MIP Modeling: from Sudoku to KenKen

The Mixed Integer Programming formulations for solving the puzzles Sudoku and KenKen have a major trait they share: the basic data structure is a binary decision variable:

Such a binary variable allows us to easily formulate so-called ‘all-different’ constructs (1). E.g. if each row \(i\) having \(n\) cells, must have different values \(1..n\) in each cell, we can write:

If we want to recover the value of a cell \((i,j)\) we can introduce a variable \(v_{i,j}\) with:

Detail: this is essentially what is sometimes called a set of accounting rows. We could also do this in post-processing outside the optimization model.

Sudoku

On the left we have a Sudoku puzzle and on the right we see the solution (2).

In Sudoku we need uniqueness along the rows, the columns and inside the 3 x 3 sub-squares. The constraints that enforce these uniqueness conditions are:

Here \(u_{s,i,j}\) contains the elements \(i,j\) for square \(s\). With this we have two issues left: How can we build up \(u_{s,i,j}\) and how can we fix the given values in the Sudoku problem.

Populate areas

We number the squares as follows:

Let’s see if we can populate \(u_{s,i,j}\) without entering data manually and without using loops. We can invent a function \(s=f(i,j)\) that takes a row and column number and returns the number of the square as follows:

Here the funny brackets indicate the ceiling function. E.g. \(f(4,7)= 3\left\lceil 4/3 \right\rceil + \left\lceil 7/3 \right\rceil –3 = 3\cdot 2+3-3=6\). This means we can populate \(u\) as in:

When we take a step back, we can consider rows and columns as just another particularly shaped block. So we add to our set \(u\) the individual rows and columns:

This means we can reduce the model equations to just:

Fixing initial values

If we use the variables \(v_{i,j}\) in the model, we can use them to fix the initial values.

If we don’t have the \(v\) variables in the model we can fix the underlying \(x\) variables:

Complete model

The input is most conveniently organized as a table:

The output looks like:

KenKen

On the left we have a KenKen puzzle and on the right we see the solution (4).

In the KenKen puzzle we need to have unique values along the rows and the columns. As we have seen in Sudoku example, this is straightforward. We can formulate this as:

As in the Sudoku example, we can combine the first two constraints into a single one:

In addition to these constraints we have a slew of restrictions related to collections of cells - these are called the cages. Each cage has a rule and a given answer. Let’s discuss them in detail:

1. n this is a single cell with a given value. This case is not present in our example above. The rule is easy: just put the number in the cell \((i,j)\): \(v_{i,j} = n\),
    
   For this example we would fix this cell to the value 1.
2. n+ indicates we add up the values in the cage and the result should be n. This is a simple linear constraint: \[\sum_{i,j|C{i,j}} v_{i,j} = n\]
3. n- is a cage with just two cells where the absolute value difference between the values of the two cells is n. I.e. \[|v_{i_1,j_1}- v_{i_2,j_2}|=n\] In (6) it is proposed to model this in a linear fashion with the help of an additional binary variable as: \[v_{i_1,j_1}- v_{i_2,j_2}=n(1 –2 \delta)\] where \(\delta \in \{0,1\}\). 
4. n* indicates we multiply all values in the cage: \[\prod_{i,j|C{i,j}} v_{i,j} = n\] In (6) a number of possible linearizations are mentioned, some of them rather complicated with such things as prime number factorizations. Let me suggest a much simpler approach using logarithms. We have: \[\begin{align}&\prod_{i,j|C{i,j}} v_{i,j} = n\\ \Longrightarrow & \sum_{i,j|C_{i,j}} \log(v_{i,j}) = \log(n)\\ \Longrightarrow & \sum_{i,j|C_{i,j}} \sum_k \log(k) x_{i,j,k} = \log(n)\end{align}\] The last constraint is linear in the decision variables \(x_{i,j,k}\)!. This simple linearization was missed in (6).
5. n÷ this operation has two cells, and indicates division and is more formally: \[\frac{v_{i_1,j_1}}{v_{i_2,j_2}}=n \>\text{or}\>\frac{v_{i_2,j_2}}{v_{i_1,j_1}}=n\] In (6) the following linearization is proposed:\[\begin{align} &v_{i_1,j_1}-n\cdot v_{i_2,j_2} \ge 0-M\cdot \delta\\&v_{i_1,j_1}-n\cdot v_{i_2,j_2} \le 0+M\cdot \delta\\&v_{i_2,j_2}-n\cdot v_{i_1,j_1} \ge 0-M\cdot (1-\delta)\\&v_{i_2,j_2}-n\cdot v_{i_1,j_1} \le 0+M\cdot (1-\delta)\\&\delta\in\{0,1\}\end{align}\] Again I would suggest a simpler linearization: \[\begin{align}&\frac{v_{i_1,j_1}}{v_{i_2,j_2}}=n \>\text{or}\>\frac{v_{i_2,j_2}}{v_{i_1,j_1}}=n\\ \Longrightarrow&\log(v_{i_1,j_1})-\log(v_{i_2,j_2})=\log(n) \> \text{or} \>\log(v_{i_2,j_2})-\log(v_{i_1,j_1})=\log(n)\\ \Longrightarrow&|\log(v_{i_1,j_1})-\log(v_{i_2,j_2})|=\log(n)\\  \Longrightarrow&\log(v_{i_1,j_1})-\log(v_{i_2,j_2})=\log(n)(1-2\delta)\\  \Longrightarrow&\sum_k \log(k) x_{i_1,j_1,k}-\sum_k \log(k) x_{i_2,j_2,k}=\log(n)(1-2\delta)\end{align}\]

Complete Model

We have now all the pieces to assemble the complete model.

[todo]

References

1. All-different and Mixed Integer Programming, http://yetanothermathprogrammingconsultant.blogspot.com/2016/05/all-different-and-mixed-integer.html
2. Sudoku, https://en.wikipedia.org/wiki/Sudoku
3. Sudoku Puzzle, http://www.nytimes.com/crosswords/game/sudoku/hard
4. KenKen, https://en.wikipedia.org/wiki/KenKen
5. KenKen Puzzle, http://www.nytimes.com/ref/crosswords/kenken.html
6. Vardges Melkonian, “An Integer Programming Model for the KenKen Problem”, American Journal of Operations Research, 2016, 6, 213-225.