Question

1. An integer is said to be perfect if the sum of its factors, including 1, is equal to the number itself. Write a Prolog predicate perfect(N, F), which determines if integer N is perfect and if so, return it’s list of factors F. If N is not perfect, the predicate fails. You may find the arithmetic operation M mod N to be useful. Test your predicate on the following.

• perfect(6, Factors) ⇒ Factors = [1, 2, 3]

• perfect(298, Factors) ⇒ no

• perfect(496, Factors) ⇒ Factors = [1, 2, 4, 8, 16, 31, 62, 124, 248]

2. Write a Prolog predicate delete(Atom, List1, List2) that deletes all occurrences, no matter how deep, of Atom in List1 and returns the result in List2. The returned list cannot contain anything in place of the deleted atoms. Test your predicate on the following.

• delete(a, [a, b, r, a, c, a, d, a, b, r, a], List) ⇒ List = [b, r, c, d, b, r]

• delete(b, [no, bs, here], List) ⇒ List = []

• delete(nest, [nest, [second, nest, level], [third, [nest], level], [[[big, nest]]]], List) ⇒ [[second, level], [third, [], level], [[[big]]]]

Answer 1

1)

program for perfect number

#include <stdio.h> 
#include <conio.h>

int main()
{
    int a, num, sum = 0 ;
    printf("Enter a number to check for the perfect number : ") ;
    scanf("%d", &num) ;
    
    for(a = 1 ; a < num ; a++)
    {
        if(num % a == 0)
        sum = sum + a ;
    }

    if (sum == num)
    printf("\n%d is a perfect number", num) ;
    else
    printf("\n%d is not a perfect number", num) ;
    getch() ;
}

2)

predicate delete(Atom, List1, List2) that deletes all occurrences, no matter how deep, of Atom in List1 and returns the result in List2. The returned list cannot contain anything in place of the deleted atoms. Test your predicate on the following.

• delete(a, [a, b, r, a, c, a, d, a, b, r, a], List) ⇒ List = [b, r, c, d, b, r]

• delete(b, [no, bs, here], List) ⇒ List = []

• delete(nest, [nest, [second, nest, level], [third, [nest], level], [[[big, nest]]]], List) ⇒ [[second, level], [third, [], level], [[[big]]]]

Success:
   [eclipse]: delete(X,[1,M,X],L), writeln((M,X,L)), fail.
   _g66 , 1 , [_g66, 1]
   _g66 , _g66 , [1, _g66]
   _g66 , _g72 , [1, _g66]
   no (more) solution.

   [eclipse]: delete(3,[1,3,5,3],L).
   L = [1, 5, 3]    More? (;)
   L = [1, 3, 5]
   yes.

   [eclipse]: delete(X,L,[a,b]), writeln((X,L)), fail.
   _g66 , [_g66, a, b]
   _g66 , [a, _g66, b]
   _g66 , [a, b, _g66]
   no (more) solution.

   delete(X,[1,2],L).   (gives X=1 L=[2]; X=2 L=[1]).
Fail:
   delete(1,[1,2,1,3],[2,3]).

Answer 2

#include <iostream> #include <cctype>using namespace std; int main(){
    int n,i=1,sum=0;
    cout << "Enter a number: ";
    cin >> n;
       while(i<n){
       if(n%i==0)
       sum=sum+i;
       i++; }
 if(sum==n)
    cout << i << " is a perfect number\n"; else
    cout << i << " is not a perfect number\n";
    system("pause"); return 0;}