Due Wednesday, 20 October, 10:10am. Please Hand it in to Hossein before the beginning of the exercise session.
A grammar has a cycle if there is a non-terminal such that .
- Show that an LL(1) grammar must have no cycles.
- Give an algorithm that eliminates cycles in a context-free grammar.
Show that the regular languages can be recognized with LL(1) parsers. Describe a process that, given a regular expression, constructs an LL(1) parser for it.
The following grammar is motivated by type definitions in Scala.
TypeDefinition := "type" Identifier "=" Type Type := Type "," Type Type := Type "=>" Type Type := "Int" | "Boolean"
- By giving at least one input which has two different parse trees prove that the grammar is ambiguous.
- Given the fact that “=>” has a lower precedence than “,” resolve the ambiguity of the grammar. That is, write a new grammar that determines the same language but is not ambiguous, such that the Type “Int ⇒ Int, Int” is parsed as function from Int to Int,Int.
The following grammar generates all the regular expressions over .
S -> S S S -> S '+' S S -> S '*' S -> '(' S ')' S -> 'a' | 'b'
- Compute the First and Follow sets for S.
- By giving two different parse trees for an input, show that the grammar is ambiguous.
- Assuming that the '*' has the highest and '+' has the lowest precedence, rewrite the grammar to an unambiguous one.
- Compute the First and Follow sets for the non-terminals of the new grammar.
- Is the new grammar LL(1)?
A grammar is called LL(k) if it needs k tokens of lookahead when parsing a sentence. Does the following grammar belongs to LL(k) for any k?
S -> AB S -> BC A -> a A -> Ba B -> Cc C -> c B -> a