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# 13.4   n queens problem

!!! question

    According to the rules of chess, a queen can attack pieces in the same row, column, or on a diagonal line. Given $n$ queens and an $n \times n$ chessboard, find arrangements where no two queens can attack each other.

As shown in Figure 13-15, when $n = 4$, there are two solutions. From the perspective of the backtracking algorithm, an $n \times n$ chessboard has $n^2$ squares, presenting all possible choices `choices`. The state of the chessboard `state` changes continuously as each queen is placed.

![Solution to the 4 queens problem](n_queens_problem.assets/solution_4_queens.png){ class="animation-figure" }

<p align="center"> Figure 13-15 &nbsp; Solution to the 4 queens problem </p>

Figure 13-16 shows the three constraints of this problem: **multiple queens cannot be on the same row, column, or diagonal**. It is important to note that diagonals are divided into the main diagonal `\` and the secondary diagonal `/`.

![Constraints of the n queens problem](n_queens_problem.assets/n_queens_constraints.png){ class="animation-figure" }

<p align="center"> Figure 13-16 &nbsp; Constraints of the n queens problem </p>

### 1. &nbsp; Row-by-row placing strategy

As the number of queens equals the number of rows on the chessboard, both being $n$, it is easy to conclude: **each row on the chessboard allows and only allows one queen to be placed**.

This means that we can adopt a row-by-row placing strategy: starting from the first row, place one queen per row until the last row is reached.

Figure 13-17 shows the row-by-row placing process for the 4 queens problem. Due to space limitations, the figure only expands one search branch of the first row, and prunes any placements that do not meet the column and diagonal constraints.

![Row-by-row placing strategy](n_queens_problem.assets/n_queens_placing.png){ class="animation-figure" }

<p align="center"> Figure 13-17 &nbsp; Row-by-row placing strategy </p>

Essentially, **the row-by-row placing strategy serves as a pruning function**, avoiding all search branches that would place multiple queens in the same row.

### 2. &nbsp; Column and diagonal pruning

To satisfy column constraints, we can use a boolean array `cols` of length $n$ to track whether a queen occupies each column. Before each placement decision, `cols` is used to prune the columns that already have queens, and it is dynamically updated during backtracking.

How about the diagonal constraints? Let the row and column indices of a cell on the chessboard be $(row, col)$. By selecting a specific main diagonal, we notice that the difference $row - col$ is the same for all cells on that diagonal, **meaning that $row - col$ is a constant value on that diagonal**.

Thus, if two cells satisfy $row_1 - col_1 = row_2 - col_2$, they are definitely on the same main diagonal. Using this pattern, we can utilize the array `diags1` shown in Figure 13-18 to track whether a queen is on any main diagonal.

Similarly, **the sum $row + col$ is a constant value for all cells on a secondary diagonal**. We can also use the array `diags2` to handle secondary diagonal constraints.

![Handling column and diagonal constraints](n_queens_problem.assets/n_queens_cols_diagonals.png){ class="animation-figure" }

<p align="center"> Figure 13-18 &nbsp; Handling column and diagonal constraints </p>

### 3. &nbsp; Code implementation

Please note, in an $n$-dimensional matrix, the range of $row - col$ is $[-n + 1, n - 1]$, and the range of $row + col$ is $[0, 2n - 2]$, thus the number of both main and secondary diagonals is $2n - 1$, meaning the length of both arrays `diags1` and `diags2` is $2n - 1$.

=== "Python"

    ```python title="n_queens.py"
    def backtrack(
        row: int,
        n: int,
        state: list[list[str]],
        res: list[list[list[str]]],
        cols: list[bool],
        diags1: list[bool],
        diags2: list[bool],
    ):
        """Backtracking algorithm: n queens"""
        # When all rows are placed, record the solution
        if row == n:
            res.append([list(row) for row in state])
            return
        # Traverse all columns
        for col in range(n):
            # Calculate the main and minor diagonals corresponding to the cell
            diag1 = row - col + n - 1
            diag2 = row + col
            # Pruning: do not allow queens on the column, main diagonal, or minor diagonal of the cell
            if not cols[col] and not diags1[diag1] and not diags2[diag2]:
                # Attempt: place the queen in the cell
                state[row][col] = "Q"
                cols[col] = diags1[diag1] = diags2[diag2] = True
                # Place the next row
                backtrack(row + 1, n, state, res, cols, diags1, diags2)
                # Retract: restore the cell to an empty spot
                state[row][col] = "#"
                cols[col] = diags1[diag1] = diags2[diag2] = False

    def n_queens(n: int) -> list[list[list[str]]]:
        """Solve n queens"""
        # Initialize an n*n size chessboard, where 'Q' represents the queen and '#' represents an empty spot
        state = [["#" for _ in range(n)] for _ in range(n)]
        cols = [False] * n  # Record columns with queens
        diags1 = [False] * (2 * n - 1)  # Record main diagonals with queens
        diags2 = [False] * (2 * n - 1)  # Record minor diagonals with queens
        res = []
        backtrack(0, n, state, res, cols, diags1, diags2)

        return res
    ```

=== "C++"

