13.4 N-Queens Problem¶
Question
According to the rules of chess, a queen can attack pieces that share the same row, column, or diagonal line. Given \(n\) queens and an \(n \times n\) chessboard, find a placement scheme such that no two queens can attack each other.
As shown in Figure 13-15, when \(n = 4\), there are two solutions that can be found. From the perspective of the backtracking algorithm, an \(n \times n\) chessboard has \(n^2\) squares, which provide all the choices choices. During the process of placing queens one by one, the chessboard state changes continuously, and the chessboard at each moment represents the state state.
Figure 13-15 Solution to the 4-queens problem
Figure 13-16 illustrates the three constraints of this problem: multiple queens cannot be in the same row, the same column, or on the same diagonal. It is worth noting that diagonals are divided into two types: the main diagonal \ and the anti-diagonal /.
Figure 13-16 Constraints of the n-queens problem
1. Row-By-Row Placement Strategy¶
Since both the number of queens and the number of rows on the chessboard are \(n\), we can easily derive a conclusion: each row of the chessboard allows and only allows exactly one queen to be placed.
This means we can adopt a row-by-row placement strategy: starting from the first row, place one queen in each row until the last row is completed.
Figure 13-17 shows the row-by-row placement process for the 4-queens problem. Due to space limitations, the figure only expands one search branch of the first row, and all schemes that do not satisfy the column constraint and diagonal constraints are pruned.
Figure 13-17 Row-by-row placement strategy
Essentially, the row-by-row placement strategy serves a pruning function, as it avoids all search branches where multiple queens appear in the same row.
2. Column and Diagonal Pruning¶
To satisfy the column constraint, we can use a boolean array cols of length \(n\) to record whether each column has a queen. Before each placement decision, we use cols to prune columns that already have queens, and dynamically update the state of cols during backtracking.
Tip
Please note that the origin of the matrix is located in the upper-left corner, where the row index increases from top to bottom, and the column index increases from left to right.
So how do we handle diagonal constraints? Consider a square on the chessboard with row and column indices \((row, col)\). If we select a specific main diagonal in the matrix, we find that all squares on that diagonal have the same difference between their row and column indices, meaning that \(row - col\) is a constant value for all squares on the main diagonal.
In other words, if two squares satisfy \(row_1 - col_1 = row_2 - col_2\), they must be on the same main diagonal. Using this pattern, we can use the array diags1 shown in Figure 13-18 to record whether there is a queen on each main diagonal.
Similarly, for all squares on an anti-diagonal, the sum \(row + col\) is a constant value. We can likewise use the array diags2 to handle anti-diagonal constraints.
Figure 13-18 Handling column and diagonal constraints
3. Code Implementation¶
Please note that in an \(n\)-dimensional square matrix, the range of \(row - col\) is \([-n + 1, n - 1]\), and the range of \(row + col\) is \([0, 2n - 2]\). Therefore, the number of both main diagonals and anti-diagonals is \(2n - 1\), meaning the length of both arrays diags1 and diags2 is \(2n - 1\).
