# POLYOMINO TILINGS

## Polyomino Tilings

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.

## L triomino¶

Area: 3.

Perimeter: 8.

Size: 2x2.

Is rectangular: no.

Is convex: yes.

Holes: 0.

Order: 2.

Square order: 12.

Odd order: 15.

Prime rectangles: 2.

## Smallest rectangle tilings¶

Smallest rectangle (2x3):

Smallest square (6x6):

Smallest odd rectangle (5x9):

## Rectangle tilings' solutions count (including symmetric)¶

Blue number (P) - strongly prime rectangle (which cannot be divided into two or more number of rectangles tileable by this set).

Green number (W) - weakly prime rectangle (which cannot be divided into two rectangles tileable by this set, but which can be divided into three or more rectangles).

Red number (C) - 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 reptiles¶

Smallest prime reptile (3Lx2):

## Reptile tilings' solutions count (including symmetric)¶

polyomino \ n²
L triomino11P4P409C108388P104574902C≥15397007360P

## Smallest tori tilings¶

Smallest torus (2x3):

Smallest square torus (3x3):

Smallest odd torus (3x3):

## Tori tilings' solutions count (including translations)¶

w \ h123456789
100
20000
30024241212
4000010810800
5000060600000
6001921925645642736273613620136206974469744
70000252252000035658035658000
80000310831080000≥500000≥5000000000
90015361536102010207776077760≥500000≥500000≥500000≥500000≥500000≥500000≥500000≥500000≥500000≥500000

## Smallest Baiocchi figures¶

Smallest Baiocchi figure (area 12):

## Formulas¶

$N(w; h)$ - 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 $A(w; h)$

$N(1; n) = T(1; n) = 0, \qquad n \geq 1 \tag{1}$

"L" triomino has width and height of $2$ thus it does not fit into $1 \times n$ rectangle. Q.E.D.

$N(3n + 1; 3m + 1) = T(3n + 1; 3m + 1) = 0, \qquad n,m \geq 0 \tag{2}$

$N(3n + 1; 3m + 2) = T(3n + 1; 3m + 2) = 0, \qquad n,m \geq 0 \tag{3}$

$N(3n + 2; 3m + 2) = T(3n + 2; 3m + 2) = 0, \qquad n,m \geq 0 \tag{4}$

Area of rectangle should be divisible by triomino area. Since $(3n + 1)(3m + 1)\equiv 1\pmod{3}$, $(3n + 1)(3m + 2)\equiv 2\pmod{3}$, $(3n + 2)(3m + 2)\equiv 1\pmod{3}$, neither of those rectangles can be tiled by "L" triomino. Q.E.D.

$N(2; n) = 2 \times N(2; n - 3), \qquad n \geq 4 \tag{5}$

If triomino tiles $2 \times n$ rectangle then all rectangle cells should filled. Top left corner (marked with dot on image) may be covered in only three ways. Only two of them are valid, because third leaves untileable hole (marked with cross on image). When placing next triomino in remaining two tilings, a $2 \times 3$ rectangle is covered. Remaining part is a $2 \times (n - 3)$ rectangle. So whole rectangle can be tiled in $2 \times N(2; n - 3)$ ways. Q.E.D.

$N(3; n) = 2 \times N(3; n - 2), \qquad n \geq 3 \tag{6}$

$N(3; 2n + 1) = T(3; 2n + 1) = 0, \qquad n \geq 0 \tag{7}$

First, prove $(6)$. If triomino tiles $3 \times n$ rectangle then all rectangle cells should filled. Top left corner (marked with dot on image) may be covered in only three ways. Only two of them are valid, because third leaves untileable hole (marked with cross on image). When filling other corner on remaining two tilings, a $2 \times 3$ rectangle is covered. Remaining part is a $3 \times (n - 2)$ rectangle. So whole rectangle can be tiled in $2 \times N(3; n - 2)$ ways. Q.E.D.

Second, prove $(7)$. Assume that triomino tiles $3 \times (2n + 1)$ rectangle. Applying formula $(6)$ $n$ times, $N(3; 2n + 1) = N(3; 2n + 1 - 2 \times n) = N(3; 1) = N(1; 3) = 0$, with final step by formula $(1)$. Contradiction, therefore triomino does not tile $3 \times (2n + 1)$ rectangle. Q.E.D.

$N(4; n) = 10 \times N(4; n - 3) - 22 \times N(4; n - 6) - 4 \times N(4; n - 9), \qquad n \geq 10 \tag{8}$

$G(N(4); x) = \frac{1 - 6x^3}{1 - 10x^3 + 22x^6 + 4x^9} \tag{9}$

$G(N(5); x) = \frac{1 - 2x^3 - 31x^6 - 40x^9 - 20x^{12}}{1 - 2x^3 - 103x^6 - 280x^9 - 380x^{12}} \tag{10}$

$G(N(6); x) = \frac{1 - 2x - 4x^2 - 2x^3 + 13x^4 + 6x^5 - 6x^6 - 6x^7}{1 - 2x - 8x^2 - 2x^3 + 43x^4 + 42x^5 - 36x^6 - 102x^7 + 44x^9 + 8x^{10} + 8x^{11}} \tag{10}$

$G(T; x; y) = \frac{x^2 y^2 \left(x^2 y+x y^2+x y+x+y\right)}{\left(1-x^3\right) \left(1-y^3\right)}-\frac{x^3 y^3 \left(1-x^2 y^2\right)}{\left(1-x^2\right) \left(1-y^2\right)} = \\ = \frac{x^9 y^9+x^9 y^7+x^9 y^5+x^7 y^9+x^7 y^6+x^6 y^7+x^6 y^6+x^6 y^5+x^6 y^4+x^6 y^3+x^6 y^2+x^5 y^9+x^5 y^6+x^4 y^6+x^4 y^3+x^3 y^6+x^3 y^4+x^3 y^2+x^2 y^6+x^2 y^3}{\left(1-x^6\right) \left(1-y^6\right)} \tag{11}$

1. Prime rectangles first have been found by Michael Reid (http://www.cflmath.com/~reid/Polyomino/l3_rect.html).
2. Smallest rectangle first have been found by Michael Reid.
3. Smallest odd rectangle first have been found by David A. Klarner, Packing a Rectangle with Congruent N-ominoes, "Journal of Combinatorial Theory 7" (1969) pp. 107-115. (http://www.sciencedirect.com/science/article/pii/S002198006980044X)