We have discussed Knight’s tour and Rat in a Maze problem in Set 1 and Set 2 respectively. Let us discuss N Queen as another example problem that can be solved using backtracking.
The N Queen is the problem of placing N chess queens on an N×N chessboard so that no two queens attack each other. For example, the following is a solution for the 4 Queen problem.
The expected output is in form of a matrix that has ‘Q’s for the blocks where queens are placed and the empty spaces are represented by ‘.’s . For example, the following is the output matrix for the above 4 queen solution.
. . Q .
Q . . .
. . . Q
. Q . .
Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.
Naive Algorithm
Generate all possible configurations of queens on board and print a configuration that satisfies the given constraints.
while there are untried configurations{ generate the next configuration if queens don't attack in this configuration then { print this configuration; }}Backtracking AlgorithmMethod 1:
The idea is to place queens one by one in different columns, starting from the leftmost column. When we place a queen in a column, we check for clashes with already placed queens. In the current column, if we find a row for which there is no clash, we mark this row and column as part of the solution. If we do not find such a row due to clashes, then we backtrack and return false.
Method 1:1) Start in the leftmost column2) If all queens are placed return true3) Try all rows in the current column. Do following for every tried row. a) If the queen can be placed safely in this row then mark this [row, column] as part of the solution and recursively check if placing queen here leads to a solution. b) If placing the queen in [row, column] leads to a solution then return true. c) If placing queen doesn't lead to a solution then unmark this [row, column] (Backtrack) and go to step (a) to try other rows.4) If all rows have been tried and nothing worked, return false to trigger backtracking.
Implementation of Backtracking solutionby method 1:
C++
/* C++ program to solve N Queen Problem using
backtracking */
#include <bits/stdc++.h>
#define N 4
using namespace std;
/* A utility function to print solution */
void printSolution(int board[N][N])
{
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++)
if(board[i][j])
cout << "Q ";
else cout<<". ";
printf("\n");
}
}
/* A utility function to check if a queen can
be placed on board[row][col]. Note that this
function is called when "col" queens are
already placed in columns from 0 to col -1.
So we need to check only left side for
attacking queens */
bool isSafe(int board[N][N], int row, int col)
{
int i, j;
/* Check this row on left side */
for (i = 0; i < col; i++)
if (board[row][i])
return false;
/* Check upper diagonal on left side */
for (i = row, j = col; i >= 0 && j >= 0; i--, j--)
if (board[i][j])
return false;
/* Check lower diagonal on left side */
for (i = row, j = col; j >= 0 && i < N; i++, j--)
if (board[i][j])
return false;
return true;
}
/* A recursive utility function to solve N
Queen problem */
bool solveNQUtil(int board[N][N], int col)
{
/* base case: If all queens are placed
then return true */
if (col >= N)
return true;
/* Consider this column and try placing
this queen in all rows one by one */
for (int i = 0; i < N; i++) {
/* Check if the queen can be placed on
board[i][col] */
if (isSafe(board, i, col)) {
/* Place this queen in board[i][col] */
board[i][col] = 1;
/* recur to place rest of the queens */
if (solveNQUtil(board, col + 1))
return true;
/* If placing queen in board[i][col]
doesn't lead to a solution, then
remove queen from board[i][col] */
board[i][col] = 0; // BACKTRACK
}
}
/* If the queen cannot be placed in any row in
this column col then return false */
return false;
}
/* This function solves the N Queen problem using
Backtracking. It mainly uses solveNQUtil() to
solve the problem. It returns false if queens
cannot be placed, otherwise, return true and
prints placement of queens in the form of 1s.
Please note that there may be more than one
solutions, this function prints one of the
feasible solutions.*/
bool solveNQ()
{
int board[N][N] = { { 0, 0, 0, 0 },
{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 } };
if (solveNQUtil(board, 0) == false) {
cout << "Solution does not exist";
return false;
}
printSolution(board);
return true;
}
// driver program to test above function
int main()
{
solveNQ();
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
C
/* C program to solve N Queen Problem using
backtracking */
#define N 4
#include <stdbool.h>
#include <stdio.h>
/* A utility function to print solution */
void printSolution(int board[N][N])
{
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++)
printf(" %d ", board[i][j]);
printf("\n");
}
}
/* A utility function to check if a queen can
be placed on board[row][col]. Note that this
function is called when "col" queens are
already placed in columns from 0 to col -1.
