CS 132 Midterm, Spring 2008: Solution
-------------------------------------

Question 1:
   A ::= x B | C y 
   B ::= y | x B 
   C ::= A y | z 
 ->
   A ::= x B | A y y | z y
   B ::= y | x B 
 -> 
   A ::= D | A y y 
   B ::= y | x B 
   D ::= x B | z y
 -> 
   A ::= D (y y)*
   B ::= y | x B 
   D ::= x B | z y
 ->
   A ::= D E
   B ::= y | x B 
   D ::= x B | z y
   E ::= epsilon | y y E

      First Follow Nullable   x     y      z
   A   x z    $       no     D E  error  error
   B   x y   y $      no     x B    y    error
   D   x z   y $      no     x B  error   z y
   E   y      $      yes    error y y E  error

   LL(1) because: 
    - A: only one choice.
    - B: not nullable and the two choices have different First sets.
    - D: not nullable and the two choices have different First sets.
    - E: nullable, and First(y y E) = {y} has an empty intersection
      with Follow(E) = {$}
    

Question 2:

A ::= z B | C y 
B ::= x C | epsilon
C ::= B A x | y 

       First Follow Nullable   x     y      z
   A   x y z   x       no     C y   C y    z B
   B   x     x y z    yes
   C

   Not LL(1) because: B is nullable and First(x C) and Follow(B)
   have x in the intersection.


Question 3:
   
 A ::= x B | C y 
 B ::= A x C 
 C ::= z B | epsilon

 proc eat(symbol a) {
   if (token==a) {
      token=next_token();  
   } 
   else {
      throw new Exception();
   }
 }

 proc main() {
   token=next_token();
   A();
   eat(EOF);
 }

 proc A() {
   if (token==x) {
      eat(x);
      B();
   }
   else if (token==y or token==z) {
      C();
      eat(y);
   } 
   else 
      throw new Exception();
 }
   
 proc B() {
   A();
   eat(x);
   C();
 }
   
 proc C():
   if (token==z) {
      eat(z); 
      B();
   }
   else
      // do nothing
 }