N Queen Problem using Branch And Bound - GeeksforGeeks.pdf

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N Queen Problem
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4 Queens Problem
8 queen problem
NQueenProblemusing
BranchAndBound
N Queen in O(n) space
Printing all solutions in N-
Queen Problem
Minimum queens required
to cover all the squares of
a chess board
The N queens puzzle is the problem of placing
N chess queens on an N×N chessboard so that no two
queens threaten each other. Thus, a solution requires
that no two queens share the same row, column, or
diagonal.
The backtracking Algorithm for N-Queen is already
discussed here. In a backtracking solution, we backtrack
when we hit a dead end. In Branch and Bound solution,
after building a partial solution, we figure out that
there is no point going any deeper as we are going to
hit a dead end.
Let’s begin by describing the backtracking solution. “The
idea is to place queens one by one in different columns,
starting from the leftmost column. When we place a
NQueenProblemusingBranch
AndBound
Read Courses Practice Jobs
DSA Branch and Bound Tutorial Backtracking Vs Branch-N-Bound 0/1 Knapsack 8 Puzzle Problem Job Assignment Problem N-Queen Problem Tra
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Number of cells a queen
can move with obstacles
on the chessboard
N-Queen Problem | Local
Search using Hill climbing
with random neighbour
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.”
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Placing 1st queen on (3, 0) and 2nd queen on (5, 1)
1. For the 1st Queen, there are total 8 possibilities as we
can place 1st Queen in any row of first column. Let’s
place Queen 1 on row 3.
2. After placing 1st Queen, there are 7 possibilities left
for the 2nd Queen. But wait, we don’t really have 7
possibilities. We cannot place Queen 2 on rows 2, 3
or 4 as those cells are under attack from Queen 1. So,
Queen 2 has only 8 – 3 = 5 valid positions left.
3. After picking a position for Queen 2, Queen 3 has
even fewer options as most of the cells in its column
are under attack from the first 2 Queens.
We need to figure out an efficient way of keeping track of
which cells are under attack. In previous solution we kept
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an 8
-by
-8 Boolean matrix and update it each time we
placed a queen, but that required linear time to update
as we need to check for safe cells.
Basically, we have to ensure 4 things:
1. No two queens share a column.
2. No two queens share a row.
3. No two queens share a top-right to left-bottom
diagonal.
4. No two queens share a top-left to bottom-right
diagonal.
Number 1 is automatic because of the way we store the
solution. For number 2, 3 and 4, we can perform updates
in O(1) time. The idea is to keep three Boolean arrays
that tell us which rows and which diagonals are
occupied.
Lets do some pre-processing first. Let’s create two N x N
matrix one for / diagonal and other one for diagonal.
Let’s call them slashCode and backslashCode
respectively. The trick is to fill them in such a way that
two queens sharing a same /
diagonal will have the same
value in matrix slashCode, and if they share same
-
diagonal, they will have the same value in
backslashCode matrix.
For an N x N matrix, fill slashCode and backslashCode
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matrix using below formula –
slashCode[row][col] = row + col
backslashCode[row][col] = row – col + (N-1)
Using above formula will result in below matrices
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The ‘N – 1’ in the backslash code is there to ensure that
the codes are never negative because we will be using
the codes as indices in an array.
Now before we place queen i on row j, we first check
whether row j is used (use an array to store row info).
Then we check whether slash code ( j + i ) or backslash
code ( j – i + 7 ) are used (keep two arrays that will tell us
which diagonals are occupied). If yes, then we have to try
a different location for queen i. If not, then we mark the
row and the two diagonals as used and recurse on queen
i + 1. After the recursive call returns and before we try
another position for queen i, we need to reset the row,
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slash code and backslash code as unused again, like in
the code from the previous notes.
