From: cdodd@acm.org   
      
   George Neuner  wrote   
   > For every LL(k) grammar where both k and the set T of tokens is fixed,   
   > there is an equivalent LL(1) grammar.  In the worst case, the   
   > equivalent LL(1) grammar will have k**T rules.   
      
   This is simply not true.  A counter-example from Kurki-Suonio   
      
      
   is:   
      
       S ::= aSA | \epsilon   
       A ::= aabS | c   
      
   This grammar (and the language) is LL(2), but there does not exist an   
   equivalent LL(1) grammar.  Here k=2 (fixed) and T = {a, b, c} (also fixed)   
      
   Chris Dodd   
   cdodd@acm.org   
      
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