find a solution to the nqueens problem using min conflicts local search algorith

Question

find a solution to the nqueens problem using min conflicts local search algorithm

import random
import sys
def nqueens(nr):
show(min_conflicts(list(range(nr)), nr), nr)

#print the solution if it exists
def show(soln, nr):
for i in range(nr):
  row = ['~'] * nr
  for col in range(nr):
   if soln[col] == nr - 1 - i:
    row[col] = 'Q'
  print(''.join(row))

#passing in number of iterations to overcome situations where min-conflicts chokes
#due to critical-ratio issues
#this should not matter in most cases
def min_conflicts(soln, nr, iters=1000000):
        #initialize queens to random positions
def random_pos(li, filt):
  return random.choice([i for i in range(nr) if filt(li[i])])

for k in range(iters):
            print (k)
            #REMOVE THE ABOVE PRINT STATEMENT WHEN YOU MAKE YOUR CHANGES#
            ## YOUR CODE HERE ##
            #AROUND 10-25 LINES OF CODE EXPECTED #
            #HINT: MAKE USE OF THE find_conflicts METHOD BELOW#
raise Exception("Incomplete solution: try more iterations.")

def find_conflicts(soln, nr):
return [hits(soln, nr, col, soln[col]) for col in range(nr)]

def hits(soln, nr, col, row):
total = 0
for i in range(nr):
  if i == col:
   continue
  if soln[i] == row or abs(i - col) == abs(soln[i] - row):
   total += 1
return total

#try running n-queens with n=64 etc.
nqueens(64)

Constraint Satisfaction using min-conflicts In this problem, you will implement a solution to the n-queens problem using min-conflicts local search algorithm. A skeleton code file is provided for you. Take a look at the output function, show (soln, nr). This function will print the solution if your code finds one in the min_conflicts method in the format shown below. Here is output from running 4-queens. And here is 10-queens.

Explanation / Answer

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

The expected output is a binary matrix which has 1s for the blocks where queens are placed. For example following is the output matrix for above 4 queen solution.

Explanation:

Naive Algorithm
Generate all possible configurations of queens on board and print a configuration that satisfies the given constraints.

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.

Program:

/* C/C++ program to solve N Queen Problem using
backtracking */
#define N 4
#include<stdio.h>

/* 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(" %d ", board[i][j]);
       printf(" ");
   }
}

/* A utility function to check if a queen can
be placed on board[row][col]. Note that this
function is called when "col" queens are
already placed in columns from 0 to col -1.
So we need to check only left side for
attacking queens */
bool isSafe(int board[N][N], int row, int col)
{
   int i, j;

   /* Check this row on left side */
   for (i = 0; i < col; i++)
       if (board[row][i])
           return false;

   /* Check upper diagonal on left side */
   for (i=row, j=col; i>=0 && j>=0; i--, j--)
       if (board[i][j])
           return false;

   /* Check lower diagonal on left side */
   for (i=row, j=col; j>=0 && i<N; i++, j--)
       if (board[i][j])
           return false;

   return true;
}

/* A recursive utility function to solve N
Queen problem */
bool solveNQUtil(int board[N][N], int col)
{
   /* 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(board, i, col) )
       {
           /* Place this queen in board[i][col] */
           board[i][col] = 1;

           /* recur to place rest of the queens */
           if ( solveNQUtil(board, col + 1) )
               return true;

           /* If placing queen in board[i][col]
           doesn't lead to a solution, then
           remove queen from board[i][col] */
           board[i][col] = 0; // BACKTRACK
       }
   }

   /* 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
Backtracking. It mainly uses solveNQUtil() 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 solveNQ()
{
   int board[N][N] = { {0, 0, 0, 0},
       {0, 0, 0, 0},
       {0, 0, 0, 0},
       {0, 0, 0, 0}
   };

   if ( solveNQUtil(board, 0) == false )
   {
   printf("Solution does not exist");
   return false;
   }

   printSolution(board);
   return true;
}

// driver program to test above function
int main()
{
   solveNQ();
   return 0;
}

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