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50 lines
3.9 KiB
50 lines
3.9 KiB
# n queens problem
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!!! question
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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.
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As shown in the figure below, 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.
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The figure below 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 `/`.
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### Row-by-row placing strategy
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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**.
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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.
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The figure below 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.
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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.
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### Column and diagonal pruning
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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.
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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**.
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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 the figure below to track whether a queen is on any main diagonal.
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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.
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### Code implementation
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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 $2n - 1$.
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```src
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[file]{n_queens}-[class]{}-[func]{n_queens}
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```
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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.
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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 $O(n^2)$**.
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