ENUMERATION OF POLYOMINOES INSCRIBED IN A RECTANGLE Alain Goupil, - - PDF document

1

ENUMERATION OF POLYOMINOES INSCRIBED IN A RECTANGLE

Alain Goupil, Hugo Cloutier, Fathallah Nouboud Université du Québec à Trois-Rivières

Polyomino inscribed Polyomino with Polyomino with in a rectangle minimal area minimal +1 area

• Problem. Enumerate polyominoes inscribed in a rectangle !

2

 Enumeration of polyominoes with minimal area

 area=b+k-1  they are disposed along a diagonal of the rectangle

First decomposition. minimal = hook × stair × hook Building blocs : hook(b,k)=1 (the corner cell is fixed) ⇒ Hook(x,y)= 1+ xbyk =1+ xy (1 x)(1 y)

b,k1

• stair(b,k) =

b +k 2 b - 1

• ⇒

Stair(x,y)= b +k 2 b - 1

• xbyk =

xy (1 x y)

b,k1

• Polyominoes on one diagonal :

⇒ Pmin,\(x,y)= 1+ xy (1 x)(1 y)

• 2

xy (1 x y)

3 Polyominoes on two diagonals: Crosses cross = hook × (hook – corner cell) except for polyominoes on one row or one column ⇒ Cross(x,y)= xy (1 x)2(1 y)2 xy2 (1 y)2 x2y (1 x)2 Inclusion-exclusion. Pmin(x,y)= Pmin,\(x,y)+Pmin,/(x,y)-cross(x,y) ⇒ Pmin(x,y)=

pmin(b,k)xbyk

b,k1

• 1+

xy (1 x )(1 y)

• 2

2xy (1 x y)

• xy

(1 x )2(1 y)2 xy2 (1 y)2 x2y (1 x )2

4 Exact formulas : Pmin(b,k)= 8 b +k 2 b - 1

• bk 2(b 1)(k 1) 6

if n = number of cells then Pmin(n) = number of polyominoes with n cells inscribed in any rectangle of perimeter 2n+2 =

pmin(b,n b +1)

b=1 n

• =

2n+2 1 2 n3 n2 +10n +4

• Pmin(z)=

pmin(n)zn

b,k1

• =

z2(1 4z +8z2 6z3 +4z4 ) (1 z )4(1 2z )

5 Second decomposition : minimal = hook × corner polyomino Corner polyominoes : inscribed polyominoes with min area and one cell in a given corner of the rectangle. pc(b,k)= 1 if b =1 or k =1 pc(b 1,k) + pc(b,k 1) +1 otherwise

• =

2 b +k 2 b - 1

• 1 for b,k 1

6

 Polyominoes with min+1 area

Benches : P is an inscribed polyomino of area min+1 ⇔ P contains exactly one bench To construct all min+1 polyominoes that contain a given bench B: 1- Fix the position of the bench B in a b × k rectangle R. 2- Complete the bench into a polyomino with area min+1 in two opposite regions; f1--f2 or f3--f4. 3- Use inclusion-exclusion and remove polyominoes that belong to both diagonals (i.e. hooks). Pmin+1(B) = f1 f2 + f3 f4

• 8t

To obtain all inscribed min+1 polyominoes : 4- sum over all benches B in the rectangle R.

7 Case 1. The bench B is in a corner. P1(t,b,k) = Corner polyomino + Hook, = 2 b +k t 2 b - 2

• + 2( t 1)

P2(t,b,k) = Corner polyomino + Hook, = 2 b +k t 2 b - 2

• + 2
• Proposition. The number g1(b,k) of polyominoes of area

min+1 inscribed in a bxk rectangle with a bench in any corner of the rectangle is

g1(b,k) = 4 p1(t,b,k) +4

t=3 k1

• + 4

p2(t,b,k) + 2k

t=3 k1

• + 4

p1(t,b,k) +4

t=3 b1

• + 4

p2(t,b,k) + 2b

t=3 b1

• = 16 b + k - 4

b - 1

• + b + k - 4

k - 1

• + 2k(2k 1) + 2b(2b 1) 72

8 Case 2. The bench is on one side of the rectangle and not in a corner.

• Proposition. The number g2(b,k) of polyominoes of area

min+1 inscribed in a b×k rectangle with a bench touching exactly one side of the rectangle is

g2(b,k) = 32 b + k - 4 b

• + b+ k - 4

k

• +

8 10 b+ k - 4 b - 2

• + b+ k - 4

b - 1

• + b+ k - 4

k - 1

• +

4 3 (b3 +k3 ) 28(b2 +k2 ) 48bk +164 3 (b +k) + 4(bk2 +b2k) +144

9 Case 3. The bench touches no side of the rectangle.

• Proposition. The number g3(b,k) of polyominoes of area

min+1 inscribed in a b×k rectangle with a bench touching no side of the rectangle is

g3(b,k) = 8 3 12

[

b + k - 4 b

• + b + k - 4

k

• +

6(b +k 6) b + k - 4 b - 1

• + b + k - 4

k - 1

• 60 b + k - 4

b - 2

• +18 b + k - 2

b - 1

• (b3 +k3 ) +15(b2 +k2 )

6(bk2 +b2k) 48bk 56(b +k) + 24

• Case 4. 2×2 benches.
• Proposition. The number

p22(b,k) of polyominoes of

area min+1 inscribed in a bxk rectangle with a bench touching no side of the rectangle is

4(b+ k - 4) if b = 2,k 3 or k = 2,b 3 8 b+ k - 4 b - 2

• + 2 b+ k - 4

b - 1

• + 2 b+ k - 4

k - 1

• 3
• if b = 3 or k = 3

8 b+ k - 4 b - 2

• +1
• (b +k 2) bk
• if b,k 4

10 All cases. Theorem. For b,k≥3, the number pmin+1(b,k)

• f

polyominoes of area min+1 inscribed in a b×k rectangle is pmin+1(b,k)= g1(b,k)+ g2(b,k)+ g3(b,k)+

p22(b,k) = 8(b +k 22) b +k 4 b - 2

• + 8(2k2 + 2kb +k 13k +13)

(k 2) b +k 4 b - 1

• + 8(2b2 + 2kb +b 13b +13)

(b 2) b +k 4 k - 1

• +48 b +k 2

b - 1

• 4

3 (b3 +k3 ) 12(b2k +bk2 ) +16(b2 +k2 ) +72bk 266 3 (b +k) +120

• Corollary. For integers n≥4, the number pmin+1(n) of

polyominoes of area n inscribed in any rectangle of perimeter 2n is given by

pmin+1(n) = pmin+1(b,n b)

b=2 n2

• = 2n 4

5 + 22 5 n

• 1

3 8n4 88n3 +430n2 902n +636

11

 Polyominoes with no loop (lattice trees)

and min+1 area.

• Corollary. The number

lmin+1(b,k) of lattice trees

inscribed in a b×k rectangle with area min+1 is

lmin+1(b,k) = fmin+1(b,k) f22(b,k)

• Corollary. For integers n ≥ 5, the number

lmin+1(n) of

lattice trees of area n inscribed in any rectangle of perimeter 2n is given by

lmin+1(n) = lmin+1(b,n b)

b=2 n2

• = 2n+1 n 1

( ) 2

3 4n4 46n3 + 227n2 473n +318

12

Next …

 More recurrences, exact formulae, generating

functions

 Minimal 3D polyominoes