So we need to check only left side for
attacking queens */
bool isSafe(int board[N][N], int row, int col)
{
int i, j;
/* Check this row on left side */
for (i = 0; i < col; i++)
if (board[row][i])
return false;
/* Check upper diagonal on left side */
for (i = row, j = col; i >= 0 && j >= 0; i--, j--)
if (board[i][j])
return false;
/* Check lower diagonal on left side */
for (i = row, j = col; j >= 0 && i < N; i++, j--)
if (board[i][j])
return false;
return true;
}
/* A recursive utility function to solve N
Queen problem */
bool solveNQUtil(int board[N][N], int col)
{
/* base case: If all queens are placed
then return true */
if (col >= N)
return true;
/* Consider this column and try placing
this queen in all rows one by one */
for (int i = 0; i < N; i++) {
/* Check if the queen can be placed on
board[i][col] */
if (isSafe(board, i, col)) {
/* Place this queen in board[i][col] */
board[i][col] = 1;
/* recur to place rest of the queens */
if (solveNQUtil(board, col + 1))
return true;
/* If placing queen in board[i][col]
doesn't lead to a solution, then
remove queen from board[i][col] */
board[i][col] = 0; // BACKTRACK
}
}
/* If the queen cannot be placed in any row in
this column col then return false */
return false;
}
/* This function solves the N Queen problem using
Backtracking. It mainly uses solveNQUtil() to
solve the problem. It returns false if queens
cannot be placed, otherwise, return true and
prints placement of queens in the form of 1s.
Please note that there may be more than one
solutions, this function prints one of the
feasible solutions.*/
bool solveNQ()
{
int board[N][N] = { { 0, 0, 0, 0 },
{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 } };
if (solveNQUtil(board, 0) == false) {
printf("Solution does not exist");
return false;
}
printSolution(board);
return true;
}
// driver program to test above function
int main()
{
solveNQ();
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
Java
/* Java program to solve N Queen Problem using
backtracking */
public class NQueenProblem {
final int N = 4;
/* A utility function to print solution */
void printSolution(int board[][])
{
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++)
System.out.print(" " + board[i][j]
+ " ");
System.out.println();
}
}
/* A utility function to check if a queen can
be placed on board[row][col]. Note that this
function is called when "col" queens are already
placeed in columns from 0 to col -1. So we need
to check only left side for attacking queens */
boolean isSafe(int board[][], int row, int col)
{
int i, j;
/* Check this row on left side */
for (i = 0; i < col; i++)
if (board[row][i] == 1)
return false;
/* Check upper diagonal on left side */
for (i = row, j = col; i >= 0 && j >= 0; i--, j--)
if (board[i][j] == 1)
return false;
/* Check lower diagonal on left side */
for (i = row, j = col; j >= 0 && i < N; i++, j--)
if (board[i][j] == 1)
return false;
return true;
}
/* A recursive utility function to solve N
Queen problem */
boolean solveNQUtil(int board[][], int col)
{
/* base case: If all queens are placed
then return true */
if (col >= N)
return true;
/* Consider this column and try placing
this queen in all rows one by one */
for (int i = 0; i < N; i++) {
/* Check if the queen can be placed on
board[i][col] */
if (isSafe(board, i, col)) {
/* Place this queen in board[i][col] */
board[i][col] = 1;
/* recur to place rest of the queens */
if (solveNQUtil(board, col + 1) == true)
return true;
/* If placing queen in board[i][col]
doesn't lead to a solution then
remove queen from board[i][col] */
board[i][col] = 0; // BACKTRACK
}
}
/* If the queen can not be placed in any row in
this column col, then return false */
return false;
}
/* This function solves the N Queen problem using
Backtracking. It mainly uses solveNQUtil () to
solve the problem. It returns false if queens
cannot be placed, otherwise, return true and
prints placement of queens in the form of 1s.
Please note that there may be more than one
solutions, this function prints one of the
feasible solutions.*/
boolean solveNQ()
{
int board[][] = { { 0, 0, 0, 0 },
{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 } };
if (solveNQUtil(board, 0) == false) {
System.out.print("Solution does not exist");
return false;
}
printSolution(board);
return true;
}
// driver program to test above function
public static void main(String args[])
{
NQueenProblem Queen = new NQueenProblem();
Queen.solveNQ();
}
}
// This code is contributed by Abhishek Shankhadhar
Python3
# Python3 program to solve N Queen
# Problem using backtracking
global N
N = 4
def printSolution(board):
for i in range(N):
for j in range(N):
print(board[i][j], end = " ")
print()
# A utility function to check if a queen can
# be placed on board[row][col]. Note that this
# function is called when "col" queens are
# already placed in columns from 0 to col -1.
# So we need to check only left side for
# attacking queens
def isSafe(board, row, col):
# Check this row on left side
for i in range(col):
if board[row][i] == 1:
return False
# Check upper diagonal on left side
for i, j in zip(range(row, -1, -1),
range(col, -1, -1)):
if board[i][j] == 1:
return False
# Check lower diagonal on left side
for i, j in zip(range(row, N, 1),
range(col, -1, -1)):
if board[i][j] == 1:
return False
return True
def solveNQUtil(board, col):
# base case: If all queens are placed
# then return true
if col >= N:
return True
# Consider this column and try placing
# this queen in all rows one by one
for i in range(N):
if isSafe(board, i, col):
# Place this queen in board[i][col]
board[i][col] = 1
# recur to place rest of the queens
if solveNQUtil(board, col + 1) == True:
