---
comments: true
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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.
{ class="animation-figure" }
Figure 13-15 Solution to the 4 queens problem
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 `/`.
{ class="animation-figure" }
Figure 13-16 Constraints of the n queens problem
### 1. 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.
{ class="animation-figure" }
Figure 13-17 Row-by-row placing strategy
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. 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.
{ class="animation-figure" }
Figure 13-18 Handling column and diagonal constraints
### 3. 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],
):
"""回溯算法：n 皇后"""
# 当放置完所有行时，记录解
if row == n:
res.append([list(row) for row in state])
return
# 遍历所有列
for col in range(n):
# 计算该格子对应的主对角线和次对角线
diag1 = row - col + n - 1
diag2 = row + col
# 剪枝：不允许该格子所在列、主对角线、次对角线上存在皇后
if not cols[col] and not diags1[diag1] and not diags2[diag2]:
# 尝试：将皇后放置在该格子
state[row][col] = "Q"
cols[col] = diags1[diag1] = diags2[diag2] = True
# 放置下一行
backtrack(row + 1, n, state, res, cols, diags1, diags2)
# 回退：将该格子恢复为空位
state[row][col] = "#"
cols[col] = diags1[diag1] = diags2[diag2] = False
def n_queens(n: int) -> list[list[list[str]]]:
"""求解 n 皇后"""
# 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
state = [["#" for _ in range(n)] for _ in range(n)]
cols = [False] * n # 记录列是否有皇后
diags1 = [False] * (2 * n - 1) # 记录主对角线上是否有皇后
diags2 = [False] * (2 * n - 1) # 记录次对角线上是否有皇后
res = []
backtrack(0, n, state, res, cols, diags1, diags2)
return res
```
=== "C++"
```cpp title="n_queens.cpp"
/* 回溯算法：n 皇后 */
void backtrack(int row, int n, vector> &state, vector>> &res, vector &cols,
vector &diags1, vector &diags2) {
// 当放置完所有行时，记录解
if (row == n) {
res.push_back(state);
return;
}
// 遍历所有列
for (int col = 0; col < n; col++) {
// 计算该格子对应的主对角线和次对角线
int diag1 = row - col + n - 1;
int diag2 = row + col;
// 剪枝：不允许该格子所在列、主对角线、次对角线上存在皇后
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// 尝试：将皇后放置在该格子
state[row][col] = "Q";
cols[col] = diags1[diag1] = diags2[diag2] = true;
// 放置下一行
backtrack(row + 1, n, state, res, cols, diags1, diags2);
// 回退：将该格子恢复为空位
state[row][col] = "#";
cols[col] = diags1[diag1] = diags2[diag2] = false;
}
}
}
/* 求解 n 皇后 */
vector>> nQueens(int n) {
// 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
vector> state(n, vector(n, "#"));
vector cols(n, false); // 记录列是否有皇后
vector diags1(2 * n - 1, false); // 记录主对角线上是否有皇后
vector diags2(2 * n - 1, false); // 记录次对角线上是否有皇后
vector>> res;
backtrack(0, n, state, res, cols, diags1, diags2);
return res;
}
```
=== "Java"
```java title="n_queens.java"
/* 回溯算法：n 皇后 */
void backtrack(int row, int n, List> state, List>> res,
boolean[] cols, boolean[] diags1, boolean[] diags2) {
// 当放置完所有行时，记录解
if (row == n) {
List> copyState = new ArrayList<>();
for (List sRow : state) {
copyState.add(new ArrayList<>(sRow));
}
res.add(copyState);
return;
}
// 遍历所有列
for (int col = 0; col < n; col++) {
// 计算该格子对应的主对角线和次对角线
int diag1 = row - col + n - 1;
int diag2 = row + col;
// 剪枝：不允许该格子所在列、主对角线、次对角线上存在皇后
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// 尝试：将皇后放置在该格子
state.get(row).set(col, "Q");
cols[col] = diags1[diag1] = diags2[diag2] = true;
// 放置下一行
backtrack(row + 1, n, state, res, cols, diags1, diags2);
// 回退：将该格子恢复为空位
state.get(row).set(col, "#");
cols[col] = diags1[diag1] = diags2[diag2] = false;
}
}
}
/* 求解 n 皇后 */
List>> nQueens(int n) {
// 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
