问题
I am dealing with the following grammar:
G = ( {S, A}, {a, b}, P, S )
P = { S -> aAb, S -> bAa,
A -> aSa,
A -> S,
A -> epsilon}
I need to find out L(G). The thing is, I figured out that the words in the grammar are of the form: starts with a and ends with b, or starts with b and ends with a, and between these letters one of the combinations : ab, ba, aaba, abaa; then the next word is formed by inserting one of these 4 combinations between the a and b in the middle..but how can I express this formally? I mean, as far as I could tell, L(A) = a^n S a^n and if w belongs to L(G), then w reversed also belongs to L(G). I tried to express it as a regular expression but failed...could anyone please help?
Thank you.
回答1:
You see that L is not regular, to prove you can use Pumping lemma or Myhill–Nerode theorem, so regular expression can't be discussed
You can notice, that since L consists just of {a,b} you can use it's power We see that language is in the form of aAb or bAa or aAa with exception that aAa can't be located at begin-end of the word
So let's use this, the only thing we miss is the combination of bAb A can generate almost everything (words |w| = 2k, and |w|>=2) but words where position of b matches position of b from the reverse
Formally
Sorry for my tex skills and my formal expression
there must be some error, because I didn't have so much time to think about this, but it can be some way how to continue, it's homework so it's fine, think about it! :)
回答2:
I tried to express it as a regular expression but failed
This language is 'probably' not regular. Context-free languages are more complex/powerful than regular expressions.
Try generating a few words and see if you can come up with a name, properties for the language. Or try to find words that are not in the language.
Some hints, a few properties you can easily see:
The smallest string is at least size 2.
The size of the string is even.
The count of b's is <= than a's.
If you take out A -> aSa, and compare a word with its reversed version, a pattern should be visible. If you include the left out rule, the pattern changes slightly...
回答3:
You are asked to find the language generated by the grammar G with productions
S → aAb | bAa
A → aSa | S | λ
First, consider small derivations beginning from the start symbol S
S ⇒1 aAb ⇒1 aaSab | aSb | ab
S ⇒1 bAa ⇒1 baSaa | bSa | ba
The difficult step is dealing with the recursion generated by the rules A → aSa, S → aAb and S → bAa. The clue to dealing with this difficulty is revealed by considering an inductive definition of the language generated by G:
1. ab ∈ L4
2. ba ∈ L4
3. w ∈ L4 → awb ∈ L4
4. w ∈ L4 → bwa ∈ L4
5. w ∈ L4 → awa ∈ L4
The rules (3)-(5) correspond to the rules A → aAa, S → aAb and S → bAa in G. It is easy to see that the inductive definition and the rules of G define the same language. The inductive definition shows that the language of G can be build incrementally in steps. Starting with the smallest strings generable in G, we build larger and larger sets of strings corresponding to the problematic rules:
L(1) = {ab, ba}
L(n + 1) = {awb, bwa, awa : w ∈ L(n)}
The set L(1) contains the smallest strings generable in G. The set L(n + 1) contains strings awb, bwa and awa for each string w ∈ L(n). That is, the strings in L(n + 1) correspond to the strings obtained by applying the rules S → aAb, S → bAa and A → aAa once to strings in L(n). All that remains is to construct the union of the L(n), which is a set:
L = ⋃ {L(n) : n ∈ ℕ}
To see that L is equivalent to the language generated by the grammar G, you can argue by induction on the length of derivations in G. Starting with the smallest strings genreable in G (i.e, ab and ba), working backwards using appropriate induction hypotheses.
回答4:
the language generated is : (a)n(b)m Were n>=m
来源：https://stackoverflow.com/questions/8123485/how-do-i-figure-out-the-language-generated-by-this-context-free-grammar