N Queen Problem using Branch And Bound
Last Updated : 11 Mar, 2024
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 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.”
Placing 1st queen on (3, 0) and 2nd queen on (5, 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.
- 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 –
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 contains or current left diagonal or current right diagonal contains any queen or not if yes return false else return true */ bool isSafe(int row ,int col ,vector<bool>rows , vector<bool>left_digonals ,vector<bool>Right_digonals) { if(rows[row] == true || left_digonals[row+col] == true || Right_digonals[col-row+N-1] == true){ return false; } return true; } // Recursive function to solve N-queen Problem bool solve(vector<vector<int>>& board ,int col ,vector<bool>rows , vector<bool>left_digonals ,vector<bool>Right_digonals) { // base Case : If all Queens are placed if(col>=N){ return true; } /* Consider this Column and move in all rows one by one */ for(int i = 0;i<N;i++) { if(isSafe(i,col,rows,left_digonals,Right_digonals) == true) { rows[i] = true; left_digonals[i+col] = true; Right_digonals[col-i+N-1] = true; board[i][col] = 1; // placing the Queen in board[i][col] /* recur to place rest of the queens */ if(solve(board,col+1,rows,left_digonals,Right_digonals) == true){ return true; } // Backtracking rows[i] = false; left_digonals[i+col] = false; Right_digonals[col-i+N-1] = false; board[i][col] = 0; // removing the Queen from board[i][col] } } return false; } int main() { // Taking input from the user cout<<"Enter the no of rows for the square Board : "; 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 occupied vector<bool>left_digonals(2*N-1,false); vector<bool>Right_digonals(2*N-1,false); bool ans = solve(board , 0, rows,left_digonals,Right_digonals); if(ans == true){ // printing the solution Board printSol(board); } else{ cout<<"Solution Does not Exist\n"; } } // This Code is Contributed by Parshant Rajput /* C++ program to solve N Queen Problem using Branch 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 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 column 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][c] = r + c, backslashCode[r][c] = 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; } // Java program to solve N Queen Problem // 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][j]); 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 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 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 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 column 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. */ static boolean solveNQueens() { int board[][] = new int[N][N]; // Helper matrices int slashCode[][] = new int[N][N]; int backslashCode[][] = new int[N][N]; // Arrays to tell us which rows are occupied boolean[] rowLookup = new boolean[N]; // Keep two arrays to tell us // which diagonals are occupied boolean slashCodeLookup[] = new boolean[2 * N - 1]; boolean backslashCodeLookup[] = new boolean[2 * N - 1]; // Initialize helper matrices for(int r = 0; r < N; r++) for(int c = 0; c < N; c++) { slashCode[r][c] = r + c; backslashCode[r][c] = r - c + 7; } if (solveNQueensUtil(board, 0, slashCode, backslashCode, rowLookup, slashCodeLookup, backslashCodeLookup) == false) { System.out.printf("Solution does not exist"); return false; } // Solution found printSolution(board); return true; } // Driver code public static void main(String[] args) { solveNQueens(); } } // 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): 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 column 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 Code solveNQueens() # This code is contributed by SHUBHAMSINGH10 using System; using System.Linq; class GFG { static int N = 8; // 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.WriteLine(); } } // A Optimized function to check if a queen // can be placed on board[row][col] static bool isSafe(int row, int col, int[, ] slashCode, int[, ] backslashCode, 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 static bool solveNQueensUtil(int[, ] board, int col, int[, ] slashCode, int[, ] backslashCode, bool[] rowLookup, 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 column col then return false return false; } /* * This function solves theN 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. */ static bool solveNQueens() { int[, ] board = new int[N, N]; // Helper matrices int[, ] slashCode = new int[N, N]; int[, ] backslashCode = new int[N, N]; // Arrays to tell us which rows are occupied bool[] rowLookup = new bool[N]; // Keep two arrays to tell us // which diagonals are occupied bool[] slashCodeLookup = new bool[2 * N - 1]; bool[] backslashCodeLookup = new bool[2 * N - 1]; // 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, rowLookup, slashCodeLookup, backslashCodeLookup)) { Console.WriteLine("Solution does not exist"); return false; } printSolution(board); return true; } // Driver code public static void Main() { solveNQueens(); } } // This code is contributed by lokeshpotta20. <script> // JavaScript program to solve N Queen Problem // using Branch and Bound let N = 8; // A utility function to print solution 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>"); } } // A Optimized function to check if a queen // can be placed on board[row][col] function isSafe(row,col,slashCode,backslashCode,rowLookup,slashCodeLookup,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 function solveNQueensUtil(board,col,slashCode, backslashCode,rowLookup,slashCodeLookup,backslashCodeLookup) { // Base case: If all queens are placed // then return true //let N = board.length; 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 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 column 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. */ 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++) { board[i][j]=0; slashCode[i][j]=0; backslashCode[i][j]=0; } } // Arrays to tell us which rows are occupied let rowLookup = new Array(N); for(let i=0;i<N;i++) rowLookup[i]=false; // Keep two arrays to tell us // which diagonals are occupied let slashCodeLookup = new Array(2 * N - 1); let backslashCodeLookup = new Array(2 * N - 1); 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][c] = r + c; backslashCode[r][c] = r - c + 7; } if (solveNQueensUtil(board, 0, slashCode, backslashCode, rowLookup, slashCodeLookup, backslashCodeLookup) == false) { document.write("Solution does not exist"); return false; } // Solution found printSolution(board); return true; } // Driver code solveNQueens(); // This code is contributed by avanitrachhadiya2155 </script> 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
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).
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