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: 4.
Square order: 4.
Prime rectangles: ≥ 1.
Smallest rectangle and 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 (4Tx4):
Smallest torus (2x4):
Smallest square torus (4x4):
Smallest known odd torus (7x12):
Smallest Baiocchi figure (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 T tetromino tiles $n\times m$ rectangles for $8 \nmid nm$.
Place numbers in rectangles' cells according to function $F(x,y)\equiv4x+4y+1\pmod{8}$, where $x$ and $y$ are cells' coordinates (zero-based). On the one hand, T 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 $\sum_{x=0}^{n-1}\sum_{y=0}^{m-1}F(x,y)$, which is not congruent to $0\pmod{8}$ for $8 \nmid nm$. 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 - T tetromino doesn't tile $n\times m$ rectangles for $8 \nmid nm$. Q.E.D.
$N(4n; 4m + 2) = T(4n; 4m + 2) = 0 \tag{2}$
$N(8n; 2m + 1) = T(8n; 2m + 1) = 0 \tag{3}$
Assume T tetromino tiles $8n \times (2m + 1)$ rectangle. Then, by joining together two rectangles along $8n$ side, one gets $8n \times (4m + 2)$ rectangle. Contradiction, as tetromino does not tile such rectangle according to formula $(2)$. Thus only assumption we made is false - T tetromino doesn't tile $8n \times (2m + 1)$ rectangle.
$T(4n; 4m) = 1 \tag{4}$
$G(T; x; y) = \frac{x^4y^4}{\left(1 - x^4\right)\left(1 - y^4\right)} \tag{5}$
By definition of generating function:
$$G(T; x; y) = \sum_{w=1}^{\infty}\sum_{h=1}^{\infty}T(w; h)x^w y^h=$$
[Splitting series by $w$ into four parts]
$$ = \sum_{w=1}^{\infty}\sum_{h=1}^{\infty}T(4w; h)x^{4w} y^h + \sum_{w=1}^{\infty}\sum_{h=1}^{\infty}T(4w + 1; h)x^{4w+1} y^h + \sum_{w=1}^{\infty}\sum_{h=1}^{\infty}T(4w + 2; h)x^{4w+2} y^h + \sum_{w=1}^{\infty}\sum_{h=1}^{\infty}T(4w + 3; h)x^{4w+3} y^h = $$
[Using formulas $(1)-(4)$]
$$ = \sum_{w=1}^{\infty}\sum_{h=1}^{\infty}T(4w; h)x^{4w} y^h +0+0+0 = \sum_{w=1}^{\infty}x^{4w}\sum_{h=1}^{\infty}T(4w; h)y^h = $$
[Splitting series by $h$ into four parts]
$$ = \sum_{w=1}^{\infty}x^{4w}\left(\sum_{h=1}^{\infty}T(4w; 4h)y^{4h} + \sum_{h=1}^{\infty}T(4w; 4h + 1)y^{4h + 1} + \sum_{h=1}^{\infty}T(4w; 4h + 2)y^{4h + 2} + \sum_{h=1}^{\infty}T(4w; 4h + 3)y^{4h + 3}\right) = $$
[Using formulas $(1)-(4)$]
$$ = \sum_{w=1}^{\infty}x^{4w}\left(\sum_{h=1}^{\infty}1\times y^{4h}+0+0+0\right) = $$
[Using infinite geometric series sum formula]
$$ = \sum_{w=1}^{\infty}x^{4w}\times\frac{y^4}{1 - y^4} = \frac{x^4}{1 - x^4}\times\frac{y^4}{1 - y^4} = \frac{x^4y^4}{\left(1 - x^4\right)\left(1 - y^4\right)}$$
Q.E.D.