return True
# If placing queen in board[i][col
# doesn't lead to a solution, then
# queen from board[i][col]
board[i][col] = 0
# if the queen can not be placed in any row in
# this column col then return false
return False
# This function solves the N Queen problem using
# Backtracking. It mainly uses solveNQUtil() to
# solve the problem. It returns false if queens
# cannot be placed, otherwise return true and
# placement of queens in the form of 1s.
# note that there may be more than one
# solutions, this function prints one of the
# feasible solutions.
def solveNQ():
board = [ [0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0] ]
if solveNQUtil(board, 0) == False:
print ("Solution does not exist")
return False
printSolution(board)
return True
# Driver Code
solveNQ()
# This code is contributed by Divyanshu Mehta
C#
// C# program to solve N Queen Problem
// using backtracking
using System;
class GFG
{
readonly int N = 4;
/* A utility function to print solution */
void printSolution(int [,]board)
{
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
Console.Write(" " + board[i, j]
+ " ");
Console.WriteLine();
}
}
/* A utility function to check if a queen can
be placed on board[row,col]. Note that this
function is called when "col" queens are already
placeed in columns from 0 to col -1. So we need
to check only left side for attacking queens */
bool isSafe(int [,]board, int row, int col)
{
int i, j;
/* Check this row on left side */
for (i = 0; i < col; i++)
if (board[row,i] == 1)
return false;
/* Check upper diagonal on left side */
for (i = row, j = col; i >= 0 &&
j >= 0; i--, j--)
if (board[i,j] == 1)
return false;
/* Check lower diagonal on left side */
for (i = row, j = col; j >= 0 &&
i < N; i++, j--)
if (board[i, j] == 1)
return false;
return true;
}
/* A recursive utility function to solve N
Queen problem */
bool solveNQUtil(int [,]board, int col)
{
/* base case: If all queens are placed
then return true */
if (col >= N)
return true;
/* Consider this column and try placing
this queen in all rows one by one */
for (int i = 0; i < N; i++)
{
/* Check if the queen can be placed on
board[i,col] */
if (isSafe(board, i, col))
{
/* Place this queen in board[i,col] */
board[i, col] = 1;
/* recur to place rest of the queens */
if (solveNQUtil(board, col + 1) == true)
return true;
/* If placing queen in board[i,col]
doesn't lead to a solution then
remove queen from board[i,col] */
board[i, col] = 0; // BACKTRACK
}
}
/* If the queen can not be placed in any row in
this column col, then return false */
return false;
}
/* This function solves the N Queen problem using
Backtracking. It mainly uses solveNQUtil () to
solve the problem. It returns false if queens
cannot be placed, otherwise, return true and
prints placement of queens in the form of 1s.
Please note that there may be more than one
solutions, this function prints one of the
feasible solutions.*/
bool solveNQ()
{
int [,]board = {{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 }};
if (solveNQUtil(board, 0) == false)
{
Console.Write("Solution does not exist");
return false;
}
printSolution(board);
return true;
}
// Driver Code
public static void Main(String []args)
{
GFG Queen = new GFG();
Queen.solveNQ();
}
}
// This code is contributed by Princi Singh
Javascript
<script>
// JavaScript program to solve N Queen
// Problem using backtracking
const N = 4
function printSolution(board)
{
for(let i = 0; i < N; i++)
{
for(let j = 0; j < N; j++)
{
document.write(board[i][j], " ")
}
document.write("</br>")
}
}
// A utility function to check if a queen can
// be placed on board[row][col]. Note that this
// function is called when "col" queens are
// already placed in columns from 0 to col -1.
// So we need to check only left side for
// attacking queens
function isSafe(board, row, col)
{
// Check this row on left side
for(let i = 0; i < col; i++){
if(board[row][i] == 1)
return false
}
// Check upper diagonal on left side
for (i = row, j = col; i >= 0 && j >= 0; i--, j--)
if (board[i][j])
return false
// Check lower diagonal on left side
for (i = row, j = col; j >= 0 && i < N; i++, j--)
if (board[i][j])
return false
return true
}
function solveNQUtil(board, col){
// base case: If all queens are placed
// then return true
if(col >= N)
return true
// Consider this column and try placing
// this queen in all rows one by one
for(let i=0;i<N;i++){
if(isSafe(board, i, col)==true){
// Place this queen in board[i][col]
board[i][col] = 1
// recur to place rest of the queens
if(solveNQUtil(board, col + 1) == true)
return true
// If placing queen in board[i][col
// doesn't lead to a solution, then
// queen from board[i][col]
board[i][col] = 0
}
}
// if the queen can not be placed in any row in
// this column col then return false
return false
}
// This function solves the N Queen problem using
// Backtracking. It mainly uses solveNQUtil() to
// solve the problem. It returns false if queens
// cannot be placed, otherwise return true and
// placement of queens in the form of 1s.
