# 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 pentomino¶

Area: 5.

Perimeter: 12.

Size: 2x4.

Is rectangular: no.

Is convex: yes.

Holes: 0.

Order: 2.

Square order: 20.

Odd order: 21.

Prime rectangles: 2.

## Smallest rectangle tilings¶

Smallest rectangle (2x5):

Smallest square (10x10):

Smallest odd rectangle (7x15):

## 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.

w \ h12345678910N>0
10
200
3000
40000
5022P044C0
6000088C0
70000000
800001616C000
9000000000
10044C01616C3232C7676C384384C386386C29762976C39523952C
110000000002052020520C5k
1200006464C0000≥1≥1C5k
13000000000≥1≥1C5k
140000128128C0000≥1≥1C5k
15088C06464C0776776C320320P≥1≥1C≥1≥1C≥1≥1Call
160000256256C0000≥1≥1C5k
17000000000≥1≥1C5k
180000512512C0000≥1≥1C5k
19000000000≥1≥1C5k
2001616C0256256C10241024C80408040C213728213728C≥1≥1C≥1≥1C≥1≥1Call
N>0x5kx5k2k5k5k5k5kall

## Smallest prime reptiles¶

Smallest prime reptile (5Lx4):

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

polyomino \ n²
L pentomino10048P1536P1080328P

## Smallest tori tilings¶

Smallest torus (2x5):

Smallest square torus (5x5):

Smallest odd torus (5x5):

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

w \ h12345678910
100
20000
3000000
400000000
5004040002602604040
6000000002500250000
7000000001401400000
8000000002150021500000000
90000000072072000000000
10003203200027602760215900215900114440114440243880243880≥500000≥500000≥500000≥500000≥500000≥500000

## Smallest Baiocchi figures¶

Smallest Baiocchi figure (area 20):

Smallest known Baiocchi figure without holes (area 40):

## 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 pentomino has minimal dimension of $2$ thus it does not fit into $1 \times n$ rectangle. Q.E.D.

$N(2; n) = 2 \times N(2; n - 5), \qquad n \geq 6 \tag{2}$

If L pentomino 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 cells (marked with crosses on image). When placing next pentomino in remaining two tilings, a $2 \times 5$ rectangle is covered. Remaining part is a $2 \times (n - 5)$ rectangle. So whole rectangle can be tiled in $2 \times N(2; n - 5)$ ways. Q.E.D.

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

If L pentomino 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 leave untileable hole (marked with cross on image). However, placing pentomino in first two ways is possible only in one way and then it also leaves untileable cells. Q.E.D.

$N(4; n) = 4 \times N(4; n - 5), \qquad n \geq 6 \tag{4}$

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

$N(5; 2n + 1) = T(5; 2n + 1) = 0, \qquad n \geq 0 \tag{6}$