Introduction
Piet Mondriaan (later Mondrian) (1872-1944) was a Dutch painter, famous for his abstract works [1].
A tiling problem is related to this [2]:
The problem can be stated as follows:
- Given is a square \((n \times n)\) area,
- Partition this square into multiple, integer-sided, non-congruent rectangles,
- Minimize the difference between the largest and the smallest rectangle.
Non-congruent means: only one \((p \times q)\) or \((q\times p)\) rectangle can be used.
Example
Consider a small problem with \(n=5\). The possible sides for each rectangle are \(1,\dots,n\). From this, we can form a set of possible rectangles (I automated this step):
We see there are 14 different rectangles we can use. Note that the rectangle with size \((5 \times 5)\) is excluded. It would completely fill the area, while we need to place multiple rectangles. Also note: we only have \((p \times q)\) but not \((q \times p)\). We cannot use both such rectangles as they are congruent. We can rotate a rectangle when placing it.
An optimal solution for this problem is:
![]()
These are rectangles \(k5, k6, k12\) with \(k5\) being rotated. Obviously, this solution is not unique.
To model this, I use a binary variable: \[ x_{k,r,i,j} = \begin{cases} 1 & \text{if rectangle $k$ (rotated if $r=\mathrm{y}$) is placed at location $(i,j)$} \\ 0 & \text{otherwise}\end{cases}\] Here \(r = \{\mathrm{n},\mathrm{y}\}\) indicates if a rectangle is rotated. Note that not all rectangles can be rotated (it does not make sense to rotate a square, or there may not be space left to rotate for a given position).
For convenience, I also introduce \[y_{k} = \begin{cases} 1 & \text{if rectangle $k$ is used} \\ 0 & \text{otherwise}\end{cases}\] This is essentially an aggration of \(x\).
In addition, I use a few more complex data structures: \[\mathit{ok}_{k,r,i,j} = \begin{cases} \mathit{Yes} & \text{if we can place rectangle $k$ at location $(i,j)$} \\ \mathit{No} & \text{otherwise}\end{cases}\] Again, \(r\) indicates if rectangle \(k\) is rotated. I implemented this as a GAMS set. You can consider it as a Boolean data structure. Furthermore: \[\mathit{cover}_{k,r,i,j,i',j'} = \begin{cases} \mathit{Yes} & \text{if cell $(i',j')$ is covered by placing rectangle $k$ at location $(i,j)$} \\ \mathit{No} & \text{otherwise}\end{cases}\]
These sets are large. The set ok looks like:
The set cover is even larger:
The reason to develop these sets is to make the equations simpler. In addition, we can develop, test and debug the calculation of these sets in isolation of the test of the model and before the equations are written.
With this, we can formulate:
When solving this model, we see:
---- 65 SET s sizes of rectangles (height or width)
s1, s2, s3, s4, s5
---- 65 SET k rectangle id
k1 , k2 , k3 , k4 , k5 , k6 , k7 , k8 , k9 , k10, k11, k12, k13, k14
---- 65 SET rec rectangles
s1 s2 s3 s4
k1 .s1 YES
k2 .s2 YES
k3 .s2 YES
k4 .s3 YES
k5 .s3 YES
k6 .s3 YES
k7 .s4 YES
k8 .s4 YES
k9 .s4 YES
k10.s4 YES
k11.s5 YES
k12.s5 YES
k13.s5 YES
k14.s5 YES
We see there are 14 different rectangles we can use. Note that the rectangle with size \((5 \times 5)\) is excluded. It would completely fill the area, while we need to place multiple rectangles. Also note: we only have \((p \times q)\) but not \((q \times p)\). We cannot use both such rectangles as they are congruent. We can rotate a rectangle when placing it.
Optimal solution
An optimal solution for this problem is:
These are rectangles \(k5, k6, k12\) with \(k5\) being rotated. Obviously, this solution is not unique.
Optimization model
To model this, I use a binary variable: \[ x_{k,r,i,j} = \begin{cases} 1 & \text{if rectangle $k$ (rotated if $r=\mathrm{y}$) is placed at location $(i,j)$} \\ 0 & \text{otherwise}\end{cases}\] Here \(r = \{\mathrm{n},\mathrm{y}\}\) indicates if a rectangle is rotated. Note that not all rectangles can be rotated (it does not make sense to rotate a square, or there may not be space left to rotate for a given position).
For convenience, I also introduce \[y_{k} = \begin{cases} 1 & \text{if rectangle $k$ is used} \\ 0 & \text{otherwise}\end{cases}\] This is essentially an aggration of \(x\).