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 diagonal and anti-diagonal corresponding to this cell
diag1 = row - col + n - 1
diag2 = row + col
# Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if not cols[col] and not diags1[diag1] and not diags2[diag2]:
# Attempt: place the queen in this 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)
# Backtrack: restore this cell to an empty cell
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 chessboard, where 'Q' represents a queen and '#' represents an empty cell
state = [["#" for _ in range(n)] for _ in range(n)]
cols = [False] * n # Record whether there is a queen in the column
diags1 = [False] * (2 * n - 1) # Record whether there is a queen on the main diagonal
diags2 = [False] * (2 * n - 1) # Record whether there is a queen on the anti-diagonal
res = []
backtrack(0, n, state, res, cols, diags1, diags2)
return res
n_queens.cpp
/* Backtracking algorithm: N queens */
void backtrack(int row, int n, vector<vector<string>> &state, vector<vector<vector<string>>> &res, vector<bool> &cols,
vector<bool> &diags1, vector<bool> &diags2) {
// When all rows are placed, record the solution
if (row == n) {
res.push_back(state);
return;
}
// Traverse all columns
for (int col = 0; col < n; col++) {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
int diag1 = row - col + n - 1;
int diag2 = row + col;
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in this 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);
// Backtrack: restore this cell to an empty cell
state[row][col] = "#";
cols[col] = diags1[diag1] = diags2[diag2] = false;
}
}
}
/* Solve N queens */
vector<vector<vector<string>>> nQueens(int n) {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
vector<vector<string>> state(n, vector<string>(n, "#"));
vector<bool> cols(n, false); // Record whether there is a queen in the column
vector<bool> diags1(2 * n - 1, false); // Record whether there is a queen on the main diagonal
vector<bool> diags2(2 * n - 1, false); // Record whether there is a queen on the anti-diagonal
vector<vector<vector<string>>> res;
backtrack(0, n, state, res, cols, diags1, diags2);
return res;
}
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 diagonal and anti-diagonal corresponding to this cell
int diag1 = row - col + n - 1;
int diag2 = row + col;
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in this 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);
// Backtrack: restore this cell to an empty cell
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 chessboard, where 'Q' represents a queen and '#' represents an empty cell
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 whether there is a queen in the column
boolean[] diags1 = new boolean[2 * n - 1]; // Record whether there is a queen on the main diagonal
boolean[] diags2 = new boolean[2 * n - 1]; // Record whether there is a queen on the anti-diagonal
List<List<List<String>>> res = new ArrayList<>();
backtrack(0, n, state, res, cols, diags1, diags2);
return res;
}
n_queens.cs
/* Backtracking algorithm: N queens */
void Backtrack(int row, int n, List<List<string>> state, List<List<List<string>>> res,
bool[] cols, bool[] diags1, bool[] diags2) {
// When all rows are placed, record the solution
if (row == n) {
List<List<string>> copyState = [];
foreach (List<string> sRow in state) {
copyState.Add(new List<string>(sRow));
}
res.Add(copyState);
return;
}
// Traverse all columns
for (int col = 0; col < n; col++) {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
int diag1 = row - col + n - 1;
int diag2 = row + col;
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in this 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);
// Backtrack: restore this cell to an empty cell
state[row][col] = "#";
cols[col] = diags1[diag1] = diags2[diag2] = false;
}
}
}
/* Solve N queens */
List<List<List<string>>> NQueens(int n) {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
List<List<string>> state = [];
for (int i = 0; i < n; i++) {
List<string> row = [];
for (int j = 0; j < n; j++) {
row.Add("#");
}
state.Add(row);
}
bool[] cols = new bool[n]; // Record whether there is a queen in the column
bool[] diags1 = new bool[2 * n - 1]; // Record whether there is a queen on the main diagonal
bool[] diags2 = new bool[2 * n - 1]; // Record whether there is a queen on the anti-diagonal
List<List<List<string>>> res = [];
Backtrack(0, n, state, res, cols, diags1, diags2);
return res;
}
n_queens.go
/* Backtracking algorithm: N queens */
func backtrack(row, n int, state *[][]string, res *[][][]string, cols, diags1, diags2 *[]bool) {
