Please share MGMT 615 assignment 1 solution idea
SOLVED CS607 Assignment 2 Solution and Discussion

Assignment No. 02
Semester: Fall 2019
CS607: Artificial Intelligence
Total Marks: 20Due Date:29/11/2019
Instructions
It should be clear that your assignment will not get any credit if:
The assignment is submitted after the due date.
The submitted assignment does not open or the file is corrupt.
The Solution is copied from any other source.
Objective
The objective of this assignment is to;
Learn and practice basic concepts of Eight Queen problem.
Basic understanding of Knowledge representation through graph.Assignment
Question No. 1
Consider the EightQueen problem to use representation and strategy on chess board. Show the positions of Individuals on chess board table and calculate the fitness values for both individuals in first iteration. The individuals (board positions) are given below:
(a) 47285421
(b) 52416782Question No. 2
Given the following information you are required to represent it by using the graph.
o Zulfiqar is Hussain’s Father
o Zahra is Hussain’s Mother
o Hussain is Zulfiqar and Zahra‘s SonYou are required to submit your solution through LMS as an MS Word document.



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.


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/* C/C++ program to solve N Queen Problem using backtracking */ #define N 4 #include <stdbool.h> #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("\n"); } } /* 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 the 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 the queen cannot be placed 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; }