// note that there may be more than one
// solutions, this function prints one of the
// feasible solutions.
function solveNQ(){
let board = [ [0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0] ]
if(solveNQUtil(board, 0) == false){
document.write("Solution does not exist")
return false
}
printSolution(board)
return true
}
// Driver Code
solveNQ()
// This code is contributed by shinjanpatra
</script>
Output
. . Q . Q . . . . . . Q . Q . .
Time Complexity: O(N!)
Auxiliary Space: O(N2)
Backtracking AlgorithmMethod 2:
The idea is to place queens one by one in different rows, starting from the topmost row. When we place a queen in a row, we check for clashes with already placed queens. In the current column, if we find a row for which there is no clash, we mark this row and column as part of the solution. If we do not find such a row due to clashes, then we backtrack and return false.
Method 2:
0) Make a board, make a space to collect all solution states.
1) Start in the topmost row.
2) Make a recursive function which takes state of board and the current row number
as its parameter.
3) Fill a queen in a safe place and use this state of board to advance to next recursive
call, add 1 to the current row. Revert the state of board after making the call.
a) Safe function checks the current column, left top diagonal and right top diagonal.
b) If no queen is present then fill else return false and stop exploring that state
and track back to the next possible solution state
4) Keep calling the function till the current row is out of bound.
5) If current row reaches the number of rows in the board then the board is filled.
6) Store the state and return.
Implementation of Backtracking solution by Method 2:
C++
#include <bits/stdc++.h>
using namespace std;
// store all the possible answers
vector<vector<string> > answer;
// print the board
void print_board()
{
for (auto& str : answer[1]) {
for (auto& letter : str)
cout << letter << " ";
cout << endl;
}
return;
}
// we need to check in three directions
// 1. in the same column above the current position
// 2. in the left top diagonal from the given cell
// 3. in the right top diagonal from the given cell
int safe(int row, int col, vector<string>& board)
{
for (int i = 0; i < board.size(); i++) {
if (board[i][col] == 'Q')
return false;
}
int i = row, j = col;
while (i >= 0 && j >= 0)
if (board[i--][j--] == 'Q')
return false;
i = row, j = col;
while (i >= 0 && j < board.size())
if (board[i--][j++] == 'Q')
return false;
return true;
}
// rec function here will fill the queens
// 1. there can be only one queen in one row
// 2. if we filled the final row in the board then row will
// be equal to total number of rows in board
// 3. push that board configuration in answer set because
// there will be more than one answers for filling the board
// with n-queens
void rec(vector<string> board, int row)
{
if (row == board.size()) {
answer.push_back(board);
return;
}
for (int i = 0; i < board.size(); i++) {
// for each position check if it is safe and if it
// safe make a recursive call with
// row+1,board[i][j]='Q' and then revert the change
// in board that is make the board[i][j]='.' again to
// generate more solutions
if (safe(row, i, board)) {
board[row][i] = 'Q';
rec(board, row + 1);
board[row][i] = '.';
}
}
return;
}
// function to solve n queens
vector<vector<string> > solveNQueens(int n)
{
string s;
for (int i = 0; i < n; i++)
s += '.';
// vector of string will make our board which is
// initially all empty
vector<string> board(n, s);
rec(board, 0);
return answer;
}
int main()
{
clock_t start, end; // this is to calculate the
// execution time for n-queens
start = clock();
// size 4x4 is taken and we can pass some other
// dimension for chess board as well
cout << solveNQueens(4).size() << endl;
end = clock();
double time_taken
= double(end - start) / double(CLOCKS_PER_SEC);
cout << time_taken << " time was taken(in miliseconds)"
<< endl;
cout << "Out of " << answer.size()
<< " solutions one is following" << endl;
print_board();
}
Java
import java.util.*;
public class NQueens
{
// store all the possible answers
static List<List<String>> answer = new ArrayList<>();
// print the board
static void print_board() {
for (String str : answer.get(1)) {
for (Character letter : str.toCharArray())
System.out.print(letter + " ");
System.out.println();
}
return;
}
// we need to check in three directions
// 1. in the same column above the current position
// 2. in the left top diagonal from the given cell
// 3. in the right top diagonal from the given cell
static boolean safe(int row, int col, List<String> board) {
for (int i = 0; i < board.size(); i++) {
if (board.get(i).charAt(col) == 'Q')
return false;
}
int i = row, j = col;
while (i >= 0 && j >= 0)
if (board.get(i--).charAt(j--) == 'Q')
return false;
i = row;
j = col;
while (i >= 0 && j < board.size())
if (board.get(i--).charAt(j++) == 'Q')
