ALGORITHM
1. Start
2. For each nonterminal A and terminal a do L(A,a):= false;
3. For each production of the form A->a or A->B do INSTALL(A,a);
4. While STACK not empty repeat step 5& 6
5. Pop top pair (B,a) from STACK;
6. For each production of the form A->B do INSTALL(A,a)
7. Stop
Algorithm For INSTALL(A,a)
1. Start
2. If L(A,a) not present do step 3 and 4.
3 . Make L(A,a)=True
4 . Push (A,a) onto stack
5 . Stop
#include<conio.h>
#include<stdio.h>
char arr[18][3] =
{
{'E','+','F'},{'E','*','F'},{'E','(','F'},{'E',')','F'},{'E','i','F'},{'E','$','F'},
{'F','+','F'},{'F','*','F'},{'F','(','F'},{'F',')','F'},{'F','i','F'},{'F','$','F'},
{'T','+','F'},{'T','*','F'},{'T','(','F'},{'T',')','F'},{'T','i','F'},{'T','$','F'},
};
char prod[6] = "EETTFF";
char res[6][3]=
{
{'E','+','T'},{'T','\0'},
{'T','*','F'},{'F','\0'},
{'(','E',')'},{'i','\0'},
};
char stack [5][2];
int top = -1;
void install(char pro,char re)
{
int i;
for(i=0;i<18;++i)
{
if(arr[i][0]==pro && arr[i][1]==re)
{
arr[i][2] = 'T';
break;
}
}
++top;
stack[top][0]=pro;
stack[top][1]=re;
}
void main()
{
int i=0,j;
char pro,re,pri=' ';
clrscr();
for(i=0;i<6;++i)
{
for(j=0;j<3 && res[i][j]!='\0';++j)
{
if(res[i][j]
=='+'||res[i][j]=='*'||res[i][j]=='('||res[i][j]==')'||res[i][j]=='i'||res[i][j]=='$')
{
install(prod[i],res[i][j]);
break;
}
}
}
while(top>=0)
{
pro = stack[top][0];
re = stack[top][1];
--top;
for(i=0;i<6;++i)
{
if(res[i][0]==pro && res[i][0]!=prod[i])
{
install(prod[i],re);
}
}
}
for(i=0;i<18;++i)
{
printf("\n\t");
for(j=0;j<3;++j)
printf("%c\t",arr[i][j]);
}
getch();
clrscr();
printf("\n\n");
for(i=0;i<18;++i)
{
if(pri!=arr[i][0])
{
pri=arr[i][0];
printf("\n\t%c -> ",pri);
}
if(arr[i][2] =='T')
printf("%c ",arr[i][1]);
}
getch();
}
OUTPUT E + T E * T E ( T E ) F E i T E $ F F + F F * F F ( T F ) F F i T F $ F T + F T * T T ( T T ) F T i T T $ F