N-Queens Problem

The 8-Queens problem aims to place 8 queens on a chessboard so that no two queens attack each other; that is, no two queens share the same row, the same column, or the same diagonal (in either direction). The N-Queens problem generalizes this: place $N$ queens on an $N\times N$ chessboard under the same conditions.

To formulate this problem using QUBO++, we use an $N\times N$ matrix $X=(x_{i,j}) of binary variables, where $x_{i,j}=1 if a queen is placed at row $i$ and column $j$, and $x_{i,j}=0$ otherwise. We impose the following constraints:

Exactly one queen in each row:

\[\begin{aligned} \sum_{j=0}^{N-1} x_{i,j}&=1 && (0\leq i\leq N-1) \end{aligned}\]

Exactly one queen in each column:

\[\begin{aligned} \sum_{i=0}^{N-1} x_{i,j}&=1 && (0\leq j\leq N) \end{aligned}\]

At most one queen on each diagonal (from top-left to bottom-right): A diagonal is characterized by $i+j=k$. We consider only diagonals of length at least 2, i.e., $k=1,2,\ldots,2N−3$, and require:

\[\begin{aligned} 0\leq \sum_{\substack{0\le i,j \le N-1\\ i+j=k}}x_{i,j}\leq 1 &&(1\leq k\leq 2N-3) \end{aligned}\]

The sum of each anti-diagonal of $X$ is 0 or 1: An anti-diagonal is characterized by $j−i=d$. We consider only anti-diagonals of length at least 2, i.e., $d=−(N−2),\ldots,(N−2)$, and require:

\[\begin{aligned} 0\leq \sum_{\substack{0\le i,j \le N-1\\ j-i=d}}x_{i,j}\leq 1 &&(-(N-2)\leq d\leq (N-2)) \end{aligned}\]

QUBO++ program

The following QUBO++ program constructs an expression representing the constraints above and then finds a feasible solution using the Easy Solver:

#include "qbpp.hpp"
#include "qbpp_easy_solver.hpp"

int main() {
  const int n = 8;
  auto x = qbpp::var("x", n, n);

  auto f = qbpp::sum(qbpp::vector_sum(x) == 1);
  f += qbpp::sum(qbpp::vector_sum(qbpp::transpose(x)) == 1);

  const int m = 2 * n - 3;
  auto a = qbpp::expr(m);
  auto b = qbpp::expr(m);

  for (int i = 0; i < m; ++i) {
    const int k = i + 1;
    for (int r = 0; r < n; ++r) {
      const int c = k - r;
      if (0 <= c && c < n) {
        a[static_cast<size_t>(i)] +=
            x[static_cast<size_t>(r)][static_cast<size_t>(c)];
      }
    }

    const int d = i - (n - 2);
    for (int r = 0; r < n; ++r) {
      const int c = r + d;
      if (0 <= c && c < n) {
        b[static_cast<size_t>(i)] +=
            x[static_cast<size_t>(r)][static_cast<size_t>(c)];
      }
    }
  }

  f += qbpp::sum(0 <= a <= 1);
  f += qbpp::sum(0 <= b <= 1);

  f.simplify_as_binary();

  auto solver = qbpp::easy_solver::EasySolver(f);
  solver.target_energy(0);
  auto sol = solver.search();
  for (size_t i = 0; i < n; i++) {
    for (size_t j = 0; j < n; j++) {
      std::cout << (sol(x[i][j]) ? "Q" : ".");
    }
    std::cout << std::endl;
  }
}

An n$\times$n matrix x of binary variables is introduced, where x[i]\[j] = 1 indicates that a queen is placed at row i and column j. The row-wise sums are computed by qbpp::vector_sum(x), which returns a vector of n expressions (one per row). The operator == is applied element-wise and produces a vector of penalty expressions; each expression becomes 0 if and only if the corresponding row sum equals 1. Finally, qbpp::sum() aggregates these penalties into a single expression that attains its minimum value 0 if and only if every row contains exactly one queen. The column constraints are constructed in the same way by applying qbpp::transpose(x).

To enforce diagonal constraints, we build two vectors of expressions, a and b, each of length m = 2*n - 3. For each index i, a[i] accumulates variables on a diagonal with a fixed value of r + c (diagonals from top-left to bottom-right), excluding diagonals of length 1. Similarly, b[i] accumulates variables on an anti-diagonal with a fixed value of c - r (diagonals from top-right to bottom-left), again excluding diagonals of length 1. The chained range comparison 0 <= a <= 1 (and similarly for b) is applied element-wise and produces penalties that become 0 if and only if each diagonal/anti-diagonal contains at most one queen. These penalties are added to f.

After converting the expression into a binary QUBO form with f.simplify_as_binary(), the Easy Solver searches for a solution with target energy 0. The resulting assignment sol is then printed as an 8-by-8 board, where Q denotes a queen and . denotes an empty square. For example, the program may produce the following output:

..Q.....
.....Q..
.......Q
.Q......
...Q....
Q.......
......Q.
....Q...

This output confirms a valid placement of eight queens, since no two queens share the same row, column, diagonal, or anti-diagonal.

Last updated: 2026.01.09