return false;
return true;
}
// rec function here will fill the queens
// 1. there can be only one queen in one row
// 2. if we filled the final row in the board then row will
// be equal to total number of rows in board
// 3. push that board configuration in answer set because
// there will be more than one answers for filling the board
// with n-queens
static void rec(List<String> board, int row) {
if (row == board.size()) {
answer.add(board);
return;
}
for (int i = 0; i < board.size(); i++)
{
// for each position check if it is safe and if it
// safe make a recursive call with
// row+1,board[i][j]='Q' and then revert the change
// in board that is make the board[i][j]='.' again to
// generate more solutions
if (safe(row, i, board)) {
List<String> temp = new ArrayList<>(board);
temp.set(row, temp.get(row).substring(0, i) + "Q" + temp.get(row).substring(i + 1));
rec(temp, row + 1);
}
}
return;
}
// function to solve n queens
static List<List<String>> solveNQueens(int n)
{
String s = new String(new char[n]).replace("\0", ".");
// vector of string will make our board which is
// initially all empty
List<String> board = new ArrayList<>();
for (int i = 0; i < n; i++)
board.add(s);
rec(board, 0);
return answer;
}
public static void main(String[] args) {
long start, end;
// this is to calculate the
// execution time for n-queens
start = System.currentTimeMillis();
// size 4x4 is taken and we can pass some other
// dimension for chess board as well
System.out.println(solveNQueens(4).size());
end = System.currentTimeMillis();
double time_taken = (end - start);
System.out.println(time_taken + " time was taken(in miliseconds)");
System.out.println("Out of " + answer.size() + " solutions one is following");
print_board();
}
}
// This code is contributed by surajrasr7277
Python3
import time
# print the board
def print_board(board, n):
for i in range(n):
for j in range(n):
print(board[i][j], end = " ")
print()
# joining '.' and 'Q'
# making combined 2D Array
#For output in desired format
def add_sol(board, ans, n):
temp = []
for i in range(n):
string = ""
for j in range(n):
string += board[i][j]
temp.append(string)
ans.append(temp)
# we need to check in three directions
# 1. in the same column above the current position
# 2. in the left top diagonal from the given cell
# 3. in the right top diagonal from the given cell
def is_safe(row, col, board, n):
x = row
y = col
#check for same upper col
while(x>=0):
if board[x][y] == "Q":
return False
else:
x -= 1
#Check for Upper Right Diagonal
x = row
y = col
while(y<n and x>=0):
if board[x][y] == "Q":
return False
else:
y += 1
x -= 1
#check for Upper Left diagonal
x = row
y = col
while(y>=0 and x>=0):
if board[x][y] == "Q":
return False
else:
x -= 1
y -= 1
return True
# function to solve n queens
# solveNQueens function here will fill the queens
# 1. there can be only one queen in one row
# 2. if we filled the final row in the board then row will
# be equal to total number of rows in board
# 3. push that board configuration in answer set because
# there will be more than one answers for filling the board
# with n-queens
def solveNQueens(row, ans, board, n):
#base Case
#Queen is depicted by "Q"
# adding solution to final answer array
if row == n:
add_sol(board, ans, n)
return
#solve 1 case and rest recursion will follow
for col in range(n):
# for each position check if it is safe and if it
# is safe make a recursive call with
# row+1, board[i][j]='Q' and then revert the change
# in board that is make the board[i][j]='.' again to
# generate more solutions
if is_safe(row, col, board, n):
# if placing Queen is safe
board[row][col] = "Q"
solveNQueens(row+1, ans, board, n)
# Backtrack
board[row][col] = "."
# Driver Code
if __name__ == "__main__":
# size 4x4 is taken and we can pass some other
# dimension for chess board as well
n = 4
# 2D array of string will make our board
# which is initially all empty
board = [["." for i in range(n)] for j in range(n)]
# store all the possible answers
ans = []
start = time.time()
solveNQueens(0, ans, board, n)
end = time.time()
time_taken = end - start
if ans == []:
print("Solution does not exist")
else:
print(len(ans))
print(f"{time_taken:.06f} time was taken(in miliseconds)")
print(f"Out Of {len(ans)} solutions one is following")
print_board(ans[0], n)
# This code is contributed by Priyank Namdeo
Output
20.000107 time was taken(in miliseconds)Out of 2 solutions one is following. . Q . Q . . . . . . Q . Q . .
Time Complexity: O(N!)
Auxiliary Space: O(N2)
Optimization in is_safe() function
The idea is not to check every element in right and left diagonal, instead use the property of diagonals:
1. The sum of i and j is constant and unique for each right diagonal, where i is the row of elements and j is the
column of elements.
2. The difference between i and j is constant and unique for each left diagonal, where i and j are row and column of element respectively.
