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Let $A/B$ = $\{ w \mid wx \in A$ for some $x \in B \}$. Show that if A is context free and B is regular, then $A/B$ is context free.

My interpretation of this is is that we need to show that if a string $wx$ is accepted by a CFG, and we know that $x$ is accepted by a regular language (and therefore is also accepted by a context-free language), then $w$ must also be accepted by a CFG.

My initial thought on how to solve this would be a proof by contradiction in which we assume that $A$ is context free, $B$ is regular, and then assume that $A/B$ is not context-free. Since $A$ is context free, we can construct an equivalent PDA that accepts $A$.

From here, my thought was to take an arbitrary $wx$ that is accepted by $A$, such that $x \in B$. We can then construct another PDA based on the first that only accepts $wx$. We could then break the PDA into two pieces: one that accepts $w$ and one that accepts $x$ (with the two pieces concatenated together). Since there then would exist a PDA that accepts just $w$, and $w$ is arbitrary insofar as $wx$ was arbitrary, $A/B$ must therefore be context-free after all (contradiction).

Will this approach work? (Is this a good general approach?) If so, how would I go about breaking the PDA that accepts $wx$ into chunks formally?

Raphael
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therealrootuser
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3 Answers3

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Imagine you have a pushdown automaton (PDA) $X$ for $A$ and a DFA $Y$ for $B$.

You want to build a PDA $Z$ for $A/B$. You can do as follows: the states $Q_Z$ of $Z$ are $Q_X\times Q_Y$. There are two phases:

Phase 1: You just read the input word and advance in $X$, states of $Y$ are ignored. This corresponds to the word $w$ of your language.

Phase 2: When you reach the end of $w$, you are allowed to perform $\epsilon$-transitions in order to reach an accepting state of $X\times Y$. So you guess a word $x$ that will continue to advance in $X$ and start to advance in $Y$, and that has to reach acceptance in both automata simultaneously. If there is an accepting run of this automaton, then the input word $w$ is in $A/B$ since you guessed the witness $x$ (and vice-versa, if $x$ exists, then it can be guessed by your automaton $Z$).

The reason why $B$ has to be regular instead of CFL is that you cannot manage two stacks at the same time, if you want to stay CFL.

Denis
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  • I followed Phase 1, and I think I am beginning to understand Phase 2. The part that I'm not sure I understand is 1) How guessing the word $x$ would work (an $\epsilon$ transition?), and 2) how the PDA ($Z$) could know that it has reached an accept state in both automata simultaneously. – therealrootuser Jan 30 '14 at 03:56
  • for guessing the word $x$, you just guess letter by letter, and advance simultenously in both automata $X$ and $Y$ according to the letter you guessed. For instance if you are in state $(p,q)$, you guess letter $a$, you go to state $(p',q')$ (and update the stack of $X$ accordingly) such that transitions in both automata are valid. A state $(p,q)$ is accepting if $p$ is accepting in $X$ and $q$ is accepting in $Y$. – Denis Jan 30 '14 at 13:38
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Here is a very short proof relying on rational transductions. Let $X$ be the alphabet. First the relation $\tau$ defined by $\tau(u) = uB$ is a rational transduction. Indeed its graph $(\Sigma_{x \in X} (x,x))^*(\{1\} \times B)$ is a rational subset of $X^* \times X^*$. This is the place where the hypothesis that $B$ is regular is mandatory. Now $$ A/B = \{u \mid \tau(u) \cap A \not= \emptyset \} = \tau^{-1}(A) $$ where $\tau^{-1}$ denotes the inverse of the relation $\tau$, which is also a rational transduction. Finally, it suffices to use the fact that if $L$ is context-free and $\sigma$ is a rational transduction, then $\sigma(L)$ is context-free.

J.-E. Pin
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Above answers are excellently written. But one other approach would be helpful I think.

We want to prove that if $L(A)$ is a $CFL$ and $L(B)$ is a regular then $L(A/B) = \{w\space|\space wx \in A,\space x\in B,\space w\in \Sigma^*\, ,\space x\in \Sigma^*\}$ is also a $CFL$.

Let's use fact that set of regular languages and set of context free languages are closed under homomorphism and inverse homomorphism.

Let's define $\overline\Sigma^* = \{\overline a\space|\space a\in\Sigma\}$

Now define homomorphism

$unbar:\space (\Sigma\space \cup \overline\Sigma)^* \rightarrow \Sigma^*$

$unbar(a) = a, \space \space if\space a\in \Sigma$

$unbar(\overline a) = a, \space \space if\space a\in \overline\Sigma$

and

$rembar:\space (\Sigma\space \cup \overline\Sigma)^* \rightarrow \Sigma^*$

$rembar(a) = a, \space \space if\space a\in \Sigma$

$rembar(\overline a) = \epsilon, \space \space if\space a\in \overline\Sigma$

Now, proof

$L_1 = unbar^{-1}(A)$ 

L_1 is CFL because A is CFL and CFL is closed under inverse homomorphism.
L_1 is set of strings in A with arbitrary letter annotated with bars.

$L_2 = unbar^{-1}(B)$ 

L_2 is a regular because regular set is also closed under inverse homomorphism.
Meaning of L_2 is similar to above.

$L_3 = L_2\space\cap\space\overline\Sigma^* $ 

Now L_3 is also regular because L_2 is regular and regular set is closed under intersection. 
L_3 is set of strings in B with every letter annotated with bar.

$L_4 = \Sigma^*L_3$

This too is regular.

$L_4$ is set of strings of form $xy, x\in \Sigma^*$ and $y$ is string in B with every letter annotated by bar.

$L(A/B) = rembar(L_1 \space\cap\space L_4)$ 

Now this is CFL because L_1 is CFL and L_4 is regular and CFL is closed under regular 
intersections.

Now this is little hard to explain but you may infer the meaning.

Inshort, $L(A/B) = rembar(unbar^{-1}(A) \space\cap\space (\Sigma^*(unbar^{-1}(B)\space\cap\overline\Sigma^*)))$

All operations used preserve Context-free-ness so L(A/B) is CFL.
Vimal Patel
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