Printing all solutions in N-Queen Problem

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 4 Queen problem.
 

In previous post, we have discussed an approach that prints only one possible solution, so now in this post, the task is to print all solutions in N-Queen Problem. Each solution contains distinct board configurations of the N-queens’ placement, where the solutions are a permutation of [1,2,3..n] in increasing order, here the number in the ith place denotes that the ith-column queen is placed in the row with that number. For the example above solution is written as [[2 4 1 3 ] [3 1 4 2 ]]. The solution discussed here is an extension of the same approach.

Backtracking Algorithm 

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.

1) Start in the leftmost column
2) If all queens are placed
    return true
3) 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 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.

A modification is that we can find whether we have a previously placed queen in a column or in left diagonal or in right diagonal in O(1) time. We can observe that 

1. For all cells in a particular left diagonal , their row + col  = constant.
2. For all cells in a particular right diagonal, their row – col + n – 1 = constant.

Let say n = 5, then we have a total of 2n-1 left and right diagonals

Let say we placed a queen at (2,0) 

(2,0) have leftDiagonal value = 2. Now we can not place another queen at (1,1) and (0,2) because both of these have same leftDiagonal value as for (2,0). Similar thing can be noticed for right diagonal as well.

Implementation:

[2 4 1 3 ][3 1 4 2 ]

Time Complexity: O(N!), where N is the size of the chessboard. 

Auxiliary Space: O(N^2)

Efficient Backtracking Approach Using Bit-Masking 

Algorithm: 
There is always only one queen in each row and each column, so idea of backtracking is to start placing queen from the leftmost column of each row and find a column where the queen could be placed without collision with previously placed queens. It is repeated from the first row till the last row. While placing a queen, it is tracked as if it is not making a collision (row-wise, column-wise and diagonally) with queens placed in previous rows. Once it is found that the queen can’t be placed at a particular column index in a row, the algorithm backtracks and change the position of the queen placed in the previous row then moves forward to place the queen in the next row. 

1. Start with three-bit vector which is used to track safe place for queen placement row-wise, column-wise and diagonally in each iteration.
2. Three-bit vector will contain information as bellow:
   • rowmask: set bit index (i) of this bit vector will indicate, the queen can’t be placed at ith column of next row.
   • ldmask: set bit index (i) of this bit vector will indicate, the queen can’t e placed at ith column of next row. It represents the unsafe column index for next row falls under left diagonal of queens placed in previous rows.
   • rdmask: set bit index (i) of this bit vector will indicate, the queen can’t be placed at ith column of next row. It represents the unsafe column index for next row falls right diagonal of queens placed in previous rows.
3. There is a 2-D (NxN) matrix (board), which will have ‘ ‘ character at all indexes in beginning and it gets filled by ‘Q’ row-by-row. Once all rows are filled by ‘Q’, the current solution is pushed into the result list.

Below is the implementation of the above approach: 

[2 4 1 3 ][3 1 4 2 ]

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Iterative backtracking solution:

In this section of the article, an iterative backtracking solution without recursion is introduced. The code will mainly utilize two nested loops and a stack to keep track of the positions of queens on the board and use this stack to backtrack.

Algorithm:

n: is the number of rows, columns, and queens to be added.

1- Create 2D-array to represent the board with each cell in the array corresponding to a box in the board
2- Create a stack to keep track of the queens' positions where the queen on the top is always the most recently added queen.
3- set counter j = 1 which will be used loop through columns in the coming steps.
4- Create an outer loop through the rows from i = 1 ----> i = n
    5- Create inner loop through columns from j  ----> j = n.
        6- in the inner body of loops check if the current cell is valid to add a queen
        if valid 
            - add the queen to the board
            - add the queen's position to the stack
            - break the inner loop
    7- after finishing or breaking the inner loop 
    a) set j = 1 (so the default is to start looping through the columns from j = 1 in the coming iteration)
    b) check if the size of the stack equals the number of the row to see if a queen was placed in current row
    if no queen was placed:
        8- back track to the last added queen by doing the following:
            a) accessing the position of the last added queen from the top of stack
            b) delete the last added queen from the board
            c) set i (counter for rows of the board) = the row of the last added queen
            d) set j = the column of last added queen + 1
    9- if i == n meaning that all rows are now visited and filled with queens
        a) print solution
        b) do step 8 again to test other solutions until the stack is empty
            
    
            

According to the below Implementation, the solutions will be printed as an n*n matrix of zeros and ones, where the cells = 1 represent the queens. 

Below is the Implementation of the above algorithm

0 1 0 0 
0 0 0 1 
1 0 0 0 
0 0 1 0 

0 0 1 0 
1 0 0 0 
0 0 0 1 
0 1 0 0 

Time Complexity: O(n² * n!), where n is the size of the board.
Space Complexity: O(n²)