Implementation of Backtracking solution(with optimization)
C++
/* C++ program to solve N Queen Problem using
backtracking */
#include<bits/stdc++.h>
using namespace std;
#define N 4
/* ld is an array where its indices indicate row-col+N-1
(N-1) is for shifting the difference to store negative
indices */
int ld[30] = { 0 };
/* rd is an array where its indices indicate row+col
and used to check whether a queen can be placed on
right diagonal or not*/
int rd[30] = { 0 };
/*column array where its indices indicates column and
used to check whether a queen can be placed in that
row or not*/
int cl[30] = { 0 };
/* A utility function to print solution */
void printSolution(int board[N][N])
{
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++)
cout<<" "<< board[i][j]<<" ";
cout<<endl;
}
}
/* A recursive utility function to solve N
Queen problem */
bool solveNQUtil(int board[N][N], int col)
{
/* base case: If all queens are placed
then return true */
if (col >= N)
return true;
/* Consider this column and try placing
this queen in all rows one by one */
for (int i = 0; i < N; i++) {
/* Check if the queen can be placed on
board[i][col] */
/* A check if a queen can be placed on
board[row][col].We just need to check
ld[row-col+n-1] and rd[row+coln] where
ld and rd are for left and right
diagonal respectively*/
if ((ld[i - col + N - 1] != 1 && rd[i + col] != 1) && cl[i] != 1) {
/* Place this queen in board[i][col] */
board[i][col] = 1;
ld[i - col + N - 1] = rd[i + col] = cl[i] = 1;
/* recur to place rest of the queens */
if (solveNQUtil(board, col + 1))
return true;
/* If placing queen in board[i][col]
doesn't lead to a solution, then
remove queen from board[i][col] */
board[i][col] = 0; // BACKTRACK
ld[i - col + N - 1] = rd[i + col] = cl[i] = 0;
}
}
/* If the queen cannot be placed in any row in
this column col then return false */
return false;
}
/* This function solves the N Queen problem using
Backtracking. It mainly uses solveNQUtil() to
solve the problem. It returns false if queens
cannot be placed, otherwise, return true and
prints placement of queens in the form of 1s.
Please note that there may be more than one
solutions, this function prints one of the
feasible solutions.*/
bool solveNQ()
{
int board[N][N] = { { 0, 0, 0, 0 },
{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 } };
if (solveNQUtil(board, 0) == false) {
cout<<"Solution does not exist";
return false;
}
printSolution(board);
return true;
}
// driver program to test above function
int main()
{
solveNQ();
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
C
/* C program to solve N Queen Problem using
backtracking */
#define N 4
#include <stdbool.h>
#include <stdio.h>
/* A utility function to print solution */
void printSolution(int board[N][N])
{
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++)
printf(" %d ", board[i][j]);
printf("\n");
}
}
/* A utility function to check if a queen can
be placed on board[row][col]. Note that this
function is called when "col" queens are
already placed in columns from 0 to col -1.
So we need to check only left side for
attacking queens */
bool isSafe(int board[N][N], int row, int col)
{
int i, j;
/* Check this row on left side */
for (i = 0; i < col; i++)
if (board[row][i])
return false;
/* Check upper diagonal on left side */
for (i = row, j = col; i >= 0 && j >= 0; i--, j--)
if (board[i][j])
return false;
/* Check lower diagonal on left side */
for (i = row, j = col; j >= 0 && i < N; i++, j--)
if (board[i][j])
return false;
return true;
}
/* A recursive utility function to solve N
Queen problem */
bool solveNQUtil(int board[N][N], int col)
{
/* base case: If all queens are placed
then return true */
if (col >= N)
return true;
/* Consider this column and try placing
this queen in all rows one by one */
for (int i = 0; i < N; i++) {
/* Check if the queen can be placed on
board[i][col] */
if (isSafe(board, i, col)) {
/* Place this queen in board[i][col] */
board[i][col] = 1;
/* recur to place rest of the queens */
if (solveNQUtil(board, col + 1))
return true;
/* If placing queen in board[i][col]
doesn't lead to a solution, then
remove queen from board[i][col] */
board[i][col] = 0; // BACKTRACK
}
}
/* If the queen cannot be placed in any row in
this column col then return false */
return false;
}
/* This function solves the N Queen problem using
Backtracking. It mainly uses solveNQUtil() to
solve the problem. It returns false if queens
cannot be placed, otherwise, return true and
prints placement of queens in the form of 1s.
