POLYOMINO TILINGS

Polyomino Tilings

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T tetromino

Area: 4.

Perimeter: 10.

Size: 2x3.

Is rectangular: no.

Is convex: yes.

Holes: 0.

Order: 4.

Square order: 4.

Odd order: ∞.

Prime rectangles: 1.

Smallest rectangle tilings

Smallest rectangle and smallest square (4x4):

No odd rectangles exist.

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 \ h1-345-789-1112N>0
1-30
4022P
5-7000
8066C08484C
9-1100000
1201818C011821182C07869678696C
13-15000000x
1605454C01664416644C052538225253822C4k
17-19000000x
200162162C0234390234390C0≥1≥1C4k
21-23000000x
240486486C033008523300852C0≥1≥1C4k
N>0x4kx4kx4k

Smallest prime reptiles

Smallest prime reptile (4Tx4):

Reptile tilings' solutions count (including symmetric)

polyomino \ n²
T tetromino10054P000≥277271568P

Smallest tori tilings

Smallest torus (2x4):

Smallest square torus (4x4):

Smallest odd torus (7x12):

Tori tilings' solutions count (including translations)

w \ h123456789
100
20000
3000000
400161600112112
50000000000
60000004244240000
700000000000000
80064640027042704160160584858481121129214492144
900000000000000128161281600
100000001549615496000000≥500000≥50000000
1100000000000000279842798400
120025625600937129371200109912109912672672≥500000≥5000003024030240

Smallest Baiocchi figures

Smallest Baiocchi figure (area 16):

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 $A(w; h)$

$N(n; m) = T(n; m) = 0, \qquad 8 \nmid nm \tag{1}$

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.

Attributions

  1. Solutions have been counted by Dmitry Grekov.
  2. Formulas have been found by Dmitry Grekov.
  3. Prime rectangles have been found by D.W. Walkup (Covering a Rectangle with T-tetrominoes, "American Mathematical Monthly 72" (1965) pp. 986-988.).

See Also

O tetrominoZ tetromino