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: ≥ 12.

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.

w \ h

1-4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

N>0

1-4

0

5

0

0

6

0

7

0

0

0

8

0

9

0

0

0

0

10

0

11

0

0

0

0

0

12

0

13

0

0

0

0

0

0

14

0

15

0

0

0

0

0

0

0

16

0

17

0

0

0

0

0

0

0

0

18

0

19

0

0

0

0

0

0

0

0

0

20

0

21

0

0

0

0

0

0

0

0

0

?

N>0

x

2k

all

2k

all

2k

all

2k

all

2k

all

?

?

?

?

?

?

Smallest prime reptiles (2Ix5, 5Xx3):

Smallest common multiple (area 20):

$N(w; h)$ - 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 domino 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, domino 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 domino 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 - domino and X pentomino don't tile $(2n + 1)\times(2m + 1)$ rectangle. Q.E.D.