// When all rows are placed, record the solution
if row == n {
newState := make([][]string, len(*state))
for i, _ := range newState {
newState[i] = make([]string, len((*state)[0]))
copy(newState[i], (*state)[i])
}
*res = append(*res, newState)
return
}
// Traverse all columns
for col := 0; col < n; col++ {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
diag1 := row - col + n - 1
diag2 := row + col
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if !(*cols)[col] && !(*diags1)[diag1] && !(*diags2)[diag2] {
// Attempt: place the queen in this cell
(*state)[row][col] = "Q"
(*cols)[col], (*diags1)[diag1], (*diags2)[diag2] = true, true, true
// Place the next row
backtrack(row+1, n, state, res, cols, diags1, diags2)
// Backtrack: restore this cell to an empty cell
(*state)[row][col] = "#"
(*cols)[col], (*diags1)[diag1], (*diags2)[diag2] = false, false, false
}
}
}
/* Solve N queens */
func nQueens(n int) [][][]string {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
state := make([][]string, n)
for i := 0; i < n; i++ {
row := make([]string, n)
for i := 0; i < n; i++ {
row[i] = "#"
}
state[i] = row
}
// Record whether there is a queen in the column
cols := make([]bool, n)
diags1 := make([]bool, 2*n-1)
diags2 := make([]bool, 2*n-1)
res := make([][][]string, 0)
backtrack(0, n, &state, &res, &cols, &diags1, &diags2)
return res
}
n_queens.swift
/* Backtracking algorithm: N queens */
func backtrack(row: Int, n: Int, state: inout [[String]], res: inout [[[String]]], cols: inout [Bool], diags1: inout [Bool], diags2: inout [Bool]) {
// When all rows are placed, record the solution
if row == n {
res.append(state)
return
}
// Traverse all columns
for col in 0 ..< n {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
let diag1 = row - col + n - 1
let diag2 = row + col
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if !cols[col] && !diags1[diag1] && !diags2[diag2] {
// Attempt: place the queen in this cell
state[row][col] = "Q"
cols[col] = true
diags1[diag1] = true
diags2[diag2] = true
// Place the next row
backtrack(row: row + 1, n: n, state: &state, res: &res, cols: &cols, diags1: &diags1, diags2: &diags2)
// Backtrack: restore this cell to an empty cell
state[row][col] = "#"
cols[col] = false
diags1[diag1] = false
diags2[diag2] = false
}
}
}
/* Solve N queens */
func nQueens(n: Int) -> [[[String]]] {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
var state = Array(repeating: Array(repeating: "#", count: n), count: n)
var cols = Array(repeating: false, count: n) // Record whether there is a queen in the column
var diags1 = Array(repeating: false, count: 2 * n - 1) // Record whether there is a queen on the main diagonal
var diags2 = Array(repeating: false, count: 2 * n - 1) // Record whether there is a queen on the anti-diagonal
var res: [[[String]]] = []
backtrack(row: 0, n: n, state: &state, res: &res, cols: &cols, diags1: &diags1, diags2: &diags2)
return res
}
n_queens.js
/* Backtracking algorithm: N queens */
function backtrack(row, n, state, res, cols, diags1, diags2) {
// When all rows are placed, record the solution
if (row === n) {
res.push(state.map((row) => row.slice()));
return;
}
// Traverse all columns
for (let col = 0; col < n; col++) {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
const diag1 = row - col + n - 1;
const diag2 = row + col;
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in this 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);
// Backtrack: restore this cell to an empty cell
state[row][col] = '#';
cols[col] = diags1[diag1] = diags2[diag2] = false;
}
}
}
/* Solve N queens */
function nQueens(n) {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
const state = Array.from({ length: n }, () => Array(n).fill('#'));
const cols = Array(n).fill(false); // Record whether there is a queen in the column
const diags1 = Array(2 * n - 1).fill(false); // Record whether there is a queen on the main diagonal
const diags2 = Array(2 * n - 1).fill(false); // Record whether there is a queen on the anti-diagonal
const res = [];
backtrack(0, n, state, res, cols, diags1, diags2);
return res;
}
n_queens.ts
/* Backtracking algorithm: N queens */
function backtrack(
row: number,
n: number,
state: string[][],
res: string[][][],
cols: boolean[],
diags1: boolean[],
diags2: boolean[]
): void {
// When all rows are placed, record the solution
if (row === n) {
res.push(state.map((row) => row.slice()));
return;
}
// Traverse all columns
for (let col = 0; col < n; col++) {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
const diag1 = row - col + n - 1;
