Re: [stack] Is this language concatenative?

On Tue, 1 May 2007 07:00 pm, William Tanksley, Jr wrote:
> Michael Nedzelsky <[email protected]> wrote:
> > > But a quotation is not a member of T*, is it? Therefore it's not a
> > > "simple program".
> >
> > It is a member if T*!
> > T* means the set of all sequences of terminal symbols, and '[', ']' are
> > terminal symbols.
>
> Oh. Then no, I don't think your definition is adequate; it admits all
> ill-formed strings, such as "]]]]a]]]". Your grammar can't be parsed.
Which grammar??
My definition is about languages, i.e. subsets of T*.

I'll try to explain by example.

Example 1.

Terminal symbols: a, b, [, ], +, (, )
Production rules:
S --> e | a | b | S S | [ S ] | (S + S)

Let l be the language which generated by this grammar.
Then we have "]]]]a]]]" \in T*, but "]]]]a]]]" \notin L.
The set of all simple programs for language L is generated by the following 
grammar
S1 --> a | b| [S] (S+S)
S --> e | a | b | S S | [ S ] | (S + S)
Examples of simple programs: a, b, [], (a+b), [a b],(+), (a+(+)), (+b)
The last three programs looks strange, but they are syntactically valid.
The language L is (syntactically) concatenative by my defintion, since all the 
three requirements are satisfied.

Example 2.

Terminal symbols: a, b, [, ], +
Production rules:
S --> e | ab | ba | a |  S S | [ S ]

Let l be the language which generated by this grammar.
The set of all simple programs for language L is generated by the following 
grammar
S1 --> a | ab| ba | [S]
S --> e | ab | ba | a |  S S | [ S ]
Examples of simple programs: a,ab, ba, [], [ab], [a]
The language L is not (syntactically) concatenative by my defintion, since the 
third requirement is not satisifed: the program
aba \in L can be factored in simple two different ways: "a" "ab" and  "a" 
"ba".

 Michael Nedzelsky



 
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