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×h rectangle (including symmetric solutions)
T(w;h)={1,N(w;h)≥10,else - tileability function, 1 if tiles rectangle, 0 otherwise
A(w;h)=(N(w;h))1wh - average number of ways to tile cell in w×h rectangle (including symmetric solutions)
G(T;x;y)=∑∞w=1∑∞h=1T(w;h)xwyh - bivariate generating function of T(w;h)
G(A;x;y)=∑∞w=1∑∞h=1A(w;h)xwyh - bivariate generating function of A(w;h)
N(2n+1;2m+1)=T(2n+1;2m+1)=0
Assume L tetromino and X pentomino tile (2n+1)×(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.