List> state = new ArrayList<>();
for (int i = 0; i < n; i++) {
List row = new ArrayList<>();
for (int j = 0; j < n; j++) {
row.add("#");
}
state.add(row);
}
boolean[] cols = new boolean[n]; // 记录列是否有皇后
boolean[] diags1 = new boolean[2 * n - 1]; // 记录主对角线上是否有皇后
boolean[] diags2 = new boolean[2 * n - 1]; // 记录次对角线上是否有皇后
List>> res = new ArrayList<>();
backtrack(0, n, state, res, cols, diags1, diags2);
return res;
}
```
=== "C#"
```csharp title="n_queens.cs"
/* 回溯算法：n 皇后 */
void Backtrack(int row, int n, List> state, List>> res,
bool[] cols, bool[] diags1, bool[] diags2) {
// 当放置完所有行时，记录解
if (row == n) {
List> copyState = [];
foreach (List sRow in state) {
copyState.Add(new List(sRow));
}
res.Add(copyState);
return;
}
// 遍历所有列
for (int col = 0; col < n; col++) {
// 计算该格子对应的主对角线和次对角线
int diag1 = row - col + n - 1;
int diag2 = row + col;
// 剪枝：不允许该格子所在列、主对角线、次对角线上存在皇后
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// 尝试：将皇后放置在该格子
state[row][col] = "Q";
cols[col] = diags1[diag1] = diags2[diag2] = true;
// 放置下一行
Backtrack(row + 1, n, state, res, cols, diags1, diags2);
// 回退：将该格子恢复为空位
state[row][col] = "#";
cols[col] = diags1[diag1] = diags2[diag2] = false;
}
}
}
/* 求解 n 皇后 */
List>> NQueens(int n) {
// 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
List> state = [];
for (int i = 0; i < n; i++) {
List row = [];
for (int j = 0; j < n; j++) {
row.Add("#");
}
state.Add(row);
}
bool[] cols = new bool[n]; // 记录列是否有皇后
bool[] diags1 = new bool[2 * n - 1]; // 记录主对角线上是否有皇后
bool[] diags2 = new bool[2 * n - 1]; // 记录次对角线上是否有皇后
List>> res = [];
Backtrack(0, n, state, res, cols, diags1, diags2);
return res;
}
```
=== "Go"
```go title="n_queens.go"
/* 回溯算法：n 皇后 */
func backtrack(row, n int, state *[][]string, res *[][][]string, cols, diags1, diags2 *[]bool) {
// 当放置完所有行时，记录解
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)
}
// 遍历所有列
for col := 0; col < n; col++ {
// 计算该格子对应的主对角线和次对角线
diag1 := row - col + n - 1
diag2 := row + col
// 剪枝：不允许该格子所在列、主对角线、次对角线上存在皇后
if !(*cols)[col] && !(*diags1)[diag1] && !(*diags2)[diag2] {
// 尝试：将皇后放置在该格子
(*state)[row][col] = "Q"
(*cols)[col], (*diags1)[diag1], (*diags2)[diag2] = true, true, true
// 放置下一行
backtrack(row+1, n, state, res, cols, diags1, diags2)
// 回退：将该格子恢复为空位
(*state)[row][col] = "#"
(*cols)[col], (*diags1)[diag1], (*diags2)[diag2] = false, false, false
}
}
}
/* 求解 n 皇后 */
func nQueens(n int) [][][]string {
// 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
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
}
// 记录列是否有皇后
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
}
```
=== "Swift"
```swift title="n_queens.swift"
/* 回溯算法：n 皇后 */
func backtrack(row: Int, n: Int, state: inout [[String]], res: inout [[[String]]], cols: inout [Bool], diags1: inout [Bool], diags2: inout [Bool]) {
// 当放置完所有行时，记录解
if row == n {
res.append(state)
return
}
// 遍历所有列
for col in 0 ..< n {
// 计算该格子对应的主对角线和次对角线
let diag1 = row - col + n - 1
let diag2 = row + col
// 剪枝：不允许该格子所在列、主对角线、次对角线上存在皇后
if !cols[col] && !diags1[diag1] && !diags2[diag2] {
// 尝试：将皇后放置在该格子
state[row][col] = "Q"
cols[col] = true
diags1[diag1] = true
diags2[diag2] = true
// 放置下一行
backtrack(row: row + 1, n: n, state: &state, res: &res, cols: &cols, diags1: &diags1, diags2: &diags2)
// 回退：将该格子恢复为空位
state[row][col] = "#"
cols[col] = false
diags1[diag1] = false
diags2[diag2] = false
}
}
}
/* 求解 n 皇后 */
func nQueens(n: Int) -> [[[String]]] {
// 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
var state = Array(repeating: Array(repeating: "#", count: n), count: n)
