8 queens problem using backtracking recurison

Question

I've been working on the 8 queens problem but I got stuck. I don't want code. I would love guidance and directions in order to understand how to solve this problem myself using backtracking recursion.

The program should enumerate all solutions to the N-queens problem by drawing the location of the queens in ASCII like the two solutions here.

My pseudocode so far is:

void queen(int n){

   for( int i = 0; i < n; i++){

       place queen[ i ] on row i;

       for(int j = 0 ; j < n ; j++){
               if( queen[ i ] is not in the same column as queen[0] through queen[ i - 1 ]  &&
                   queen[ i ] is not on the same major diagonal with queen[0] through queen[ i -1 ]  &&
                   queen[ i ] is not on the same minor diagonal with queen[0] through queen[ i -1 ]  ) {
                              print 'Q ';
                   }
               else{
                              print '* ';
                   }

               System.out.println();
         }

         System.out.println();

  }

}

There is no any backtracking recursion in my pseudocode because I don't know how to do it.

Any help is greatly appreciated.No code, please.

(Update in response to Nemo):

solver(int n, Board b){
    for(int i = 0; i < b.length; i++){
       place queen in column i;
       for(int j = 0; j < b.length; j++){
           change b;
           solver(n+1,changed b); 
       }
    }
}

Is it correct?

(Update 2):

 solver8(board /* with queens presented in first 7 columns */){
    // place a queen in the 8th column;
    for(each possible placement of the queen in column 8 
        or in other words int i = 0; i < board.length; i++ ){
             place the queen and print the board
    }
}


 solver7(board /* with queens presented in first 6 columns */){
    // place a queen in the 7th column;
    for(each possible placement of the queen in column 7 
        or in other words int i = 0; i < board.length; i++ ){
             solver8(board with queens placed in first 7 columns);
    }
}


 solver6(board /* with queens presented in first 5 columns */ ){
    // place a queen in the 6th column;
    for(each possible placement of the queen in column 6 
        or in other words int i = 0; i < board.length; i++ ){
             solver7(board with queens presented in first 6 columns);
    }
}

and so on until

 solver1(1, empty board){
     for(int i = 0; i < board.length; i++){
        place queen in row[i] of column 1;
        solver2(board with queen in row[i] of column 1);
      }
}

Update 3 (Edited):

private int numberOfQueens = 8;
solver(int n, Board b){

        for(int r = 0; r < b.length; r++){

               place queen in row[r] of column[n];

               if(n == numberOfQueens){
                    print the board;
                    return;
                }
                else{
                    solver(n+1, board with queen in row[r] of column[n]);
                }
           }
     }
}

Answer 1

The purpose of using recursion for these kinds of problems is that they allow you to think in terms of "I have now placed k queens; how can I place the remaining ones if the total number of queens is n?" So the recursive function should take two parameters: the target number of queens, and the number of queens placed so far. When writing the function, your goal is first and foremost to try out different ways of placing the k th queen. But when you have selected a possible placement and found it to be valid, you need to place the remaining n - k - 1 queens. How can we do this? The answer: recursion! Call the function (from within itself) with the parameter k - 1 to indicate that you want to place the remaining k - 1 queens. Whenever you exhaust all possibilities (or find that none are possible), simply return from the function - you will then get back to the previous function call (e.g. the one that tries to place the k th queen).

Edit: You will also need to create a two-dimensional array to represent the current state of your board; this array must either be sent as an additional parameter to the recursive function, or be kept as a field of the class that contains the method.

As for the backtracking, that is accomplished simply by making sure that the function that gets called with k + 1 removes the k + 1 th queen from the board before returning; this essentially says "We've now (unsuccessfully) tried all ways of placing the remainder of the queens - based on the positions of the k queens that have already been placed. None of them succeeded, so please adjust the positions of the first k queens (which will be done by the function that was called with k, and the function which called that function, and so on) and try again."