Context-free Languages closed under Reversal

Question

In class this week we've been learning about the CFLs and their closure properties. I've seen proofs for union, intersection and compliment but for reversal my lecturer just said its closed. I wanted to see the proof so I've been searching for the past few days but all I've found is most people just say that to reverse the productions is enough to prove it. Those that do go a little more formal just state there is an easy inductive proof you can give. Can anyone provide me with some more information/hints about the inductive proof? Try as I might I can't come up with it.

Answer 1

There is another way to look at this problem.

Consider that the Language $L$ is a CFL. This means that there is a grammar $G=\{N,\sum,P,S\}$ that satisfies the CFL. We can assume that this is in Chomsky Normal Form.

If $\epsilon$ is part of the language, trivially $\epsilon^R$ is also part of the language. Now for every production of the form $P_1 \longrightarrow AB$, replace it with, $P_1 \longrightarrow BA$ and for the productions of the form $P_1 \longrightarrow a$, where $a \in \sum$, leave it the same.

From the parse tree of the derived string, it is easy to see that the language derived will be exactly the reverse of the initial language as the construction mirrors the original parse tree.

Answer 2

Your sources are right, and I am afraid there is only little to add, except formalism. Indeed it is enough to reverse all productions to reverse the language. I denote the reverse (mirror) of string $w$ by $w^R$.

If $G$ is a grammar, let $H$ be its reverse, so for production $A\to w$ in $G$ we have $A\to w^R$ in $H$.

Then by induction we show that original grammar $G$ generates a string iff the reverse grammar $H$ generates the reverse of the string. Formally $A\Rightarrow_G^*w$ iff $A\Rightarrow_H^*w^R$.

• (basis) In zero steps we have $A\Rightarrow_G^0 A$ iff $A\Rightarrow_H^0 A$.
• (induction) Assuming $A\Rightarrow_G^*w_1Bw_2$ iff $A\Rightarrow_H^*w_2^RBw_1^R$ we can apply any production $B\to u$ in $G$ (and in $H$ in reverse) and obtain $A\Rightarrow_G^*w_1uw_2$ and $A\Rightarrow_H^*w_2^Ru^Rw_1^R$ respectively, where indeed $w_2^Ru^Rw_1^R$ is the reverse of $w_1uw_2$.

This is a very condensed proof, but contains all necessary ingredients. Again, a derivation of the reverse grammar is the reverse of the original one. This is especially clear when looking at the two derivation trees: they are just reversed (or mirrored).

Answer 3

First off. CFL's are not closed under intersection or complement (or difference for that matter). They are closed under Union, Concatenation, Kleene star closure, substitution, homomorphism, inverse homomorphism, and reversal. NOTE: The two homomorphism's are usually not covered in an intro Computer Theory course.

To prove reversal, Let L be a CFL, with grammar G=(V,T,P,S). Let L^R be the reverse of L, such that the Grammar is G^R = (V,T,P^R,S). That is, reverse every production.

Ex. P -> AB would become P -> BA

Since G^R is a CFG, therefore L(G^R) is a CFL.