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.
Area: 4.
Perimeter: 10.
Size: 2x3.
Is rectangular: no.
Is convex: yes.
Holes: 0.
Order: 2.
Square order: 4.
Prime rectangles: ≥ 2.
Some facts:
Smallest rectangle (2x4):
Smallest square (4x4):
No odd rectangles exist.
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.
Smallest prime reptile (4Lx2):
Smallest torus (2x4):
Smallest square torus (4x4):
Smallest Baiocchi figure (area 8):
Smallest Baiocchi figure without holes (area 16):
$N(w; h)$ - 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 $A(w; h)$
$N(n; m) = T(n; m) = 0, \qquad 8\nmid nm \tag{1}$
Assume L tetromino tiles $n\times m$ rectangles for $nm\not\equiv 0\pmod{8}$.
Place numbers in rectangles' cells according to function $F(x,y)\equiv4x+1\pmod{8}$, where $x$ and $y$ are cells' coordinates (zero-based). On the one hand, L tetromino, no matter how placed, covers sum congruent to $0\pmod{8}$. Then sum covered by all tetrominoes is also congruent to $0\pmod{8}$. On the other hand, rectangle covers sum congruent to $\sum_{x=0}^{n-1}\sum_{y=0}^{m-1}\left(4x+1\right)$, which is not congruent to $0\pmod{8}$ for $nm\not\equiv 0\pmod{8}$. Contradiction, as tetromino tiles this rectangle and thus sum covered by all tetrominoes should be equal to sum covered by rectangle. Thus only assumption we made is false - L tetromino doesn't tile $n\times m$ rectangles for $nm\not\equiv 0\pmod{8}$. Q.E.D.
$T(n; m) = 1, \qquad 8\mid nm \tag{2}$
