Suppose we have a solution to the n_Queens problem instance in which n = 4. Can we extend the solution to find a solution to the problem instance n = 5? Can we then use the solution for n = 4 and n = 5 to construct a solution to the instance n = 6 and continue this dynamic programming approach to find a solution to any instance in which n > 4? Justify your answer. Solution solution package com.qp.queens; import java.util.Scanner; public class QueensProblem { public static void possibleChoices(int k) { int[] choices = new int[k]; possibleChoices (choices, 0); } public static void possibleChoices(int[] queen, int no) { int n = queen.length; if (no == n) display(queen); else { for (int i = 0; i < n; i++) { queen[no] = i; if (isConsistentApproach(queen, no)) possibleChoices(queen, no+1); } } } public static void display(int[] queen) { int z = queen.length; for (int k = 0; k < z; k++) { for (int j = 0; j < z; j++) { if (queen[k] == j) System.out.print(\"p \"); else System.out.print(\"* \"); } System.out.println(); } System.out.println(); } public static boolean isConsistentApproach(int[] queen, int no) { for (int k = 0; k < no; k++) { if (queen[k] == queen[no]) return false; // same column if ((queen[k] - queen[no]) == (no - k)) return false; // same major diagonal if ((queen[no] - queen[k]) == (no - k)) return false; // same minor diagonal } return true; } public static void main(String... arguments) { Scanner scan=new Scanner(System.in); System.out.println(\"enter the instance you want\"); int choice=scan.nextInt(); QueensProblem.possibleChoices(choice); } } output enter the instance you want 5 p * * * * * * p * * * * * * p * p * * * * * * p * p * * * * * * * p * * p * * * * * * * p * * p * * * p * * * * * * p * p * * * * * * p * * * * * * p * p * * * * * * * p * * p * * p * * * * * * * p * * * p * * p * * * * * * * p * * p * * * * * * * p * * p * * * * * * p * p * * * * * * p * p * * * * * * * p * p * * * * * * p * * * * * * p * p * * * * * * p * * p * * * * * * * p * * p * * p * * * * * * * * p * p * * * * * * p * p * * * * * * p * * * * * * p * * p * * p * * * * * * * p * * p * * * enter the instance you want 6 * p * * * * * * * p * * * * * * * p p * * * * * * * p * * * * * * * p * * * p * * * * * * * * p * p * * * * * * * * p * p * * * * * * * * p * * * * * p * * p * * * * * * * * * p * * p * * * * * * * * * p * * p * * * * * * * p * * * p * * * p * * * * * * * * * * p * * * p * * * p * * * * enter the instance you want 7 p * * * * * * * * p * * * * * * * * p * * * * * * * * p * p * * * * * * * * p * * * * * * * * p * p * * * * * * * * * p * * * * * * * * * p * * p * * * * * * * * * p * * p * * * * * * * * * p * * p * * * * * * * * * * p * * * p * * * * * * * * * * p * * * p * * * * * * * * * * p * * * p * * * p * * * * * * * * * * * p * * * * p * * * * p * * * * * * * * * * * p * * * * p * * * * p * * * * * p * * * * * * * * p * * * p * * * * * * * * * * * * p * * * * p * * * * p * * * * * * * * * p * * p * * * * * * * * p * * * * * * * * p.

1. Suppose we have a solution to the n_Queens problem instance in which n = 4. Can we extend the
solution to find a solution to the problem instance n = 5? Can we then use the solution for n = 4
and n = 5 to construct a solution to the instance n = 6 and continue this dynamic programming
approach to find a solution to any instance in which n > 4? Justify your answer.
Solution
solution
package com.qp.queens;
import java.util.Scanner;
public class QueensProblem {
public static void possibleChoices(int k) {
int[] choices = new int[k];
possibleChoices (choices, 0);
}
public static void possibleChoices(int[] queen, int no) {
int n = queen.length;
if (no == n) display(queen);
else {
for (int i = 0; i < n; i++) {
queen[no] = i;
if (isConsistentApproach(queen, no)) possibleChoices(queen, no+1);
}
}
}
public static void display(int[] queen) {
int z = queen.length;
for (int k = 0; k < z; k++) {
for (int j = 0; j < z; j++) {
if (queen[k] == j) System.out.print("p ");
else System.out.print("* ");

