In [1] the famous \(n\)-queens problem was discussed: how many queens can we place on a chess-board such that non of them are attacked by others. Interestingly, we had some non-obvious problem with Cplex enumerating the different configurations (due to tolerance issues).
A different problem is the Minimum Dominating Set of Queens Problem:
To develop is Mixed Integer Programming model, we introduce the familiar binary variables
\[x_{i,j} = \begin{cases} 1 & \text{if square \((i,j)\) is occupied by a queen}\\ 0 & \text{otherwise}\end{cases}\]
A square \((i,j)\) is dominated if:
We can write these conditions in one big constraint. So we can write:
For an \(8\times 8\) board, we need 5 queens:
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Note that this formulation actually does a bit of double counting. E.g. when we look at the constraint for cell \((2,3)\) we see:
We see that x(2,3) has a coefficient 4. This is harmless. With a little bit more careful modeling we can get rid of this coefficient 4. A complicated way would be: \[ \sum_{j'}x_{i,j'} + \sum_{i'|i'\ne i} x_{i',j} + \sum_{i',j'|i'-j'=i-j,i'\ne i, j'\ne j} x_{i',j'} +\sum_{i',j'|i'+j'=i+j,i'\ne i, j'\ne j} x_{i',j'} \ge 1 \>\> \forall i,j\] Simpler is just to subtract \(3 x_{i,j}\): \[\sum_{j'} x_{i,j'} + \sum_{i'} x_{i',j} + \sum_{i',j'|i'-j'=i-j} x_{i',j'} +\sum_{i',j'|i'+j'=i+j} x_{i',j'}-3x_{i,j} \ge 1\>\> \forall i,j\]
Now we see:
When we ask: how many different solution are there?, we use again the Cplex solution pool. And again, we have our tolerance problem:
As the objective is integer valued, in some sense an absolute gap tolerance of 0.5 seems the safest. With this we find the correct number of solutions. See [1] for a more in-depth discussion of this tolerance problem.
A different problem is the Minimum Dominating Set of Queens Problem:
Find the minimum number of queens needed such that they dominate each square.Here dominate means: either a square is under attack by at least one queen, or it is occupied by a queen.
To develop is Mixed Integer Programming model, we introduce the familiar binary variables
\[x_{i,j} = \begin{cases} 1 & \text{if square \((i,j)\) is occupied by a queen}\\ 0 & \text{otherwise}\end{cases}\]
A square \((i,j)\) is dominated if:
- \(x_{i,j}=1\): a queen is located here. This case will be covered by the following other cases, so we don't have to worry about this.
- There is a queen in row \(i\): \[\sum_{j'} x_{i,j'} \ge 1\]
- There is a queen in column \(j\):\[\sum_{i'} x_{i',j} \ge 1\]
- There is a queen in the same diagonal. The diagonal is described by \((i',j')\) such that \(i'-j'=i-j\). So we have: \[\sum_{i',j'|i'-j'=i-j} x_{i',j'}\ge 1\]
- There is a queen in the same anti-diagonal. The anti-diagonal is described by \((i',j')\) such that \(i'+j'=i+j\). So we have: \[\sum_{i',j'|i'+j'=i+j} x_{i',j'}\ge 1\]
We can write these conditions in one big constraint. So we can write:
| Dominated Queens Problem |
|---|
| \[\begin{align}\min& \>\color{DarkRed}z=\sum_{i,j} \color{DarkRed} x_{i,j}\\ & \sum_{j'} \color{DarkRed}x_{i,j'} + \sum_{i'} \color{DarkRed}x_{i',j} + \sum_{i',j'|i'-j'=i-j} \color{DarkRed}x_{i',j'} +\sum_{i',j'|i'+j'=i+j} \color{DarkRed}x_{i',j'} \ge 1 && \forall i,j \\ & \color{DarkRed}x_{i,j}\in\{0,1\} \end{align}\] |
For an \(8\times 8\) board, we need 5 queens:
Note that this formulation actually does a bit of double counting. E.g. when we look at the constraint for cell \((2,3)\) we see:
e(2,3).. x(1,2) + x(1,3) + x(1,4) + x(2,1) + x(2,2) +4*x(2,3) + x(2,4)
+ x(2,5) + x(2,6) + x(2,7) + x(2,8) + x(3,2) + x(3,3) + x(3,4) + x(4,1)
+ x(4,3) + x(4,5) + x(5,3) + x(5,6) + x(6,3) + x(6,7) + x(7,3) + x(7,8)
+ x(8,3) =G= 1 ; (LHS = 0, INFES = 1 ****)
We see that x(2,3) has a coefficient 4. This is harmless. With a little bit more careful modeling we can get rid of this coefficient 4. A complicated way would be: \[ \sum_{j'}x_{i,j'} + \sum_{i'|i'\ne i} x_{i',j} + \sum_{i',j'|i'-j'=i-j,i'\ne i, j'\ne j} x_{i',j'} +\sum_{i',j'|i'+j'=i+j,i'\ne i, j'\ne j} x_{i',j'} \ge 1 \>\> \forall i,j\] Simpler is just to subtract \(3 x_{i,j}\): \[\sum_{j'} x_{i,j'} + \sum_{i'} x_{i',j} + \sum_{i',j'|i'-j'=i-j} x_{i',j'} +\sum_{i',j'|i'+j'=i+j} x_{i',j'}-3x_{i,j} \ge 1\>\> \forall i,j\]
Now we see:
e(2,3).. x(1,2) + x(1,3) + x(1,4) + x(2,1) + x(2,2) + x(2,3) + x(2,4) + x(2,5)
+ x(2,6) + x(2,7) + x(2,8) + x(3,2) + x(3,3) + x(3,4) + x(4,1) + x(4,3)
+ x(4,5) + x(5,3) + x(5,6) + x(6,3) + x(6,7) + x(7,3) + x(7,8) + x(8,3)
=G= 1 ; (LHS = 0, INFES = 1 ****)
When we ask: how many different solution are there?, we use again the Cplex solution pool. And again, we have our tolerance problem:
| pool.absgap | Number of solutions |
|---|---|
| 0 | 2 |
| 0.5 | 4860 |
As the objective is integer valued, in some sense an absolute gap tolerance of 0.5 seems the safest. With this we find the correct number of solutions. See [1] for a more in-depth discussion of this tolerance problem.
References
- Chess and solution pool, http://yetanothermathprogrammingconsultant.blogspot.com/2018/11/chess-and-solution-pool.html.