Composable Art:Objects that can be

arranged in many ways

Marc van KreveldInstitute of Information and

Computing SciencesUtrecht University

Composable art objects

• Not ‘fixed’• Several pieces• Changeable• Not kinetic

Example: composable tapestry

• Rods hanging by hooks

• 54 pieces

Example: composable painting

• 6 panels in box• Box has 6 slots• Panels have holes

Example: composable painting

Facts for the painting

• 5 panels, each in some color, have holes– Square, circle, triangle, ellipse, and moon– Irregular of size and position

• 1 panel has no holes but 4 colors (background panel)

• Size 100 x 50 cm

Facts for the painting

• Box of size 155 x 55 x 8 cm with– 4 slots to keep panels– Interior is black (possible background color)– Opening in front of 45 x 45 cm

Needed for counting

• Definition of what to count (what is different)

• Example (painting):– Which panels are used– Which order– Which orientation– Every 10 cm sliding (sufficiently different)

Nice compositions have 2 or 3 panels with holes

Counting the painting

• Choose 1 out of 5 background colors• Choose 2 or 3 from 5 panels• For each panel choose

– 1 out of 4 orientations– 1 out of 6 sliding positions

57,60056422

5 22

4,147,2005643!

3

5 33

Town and Forest

• 12 pieces of hardboard• 1 shelf with 3 slots

Town and Forest

• Slots have lengths 160, 170, 180 cm• Pieces have lengths 15 - 38 cm• Two compositions are different if:

– any piece is in a different slot, or– the order in a slot is different

Count of Town and Forest

• If slots are long enough and there’s only one slot: N1(k) = k! compositions

• Two slots:

• Three slots:

1)!(kk!i)(kNi!i

k(k)N

k

0i1

k

0i2

k

0i232

21

23 1kkk!i)(kNi!i

k(k)N

Count of Town and Forest

• Also: k pieces in j slots consider (k+j-1)! permutations of k pieces and j-1 separators

first slot,this order

second slot,this order

separators

divide by (j-1)! permutations of the separators (k+j-1)!

(j-1)!

k=13, j=4

Count of Town and Forest

• If slots are not long enough:– Consider all 312 = 531,441 assignments to

sets S1, S2, and S3

– If a slot length, for i=1,2,3 then count

S1! S2! S3!

• There were 471,902 assignments o.k. and 19,497,542,400 compositions

a S i

Composable tapestry

• 3 x 3 rods of 125 cm (4 hooks);3 x 3 rods of 85 cm;3 x 3 rods of 45 cm

• 3 x 9 rods of 5 cm (1 hook)

• Aluminum, wood, black / square, round, diamond

Count of the composable tapestry

• Only rectangle hangings, four-wide, no wide holes

• Count color assignments and silhouettes separately

Counting color assignments

• Given any silhouette, there are

color assignments

( )279

( )189

( )93 ( )6

3[ ]3

Counting silhouettes

• We’ve got rows, columns, and rods

• What recurrence?

Counting silhouettes

• Any level consists of– One 125 cm rod– One 85 cm rod and

one 5 cm rod– Two 45 cm rods– One 45 cm rod and

two 5 cm rods– Four 5 cm rods

(1)

(2)

(3)

(1)

(1)

Counting silhouettes

• Suppose that A(j,k,m,n) is the no. of silhouettes for– j 125 cm rods,– k 85 cm rods,– m 45 cm rods,– n 5 cm rods

A(j,k,m,n) = A(j-1,k,m,n) + 2 A(j,k-1,m,n-1) +

A(j,k,m-2,n) + 3 A(j,k,m-1,n-2) + A(j,k,m,n-4)

Counting silhouettes

A(j,k,m,n) = A(j-1,k,m,n) + 2 A(j,k-1,m,n-1) +

A(j,k,m-2,n) + 3 A(j,k,m-1,n-2) + A(j,k,m,n-4)

A(0,0,0,0) = 1

A(., ., ., .) = 0 if any negative

B(j,k,m,n) = B(j-1,k,m,n) + B(j,k-1,m,n-1) +

B(j,k,m-2,n) + B(j,k,m-1,n-2) + B(j,k,m,n-4)