var cols = Array(repeating: false, count: n) // 记录列是否有皇后
var diags1 = Array(repeating: false, count: 2 * n - 1) // 记录主对角线上是否有皇后
var diags2 = Array(repeating: false, count: 2 * n - 1) // 记录次对角线上是否有皇后
var res: [[[String]]] = []
backtrack(row: 0, n: n, state: &state, res: &res, cols: &cols, diags1: &diags1, diags2: &diags2)
return res
}
```
=== "JS"
```javascript title="n_queens.js"
/* 回溯算法：n 皇后 */
function backtrack(row, n, state, res, cols, diags1, diags2) {
// 当放置完所有行时，记录解
if (row === n) {
res.push(state.map((row) => row.slice()));
return;
}
// 遍历所有列
for (let col = 0; col < n; col++) {
// 计算该格子对应的主对角线和次对角线
const diag1 = row - col + n - 1;
const diag2 = row + col;
// 剪枝：不允许该格子所在列、主对角线、次对角线上存在皇后
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// 尝试：将皇后放置在该格子
state[row][col] = 'Q';
cols[col] = diags1[diag1] = diags2[diag2] = true;
// 放置下一行
backtrack(row + 1, n, state, res, cols, diags1, diags2);
// 回退：将该格子恢复为空位
state[row][col] = '#';
cols[col] = diags1[diag1] = diags2[diag2] = false;
}
}
}
/* 求解 n 皇后 */
function nQueens(n) {
// 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
const state = Array.from({ length: n }, () => Array(n).fill('#'));
const cols = Array(n).fill(false); // 记录列是否有皇后
const diags1 = Array(2 * n - 1).fill(false); // 记录主对角线上是否有皇后
const diags2 = Array(2 * n - 1).fill(false); // 记录次对角线上是否有皇后
const res = [];
backtrack(0, n, state, res, cols, diags1, diags2);
return res;
}
```
=== "TS"
```typescript title="n_queens.ts"
/* 回溯算法：n 皇后 */
function backtrack(
row: number,
n: number,
state: string[][],
res: string[][][],
cols: boolean[],
diags1: boolean[],
diags2: boolean[]
): void {
// 当放置完所有行时，记录解
if (row === n) {
res.push(state.map((row) => row.slice()));
return;
}
// 遍历所有列
for (let col = 0; col < n; col++) {
// 计算该格子对应的主对角线和次对角线
const diag1 = row - col + n - 1;
const diag2 = row + col;
// 剪枝：不允许该格子所在列、主对角线、次对角线上存在皇后
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// 尝试：将皇后放置在该格子
state[row][col] = 'Q';
cols[col] = diags1[diag1] = diags2[diag2] = true;
// 放置下一行
backtrack(row + 1, n, state, res, cols, diags1, diags2);
// 回退：将该格子恢复为空位
state[row][col] = '#';
cols[col] = diags1[diag1] = diags2[diag2] = false;
}
}
}
/* 求解 n 皇后 */
function nQueens(n: number): string[][][] {
// 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
const state = Array.from({ length: n }, () => Array(n).fill('#'));
const cols = Array(n).fill(false); // 记录列是否有皇后
const diags1 = Array(2 * n - 1).fill(false); // 记录主对角线上是否有皇后
const diags2 = Array(2 * n - 1).fill(false); // 记录次对角线上是否有皇后
const res: string[][][] = [];
backtrack(0, n, state, res, cols, diags1, diags2);
return res;
}
```
=== "Dart"
```dart title="n_queens.dart"
/* 回溯算法：n 皇后 */
void backtrack(
int row,
int n,
List> state,
List>> res,
List cols,
List diags1,
List diags2,
) {
// 当放置完所有行时，记录解
if (row == n) {
List> copyState = [];
for (List sRow in state) {
copyState.add(List.from(sRow));
}
res.add(copyState);
return;
}
// 遍历所有列
for (int col = 0; col < n; col++) {
// 计算该格子对应的主对角线和次对角线
int diag1 = row - col + n - 1;
int diag2 = row + col;
// 剪枝：不允许该格子所在列、主对角线、次对角线上存在皇后
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// 尝试：将皇后放置在该格子
state[row][col] = "Q";
cols[col] = true;
diags1[diag1] = true;
diags2[diag2] = true;
// 放置下一行
backtrack(row + 1, n, state, res, cols, diags1, diags2);
// 回退：将该格子恢复为空位
state[row][col] = "#";
cols[col] = false;
diags1[diag1] = false;
diags2[diag2] = false;
}
}
}
/* 求解 n 皇后 */
List>> nQueens(int n) {
// 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
List> state = List.generate(n, (index) => List.filled(n, "#"));