Please note that there may be more than one
solutions, this function prints one of the
feasible solutions.*/
bool solveNQ()
{
int board[N][N] = { { 0, 0, 0, 0 },
{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 } };
if (solveNQUtil(board, 0) == false) {
printf("Solution does not exist");
return false;
}
printSolution(board);
return true;
}
// driver program to test above function
int main()
{
solveNQ();
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
Java
/* Java program to solve N Queen Problem
using backtracking */
import java.util.*;
class GFG
{
static int N = 4;
/* ld is an array where its indices indicate row-col+N-1
(N-1) is for shifting the difference to store negative
indices */
static int []ld = new int[30];
/* rd is an array where its indices indicate row+col
and used to check whether a queen can be placed on
right diagonal or not*/
static int []rd = new int[30];
/*column array where its indices indicates column and
used to check whether a queen can be placed in that
row or not*/
static int []cl = new int[30];
/* A utility function to print solution */
static void printSolution(int board[][])
{
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
System.out.printf(" %d ", board[i][j]);
System.out.printf("\n");
}
}
/* A recursive utility function to solve N
Queen problem */
static boolean solveNQUtil(int board[][], int col)
{
/* base case: If all queens are placed
then return true */
if (col >= N)
return true;
/* Consider this column and try placing
this queen in all rows one by one */
for (int i = 0; i < N; i++)
{
/* Check if the queen can be placed on
board[i][col] */
/* A check if a queen can be placed on
board[row][col].We just need to check
ld[row-col+n-1] and rd[row+coln] where
ld and rd are for left and right
diagonal respectively*/
if ((ld[i - col + N - 1] != 1 &&
rd[i + col] != 1) && cl[i] != 1)
{
/* Place this queen in board[i][col] */
board[i][col] = 1;
ld[i - col + N - 1] =
rd[i + col] = cl[i] = 1;
/* recur to place rest of the queens */
if (solveNQUtil(board, col + 1))
return true;
/* If placing queen in board[i][col]
doesn't lead to a solution, then
remove queen from board[i][col] */
board[i][col] = 0; // BACKTRACK
ld[i - col + N - 1] =
rd[i + col] = cl[i] = 0;
}
}
/* If the queen cannot be placed in any row in
this column col then return false */
return false;
}
/* This function solves the N Queen problem using
Backtracking. It mainly uses solveNQUtil() to
solve the problem. It returns false if queens
cannot be placed, otherwise, return true and
prints placement of queens in the form of 1s.
Please note that there may be more than one
solutions, this function prints one of the
feasible solutions.*/
static boolean solveNQ()
{
int board[][] = {{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 }};
if (solveNQUtil(board, 0) == false)
{
System.out.printf("Solution does not exist");
return false;
}
printSolution(board);
return true;
}
// Driver Code
public static void main(String[] args)
{
solveNQ();
}
}
// This code is contributed by Princi Singh
Python3
""" Python3 program to solve N Queen Problem using
backtracking """
N = 4
""" ld is an array where its indices indicate row-col+N-1
(N-1) is for shifting the difference to store negative
indices """
ld = [0] * 30
""" rd is an array where its indices indicate row+col
and used to check whether a queen can be placed on
right diagonal or not"""
rd = [0] * 30
"""column array where its indices indicates column and
used to check whether a queen can be placed in that
row or not"""
cl = [0] * 30
""" A utility function to print solution """
def printSolution(board):
for i in range(N):
for j in range(N):
print(board[i][j], end = " ")
print()
""" A recursive utility function to solve N
Queen problem """
def solveNQUtil(board, col):
""" base case: If all queens are placed
then return True """
if (col >= N):
return True
""" Consider this column and try placing
this queen in all rows one by one """
for i in range(N):
""" Check if the queen can be placed on board[i][col] """
""" A check if a queen can be placed on board[row][col].
We just need to check ld[row-col+n-1] and rd[row+coln]
where ld and rd are for left and right diagonal respectively"""
if ((ld[i - col + N - 1] != 1 and
rd[i + col] != 1) and cl[i] != 1):
""" Place this queen in board[i][col] """
board[i][col] = 1
ld[i - col + N - 1] = rd[i + col] = cl[i] = 1
""" recur to place rest of the queens """
if (solveNQUtil(board, col + 1)):
return True
""" If placing queen in board[i][col]
doesn't lead to a solution,
then remove queen from board[i][col] """
board[i][col] = 0 # BACKTRACK
ld[i - col + N - 1] = rd[i + col] = cl[i] = 0
""" If the queen cannot be placed in
any row in this column col then return False """
return False
""" This function solves the N Queen problem using
Backtracking. It mainly uses solveNQUtil() to
solve the problem. It returns False if queens
cannot be placed, otherwise, return True and
prints placement of queens in the form of 1s.
Please note that there may be more than one
solutions, this function prints one of the
feasible solutions."""
