Constructing a linear grammar for the language

Question

I find difficulties in constructing a Grammar for the language especially with linear grammar.

Can anyone please give me some basic tips/methodology where i can construct the grammar for any language ? thanks in advance

I have a doubt whether the answer for this question "Construct a linear grammar for the language: is right

L ={a^n b c^n | n belongs to Natural numbers}

Solution:

Right-Linear Grammar :

S--> aS | bA

A--> cA | ^

Left-Linear Grammar:

S--> Sc | Ab

A--> Aa | ^

Answer 1

As pointed out in the comments, these grammars are wrong since they generate strings not in the language. Here's a derivation of abcc in both grammars:

S -> aS -> abA -> abcA -> abccA -> abcc
S -> Sc -> Scc -> Abcc -> Aabcc -> abcc

Also as pointed out in the comments, there is a simple linear grammar for this language, where a linear grammar is defined as having at most one nonterminal symbol in the RHS of any production:

S -> aSc | b

There are some general rules for constructing grammars for languages. These are either obvious simple rules or rules derived from closure properties and the way grammars work. For instance:

1. if L = {a} for an alphabet symbol a, then S -> a is a gammar for L.
2. if L = {e} for the empty string e, then S -> e is a grammar for L.
3. if L = R U T for languages R and T, then S -> S' | S'' along with the grammars for R and T are a grammar for L if S' is the start symbol of the grammar for R and S'' is the start symbol of the grammar for T.
4. if L = RT for languages R and T, then S = S'S'' is a grammar for L if S' is the start symbol of the grammar for R and S'' is the start symbol of the grammar for T.
5. if L = R* for language R, then S = S'S | e is a grammar for L if S' is the start symbol of the grammar for R.

Rules 4 and 5, as written, do not preserve linearity. Linearity can be preserved for left-linear and right-linear grammars (since those grammars describe regular languages, and regular languages are closed under these kinds of operations); but linearity cannot be preserved in general. To prove this, an example suffices:

R -> aRb | ab
T -> cTd | cd

L  = RT = a^n b^n c^m d^m, 0 < a,b,c,d
L' = R* = (a^n b^n)*, 0 < a,b

Suppose there were a linear grammar for L. We must have a production for the start symbol S that produces something. To produce something, we require a string of terminal and nonterminal symbols. To be linear, we must have at most one nonterminal symbol. That is, our production must be of the form

S := xYz

where x is a string of terminals, Y is a single nonterminal, and z is a string of terminals. If x is non-empty, reflection shows the only useful choice is a; anything else fails to derive known strings in the language. Similarly, if z is non-empty, the only useful choice is d. This gives four cases:

1. x empty, z empty. This is useless, since we now have the same problem to solve for nonterminal Y as we had for S.
2. x = a, z empty. Y must now generate exactly a^n' b^n' b c^m d^m where n' = n - 1. But then the exact same argument applies to the grammar whose start symbol is Y.
3. x empty, z = d. Y must now generate exactly a^n b^n c c^m' d^m' where m' = m - 1. But then the exact same argument applies to the grammar whose start symbol is Y.
4. x = a, z = d. Y must now generate exactly a^n' b^n' bc c^m' d^m' where n' and m' are as in 2 and 3. But then the exact same argument applies to the grammar whose start symbol is Y.

None of the possible choices for a useful production for S is actually useful in getting us closer to a string in the language. Therefore, no strings are derived, a contradiction, meaning that the grammar for L cannot be linear.

Suppose there were a grammar for L'. Then that grammar has to generate all the strings in (a^n b^n)R(a^m b^m), plus those in e + R. But it can't generate the ones in the former by the argument used above: any production useful for that purpose would get us no closer to a string in the language.