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How to prove left-recursive grammar is not in LL(1) using parsing table


How to prove left-recursive grammar is not in LL(1) using parsing table

By : user2952571
Date : November 19 2020, 01:01 AM
should help you out First of all you are finding FIRST and FOLLOW over the grammar in which you have removed left recursion. Therefore surely if you try to create LL(1) parsing table there won't be any 2 entries as left recursion is removed and grammar is unambiguous.
Grammar[ S->SA|A A->a ] is not LL(1) as left recursion exists. To prove it by constructing LL(1) parsing table you need to find FIRST and FOLLOW on this grammar only without modifying it.
code :
      a
------------ 
S   S->SA
    S->A
-------------
A   A->a 


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Left-factoring a non-left-recursive grammar to make it LL(1)

Left-factoring a non-left-recursive grammar to make it LL(1)


By : PatPawlowski
Date : March 29 2020, 07:55 AM
Hope that helps No, this grammar is not LL(1). At least, the last rules group is not left factored, since FIRST( Expr) and FIRST( Expr, Arglist) do interstect.
Is this grammar left recursive?

Is this grammar left recursive?


By : user3369638
Date : March 29 2020, 07:55 AM
this will help I assume you start with E. Both of E’s alternatives start with a T. Both of T’s alternatives start with F. Both of F’s alternatives start with a terminal symbol. Thus, the grammar is not left recursive.
Parsing left recursive grammar

Parsing left recursive grammar


By : punk calvin
Date : March 29 2020, 07:55 AM
around this issue If you take a look at modifierList it is essentially requiring at least one modifier. We don't have to do much to get rid of the left recursion.
code :
modifierList ::= modifier [ modifierList ] 
scopedClauseTail ::= booleanGroup searchClause [ scopedClauseTail ] 
scopedClause ::= searchClause [ scopedClauseTail ] 
Why can't a LL grammar be left-recursive?

Why can't a LL grammar be left-recursive?


By : Adrian
Date : March 29 2020, 07:55 AM
will help you OK, I figure it out, if a grammar contains left-recursive production, like:
code :
S->SA
S->B
Parsing left-recursive grammar in sum-type inifinite recursion

Parsing left-recursive grammar in sum-type inifinite recursion


By : Ezsrac
Date : March 29 2020, 07:55 AM
may help you . Though you cannnot use chainl1 here, You can define chainl1-like combinator like this:
code :
leftRec :: (Stream s m t)
        => ParsecT s u m a -> ParsecT s u m (a -> a) -> ParsecT s u m a
leftRec p op = rest =<< p
  where
    rest x = do f <- op
                rest (f x)
          <|> return x
lvalueParser :: Parser LValue
lvalueParser = leftRec idParser (recordAccessModifier <|> arraySubscriptModifier)
  where
    idParser = Id <$> atomParser
    recordAccessModifier = do
      a <- char '.' *> atomParser
      return (\l -> RecordAccess l a)
    arraySubscriptModifier = do
      e <- between (char '[') (char ']') expParser
      return (\l -> ArraySubscript l e)
main = parseTest lvalueParser "x.y[2].z"
-- => RecordAccess (ArraySubscript (RecordAccess (Id 'x') 'y') (ENat 2)) 'z'
type Atom = Char
atomParser :: Parser Atom
atomParser = letter <?> "atom"

data Expression = ENat Int
  deriving Show
expParser :: Parser Expression
expParser = (ENat . read) <$> many digit
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