2. }
System.out.println();
}
System.out.println();
}
public static boolean isConsistentApproach(int[] queen, int no) {
for (int k = 0; k < no; k++) {
if (queen[k] == queen[no]) return false; // same column
if ((queen[k] - queen[no]) == (no - k)) return false; // same major diagonal
if ((queen[no] - queen[k]) == (no - k)) return false; // same minor diagonal
}
return true;
}
public static void main(String... arguments) {
Scanner scan=new Scanner(System.in);
System.out.println("enter the instance you want");
int choice=scan.nextInt();
QueensProblem.possibleChoices(choice);
}
}
output
enter the instance you want
5
p * * * *
* * p * *
* * * * p
* p * * *
* * * p *

3. p * * * *
* * * p *
* p * * *
* * * * p
* * p * *
* p * * *
* * * p *
p * * * *
* * p * *
* * * * p
* p * * *
* * * * p
* * p * *
p * * * *
* * * p *
* * p * *
p * * * *
* * * p *
* p * * *
* * * * p
* * p * *
* * * * p
* p * * *
* * * p *
p * * * *
* * * p *
p * * * *
* * p * *
* * * * p
* p * * *
* * * p *
* p * * *
* * * * p
* * p * *
p * * * *
* * * * p

4. * p * * *
* * * p *
p * * * *
* * p * *
* * * * p
* * p * *
p * * * *
* * * p *
* p * * *
enter the instance you want
6
* p * * * *
* * * p * *
* * * * * p
p * * * * *
* * p * * *
* * * * p *
* * p * * *
* * * * * p
* p * * * *
* * * * p *
p * * * * *
* * * p * *
* * * p * *
p * * * * *
* * * * p *
* p * * * *
* * * * * p
* * p * * *
* * * * p *
* * p * * *
p * * * * *
* * * * * p
* * * p * *
* p * * * *
enter the instance you want

5. 7
p * * * * * *
* * p * * * *
* * * * p * *
* * * * * * p
* p * * * * *
* * * p * * *
* * * * * p *
p * * * * * *
* * * p * * *
* * * * * * p
* * p * * * *
* * * * * p *
* p * * * * *
* * * * p * *
p * * * * * *
* * * * p * *
* p * * * * *
* * * * * p *
* * p * * * *
* * * * * * p
* * * p * * *
p * * * * * *
* * * * * p *
* * * p * * *
* p * * * * *
* * * * * * p
* * * * p * *
* * p * * * *
* p * * * * *
* * * p * * *
p * * * * * *
* * * * * * p
* * * * p * *
* * p * * * *
* * * * * p *

6. * p * * * * *
* * * p * * *
* * * * * p *
p * * * * * *
* * p * * * *
* * * * p * *
* * * * * * p
* p * * * * *
* * * * p * *
p * * * * * *
* * * p * * *
* * * * * * p
* * p * * * *
* * * * * p *
* p * * * * *
* * * * p * *
* * p * * * *
p * * * * * *
* * * * * * p
* * * p * * *
* * * * * p *
* p * * * * *
* * * * p * *
* * * * * * p
* * * p * * *
p * * * * * *
* * p * * * *
* * * * * p *
* p * * * * *
* * * * * p *
* * p * * * *
* * * * * * p
* * * p * * *
p * * * * * *
* * * * p * *
* p * * * * *

7. * * * * * * p
* * * * p * *
* * p * * * *
p * * * * * *
* * * * * p *
* * * p * * *
* * p * * * *
p * * * * * *
* * * * * p *
* p * * * * *
* * * * p * *
* * * * * * p
* * * p * * *
* * p * * * *
p * * * * * *
* * * * * p *
* * * p * * *
* p * * * * *
* * * * * * p
* * * * p * *
* * p * * * *
* * * * p * *
* * * * * * p
* p * * * * *
* * * p * * *
* * * * * p *
p * * * * * *
* * p * * * *
* * * * * p *
* p * * * * *
* * * * p * *
p * * * * * *
* * * p * * *
* * * * * * p
* * p * * * *
* * * * * * p

