N-Queens Solutions C++

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So I need help with the classic N-Queens problem.

The command to run the program will be: nqueens N k - where N is the size of the table (N x N) and k is the number of solutions

So for example if I were to run the program by typing nqueens 4 1 the following would be printed out.

_ Q _ _

_ _ _ Q

Q _ _ _

_ _ Q _

However, I cannot figure out how to handle more than 1 solution? How can I determine more than just one solution for this problem?

What I have so far is below:

#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <vector>


using namespace std;

class Board
{
private:
    bool** spaces;
    int size;

public:
    Board(int size)
    {
        this->size = size;
        spaces = new bool*[size];
        for (int i = 0; i < size; ++i)
        {
            spaces[i] = new bool[size];
        }
    }

    bool isSafe(int row, int column, vector<int>& state)
    {
       //check row conflict
       //no need to check for column conflicts
       //because the board is beimg filled column
       //by column
       for(int i = 0; i < column; ++i)
       {
          if(state[i] == row)
             return false;
       }

       //check for diagonal conflicts
       int columnDiff = 0;
       int rowDiff = 0;
       for(int i = 0; i < column; ++i)
       {
          columnDiff = abs(column - i);
          rowDiff = abs(row - state[i]);
          if(columnDiff == rowDiff)
             return false;
       }

       return true;

    }

    int getSize()
    {
        return size;
    }

    bool getSpace(int x, int y)
    {
        return spaces[y][x];
    }

    void setSpace(int x, int y, bool value)
    {
        spaces[y][x] = value;
    }

    Board(Board& other)
    {
        this->size = other.getSize();
        spaces = new bool*[size];
        for (int i = 0; i < size; ++i)
        {
            spaces[i] = new bool[size];
            for (int j = 0; j < size; ++j)
            {
                spaces[i][j] = other.getSpace(j, i);
            }
        }
    }

    void backtrack(vector<int>& state, int board_size)
  {
     int row = 0;
     int column = 0;
     bool backtrack = false;

     while(column < board_size)
     {
        while(row < board_size)
        {
           if(backtrack)
           {
              row = state[column] + 1;
              if(row == board_size)
              {
                 column--; //backtrack more
                 backtrack = true;
                 row = 0;
                 break;
              }

              backtrack = false;
           }

           if(isSafe(row, column, state))
           {
              state[column] = row;
              row = 0;
              column++; //advance
              backtrack = false;
              break;
           }

           else
           {
              if(row == (board_size - 1))
              {
                 column--; //backtrack
                 backtrack = true;
                 row = 0;
              }
              else
              {
                 row++;
              }
           }
        }
     }
  }
};

int print_solutions(Board *board, vector<int>& state, int board_size)
{
   for(int i=0; i < board_size; ++i)
   {
      for(int j=0; j < board_size; ++j)
      {
         if(state[i] == j)
            cout << 'Q' << " ";
         else
            cout << '_' << " ";
      }

      cout << endl;
   }
}   

int main(int argc, char* argv[])
{
    if (argc < 2)
    {
        cout << "Usage: nqueens [Board Size] [Number of Solutions]" << endl;
    return 1;
    }

    int board_size = atoi(argv[1]);
    //int solution_count = atoi(argv[2]);
    vector<int> state;
    state.resize(board_size);

    Board *my_board = new Board(board_size);
    my_board->backtrack(state, board_size);

    print_solutions(my_board, state, board_size);

    return 0;
}

Printing all solutions in N-Queen Problem, 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. 3) If all rows have been tried and  So I need help with the classic N-Queens problem. The command to run the program will be: nqueens N k - where N is the size of the table (N x N) and k is the number of solutions. So for example if I were to run the program by typing nqueens 4 1 the following would be printed out. _ Q _ _ _ _ _ Q. Q _ _ _ _ _ Q _

Here is my brute force recursive solution. It is not optimal one (without backtracking), but works reasonable good for chess boards up to 14 x 14. In the recursive method queensSolution the first argument is the size of chess board. The second argument encodes actual position of the queens.

For example, vector describing the position on the picture would be {1, 3, 0, 2}. This means: in the first row (count starts with 0, so it is first element of the vector) a queen is on position 1 (second quadrat from left), in the second row (second element in the vector) a queen is placed on position 3 (the last quadrat of the row) etc.

