Implementation of Unification algorithm

unification.c

 

 

 #include<stdio.h>
int no_of_pred;
int no_of_arg[10];
int i,j;
char nouse;
char predicate[10];
char argument[10][10];


void unify();
void display();
void chk_arg_pred();


   void main()
   {
   char ch;
   do{
   

 
            printf("\t=========PROGRAM FOR UNIFICATION=========\n");
            printf("\nEnter Number of Predicates:- [ ]\b\b");
            scanf("%d",&no_of_pred);

            for(i=0;i<no_of_pred;i++)
            {
            scanf("%c",&nouse);    //to accept "enter" as a character
            printf("\nEnter Predicate %d:-[ ]\b\b",i+1);
            scanf("%c",&predicate[i]);
            printf("\n\tEnter No.of Arguments for Predicate %c:-[ ]\b\b",predicate[i]);
            scanf("%d",&no_of_arg[i]);
                        for(j=0;j<no_of_arg[i];j++)
                        {
                         scanf("%c",&nouse);
                         printf("\n\tEnter argument %d:( )\b\b",j+1);
                         scanf("%c",&argument[i][j]);
                        }
            }

            display();
            chk_arg_pred();
            
            
            printf("Do you want to continue(y/n):   ");
            scanf("%c",&ch);
       }while(ch == 'y');
   }



   void display()
   {
       printf("\n\t=======PREDICATES ARE======");
            for(i=0;i<no_of_pred;i++)
            {
             printf("\n\t%c(",predicate[i]);
                        for(j=0;j<no_of_arg[i];j++)
                        {
                        printf("%c",argument[i][j]);
                        if(j!=no_of_arg[i]-1)
                                    printf(",");
                        }
             printf(")");
            }
   }

   void chk_arg_pred()
   {
   int pred_flag=0;
   int arg_flag=0;

   /*======Checking Prediactes========*/
            for(i=0;i<no_of_pred-1;i++)
            {
                        if(predicate[i]!=predicate[i+1])
                        {
                        printf("\nPredicates not same..");
                        printf("\nUnification cannot progress!");
                        pred_flag=1;
                        break;
                        }
            }
   /*=====Chking No of Arguments====*/
     if(pred_flag!=1)
     {
            for(i=0;i<no_of_arg[i]-1;i++)
            {
                        if(no_of_arg[i]!=no_of_arg[i+1])
                        {
                        printf("\nArguments Not Same..!");
                        arg_flag=1;
                        break;
                        }
            }
     }
            if(arg_flag==0&&pred_flag!=1)
                        unify();

   }
/*==========UNIFY FUNCTION=========*/
   void unify()
   {
            int flag=0;
            for(i=0;i<no_of_pred-1;i++)
            {
                 for(j=0;j<no_of_arg[i];j++)
                 {
                        if(argument[i][j]!=argument[i+1][j])
                        {
                          if(flag==0)
                          printf("\n\t======SUBSTITUTION IS======");
                        printf("\n\t%c/%c",argument[i+1][j],argument[i][j]);
                         flag++;
                        }
                }
            }
            if(flag==0)
            {          printf("\nArguments are Identical...");
                        printf("\nNo need of Substitution\n");
            }
   }



/*
dypcoe@dypcoe-OptiPlex-390:~$ gcc u.c
dypcoe@dypcoe-OptiPlex-390:~$ ./a.out
 =========PROGRAM FOR UNIFICATION=========

case 1 
when args are same 


Enter Number of Predicates:- [2]

Enter Predicate 1:-[a]

 Enter No.of Arguments for Predicate a:-[2]

 Enter argument 1:(5)

 Enter argument 2:(x)

Enter Predicate 2:-[a]

 Enter No.of Arguments for Predicate a:-[2]

 Enter argument 1:(3)

 Enter argument 2:(y)

 =======PREDICATES ARE======
 a(5,x)
 a(3,y)
 ======SUBSTITUTION IS======
 3/5


case 2
when args are not same


dypcoe@dypcoe-OptiPlex-390:~$ gcc u.c
dypcoe@dypcoe-OptiPlex-390:~$ ./a.out
 =========PROGRAM FOR UNIFICATION=========

Enter Number of Predicates:- [2]

Enter Predicate 1:-[a]

 Enter No.of Arguments for Predicate a:-[2]

 Enter argument 1:(5)

 Enter argument 2:(x)

Enter Predicate 2:-[a]

 Enter No.of Arguments for Predicate a:-[3]

 Enter argument 1:(3)

 Enter argument 2:(x)

 Enter argument 3:(y)

 =======PREDICATES ARE======
 a(5,x)
 a(3,x,y)
Arguments Not Same..!Do you want to continue(y/n):   dypcoe@dypcoe-OptiPlex-390:~$ gcc u.c
dypcoe@dypcoe-OptiPlex-390:~$ 

case 3
when args are identical

nter Number of Predicates:- [2]

Enter Predicate 1:-[a]

 Enter No.of Arguments for Predicate a:-[2]

 Enter argument 1:(3)

 Enter argument 2:(7)

Enter Predicate 2:-[a]

 Enter No.of Arguments for Predicate a:-[2]

 Enter argument 1:(3)

 Enter argument 2:(7)

 =======PREDICATES ARE======
 a(3,7)
 a(3,7)
Arguments are Identical...
No need of Substitution


*/