8. * p * * * * *
* * * p * * *
* * * * * p *
p * * * * * *
* * * * p * *
* * p * * * *
* * * * * * p
* * * p * * *
p * * * * * *
* * * * p * *
* p * * * * *
* * * * * p *
* * * p * * *
p * * * * * *
* * p * * * *
* * * * * p *
* p * * * * *
* * * * * * p
* * * * p * *
* * * p * * *
p * * * * * *
* * * * p * *
* p * * * * *
* * * * * p *
* * p * * * *
* * * * * * p
* * * p * * *
* p * * * * *
* * * * * * p
* * * * p * *
* * p * * * *
p * * * * * *
* * * * * p *
* * * p * * *
* * * * * p *
p * * * * * *

9. * * p * * * *
* * * * p * *
* * * * * * p
* p * * * * *
* * * p * * *
* * * * * * p
* * p * * * *
* * * * * p *
* p * * * * *
* * * * p * *
p * * * * * *
* * * p * * *
* * * * * * p
* * * * p * *
* p * * * * *
* * * * * p *
p * * * * * *
* * p * * * *
* * * * p * *
p * * * * * *
* * * p * * *
* * * * * * p
* * p * * * *
* * * * * p *
* p * * * * *
* * * * p * *
p * * * * * *
* * * * * p *
* * * p * * *
* p * * * * *
* * * * * * p
* * p * * * *
* * * * p * *
* p * * * * *
* * * * * p *
* * p * * * *

10. * * * * * * p
* * * p * * *
p * * * * * *
* * * * p * *
* * p * * * *
p * * * * * *
* * * * * p *
* * * p * * *
* p * * * * *
* * * * * * p
* * * * p * *
* * * * * * p
* p * * * * *
* * * p * * *
* * * * * p *
p * * * * * *
* * p * * * *
* * * * p * *
* * * * * * p
* p * * * * *
* * * * * p *
* * p * * * *
p * * * * * *
* * * p * * *
* * * * * p *
p * * * * * *
* * p * * * *
* * * * p * *
* * * * * * p
* p * * * * *
* * * p * * *
* * * * * p *
* p * * * * *
* * * * p * *
p * * * * * *
* * * p * * *

11. * * * * * * p
* * p * * * *
* * * * * p *
* * p * * * *
p * * * * * *
* * * p * * *
* * * * * * p
* * * * p * *
* p * * * * *
* * * * * p *
* * p * * * *
* * * * p * *
* * * * * * p
p * * * * * *
* * * p * * *
* p * * * * *
* * * * * p *
* * p * * * *
* * * * * * p
* * * p * * *
p * * * * * *
* * * * p * *
* p * * * * *
* * * * * p *
* * * p * * *
* p * * * * *
* * * * * * p
* * * * p * *
* * p * * * *
p * * * * * *
* * * * * p *
* * * p * * *
* * * * * * p
p * * * * * *
* * p * * * *
* * * * p * *

12. * p * * * * *
* * * * * * p
* p * * * * *
* * * p * * *
* * * * * p *
p * * * * * *
* * p * * * *
* * * * p * *
* * * * * * p
* * p * * * *
* * * * * p *
* p * * * * *
* * * * p * *
p * * * * * *
* * * p * * *
* * * * * * p
* * * p * * *
p * * * * * *
* * * * p * *
* p * * * * *
* * * * * p *
* * p * * * *
* * * * * * p
* * * * p * *
* * p * * * *
p * * * * * *
* * * * * p *
* * * p * * *
* p * * * * *
if no is increasing the no of possible outcomes are increasing.so some cases it ends up with
infiniteloop.no problem if no is greaterthan 4.