Below is the implementation of above idea –
This code Takes the Dynamic Input:
C++
#include<bits/stdc++.h>
using namespace std;
int N;
// function for printing the solution
void printSol(vector<vector<int>>board)
{
for(int i = 0;i<N;i++){
for(int j = 0;j<N;j++){
cout<<board[i][j]<<" ";
}
cout<<"n";
}
}
/* Optimized isSafe function
isSafe function to check if current row conta
yes return false
else return true
*/
bool isSafe(int row ,int col ,vector<bool>ro
{
if(rows[row] == true || left_digonals[row
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return false;
}
return true;
}
// Recursive function to solve N-queen Proble
bool solve(vector<vector<int>>& board ,int c
{
// base Case : If all Queens are place
if(col>=N){
return true;
}
/* Consider this Column and move in al
for(int i = 0;i<N;i++)
{
if(isSafe(i,col,rows,left_digonals
{
rows[i] = true;
left_digonals[i+col] = true;
Right_digonals[col-i+N-1] = tr
board[i][col] = 1; // placin
/* recur to place rest of the
if(solve(board,col+1,rows,left
return true;
}
// Backtracking
rows[i] = false;
left_digonals[i+col] = false;
Right_digonals[col-i+N-1] = fa
board[i][col] = 0; // remov
}
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}
return false;
}
int main()
{
// Taking input from the user
cout<<"Enter the no of rows for the square
cin>>N;
// board of size N*N
vector<vector<int>>board(N,vector<int>(N,0
// array to tell which rows are occupied
vector<bool>rows(N,false);
// arrays to tell which diagonals are occu
vector<bool>left_digonals(2*N-1,false);
vector<bool>Right_digonals(2*N-1,false);
bool ans = solve(board , 0, rows,left_di
if(ans == true){
// printing the solution Board
printSol(board);
}
else{
cout<<"Solution Does not Existn";
}
}
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C
// This Code is Contributed by Parshant Rajp
/* C++ program to solve N Queen Problem using
and Bound */
#include<stdio.h>
#include<string.h>
#include<stdbool.h>
#define N 8
/* 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("%2d ", board[i][j]);
printf("n");
}
}
/* A Optimized function to check if a queen c
be placed on board[row][col] */
bool isSafe(int row, int col, int slashCode[
int backslashCode[N][N], bool row
bool slashCodeLookup[], bool backslash
{
if (slashCodeLookup[slashCode[row][col]]
backslashCodeLookup[backslashCode[row
rowLookup[row])
return false;
return true;
}
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/* A recursive utility function
to solve N Queen problem */
bool solveNQueensUtil(int board[N][N], int c
int slashCode[N][N], int backslashCode[N
bool rowLoo
bool slashCodeLoo
bool backslashCode
{
/* 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 queen can be placed on
board[i][col] */
if ( isSafe(i, col, slashCode,
backslashCode, rowLookup
slashCodeLookup, backslashCodeLooku
{
/* Place this queen in board[i][c
board[i][col] = 1;
rowLookup[i] = true;
slashCodeLookup[slashCode[i][col
backslashCodeLookup[backslashCode
/* recur to place rest of the que
if ( solveNQueensUtil(board, col
slashCode,
rowLookup, slashCodeLookup, back
return true;
/* If placing queen in board[i][c
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doesn't lead to a solution, then
/* Remove queen from board[i][col
board[i][col] = 0;
rowLookup[i] = false;
}
}
/* If queen can not be place in any row i
this column col then return false */
return false;
}
/* This function solves the N Queen problem u
Branch and Bound. It mainly uses solveNQueens
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 solveNQueens()
{
int board[N][N];
memset(board, 0, sizeof board);
// helper matrices
int slashCode[N][N];
int backslashCode[N][N];
// arrays to tell us which rows are occup
bool rowLookup[N] = {false};
//keep two arrays to tell us
// which diagonals are occupied
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Java
bool slashCodeLookup[2*N - 1] = {false};
bool backslashCodeLookup[2*N - 1] = {fal