List cols = List.filled(n, false); // 记录列是否有皇后
List diags1 = List.filled(2 * n - 1, false); // 记录主对角线上是否有皇后
List diags2 = List.filled(2 * n - 1, false); // 记录次对角线上是否有皇后
List>> res = [];
backtrack(0, n, state, res, cols, diags1, diags2);
return res;
}
```
=== "Rust"
```rust title="n_queens.rs"
/* 回溯算法：n 皇后 */
fn backtrack(
row: usize,
n: usize,
state: &mut Vec>,
res: &mut Vec>>,
cols: &mut [bool],
diags1: &mut [bool],
diags2: &mut [bool],
) {
// 当放置完所有行时，记录解
if row == n {
let mut copy_state: Vec> = Vec::new();
for s_row in state.clone() {
copy_state.push(s_row);
}
res.push(copy_state);
return;
}
// 遍历所有列
for col in 0..n {
// 计算该格子对应的主对角线和次对角线
let diag1 = row + n - 1 - col;
let diag2 = row + col;
// 剪枝：不允许该格子所在列、主对角线、次对角线上存在皇后
if !cols[col] && !diags1[diag1] && !diags2[diag2] {
// 尝试：将皇后放置在该格子
state.get_mut(row).unwrap()[col] = "Q".into();
(cols[col], diags1[diag1], diags2[diag2]) = (true, true, true);
// 放置下一行
backtrack(row + 1, n, state, res, cols, diags1, diags2);
// 回退：将该格子恢复为空位
state.get_mut(row).unwrap()[col] = "#".into();
(cols[col], diags1[diag1], diags2[diag2]) = (false, false, false);
}
}
}
/* 求解 n 皇后 */
fn n_queens(n: usize) -> Vec>> {
// 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
let mut state: Vec> = Vec::new();
for _ in 0..n {
let mut row: Vec = Vec::new();
for _ in 0..n {
row.push("#".into());
}
state.push(row);
}
let mut cols = vec![false; n]; // 记录列是否有皇后
let mut diags1 = vec![false; 2 * n - 1]; // 记录主对角线上是否有皇后
let mut diags2 = vec![false; 2 * n - 1]; // 记录次对角线上是否有皇后
let mut res: Vec>> = Vec::new();
backtrack(
0,
n,
&mut state,
&mut res,
&mut cols,
&mut diags1,
&mut diags2,
);
res
}
```
=== "C"
```c title="n_queens.c"
/* 回溯算法：n 皇后 */
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]) {
// 当放置完所有行时，记录解
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;
}
// 遍历所有列
for (int col = 0; col < n; col++) {
// 计算该格子对应的主对角线和次对角线
int diag1 = row - col + n - 1;
int diag2 = row + col;
// 剪枝：不允许该格子所在列、主对角线、次对角线上存在皇后
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
// 尝试：将皇后放置在该格子
state[row][col] = 'Q';
cols[col] = diags1[diag1] = diags2[diag2] = true;
// 放置下一行
backtrack(row + 1, n, state, res, resSize, cols, diags1, diags2);
// 回退：将该格子恢复为空位
state[row][col] = '#';
cols[col] = diags1[diag1] = diags2[diag2] = false;
}
}
}
/* 求解 n 皇后 */
char ***nQueens(int n, int *returnSize) {
char state[MAX_SIZE][MAX_SIZE];
// 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
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}; // 记录列是否有皇后
bool diags1[2 * MAX_SIZE - 1] = {false}; // 记录主对角线上是否有皇后
bool diags2[2 * MAX_SIZE - 1] = {false}; // 记录次对角线上是否有皇后
char ***res = (char ***)malloc(sizeof(char **) * MAX_SIZE);
*returnSize = 0;
backtrack(0, n, state, res, returnSize, cols, diags1, diags2);
return res;
}
```
=== "Kotlin"
```kotlin title="n_queens.kt"
/* 回溯算法：n 皇后 */
fun backtrack(
row: Int,
n: Int,
state: MutableList>,
res: MutableList>?>,
cols: BooleanArray,
diags1: BooleanArray,
diags2: BooleanArray
) {
// 当放置完所有行时，记录解
if (row == n) {
val copyState = mutableListOf>()
for (sRow in state) {
copyState.add(sRow.toMutableList())
}
res.add(copyState)
return
}
// 遍历所有列
for (col in 0..>?> {
// 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
val state = mutableListOf>()
for (i in 0..()
for (j in 0..>?>()
backtrack(0, n, state, res, cols, diags1, diags2)
return res
}
```
=== "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}
```
??? pythontutor "Code Visualization"
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 $O(n^2)$**.