Here is a difficult Sudoku variant designed by Aad van de Wetering. Aad is a former colleague of mine.
The standard Sudoku rules apply:
In addition, we have the following restriction:
To solve this you can follow this video [1]:
During post-processing, we can calculate the completed grid using the optimal values of \(x\): \[v_{i,j} = \sum_k k \cdot x^*_{i,j,k}\]
We need to introduce the concept of "neighbors". We can define the set: \[nb_{i,j,i',j'} \> \text{exists iff cells $(i,j)$ and $(i',j')$ are neighbors}\] I excluded symmetric cases. To be precise: in GAMS, I populated this set with:
We can introduce the post-processing step directly into the MIP and then enforce our difference constraint on the variables \(v_{i,j}\). This can look like \[\begin{align} & v_{i,j} = \sum_k k \cdot x_{i,j,k} && \text{cell values as constraint} \\ & -5 \le v_{i,j} - v_{i',j'} \le 5 && \text{for neigboring cells $(i,j)$ and $(i',j')$}\end{align}\]
With this, we can formulate:
This model solves very easily. The presolver can solve this model completely, so we don't even have to start the branch-and-bound phase. As a result, we can make a much shorter movie:
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The solution of this model looks like:
A proper Sudoku puzzle has one unique solution. We can verify this by adding a cut to the problem after solving it. The cut should forbid the current solution (and allow all other solutions). In this case, the cut can look like \[\sum_{i,j,k} x^*_{i,j,k} x_{i,j,k}\le 9^2-1\] where \(x^*\) is the previously found solution. Indeed, this puzzle has a single, unique solution: after adding the cut, the problem becomes (integer) infeasible.
 |
| Problem |
- Each cell has an integer value between 1 and 9
- In each row, column and sub-block cells must be unique
- Some cells have a predefined value
In addition, we have the following restriction:
- The maximum difference between the value of two neighboring cells (orthogonal, that is horizontal or vertical) is 5
To solve this you can follow this video [1]:
Of course, we want to try to solve this as a Mixed-Integer Programming model.
Review: a linear integer programming model for standard Sudoku
When we want to solve Sudokus, the easiest approach is to define the following binary decision variables [2]: \[x_{i,j,k} = \begin{cases} 1 & \text{if cell $(i,j)$ has value $k$} \\ 0 & \text{otherwise}\end{cases}\] Here \(k \in \{1,\dots,9\}\). We have 27 areas we need to check for unique values: rows, columns and sub-blocks. We can organize this as a set: \[u_{a,i,j}\>\text{exists if and only if area $a$ contains cell $(i,j)$}\] This is data. We also have a set of given cells, which we can fix. The resultimg MIP model can look like [2]:
| Mixed-Integer Programming Model for standard Sudoku |
|---|
| \[\begin{align}\min\>& 0 && && \text{Dummy objective}\\ & \sum_k \color{darkred}x_{i,j,k} = 1 &&\forall i,j && \text{One $k$ in each cell}\\ & \sum_{i,j|\color{darkblue}u_{a,i,j}} \color{darkred}x_{i,j,k} = 1 && \forall a,k && \text{Unique values in each area}\\ & \color{darkred}x_{i,j,k} = 1 && \text{where } k=\color{darkblue}{\mathit{Given}}_{i,j} &&\text{Fix given values}\\ &\color{darkred}x_{i,j,k} \in \{0,1\} \end{align}\] |
Linear integer programming model for Sudoku variant
alias (i,ii),(j,jj); set nb(i,j,ii,jj) 'neighbors' ; nb(i,j,i-1,j) = yes; nb(i,j,i,j-1) = yes; |
We can introduce the post-processing step directly into the MIP and then enforce our difference constraint on the variables \(v_{i,j}\). This can look like \[\begin{align} & v_{i,j} = \sum_k k \cdot x_{i,j,k} && \text{cell values as constraint} \\ & -5 \le v_{i,j} - v_{i',j'} \le 5 && \text{for neigboring cells $(i,j)$ and $(i',j')$}\end{align}\]
With this, we can formulate:
| Mixed-Integer Programming Model for Sudoku Variant |
|---|
| \[\begin{align}\min\>& 0 && && \text{Dummy objective}\\ & \sum_k \color{darkred}x_{i,j,k} = 1 &&\forall i,j && \text{One $k$ in each cell}\\ & \sum_{i,j|\color{darkblue}u_{a,i,j}} \color{darkred}x_{i,j,k} = 1 && \forall a,k && \text{Unique values in each area}\\ &\color{darkred}v_{i,j} =\sum_k \color{darkblue}k\cdot\color{darkred}x_{i,j,k} && \forall i,j &&\text{Value of cell}\\ & -5 \le \color{darkred}v_{i,j} - \color{darkred}v_{i',j'} \le 5 &&\forall i,j,i',j'|\color{darkblue}{nb}_{i,j,i',j'} && \text{Neighbors differ by max 5} \\ & \color{darkred}x_{i,j,k} = 1 && \text{where } k=\color{darkblue}{\mathit{Given}}_{i,j} &&\text{Fix given values}\\&\color{darkred}x_{i,j,k} \in \{0,1\} \\ &\color{darkred}v_{i,j} \in \{1,\dots,9\} \end{align}\] |
This model solves very easily. The presolver can solve this model completely, so we don't even have to start the branch-and-bound phase. As a result, we can make a much shorter movie:
The solution of this model looks like:
---- 80 VARIABLE v.L value of each cell
c1 c2 c3 c4 c5 c6 c7 c8 c9
r1 249675831
r2 385149672
r3 761238954
r4 896314527
r5 573862149
r6 124597386
r7 612483795
r8 437951268
r9 958726413
Unique solution
A proper Sudoku puzzle has one unique solution. We can verify this by adding a cut to the problem after solving it. The cut should forbid the current solution (and allow all other solutions). In this case, the cut can look like \[\sum_{i,j,k} x^*_{i,j,k} x_{i,j,k}\le 9^2-1\] where \(x^*\) is the previously found solution. Indeed, this puzzle has a single, unique solution: after adding the cut, the problem becomes (integer) infeasible.
References
- A Sudoku With Just 11 Digits = Dutch Genius!, https://www.youtube.com/watch?v=ZU5fSDHJq8k
- MIP Modeling: from Sudoku to KenKen via Logarithms, https://yetanothermathprogrammingconsultant.blogspot.com/2016/10/mip-modeling-from-sudoku-to-kenken.html