// initialize helper matrices
for (int r = 0; r < N; r++)
for (int c = 0; c < N; c++) {
slashCode[r] = r + c,
backslashCode[r] = r - c + 7;
}
if (solveNQueensUtil(board, 0,
slashCode, backslas
rowLookup, slashCodeLookup, backslashCo
{
printf("Solution does not exist");
return false;
}
// solution found
printSolution(board);
return true;
}
// Driver program to test above function
int main()
{
solveNQueens();
return 0;
}
// Java program to solve N Queen Problem
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// using Branch and Bound
import java.io.*;
import java.util.Arrays;
class GFG{
static int N = 8;
// A utility function to print solution
static void printSolution(int board[][])
{
int N = board.length;
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
System.out.printf("%2d ", board[i
System.out.printf("n");
}
}
// A Optimized function to check if a queen
// can be placed on board[row][col]
static boolean isSafe(int row, int col,
int slashCode[][],
int backslashCode[][],
boolean rowLookup[],
boolean slashCodeLookup
boolean backslashCodeLo
{
rowLookup[row])
return false;
return true;
}
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// A recursive utility function to
// solve N Queen problem
static boolean solveNQueensUtil(
int board[][], int col, int slashCode[][
int backslashCode[][], boolean rowLookup
boolean slashCodeLookup[],
boolean backslashCodeLookup[])
{
// Base case: If all queens are placed
// then return true
int N = board.length;
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 queen can be placed on bo
if (isSafe(i, col, slashCode, backsl
rowLookup, slashCodeLookup
backslashCodeLookup))
{
// Place this queen in board[i][c
board[i][col] = 1;
// recur to place rest of the que
if (solveNQueensUtil(
board, col + 1, slashCode
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slashCodeLookup,
return true;
// If placing queen in board[i][c
// lead to a solution, then backt
// Remove queen from board[i][col
board[i][col] = 0;
}
}
// If queen can not be place in any row
// in this column col then return false
return false;
}
/*
* This function solves the N Queen problem u
* and Bound. It mainly uses solveNQueensUtil
* the problem. It returns false if queens ca
* placed, otherwise return true and prints p
* queens in the form of 1s. Please note that
* be more than one solutions, this function
* of the feasible solutions.
*/
static boolean solveNQueens()
{
int board[][] = new int[N][N];
// Helper matrices
int slashCode[][] = new int[N][N];
int backslashCode[][] = new int[N][N];
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// Arrays to tell us which rows are occup
boolean[] rowLookup = new boolean[N];
// Keep two arrays to tell us
boolean slashCodeLookup[] = new boolean[
boolean backslashCodeLookup[] = new bool
// Initialize helper matrices
for(int r = 0; r < N; r++)
for(int c = 0; c < N; c++)
{
slashCode[r] = r + c;
}
if (solveNQueensUtil(board, 0, slashCode
backslashCode, rowLo
slashCodeLookup,
backslashCodeLookup
{
System.out.printf("Solution does not
return false;
}
// Solution found
return true;
}
// Driver code
public static void main(String[] args)
{
solveNQueens();
}
}
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Python3
// This code is contributed by sujitmeshram
""" Python3 program to solve N Queen Problem
using Branch or Bound """
N = 8
""" 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 Optimized function to check if
a queen can be placed on board[row][col] """
def isSafe(row, col, slashCode, backslashCode
rowLookup, slashCodeLookup,
backslashCodeLookup):
rowLookup[row]):
return False
return True
""" A recursive utility function
to solve N Queen problem """
def solveNQueensUtil(board, col, slashCode, b
rowLookup, slashCodeLook
backslashCodeLookup):
""" base case: If all queens are
placed then return True """
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if(col >= N):
return True
for i in range(N):
