N Queen Problem using Branch And Bound

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.

Backtracking Algorithm for N-Queen is already discussed here. In 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 backtracking solution. “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.”

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.
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.
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 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 matrix using below formula –

slashCode[row][col] = row + col
backslashCode[row][col] = row – col + (N-1)

Using above formula will result in below matrices

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, slash code and backslash code as unused again, like in the code from the previous notes.

Below is the implementation of above idea –

C++

/* C++ program to solve N Queen Problem using Branch 
   and Bound */
#include<stdio.h> 
#include<string.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 can 
be placed on board[row][col] */
bool isSafe(int row, int col, int slashCode[N][N], 
            int backslashCode[N][N], bool rowLookup[], 
            bool slashCodeLookup[], bool backslashCodeLookup[] ) 
{ 
    if (slashCodeLookup[slashCode[row][col]] || 
        backslashCodeLookup[backslashCode[row][col]] || 
        rowLookup[row]) 
    return false; 
  
    return true; 
} 
  
/* A recursive utility function to solve N Queen problem */
bool solveNQueensUtil(int board[N][N], int col, 
    int slashCode[N][N], int backslashCode[N][N], bool rowLookup[N], 
    bool slashCodeLookup[], bool backslashCodeLookup[] ) 
{ 
    /* 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, backslashCodeLookup) ) 
        { 
            /* Place this queen in board[i][col] */
            board[i][col] = 1; 
            rowLookup[i] = true; 
            slashCodeLookup[slashCode[i][col]] = true; 
            backslashCodeLookup[backslashCode[i][col]] = true; 
  
            /* recur to place rest of the queens */
            if ( solveNQueensUtil(board, col + 1, slashCode, backslashCode, 
                    rowLookup, slashCodeLookup, backslashCodeLookup) ) 
                return true; 
  
            /* If placing queen in board[i][col] 
            doesn't lead to a solution, then backtrack */
  
            /* Remove queen from board[i][col] */
            board[i][col] = 0; 
            rowLookup[i] = false; 
            slashCodeLookup[slashCode[i][col]] = false; 
            backslashCodeLookup[backslashCode[i][col]] = false; 
        } 
    } 
  
    /* If queen can not be place in any row in 
        this colum col then return false */
    return false; 
} 
  
/* This function solves the N Queen problem using 
Branch and Bound. It mainly uses solveNQueensUtil() 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 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 occupied 
    bool rowLookup[N] = {false}; 
  
    //keep two arrays to tell us which diagonals are occupied 
    bool slashCodeLookup[2*N - 1] = {false}; 
    bool backslashCodeLookup[2*N - 1] = {false}; 
  
    // 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, backslashCode, 
      rowLookup, slashCodeLookup, backslashCodeLookup) == false ) 
    { 
        printf("Solution does not exist"); 
        return false; 
    } 
  
    // solution found 
    printSolution(board); 
    return true; 
} 
  
// driver program to test above function 
int main() 
{ 
    solveNQueens(); 
  
    return 0; 
} 

Python3

""" Python3 program to solve N Queen Problem  
using Branch or Bound """
  
N = 8
  
""" A utility function to prsolution """
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): 
    if (slashCodeLookup[slashCode[row][col]] or
        backslashCodeLookup[backslashCode[row][col]] or
        rowLookup[row]): 
        return False
    return True
  
""" A recursive utility function to solve N Queen problem """
def solveNQueensUtil(board, col, slashCode, backslashCode,  
                     rowLookup, slashCodeLookup, 
                     backslashCodeLookup): 
                          
    """ base case: If all queens are placed then return True """
    if(col >= N): 
        return True
    for i in range(N): 
        if(isSafe(i, col, slashCode, backslashCode,  
                  rowLookup, slashCodeLookup, 
                  backslashCodeLookup)): 
                      
            """ Place this queen in board[i][col] """
            board[i][col] = 1
            rowLookup[i] = True
            slashCodeLookup[slashCode[i][col]] = True
            backslashCodeLookup[backslashCode[i][col]] = True
              
            """ recur to place rest of the queens """
            if(solveNQueensUtil(board, col + 1, slashCode, backslashCode,  
                                rowLookup, slashCodeLookup,  
                                backslashCodeLookup)): 
                return True
              
            """ If placing queen in board[i][col]  
            doesn't lead to a solution,then backtrack """
              
            """ Remove queen from board[i][col] """
            board[i][col] = 0
            rowLookup[i] = False
            slashCodeLookup[slashCode[i][col]] = False
            backslashCodeLookup[backslashCode[i][col]] = False
              
    """ If queen can not be place in any row in  
    this colum col then return False """
    return False
  
""" This function solves the N Queen problem using  
Branch or Bound. It mainly uses solveNQueensUtil()to  
solve the problem. It returns False if queens  
cannot be placed,otherwise return True or  
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 occupied  
    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 + 7
      
    if(solveNQueensUtil(board, 0, slashCode, backslashCode,  
                        rowLookup, slashCodeLookup,  
                        backslashCodeLookup) == False): 
        print("Solution does not exist") 
        return False
          
    # solution found  
    printSolution(board) 
    return True
  
# Driver Cde 
solveNQueens() 
  
# This code is contributed by SHUBHAMSINGH10 

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

Performance:
When run on local machines for N = 32, the backtracking solution took 659.68 seconds while above branch and bound solution took only 119.63 seconds. The difference will be even huge for larger values of N.

References :
https://en.wikipedia.org/wiki/Eight_queens_puzzle
www.cs.cornell.edu/~wdtseng/icpc/notes/bt2.pdf

This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

C++

Python3

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