    ```cpp title="n_queens.cpp"
    [class]{}-[func]{backtrack}

    [class]{}-[func]{nQueens}
    ```

=== "Java"

    ```java title="n_queens.java"
    /* Backtracking algorithm: n queens */
    void backtrack(int row, int n, List<List<String>> state, List<List<List<String>>> res,
            boolean[] cols, boolean[] diags1, boolean[] diags2) {
        // When all rows are placed, record the solution
        if (row == n) {
            List<List<String>> copyState = new ArrayList<>();
            for (List<String> sRow : state) {
                copyState.add(new ArrayList<>(sRow));
            }
            res.add(copyState);
            return;
        }
        // Traverse all columns
        for (int col = 0; col < n; col++) {
            // Calculate the main and minor diagonals corresponding to the cell
            int diag1 = row - col + n - 1;
            int diag2 = row + col;
            // Pruning: do not allow queens on the column, main diagonal, or minor diagonal of the cell
            if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
                // Attempt: place the queen in the cell
                state.get(row).set(col, "Q");
                cols[col] = diags1[diag1] = diags2[diag2] = true;
                // Place the next row
                backtrack(row + 1, n, state, res, cols, diags1, diags2);
                // Retract: restore the cell to an empty spot
                state.get(row).set(col, "#");
                cols[col] = diags1[diag1] = diags2[diag2] = false;
            }
        }
    }

    /* Solve n queens */
    List<List<List<String>>> nQueens(int n) {
        // Initialize an n*n size chessboard, where 'Q' represents the queen and '#' represents an empty spot
        List<List<String>> state = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            List<String> row = new ArrayList<>();
            for (int j = 0; j < n; j++) {
                row.add("#");
            }
            state.add(row);
        }
        boolean[] cols = new boolean[n]; // Record columns with queens
        boolean[] diags1 = new boolean[2 * n - 1]; // Record main diagonals with queens
        boolean[] diags2 = new boolean[2 * n - 1]; // Record minor diagonals with queens
        List<List<List<String>>> res = new ArrayList<>();

        backtrack(0, n, state, res, cols, diags1, diags2);

        return res;
    }
    ```

=== "C#"

    ```csharp title="n_queens.cs"
    [class]{n_queens}-[func]{Backtrack}

    [class]{n_queens}-[func]{NQueens}
    ```

=== "Go"

    ```go title="n_queens.go"
    [class]{}-[func]{backtrack}

    [class]{}-[func]{nQueens}
    ```

=== "Swift"

    ```swift title="n_queens.swift"
    [class]{}-[func]{backtrack}

    [class]{}-[func]{nQueens}
    ```

=== "JS"

    ```javascript title="n_queens.js"
    [class]{}-[func]{backtrack}

    [class]{}-[func]{nQueens}
    ```

=== "TS"

    ```typescript title="n_queens.ts"
    [class]{}-[func]{backtrack}

    [class]{}-[func]{nQueens}
    ```

=== "Dart"

    ```dart title="n_queens.dart"
    [class]{}-[func]{backtrack}

    [class]{}-[func]{nQueens}
    ```

=== "Rust"

    ```rust title="n_queens.rs"
    [class]{}-[func]{backtrack}

    [class]{}-[func]{n_queens}
    ```

=== "C"

    ```c title="n_queens.c"
    [class]{}-[func]{backtrack}

    [class]{}-[func]{nQueens}
    ```

=== "Kotlin"

    ```kotlin title="n_queens.kt"
    [class]{}-[func]{backtrack}

    [class]{}-[func]{nQueens}
    ```

=== "Ruby"

    ```ruby title="n_queens.rb"
    [class]{}-[func]{backtrack}

    [class]{}-[func]{n_queens}
    ```

=== "Zig"

    ```zig title="n_queens.zig"
    [class]{}-[func]{backtrack}

    [class]{}-[func]{nQueens}
    ```

Placing $n$ queens row-by-row, considering column constraints, from the first row to the last row there are $n$, $n-1$, $\dots$, $2$, $1$ choices, using $O(n!)$ time. When recording a solution, it is necessary to copy the matrix `state` and add it to `res`, with the copying operation using $O(n^2)$ time. Therefore, **the overall time complexity is $O(n! \cdot n^2)$**. In practice, pruning based on diagonal constraints can significantly reduce the search space, thus often the search efficiency is better than the above time complexity.

Array `state` uses $O(n^2)$ space, and arrays `cols`, `diags1`, and `diags2` each use $O(n)$ space. The maximum recursion depth is $n$, using $O(n)$ stack space. Therefore, **the space complexity is $O(n^2)$**.