const diag2 = row + col;
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in this 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);
// Backtrack: restore this cell to an empty cell
state[row][col] = '#';
cols[col] = diags1[diag1] = diags2[diag2] = false;
}
}
}
/* Solve N queens */
function nQueens(n: number): string[][][] {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
const state = Array.from({ length: n }, () => Array(n).fill('#'));
const cols = Array(n).fill(false); // Record whether there is a queen in the column
const diags1 = Array(2 * n - 1).fill(false); // Record whether there is a queen on the main diagonal
const diags2 = Array(2 * n - 1).fill(false); // Record whether there is a queen on the anti-diagonal
const res: string[][][] = [];
backtrack(0, n, state, res, cols, diags1, diags2);
return res;
}
n_queens.dart
/* Backtracking algorithm: N queens */
void backtrack(
int row,
int n,
List<List<String>> state,
List<List<List<String>>> res,
List<bool> cols,
List<bool> diags1,
List<bool> diags2,
) {
// When all rows are placed, record the solution
if (row == n) {
List<List<String>> copyState = [];
for (List<String> sRow in state) {
copyState.add(List.from(sRow));
}
res.add(copyState);
return;
}
// Traverse all columns
for (int col = 0; col < n; col++) {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
int diag1 = row - col + n - 1;
int diag2 = row + col;
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in this cell
state[row][col] = "Q";
cols[col] = true;
diags1[diag1] = true;
diags2[diag2] = true;
// Place the next row
backtrack(row + 1, n, state, res, cols, diags1, diags2);
// Backtrack: restore this cell to an empty cell
state[row][col] = "#";
cols[col] = false;
diags1[diag1] = false;
diags2[diag2] = false;
}
}
}
/* Solve N queens */
List<List<List<String>>> nQueens(int n) {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
List<List<String>> state = List.generate(n, (index) => List.filled(n, "#"));
List<bool> cols = List.filled(n, false); // Record whether there is a queen in the column
List<bool> diags1 = List.filled(2 * n - 1, false); // Record whether there is a queen on the main diagonal
List<bool> diags2 = List.filled(2 * n - 1, false); // Record whether there is a queen on the anti-diagonal
List<List<List<String>>> res = [];
backtrack(0, n, state, res, cols, diags1, diags2);
return res;
}
n_queens.rs
/* Backtracking algorithm: N queens */
fn backtrack(
row: usize,
n: usize,
state: &mut Vec<Vec<String>>,
res: &mut Vec<Vec<Vec<String>>>,
cols: &mut [bool],
diags1: &mut [bool],
diags2: &mut [bool],
) {
// When all rows are placed, record the solution
if row == n {
res.push(state.clone());
return;
}
// Traverse all columns
for col in 0..n {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
let diag1 = row + n - 1 - col;
let diag2 = row + col;
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if !cols[col] && !diags1[diag1] && !diags2[diag2] {
// Attempt: place the queen in this cell
state[row][col] = "Q".into();
(cols[col], diags1[diag1], diags2[diag2]) = (true, true, true);
// Place the next row
backtrack(row + 1, n, state, res, cols, diags1, diags2);
// Backtrack: restore this cell to an empty cell
state[row][col] = "#".into();
(cols[col], diags1[diag1], diags2[diag2]) = (false, false, false);
}
}
}
/* Solve N queens */
fn n_queens(n: usize) -> Vec<Vec<Vec<String>>> {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
let mut state: Vec<Vec<String>> = vec![vec!["#".to_string(); n]; n];
let mut cols = vec![false; n]; // Record whether there is a queen in the column
let mut diags1 = vec![false; 2 * n - 1]; // Record whether there is a queen on the main diagonal
let mut diags2 = vec![false; 2 * n - 1]; // Record whether there is a queen on the anti-diagonal
let mut res: Vec<Vec<Vec<String>>> = Vec::new();
backtrack(
0,
n,
&mut state,
&mut res,
&mut cols,
&mut diags1,
&mut diags2,
);
res
}
n_queens.c
/* Backtracking algorithm: N queens */
void backtrack(int row, int n, char state[MAX_SIZE][MAX_SIZE], char ***res, int *resSize, bool cols[MAX_SIZE],
bool diags1[2 * MAX_SIZE - 1], bool diags2[2 * MAX_SIZE - 1]) {
// When all rows are placed, record the solution
if (row == n) {
res[*resSize] = (char **)malloc(sizeof(char *) * n);
for (int i = 0; i < n; ++i) {
res[*resSize][i] = (char *)malloc(sizeof(char) * (n + 1));
strcpy(res[*resSize][i], state[i]);
}
(*resSize)++;
return;
}
// Traverse all columns
for (int col = 0; col < n; col++) {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
int diag1 = row - col + n - 1;
int diag2 = row + col;
