NOTE: The midterm exam is on Thursday, March 29.

The exam will cover the material on expression grammars and recursive descent parsers from the following zip file.

• Grammars-and-Parsing.zip

Problem 2:

For each of the following expressions, rewrite the expression in both prefix and postfix notation, and draw an abstract syntax tree (i.e., an expression tree) for the expression.
(a) a * b + c
(b) a * (b + c)
(c) a * b + c * d
(d) a * (b + c) * d

Problem 3:

Describe in English the language defined by the following grammar. (Note: Do not use English to describe the grammar, describe the language defined by the grammar.)

         <S> -> <A> <B> <C>
         <A> -> 'a' <A> | 'a'
         <B> -> 'b' <B> | 'b'
         <C> -> 'c' <C> | 'c'

Problem 4:

Consider the following grammar.

         <S> -> <A> 'a' <B> 'b'
         <A> -> <A> 'b' | 'b'
         <B> -> 'a' <B> | 'a'

Which of the following words are in the language generated by this grammar?
(a) baab
(b) bbbab
(c) bbaaaaa
(d) bbaab

Problem 5:

Consider the following grammar.

         <S> -> 'a' <S> 'c' <B> | <A> | 'b'
         <A> -> 'c' <A> | 'c'
         <B> -> <A> | 'd'

Which of the following words are in the language generated by this grammar?
(a) abcd
(b) acccbd
(c) acccbcc
(d) acd
(e) accc

Problem 6:

For each of the following parts, define a grammar for the words, over the alphabet containing just 0 and 1, that are described.
(a) All words ending with a 1.
(b) All words of length exactly 8.
(c) All words that contain the string 0101.
(d) All words with odd length. Show how to parse the word 00100 using your grammar.

Problem 7:

Show that the grammar

   expr ->  '-' expr
         |  expr '-' ( 'x' | 'y' | 'z' )
         |  'x' | 'y' | 'z'

is ambiguous by finding a string that has two different parse trees.

Problem 8:

The following grammar is motivated by declarations in C.

Declaration ::= Type Declarator ';'
Type ::= 'int' | 'char'
Declarator ::= '*' Declarator
            |  Declarator '[' Digit ']'
            |  Declarator '(' Type ')'
            |  '(' Declarator ')'
            |  Letter
Letter ::= 'a' | 'b' | ... | 'y' | 'z'
Digit  ::= '1' | '2' | ... | '8' | '9'

(a) Give 5 examples of strings generated by this grammar.
(b) Show that this grammar is ambiguous.

Problem 9:

Consider the following grammar for postfix expressions.

     Postfix -> Postfix Postfix BinOp
             |  Postfix UnOp
             |  Number

       BinOp -> '+' | '-' | '*' | '/'

       UnOp  -> 'neg' | 'sqrt'

      Number -> any valid Java double

Part (a) Draw the parse tree (not the expression tree) for the expression

   4 5 + sqrt 3 *

Part (b) Explain why you cannot write a recursive descent parser for this grammar.

Part (c) Draw the abstract syntax tree for the expression

   4 5 + sqrt 3 *

Problem 10:

Consider the following grammar for postfix expressions. (The symbol EMPTY in the grammar means that we produce an empty string.)

     Postfix -> Number Postfix2

    Postfix2 -> Postfix BinOp Postfix2
             |  UnOp Postfix2
             |  EMPTY

       BinOp -> '+' | '-' | '*' | '/'

       UnOp  -> 'neg' | 'sqrt'

      Number -> any valid Java double

Part (a) Draw the parse tree (not the expression tree) for the expression

   4 5 + sqrt 3 *

Part (b) Explain why the abstract syntax tree for the expression

   4 5 + sqrt 3 *

is the same as it was for the previous problem, even though the grammar has changed.

Part (c) Explain why you can write a recursive descent parser for this grammar.

Part (d) Write the Java code for the getPostfix() and getPostfix2() methods that would implement a recognizing parser (assume the usual interface to the Tokenizer).

   private static void getPostfix(Tokenizer tokens)
   {




   }//getPostfix()
   private static void getPostfix2(Tokenizer tokens)
   {




   }//getPostfix2()

Problem 11:

Suppose that a grammar for some language has the following production, P3. Use Java to write the function that parses this production in a recognizing recusive descent parser. Assume the usual interface to the Tokenizer class.

   P3  -> '*'  P4
        | '@'  P4  P5
        | '#'  P5 ( '#' P5 )*
        |  P6