Re: Question re the (non-)equivalence of Z -> z and Z -> z e (e the empty string)

On 08/10/2004 04:35 PM, Kenn Heinrich wrote:
[snip]
> I'd argue with your use of e. In automata theory, e is not usually
> interpreted as a symbol in the alphabet Vt. Rather, it's the identity
> element you add to Vt+ (the free semigroup over Vt) to get Vt* (free
> monoid). So by definition, (x e) = (x). e may be a string in a


Kenn,


Is there something like an identity element for the alternative
operator, |. I'm thinking that in (x e) the space is the implicit
multiplication operator with identity e, but I was wondering
if there were something similar for the addition (i.e. | )
operator. In addition, could this be what parsers are "in theory"
using when the lexer's return # (as in yacc) to signal the end of input?


-Larry

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