def solveNQ():
board = [[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0]]
if (solveNQUtil(board, 0) == False):
printf("Solution does not exist")
return False
printSolution(board)
return True
# Driver Code
solveNQ()
# This code is contributed by SHUBHAMSINGH10
C#
/* C# program to solve N Queen Problem
using backtracking */
using System;
class GFG
{
static int N = 4;
/* ld is an array where its indices indicate row-col+N-1
(N-1) is for shifting the difference to store negative
indices */
static int []ld = new int[30];
/* rd is an array where its indices indicate row+col
and used to check whether a queen can be placed on
right diagonal or not*/
static int []rd = new int[30];
/*column array where its indices indicates column and
used to check whether a queen can be placed in that
row or not*/
static int []cl = new int[30];
/* A utility function to print solution */
static void printSolution(int [,]board)
{
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
Console.Write(" {0} ", board[i, j]);
Console.Write("\n");
}
}
/* A recursive utility function to solve N
Queen problem */
static bool solveNQUtil(int [,]board, int col)
{
/* base case: If all queens are placed
then return true */
if (col >= N)
return true;
/* Consider this column and try placing
this queen in all rows one by one */
for (int i = 0; i < N; i++)
{
/* Check if the queen can be placed on
board[i,col] */
/* A check if a queen can be placed on
board[row,col].We just need to check
ld[row-col+n-1] and rd[row+coln] where
ld and rd are for left and right
diagonal respectively*/
if ((ld[i - col + N - 1] != 1 &&
rd[i + col] != 1) && cl[i] != 1)
{
/* Place this queen in board[i,col] */
board[i, col] = 1;
ld[i - col + N - 1] =
rd[i + col] = cl[i] = 1;
/* recur to place rest of the queens */
if (solveNQUtil(board, col + 1))
return true;
/* If placing queen in board[i,col]
doesn't lead to a solution, then
remove queen from board[i,col] */
board[i, col] = 0; // BACKTRACK
ld[i - col + N - 1] =
rd[i + col] = cl[i] = 0;
}
}
/* If the queen cannot be placed in any row in
this column col then return false */
return false;
}
/* This function solves the N Queen problem using
Backtracking. It mainly uses solveNQUtil() to
solve the problem. It returns false if queens
cannot be placed, otherwise, return true and
prints placement of queens in the form of 1s.
Please note that there may be more than one
solutions, this function prints one of the
feasible solutions.*/
static bool solveNQ()
{
int [,]board = {{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 }};
if (solveNQUtil(board, 0) == false)
{
Console.Write("Solution does not exist");
return false;
}
printSolution(board);
return true;
}
// Driver Code
public static void Main(String[] args)
{
solveNQ();
}
}
// This code is contributed by Rajput-Ji
Javascript
<script>
// JavaScript code to implement the approach
let N = 4;
/* ld is an array where its indices indicate row-col+N-1
(N-1) is for shifting the difference to store negative
indices */
let ld = new Array(30);
/* rd is an array where its indices indicate row+col
and used to check whether a queen can be placed on
right diagonal or not*/
let rd = new Array(30);
/*column array where its indices indicates column and
used to check whether a queen can be placed in that
row or not*/
let cl = new Array(30);
/* A utility function to print solution */
function printSolution( board)
{
for (let i = 0; i < N; i++)
{
for (let j = 0; j < N; j++)
document.write(board[i][j] + " ");
document.write("<br/>");
}
}
/* A recursive utility function to solve N
Queen problem */
function solveNQUtil(board, col)
{
/* base case: If all queens are placed
then return true */
if (col >= N)
return true;
/* Consider this column and try placing
this queen in all rows one by one */
for (let i = 0; i < N; i++)
{
/* Check if the queen can be placed on
board[i][col] */
/* A check if a queen can be placed on
board[row][col].We just need to check
ld[row-col+n-1] and rd[row+coln] where
ld and rd are for left and right
diagonal respectively*/
if ((ld[i - col + N - 1] != 1 &&
rd[i + col] != 1) && cl[i] != 1)
{
/* Place this queen in board[i][col] */
board[i][col] = 1;
ld[i - col + N - 1] =
rd[i + col] = cl[i] = 1;
/* recur to place rest of the queens */
if (solveNQUtil(board, col + 1))
return true;
/* If placing queen in board[i][col]
doesn't lead to a solution, then
remove queen from board[i][col] */
board[i][col] = 0; // BACKTRACK
ld[i - col + N - 1] =
rd[i + col] = cl[i] = 0;
}
}
/* If the queen cannot be placed in any row in
this column col then return false */
return false;
}
/* This function solves the N Queen problem using
Backtracking. It mainly uses solveNQUtil() to
solve the problem. It returns false if queens
cannot be placed, otherwise, return true and
prints placement of queens in the form of 1s.
Please note that there may be more than one
solutions, this function prints one of the
feasible solutions.*/
function solveNQ()
{
let board = [[ 0, 0, 0, 0 ],
[ 0, 0, 0, 0 ],
[ 0, 0, 0, 0 ],
[ 0, 0, 0, 0 ]];
if (solveNQUtil(board, 0) == false)
{
document.write("Solution does not exist");
return false;
}
printSolution(board);
return true;
}
// Driver code
solveNQ();
// This code is contributed by sanjoy_62.
</script>
Output
0 0 1 0 1 0 0 0 0 0 0 1 0 1 0 0
Time Complexity: O(N!)
Auxiliary Space: O(N)
Printing all solutions in N-Queen Problem
Sources:
http://see.stanford.edu/materials/icspacs106b/H19-RecBacktrackExamples.pdf
http://en.literateprograms.org/Eight_queens_puzzle_%28C%29
http://en.wikipedia.org/wiki/Eight_queens_puzzle
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