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in this cell
state[row][col] = 'Q';
cols[col] = diags1[diag1] = diags2[diag2] = true;
// Place the next row
backtrack(row + 1, n, state, res, resSize, cols, diags1, diags2);
// Backtrack: restore this cell to an empty cell
state[row][col] = '#';
cols[col] = diags1[diag1] = diags2[diag2] = false;
}
}
}
/* Solve N queens */
char ***nQueens(int n, int *returnSize) {
char state[MAX_SIZE][MAX_SIZE];
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
state[i][j] = '#';
}
state[i][n] = '\0';
}
bool cols[MAX_SIZE] = {false}; // Record whether there is a queen in the column
bool diags1[2 * MAX_SIZE - 1] = {false}; // Record whether there is a queen on the main diagonal
bool diags2[2 * MAX_SIZE - 1] = {false}; // Record whether there is a queen on the anti-diagonal
char ***res = (char ***)malloc(sizeof(char **) * MAX_SIZE);
*returnSize = 0;
backtrack(0, n, state, res, returnSize, cols, diags1, diags2);
return res;
}
n_queens.kt
/* Backtracking algorithm: N queens */
fun backtrack(
row: Int,
n: Int,
state: MutableList<MutableList<String>>,
res: MutableList<MutableList<MutableList<String>>?>,
cols: BooleanArray,
diags1: BooleanArray,
diags2: BooleanArray
) {
// When all rows are placed, record the solution
if (row == n) {
val copyState = mutableListOf<MutableList<String>>()
for (sRow in state) {
copyState.add(sRow.toMutableList())
}
res.add(copyState)
return
}
// Traverse all columns
for (col in 0..<n) {
// Calculate the main diagonal and anti-diagonal corresponding to this cell
val diag1 = row - col + n - 1
val diag2 = row + col
// Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// Attempt: place the queen in this cell
state[row][col] = "Q"
diags2[diag2] = true
diags1[diag1] = diags2[diag2]
cols[col] = diags1[diag1]
// Place the next row
backtrack(row + 1, n, state, res, cols, diags1, diags2)
// Backtrack: restore this cell to an empty cell
state[row][col] = "#"
diags2[diag2] = false
diags1[diag1] = diags2[diag2]
cols[col] = diags1[diag1]
}
}
}
/* Solve N queens */
fun nQueens(n: Int): MutableList<MutableList<MutableList<String>>?> {
// Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
val state = mutableListOf<MutableList<String>>()
for (i in 0..<n) {
val row = mutableListOf<String>()
for (j in 0..<n) {
row.add("#")
}
state.add(row)
}
val cols = BooleanArray(n) // Record whether there is a queen in the column
val diags1 = BooleanArray(2 * n - 1) // Record whether there is a queen on the main diagonal
val diags2 = BooleanArray(2 * n - 1) // Record whether there is a queen on the anti-diagonal
val res = mutableListOf<MutableList<MutableList<String>>?>()
backtrack(0, n, state, res, cols, diags1, diags2)
return res
}
n_queens.rb
### Backtracking: n queens ###
def backtrack(row, n, state, res, cols, diags1, diags2)
# When all rows are placed, record the solution
if row == n
res << state.map { |row| row.dup }
return
end
# Traverse all columns
for col in 0...n
# Calculate the main diagonal and anti-diagonal corresponding to this cell
diag1 = row - col + n - 1
diag2 = row + col
# Pruning: do not allow queens to exist in the column, main diagonal, and anti-diagonal of this cell
if !cols[col] && !diags1[diag1] && !diags2[diag2]
# Attempt: place the queen in this 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)
# Backtrack: restore this cell to an empty cell
state[row][col] = "#"
cols[col] = diags1[diag1] = diags2[diag2] = false
end
end
end
### Solve n queens ###
def n_queens(n)
# Initialize an n*n chessboard, where 'Q' represents a queen and '#' represents an empty cell
state = Array.new(n) { Array.new(n, "#") }
cols = Array.new(n, false) # Record whether there is a queen in the column
diags1 = Array.new(2 * n - 1, false) # Record whether there is a queen on the main diagonal
diags2 = Array.new(2 * n - 1, false) # Record whether there is a queen on the anti-diagonal
res = []
backtrack(0, n, state, res, cols, diags1, diags2)
res
end
Placing \(n\) queens row by row, considering the column constraint, 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, and the copy operation uses \(O(n^2)\) time. Therefore, the overall time complexity is \(O(n! \cdot n^2)\). In practice, pruning based on diagonal constraints can also significantly reduce the search space, so the search efficiency is often better than the time complexity mentioned above.
The array state uses \(O(n^2)\) space, and the arrays cols, diags1, and diags2 each use \(O(n)\) space. The maximum recursion depth is \(n\), using \(O(n)\) stack frame space. Therefore, the space complexity is \(O(n^2)\).