The third argument contains vector that will contain all solution positions (encoded as vector as described above). The help method intersect checks whether there is a collision between a new queen that will be placed on position {queens.size(), x}. Collisions occur when the new queen is on the same 'column' (x position) as any of existing queens or whether the distances between x positions and y positions of the new queen with any of existing queens are equal (diagonal position). We do not have to check whether the new queen will be placed in a row (y) where some other queen is already placed, because every time we add an element to the queens vector we put it in new row.

#include<vector>
using namespace std;

bool intersect(const vector<int> &queens, int x) {
    int y = queens.size();
    for (int i = 0; i < y; i++) {
        if ((abs(queens[i] - x) == 0) || (abs(queens[i] - x) == y - i))
            return true;
    }
    return false;
}

void queensSolution(const int dim, vector<int> queens, vector<vector<int>> &res) {
    if (queens.size() >= dim) {
        res.push_back(queens);
        queens.clear();
        return;
    }
    for (int i = 0; i < dim; i++) {
        if (!intersect(queens, i)) {
            queens.push_back(i);
            queensSolution(dim, queens, res);
            queens.pop_back();
        }
    }
}

For example, to find all solutions for 4 x 4 chessboard, do following:

int main(){
  int dim = 4;
  vector<int>queens;
  vector<vector<int>> result;
  queensSolution(dim, queens, result);
}

After queensSolution returns, vector result contains two vectors: {1, 3, 0, 2} and {2, 0, 3, 1}.

C Program for 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, following is a solution for  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, following is a solution for 4 Queen problem.

You can use a recursive approach for solving it. I have written an article about this: Recursion tutorial: N-queens in C. To obtain all the solutions, simply run the recursion without terminating for the first solution found.

There is also a heuristic solution available here: Eight queen puzzle.

N Queens Problem in C Using Backtracking, In this tutorial I am sharing the C program to find solution for N Queens problem using backtracking. Below animation shows the solution for 8 queens problem  therefore, for N=4, only two solutions are possible. Code implementation in C++ N queens problem //program to solve the n queen problem //grid[][] is represent the 2-d array with value(0 and 1) for grid[i][j]=1 means queen i are placed at j column. //we can take any number of queen , for this time we take the atmost 10 queen (grid[10][10]).

Check this gist. It's a simple recursive function that returns all the solutions. It works by placing each time a queen a the next row. The method is_safechecks if it is safe to place a queen at column col in the next row. The solution are vector where the index i is the row and the value at that index is the column. Each time a queen is placed successfully, the vector grows by one element and is added to the set of candidate solutions which are returned back up in the recursive calls stack.

Print all possible solutions to N Queens problem, If current configuration doesn't result in a solution, we backtrack. Before exploring any square, we ignore the square if two queens threaten each other. C++; Java  N Queens Problem is a famous puzzle in which n-queens are to be placed on a nxn chess board such that no two queens are in the same row, column or diagonal. In this tutorial I am sharing the C program to find solution for N Queens problem using backtracking. Below animation shows the solution for 8 queens problem using backtracking.

Eight queens puzzle, The eight queens puzzle has 92 distinct solutions. If n > 1, it is not possible for a solution to be equivalent to its own reflection because that would require two queens to be facing each other. a, b, c, d, e, f, g, h. 8. If a queen is not in A and B and C, all Solutions is Unique. Judgment value is investigated when that is not right. 90-degree rotation. (A) 180-degree rotation. (B) 270-degree rotation. (C) 3. Total Solutions from Unique Solutions If first queen is in the corner. Total Solutions = Unique Solutions X 8. If first queen is inside.

What is the type of algorithm used in solving the 8 Queens problem , 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. 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. For example, for standard 8 × 8 chessboard below is one such …

8 Queens Problem using Backtracking, Which type of algorithm is used to solve the 8 queens problem? The N -Queens Counter programs have been used to find the number of queen placements and the number of N -queens solutions for values of N between 1 and 15 with the Go, Python, and R implementations, and values of N between 1 and 19 with the C implementation.

Comments
  • You might find this useful: igorsevo.com/…
  • Well, argc < 2 should probably be argc < 3 since you expect two parameters, then you would likely want to actually use the number of solutions parameter in some way.