In addition, I use a few more complex data structures: \[\mathit{ok}_{k,r,i,j} = \begin{cases} \mathit{Yes} & \text{if we can place rectangle $k$ at location $(i,j)$} \\ \mathit{No} & \text{otherwise}\end{cases}\] Again, \(r\) indicates if rectangle \(k\) is rotated. I implemented this as a GAMS set. You can consider it as a Boolean data structure. Furthermore: \[\mathit{cover}_{k,r,i,j,i',j'} = \begin{cases} \mathit{Yes} & \text{if cell $(i',j')$ is covered by placing rectangle $k$ at location $(i,j)$} \\ \mathit{No} & \text{otherwise}\end{cases}\]
These sets are large. The set ok looks like:
---- 103 SET ok usable position
INDEX 1 = k1
j1 j2 j3 j4 j5
n.i1 YES YES YES YES YES
n.i2 YES YES YES YES YES
n.i3 YES YES YES YES YES
n.i4 YES YES YES YES YES
n.i5 YES YES YES YES YES
INDEX 1 = k2
j1 j2 j3 j4 j5
n.i1 YES YES YES YES YES
n.i2 YES YES YES YES YES
n.i3 YES YES YES YES YES
n.i4 YES YES YES YES YES
y.i1 YES YES YES YES
y.i2 YES YES YES YES
y.i3 YES YES YES YES
y.i4 YES YES YES YES
y.i5 YES YES YES YES
...
INDEX 1 = k13
j1 j2 j3
n.i1 YES YES YES
y.i1 YES
y.i2 YES
y.i3 YES
INDEX 1 = k14
j1 j2
n.i1 YES YES
y.i1 YES
y.i2 YES
The set cover is even larger:
---- 103 SET cover coverage of cells
INDEX 1 = k1 INDEX 2 = n INDEX 3 = i1 INDEX 4 = j1
j1
i1 YES
INDEX 1 = k1 INDEX 2 = n INDEX 3 = i1 INDEX 4 = j2
j2
i1 YES
...
INDEX 1 = k14 INDEX 2 = y INDEX 3 = i1 INDEX 4 = j1
j1 j2 j3 j4 j5
i1 YES YES YES YES YES
i2 YES YES YES YES YES
i3 YES YES YES YES YES
i4 YES YES YES YES YES
INDEX 1 = k14 INDEX 2 = y INDEX 3 = i2 INDEX 4 = j1
j1 j2 j3 j4 j5
i2 YES YES YES YES YES
i3 YES YES YES YES YES
i4 YES YES YES YES YES
i5 YES YES YES YES YES
The reason to develop these sets is to make the equations simpler. In addition, we can develop, test and debug the calculation of these sets in isolation of the test of the model and before the equations are written.
With this, we can formulate:
| MIP Model |
|---|
| \[\begin{align}\min\> & \color{darkred}u - \color{darkred}\ell\\ & \sum_{k,r,i,j|\color{darkblue}{\mathit{cover}}(k,r,i,j,i',j')}\color{darkred}x_{k,r,i',j'}=1 && \forall i',j'&&\text{cover cell $(i',j')$}\\ &\color{darkred}y_k = \sum_{r,i,j|\color{darkblue}{\mathit{ok}}(k,r,i,j)}\color{darkred}x_{k,r,i,j}&& \forall k && \text{select max one type of rectangle} \\ & \color{darkred}\ell \le \color{darkblue}{\mathit{area}}_k \cdot\color{darkred}y_k + \color{darkblue}M (1-\color{darkred}y_k) && \forall k && \text{bound smallest area}\\ & \color{darkred}u \ge \color{darkblue}{\mathit{area}}_k \cdot\color{darkred}y_k && \forall k && \text{bound largest area}\\ & \sum_k \color{darkblue}{\mathit{area}}_k \cdot\color{darkred}y_k = \color{darkblue} n^2 && && \text{extra constraint}\\& \color{darkblue}M = \max_k \color{darkblue}{\mathit{area}}_k\end{align}\] |
When solving this model, we see:
---- 146 VARIABLE x.L place tile
j1 j4
k5 .y.i4 1
k6 .n.i1 1
k12.n.i1 1
---- 149 VARIABLE y.L tile is used
k5 1, k6 1, k12 1
---- 157 PARAMETER v tile numbers
j1 j2 j3 j4 j5
i1 6661212
i2 6661212
i3 6661212
i4 5551212
i5 5551212
---- 159 VARIABLE smallest.L = 6.000 smallest used tile
VARIABLE largest.L = 10.000 largest used tile
VARIABLE z.L = 4.000 objective
One remaining question is if we can use some symmetry-breaking constraint to speed up solution times.
A larger problem
It looks like \(n=19\) is interesting [3]. This is about the limit that we can solve comfortably (a few hours to global optimality) with a MIP solver.
 |
| An optimal 19x19 problem |
The colors are not part of the model: they were added by hand to make the picture more Mondrianic.
A list of optimal objective values can be found in [4].
References
- Piet Mondrian, https://en.wikipedia.org/wiki/Piet_Mondrian
- Mondrian Puzzle -- Numberphile, https://www.youtube.com/watch?v=49KvZrioFB0
- Brady Haran, Mondrian Art Puzzle, https://www.bradyharanblog.com/blog/mondrian-art-puzzle
- A276523 in The Online Encyclopedia of Sequences, https://oeis.org/A276523