CS 132 Fall 2003, Midterm Solution
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Question 1
----------
Here follows two somewhat different solutions (labeled A and B).

Solution A:       
   E ::= ( E ) S  |  i S
   S ::= . i ( E ) S  |  ? E : E  |  epsilon

      First     Follow
   E    ( i      $ ) :
   S  . ? eps    $ ) :

       .      ?     :    (     )    i   $
   E                     (E)S       iS
   S   .i(E)S ?E:E  eps        eps      eps

Solution B:
   E ::= ( E ) D Q  |  i D Q
   Q ::= epsilon  |  ? E : E
   D ::= epsilon  |  . i ( E ) D

      First     Follow
   E   ( i       $ ) :
   Q   ? eps     $ ) : 
   D   . eps     $ ) : ?

       .      ?     :    (     )    i   $
   E                     (E)DQ      iDQ
   Q          ?E:E  eps        eps      eps
   D   .i(E)D eps   eps        eps      eps
   

Question 2
----------
   Fi(A) = Fi(B) union Fi(C) union {y} = {x,y}
   Fi(B) = {y} union {eps}             = {y,eps}
   Fi(C) = Fi(A) union {x}             = {x,y}
   Fo(A) = {y} union Fo(B)             = {x,y}
   Fo(B) = Fi(C)                       = {x,y}
   Fo(C)                               = {x}  

   Not LL(1) because the two productions for A
   having overlapping First sets.


Question 3
----------
Here is a sketch of the Java code:

   void E() {
      B();
      A();
      if ( token != EOF ) error();
   }
   
   void A() {
      if ( token == '&' ) {
         get-next-token();
         B();
         A();
      }
   }
   
   void B() {
      if ( token == "true" || token == "false" ) {
         get-next-token();
      else
         error();
   }