Question

Consider the following grammar:

A -> aB | b | cBB

B -> aB | bA | aBb

C -> aaA | b | caB

1. Perform the pairwise disjoint test for the grammar.
2. Rewrite the above grammar so that all grammar rules pass the pairwise disjoint test.
3. Suppose lex() is the lexical analyzer which gets the next lexeme and puts its token code in the global variable nextToken. And suppose the token codes for terminals a, b, and c are A_CODE, B_CODE, and C_CODE, respectively. Sketch a subprogram in a recursive-descent LL parser to parse the nonterminal A in the grammar you defined above in the question 2.

Answer 1

Ans:-

   Given,

A -> aB | b | cBB

B -> aB | bA | aBb

C -> aaA | b | caB

let A, the first sets of each RHS of each nonterminal

a] A -> aB | b | cBB

FIRST(aB) = a

FIRST(b) = b

FIRST(cBB) = c

PASSED

b] B -> aB | bA | aBb

FIRST(aB) = {a}

FIRST(bA) = {b}

FIRST(aBb) = {a}

FAILED

c] C -> aaA | b | caB

FIRST(aaA) = {a}

FIRST(b) = {b}

FIRST(caB) = {c}

PASSED

* PROGRAM:

#include <stdio.h>

void A();
void B();
void C();
char input[100];
int i,error;
int main()
{
printf("/n Enter the inputs:");
   A();
  
}

A()
{
   if (input(i) == 'a''B') {
   i++;
       B();
   }
   else if(input(i) == '|')
   {
   i++;
   }
   else if(input(i) == 'b')
   {
   i++;
   }
   else if(input(i) == '|')
   {
   i++;
   }
   else if(input(i) == 'c''B''B')
   {
   i++;
   B();
   }
}

B()
{
   if (input(i) == 'a''B') {
   i++;
       B();
   }
   else if(input(i) == '|')
   {
   i++;
   }
   else if(input(i) == 'b''A')
   {
   i++;
   A();
   }
   else if(input(i) == '|')
   {
   i++;
   }
   else if(input(i) == 'a''B''b')
   {
   i++;
   B();
   }
}

C()
{
   if (input(i) == 'a''a''A') {
   i++;
       A();
   }
   else if(input(i) == '|')
   {
   i++;
   }
   else if(input(i) == 'b')
   {
   i++;
  
   }
   else if(input(i) == '|')
   {
   i++;
   }
   else if(input(i) == 'c''a''B')
   {
   i++;
   B();
   }
}