# How to give a context free grammar for any given language ie: a^n (ba)^m a^n

i am trying to understand Context free grammar and generate a CFG for any given language.

when you're given a language , what is the best way to generate a CFG from it? are there any steps to follow to help you create a CFG for any language? is there ways of breaking down the language to make it simpler so that it helps you generate the CFG.

for example if i was given L={a^n (ba)^m a^n | n,m >=0 }

does that mean i need to have equal number of a's on both side, and have zero or more ba in between the a's ?

S---> aSa |ε

X---> baX |ε

any helps is appreciated.

• Possible duplicate of How to prove that a language is context-free? – Hendrik Jan Dec 17 '18 at 23:44
• my question is not about whether the language is context free or not. i am just finding it difficult give a CFG for any given language. there must be a way of breaking down languages into smaller parts to make it easy when generating a CFG for it. that is what i am asking here. and i want to understand it by going through some examples with people who know it well. – chelseablue Dec 17 '18 at 23:51
• I think my cfg was not working . and i come up with a better answer , which i think it is working : S ----> aSa | X X ----> baX | ε – chelseablue Dec 17 '18 at 23:52
• well, in an answer to the "How to prove ..." reference question it is explained that the grammar can indeed be made from standard components once the "nesting structure" of the strings is found. In your case the nesting is from the outside, so I would start with $S\to a S a$. – Hendrik Jan Dec 17 '18 at 23:58
• @Hendrik Jan thanks for your hints and answer, "Nesting" , that is the sort of answer i was looking for . can you please elaborate on that more in relation to the language or any other language. how does that work? and yes as for my previous answer . i came up with a different solution which is : S ----> aSa | X X ----> baX | ε – chelseablue Dec 18 '18 at 0:07

You ask about the difference between language like $$L_1 = \{ab^na^nc \mid \dots \}$$ and $$L_2 = \{a^n (ab)^m a^n \mid \dots \}$$. In both cases the parts that are iterated the same $$n$$ number of times have to be generated in parallel. Productions like $$Y\to b Y a$$ are good for that.
In case of $$L_1$$ the $$a\dots c$$ pair is around the repeat $$b^nc^n$$ and can be generated before $$S\to a X c$$, but also independently as $$S\to AXC, A\to a, C\to c$$ . In case of $$L_2$$ the $$(ab)^m$$ is inside the repeat $$a^n\dots a^n$$ and has to be generated after, like your own solution $$S\to aSa, S\to X, X\to abX\mid \varepsilon$$.
• @chelseablue Not quite. Your grammar would generate $aabbaacc$, which isn't in $L_1$. Try $S\rightarrow aTc, T\rightarrow bTa\mid\epsilon$. – Rick Decker Dec 19 '18 at 16:13