#recursive calls for A(9,9,9,27) is B(9,9,9,27)= 7 • 1014

Counting silhouettes

A[j,k,m,n] = A[j-1,k,m,n] + 2 A[j,k-1,m,n-1] +

A[j,k,m-2,n] + 3 A[j,k,m-1,n-2] + A[j,k,m,n-4]

A[0,0,0,0] = 1

A(., ., ., .) = 0 if any negative

Dynamic programming: 4-dimensional table with 28,000 entries to get A[9,9,9,27]

Counting silhouettes

• A(1,1,1,3) = 36• A(2,2,2,6) = 3960• A(3,3,3,9) = 604,800• A(4,4,4,12) = 1.1 • 108

• A(5,5,5,15) = 2.1 • 1010

• A(6,6,6,18) = 4.4 • 1012

• A(7,7,7,21) = 9.4 • 1014

• A(8,8,8,24) = 2.1 • 1017

• A(9,9,9,27) = 4.8 • 1019

Diago

• 20 rods; 4 x 5 lengths• Logging cabin principle• 9 protrusions

Counting Diago

• In how many way can we put 20 rods of lengths 2, 3, 4, 5, and 6 on a 10 by 10 grid?

Counting Diago

• Assume all pieces are horizontal• Let grid size be Z x Z• What recurrence?

Let A(i,j,k,m,n,r,c) be the # of ways to placei, j, k, m, and n rods of lengths 2, 3, 4, 5, and 6 starting at row r and column c on a grid[0..Z-1] x [0..Z-1]

Counting Diago

r

cZ-1(0,0)

A(i,j,k,m,n,r,c) = A(i,j-1,k,m,n,r,c+3)

or A(i,j-1,k,m,n,r+1,0)(or 1 or 0)

3-rod

Counting Diago

A(i,j,k,m,n,r,c) =

#ways if we place a 2-rod at (r,c)

#ways if we place a 3-rod at (r,c)

#ways if we place a 4-rod at (r,c)

#ways if we place a 5-rod at (r,c)

#ways if we place a 6-rod at (r,c)

#ways if we place no rod at (r,c)

if j > 0 and c Z-4, then A(i,j-1,k,m,n,r,c+3), etc.

Counting Diago

• Both horizontal and vertical:– Consider all 55 = 3125 ways to select

a horizontal and a vertical subset of the 4+4+4+4+4 rods

#comp = A(i,j,k,m,n,0,0) • A(4-i,4-j,4-k,4-m,4-n,0,0)0 i,j,k,m,n 4

Counts for Diago

• 6 x 6 grid: 0• 7 x 7 grid: 2.1 • 1016

• 8 x 8 grid: 6.5 • 1023

• 9 x 9 grid: 3.1 • 1028

• 10 x 10 grid: 9.9 • 1031

• 11 x 11 grid: 6.7 • 1034

• 12 x 12 grid: 1.6 • 1037

Counts for Symmetric Diago

• 6 x 6 grid: 0• 8 x 8 grid: 91,344• 10 x 10 grid: 7.3 • 106

• 12 x 12 grid: 1.3 • 108

A(1,1,1,1,1, Z/2, 0)

Z

Z/2

Symmetric Diago

Z

Z/2

Same for 2 lines symmetry

Refinements

• Some compositions are not connected• Some compositions may not be

supported by the protrusions

Cannot use recurrences to accommodate such constraints

Monte Carlo Counting

• Dynamic programming gives the sub-counts for A(4,4,4,4,4,0,0)

A[4,4,4,4,4,0,0] =

A[3,4,4,4,4,0,2]

A[4,3,4,4,4,0,3]

A[4,4,3,4,4,0,4]

A[4,4,4,3,4,0,5]

A[4,4,4,4,3,0,6]

A[4,4,4,4,4,0,1]

Monte Carlo Counting

• Choose one previous count uniformly, weighted. Then repeat

we can generate a composition uniformly at random in Z x Z steps

• Testing a (random) composition for connectedness or support is easy Monte Carlo simulation

this can also be used for the composable tapestry

More composable tapestries

to be counted

Random Generation from Dynamic Programming

• Used before for:– random string generation (McKenzie

’97)– random alignment of strings (Allison

’94)– random outerplanar graphs (Bodirski

and Kang ’03)– ……

A second Diago

GeoMecca

Composable Tower & Facade

Matisse’s leaves

Future research

• Automated analysis of nice compositions

• Automated generation of nice compositions

• Design new composable art objects