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* Retranslate the complete English edition from Simplified Chinese * Polish three English explanations after reader-focused review
56 lines
2.0 KiB
JavaScript
56 lines
2.0 KiB
JavaScript
/**
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* File: n_queens.js
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* Created Time: 2023-05-13
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* Author: Justin (xiefahit@gmail.com)
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*/
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/* Backtracking: n-queens */
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function backtrack(row, n, state, res, cols, diags1, diags2) {
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// Record a solution when a queen has been placed in every row
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if (row === n) {
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res.push(state.map((row) => row.slice()));
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return;
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}
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// Iterate over all columns
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for (let col = 0; col < n; col++) {
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// Calculate the main-diagonal and anti-diagonal indices for this square
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const diag1 = row - col + n - 1;
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const diag2 = row + col;
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// Prune: the square must have no queen in its column, main diagonal, or anti-diagonal
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if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
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// Try: place a queen on this square
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state[row][col] = 'Q';
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cols[col] = diags1[diag1] = diags2[diag2] = true;
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// Place a queen in the next row
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backtrack(row + 1, n, state, res, cols, diags1, diags2);
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// Backtrack: restore this square to empty
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state[row][col] = '#';
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cols[col] = diags1[diag1] = diags2[diag2] = false;
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}
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}
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}
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/* Solve the n-queens problem */
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function nQueens(n) {
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// Initialize an n*n board, where 'Q' represents a queen and '#' represents an empty square
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const state = Array.from({ length: n }, () => Array(n).fill('#'));
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const cols = Array(n).fill(false); // Track whether a column contains a queen
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const diags1 = Array(2 * n - 1).fill(false); // Track whether a main diagonal contains a queen
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const diags2 = Array(2 * n - 1).fill(false); // Track whether an anti-diagonal contains a queen
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const res = [];
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backtrack(0, n, state, res, cols, diags1, diags2);
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return res;
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}
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// Driver code
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const n = 4;
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const res = nQueens(n);
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console.log(`Input board side length: ${n}`);
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console.log(`Number of queen arrangements: ${res.length}`);
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res.forEach((state) => {
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console.log('--------------------');
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state.forEach((row) => console.log(row));
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});
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