if(isSafe(i, col, slashCode, backslas
backslashCodeLookup)):
""" Place this queen in board[i]
board[i][col] = 1
rowLookup[i] = True
""" recur to place rest of the qu
if(solveNQueensUtil(board, col +
slashCode, ba
rowLookup, sl
backslashCode
return True
""" If placing queen in board[i]
doesn't lead to a solution,then b
""" Remove queen from board[i][co
board[i][col] = 0
rowLookup[i] = False
""" If queen can not be place in any row
this column col then return False """
return False
""" This function solves the N Queen problem
Branch or Bound. It mainly uses solveNQueensU
solve the problem. It returns False if queens
cannot be placed,otherwise return True or
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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 solveNQueens():
board = [[0 for i in range(N)]
for j in range(N)]
# helper matrices
slashCode = [[0 for i in range(N)]
for j in range(N)]
backslashCode = [[0 for i in range(N)]
for j in range(N)]
# arrays to tell us which rows are occupi
rowLookup = [False] * N
# keep two arrays to tell us
# which diagonals are occupied
x = 2 * N - 1
slashCodeLookup = [False] * x
backslashCodeLookup = [False] * x
# initialize helper matrices
for rr in range(N):
for cc in range(N):
slashCode[rr][cc] = rr + cc
backslashCode[rr][cc] = rr - cc
if(solveNQueensUtil(board, 0, slashCode,
rowLookup, slashCodeL
backslashCodeLookup)
print("Solution does not exist")
return False
# solution found
printSolution(board)
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C#
return True
# Driver Code
solveNQueens()
# This code is contributed by SHUBHAMSINGH10
using System;
using System.Linq;
class GFG {
static int N = 8;
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.WriteLine();
}
}
static bool isSafe(int row, int col, int[,
int[, ] backslashCode,
bool[] rowLookup,
bool[] slashCodeLookup,
bool[] backslashCodeLook
{
if (slashCodeLookup[slashCode[row, col]]
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|| backslashCodeLookup[backslashCode
|| rowLookup[row])
return false;
return true;
}
static bool solveNQueensUtil(int[, ] board
int[, ] slashC
int[, ] backsl
bool[] rowLook
bool[] slashCo
bool[] backsla
{
// then return true
if (col >= N)
return true;
for (int i = 0; i < N; i++) {
// Check if queen can be placed on boar
if (isSafe(i, col, slashCode, backslash
rowLookup, slashCodeLookup,
backslashCodeLookup)) {
// Place this queen in board[i][col]
board[i, col] = 1;
slashCodeLookup[slashCode[i, col]] =
backslashCodeLookup[backslashCode[i,
= true;
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// recur to place rest of the queens
board, col + 1, slashCode,
backslashCode, rowLookup,
slashCodeLookup,
return true;
// If placing queen in board[i][col]
// lead to a solution, then backtrack
// Remove queen from board[i][col]
board[i, col] = 0;
slashCodeLookup[slashCode[i, col]] =
backslashCodeLookup[backslashCode[i,
= false;
}
}
return false;
}
/*
* This function solves theN Queen proble
* and Bound. It mainly uses solveNQueens
* the problem. It returns false if queen
* placed, otherwise return true and prin
* queens in the form of 1s. Please note
* be more than one solutions, this funct
*/
static bool solveNQueens()
{
int[, ] board = new int[N, N];
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// Helper matrices
int[, ] slashCode = new int[N, N];
int[, ] backslashCode = new int[N, N];
bool[] rowLookup = new bool[N];
bool[] slashCodeLookup = new bool[2 * N
bool[] backslashCodeLookup = new bool[2
// Initialize helper arrays
for (int r = 0; r < N; r++) {
for (int c = 0; c < N; c++) {
slashCode[r, c] = r + c;
backslashCode[r, c] = r - c + N - 1;
}
}
if (!solveNQueensUtil(board, 0, slashCode
backslashCode, rowL
slashCodeLookup,
backslashCodeLookup
Console.WriteLine("Solution does not ex
return false;
}
return true;
}
// Driver code
public static void Main() { solveNQueens()
}
// This code is contributed by lokeshpotta20
