Chess and solution pool

The \(n\)-queens problem is a popular example of a "chessboard non-attacking problem" [1,2]:

• On an \(n \times n\) chessboard place as many queens as possible such that none of them is attacked
• Given this, how many different ways can we place those queens?

We use the usual \(8 \times 8\) board.

Rooks

A related, simpler problem is about placing rooks. The optimization problem is very simple: it is an assignment problem:

n-Rooks Problem
\[\begin{align}\max& \sum_{i,j} \color{DarkRed} x_{i,j}\\ & \sum_j \color{DarkRed}x_{i,j} \le 1 && \forall i\\& \sum_i \color{DarkRed}x_{i,j} \le 1 && \forall j\\ & \color{DarkRed}x_{i,j}\in\{0,1\} \end{align}\]

For an \(8 \times 8\) board it is obvious that the number of rooks we can place is 8. Here we show two configurations:



We can easily answer: how many alternative configuration with 8 rooks can we find? This is \[n! = 8! = 40,320\] Basically, w enumerate all row and column permutations of the diagonal above. We can use the solution pool in Cplex and Gurobi to find all of them. Here I use Cplex and the performance is phenomenal: all 40,320 solutions are found in 16 seconds.

Bishops

The model for placing bishops is slightly more complicated: we need to check the main diagonals and the anti-diagonals. For the main diagonal we have the rule: \[i=j\] or for all of them \[i=j+k\] for \(k=-6,\dots,6\). The anti-diagonals can be described as: \[i+j=k\] for \(k=3,\dots,15\). We can illustrate this with:



The first picture has cell values \(a_{i,j}=i-j\) and the second one has cell values \(a_{i,j}=i+j\). So with this, our model can look like:

n-Bishops Problem
\[\begin{align}\max& \sum_{i,j} \color{DarkRed} x_{i,j}\\ & \sum_{i-j=k} \color{DarkRed}x_{i,j} \le 1 && k=-(n-2),\dots,(n-2)\\& \sum_{i+j=k} \color{DarkRed}x_{i,j} \le 1 && k=3,\dots,2n-1\\ & \color{DarkRed}x_{i,j}\in\{0,1\} \end{align}\]

Note that these constraints translate directly into GAMS:

bishop_main(k1).. sum((i,j)$(v(i)-v(j)=v(k1)),x(i,j)) =l= 1;
bishop_anti(k2).. sum((i,j)$(v(i)+v(j)=v(k2)),x(i,j)) =l= 1;

Two possible solutions are:

We can accommodate 14 bishops. There are 256 different solutions (according to Cplex).

Queens

The \(n\) queens problem is now very simple: we just need to combine the two previous models.

n-Queens Problem
\[\begin{align}\max& \sum_{i,j} \color{DarkRed} x_{i,j}\\ & \sum_j \color{DarkRed}x_{i,j} \le 1 && \forall i\\& \sum_i \color{DarkRed}x_{i,j} \le 1 && \forall j\\ & \sum_{i-j=k} \color{DarkRed}x_{i,j} \le 1 && k=-(n-2),\dots,(n-2)\\& \sum_{i+j=k} \color{DarkRed}x_{i,j} \le 1 && k=3,\dots,2n-1\\ & \color{DarkRed}x_{i,j}\in\{0,1\} \end{align}\]

A solution can look like:

We can place 8 queens and there are 92 solutions.

Notes:

• Cplex delivered 91 solutions. This seems to be a tolerance issue. I used a solution pool absolute gap of zero. As the objective jumps by one, it is safe to set the absolute gap to 0.01. With this gap we found all 92 solutions. Is this a bug? Maybe. Possibly. Probably. This 92-th solution is not supposed to be cut off.
• Many of the 92 solutions are the result of simple symmetric operations, such as rotating the board, or reflection [2]. The number of different solutions after ignoring these symmetries is just 12. The GAMS model in [3] finds those.
• The model in [3] handles the bishop constraints differently, by calculating an offset and writing \[\sum_i x_{i,i +\mathit{sh}_s} \le 1, \>\forall s\] I somewhat prefer our approach.  

Kings

Kings are handled quite cleverly in [1]. They observe that there cannot be more than one king in each block \[\begin{matrix} x_{i,j} & x_{i,j+1}\\ x_{i+1,j} & x_{i+1,j+1}\end{matrix}\] Note that we only have to look forward (and downward) as a previous block would have covered things to the left (or above). The model can look like:

n-Kings Problem
\[\begin{align}\max& \sum_{i,j} \color{DarkRed} x_{i,j}\\ & \color{DarkRed}x_{i,j}+\color{DarkRed}x_{i+1,j}+\color{DarkRed}x_{i,j+1}+\color{DarkRed}x_{i+1,j+1}\le 1 && \forall i,j\le n-1\\ & \color{DarkRed}x_{i,j}\in\{0,1\} \end{align}\]

Two solutions are:



We can place 16 kings, and there are a lot of possible configurations: 281,571 (says Cplex).

Knights

To place knights we set up a set \(\mathit{jump}_{i,j,i',j'}\) indicating if we can jump from cell \((i,j)\) to cell \((i',j')\). We only need to look forward, so for each \((i,j)\) we need to consider four cases: \[\begin{matrix} & & x_{i-2,j+1} & \\ & &  & x_{i-1,j+2} \\ & x_{i,j} &&\\ &&&x_{i+1,j+2}\\ & & x_{i+2,j+1}& \end{matrix}\] Note that near the border we may have fewer than four cases. In GAMS we can populate the set \(\mathit{jump}\) straightforwardly:

jump(i,j,i-2,j+1) = yes;
jump(i,j,i-1,j+2) = yes;
jump(i,j,i+1,j+2) = yes;
jump(i,j,i+2,j+1) = yes;

The model can look like:

n-Knights Problem
\[\begin{align}\max& \sum_{i,j} \color{DarkRed} x_{i,j}\\ & \color{DarkRed}x_{i,j}+\color{DarkRed}x_{i',j'}\le 1 && \forall i,j,i',j'|\mathit{jump}_{i,j,i',j'}\\ & \color{DarkRed}x_{i,j}\in\{0,1\} \end{align}\]

We can place 32 knights. There are only two different solutions:

References