) \geq 1 \\ 0, & \text{else} \end{cases}$ - tileability function, $1$ if tiles rectangle, $0$ otherwise
$A(w; h) = \left(N(w; h)\right)^{\frac{1}{wh}}$ - average number of ways to tile cell in $w\times h$ rectangle (including symmetric solutions)
$G(T; x; y) = \sum_{w=1}^{\infty}\sum _{h=1}^{\infty}T(w; h)x^wy^h$ - bivariate generating function of $T(w; h)$
$G(A; x; y) = \sum_{w=1}^{\infty}\sum _{h=1}^{\infty}A(w; h)x^wy^h$ - bivariate generating function of $A(w; h)$
$N(2n + 1; 2m + 1) = T(2n + 1; 2m + 1) = 0 \tag{1}$
Assume L tetromino and X pentomino tile $(2n + 1)\times(2m + 1)$ rectangle. Place numbers in rectangle's cells according to function $F(x,y)\equiv(-1)^{x+y}\pmod{3}$, where $x$ and $y$ are cells' coordinates (zero-based). On the one hand, L tetromino and X pentomino, no matter how placed, cover sum congruent to $0\pmod{3}$. Then sum covered by all polyominoes is also congruent to $0\pmod{3}$. On the other hand, rectangle covers sum congruent to $\sum_{x=0}^{(2n+1)-1}\sum_{y=0}^{(2m+1)-1}(-1)^{x+y}\equiv1\pmod{3}$. Contradiction, as L tetromino and X pentomino tile this rectangle and thus sum covered by all polyominoes should be equal to sum covered by rectangle. Thus only assumption we made is false - L tetromino and X pentomino don't tile $(2n + 1)\times(2m + 1)$ rectangle. Q.E.D.