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Javascript
// JavaScript program to solve N Queen Proble
// using Branch and Bound
let N = 8;
function printSolution(board)
{
let N = board.length;
for(let i = 0; i < N; i++)
{
for(let j = 0; j < N; j++)
document.write(board[i][j]+" ");
document.write("<br>");
}
}
function isSafe(row,col,slashCode,backslashCo
{
rowLookup[row])
return false;
return true;
}
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function solveNQueensUtil(board,col,slashCode
backslashCode,rowLookup,slashCodeLookup,backs
{
// then return true
//let N = board.length;
if (col >= N)
return true;
for(let i = 0; i < N; i++)
{
// Check if queen can be placed on bo
if (isSafe(i, col, slashCode, backsl
{
// Place this queen in board[i][c
board[i][col] = 1;
// recur to place rest of the que
board, col + 1, slashCode
slashCodeLookup,
return true;
// If placing queen in board[i][c
// lead to a solution, then backt
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// Remove queen from board[i][col
board[i][col] = 0;
}
}
return false;
}
/*
* This function solves the N Queen problem u
* and Bound. It mainly uses solveNQueensUtil
* the problem. It returns false if queens ca
* placed, otherwise return true and prints p
* queens in the form of 1s. Please note that
* be more than one solutions, this function
*/
function solveNQueens()
{
let board = new Array(N);
// Helper matrices
let slashCode = new Array(N);
let backslashCode = new Array(N);
for(let i=0;i<N;i++)
{
board[i]=new Array(N);
slashCode[i]=new Array(N);
backslashCode[i]=new Array(N);
for(let j=0;j<N;j++)
{
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board[i][j]=0;
slashCode[i][j]=0;
backslashCode[i][j]=0;
}
}
let rowLookup = new Array(N);
for(let i=0;i<N;i++)
rowLookup[i]=false;
let slashCodeLookup = new Array(2 * N -
let backslashCodeLookup = new Array(2 * N
for(let i=0;i<2*N-1;i++)
{
slashCodeLookup[i]=false;
backslashCodeLookup[i]=false;
}
// Initialize helper matrices
for(let r = 0; r < N; r++)
for(let c = 0; c < N; c++)
{
slashCode[r] = r + c;
}
if (solveNQueensUtil(board, 0, slashCode
backslashCode, rowLo
slashCodeLookup,
backslashCodeLookup
{
document.write("Solution does not exi
return false;
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Learn Data Structures & Algorithms with GeeksforGeeks
Input:
Enter the no of rows for the square Board : 8
Output :
1 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0
0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 1
0 1 0 0 0 0 0 0
0 0 0 1 0 0 0 0
0 0 0 0 0 1 0 0
0 0 1 0 0 0 0 0
}
// Solution found
return true;
}
// Driver code
solveNQueens();
// This code is contributed by avanitrachhadi
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The time and space complexity of the N-Queen
problem solver implemented in the provided code are:
Time Complexity: The time complexity of the solver
algorithm is O(N!), where N is the number of rows and
columns in the square board. This is because for each
column, the algorithm tries to place a queen in each row
and then recursively tries to place the queens in the
remaining columns. The number of possible
combinations of queen placements in the board is N!
since there can be only one queen in each row and each
column.
Space Complexity: The space complexity of the solver
algorithm is O(N^2), where N is the number of rows and
columns in the square board. This is because we are
using a 2D vector to represent the board, which takes up
N^2 space. Additionally, we are using three boolean
arrays to keep track of the occupied rows and diagonals,
which take up 2N-1 space each. Therefore, the total
space complexity is O(N^2 + 6N – 3), which is equivalent
to O(N^2).
References :
https://en.wikipedia.org/wiki/Eight_queens_puzzle
www.cs.cornell.edu/~wdtseng/icpc/notes/bt2.pdf
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Last Updated : 20 Sep, 2023
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