Let $G$ be a context-free grammar. A string of terminals and nonterminals of $G$ is said to be a sentential form of $G$ if you can obtain it by applying productions of $G$ zero or more times to the start symbol of $S$. Let $\operatorname{SF}(G)$ be the set of sentential forms of $G$.

Let $\alpha \in \operatorname{SF}(G)$ and let $\beta$ be a substring of $\alpha$ - we call $\beta$ a fragment of $\operatorname{SF}(G)$. Now let

$\operatorname{Before}(\beta) = \{ \gamma \ |\ \exists \delta . \gamma \beta \delta \in \operatorname{SF}(G) \}$


$\operatorname{After}(\beta) = \{ \delta \ |\ \exists \gamma . \gamma \beta \delta \in \operatorname{SF}(G) \}$.

Are $\operatorname{Before}(\beta)$ and $\operatorname{After}(\beta)$ context-free languages? What if $G$ is unambiguous? If $G$ is unambiguous, are $\operatorname{Before}(\beta)$ and $\operatorname{After}(\beta)$ also describable by an unambiguous context-free language?

This is a followup to my earlier question, after an earlier attempt to make my question easier to answer failed. A negative answer will make the encompassing question I'm working on very hard to answer.


2 Answers 2


Let us get a feeling for $\operatorname{Before}(\beta)$ and $\operatorname{After}(\beta)$ first. Consider a derivation tree which contains $\beta$; "contains" here means that you can cut away subtrees so that $\beta$ is a subword of the tree's front. Then, the before (after) set are all potential fronts of the tree's part left (right) of $\beta$:

tree with before and after sets

So we have to build a grammar for the horizontally lined (vertically lined) part of the tree. That seems easy enough as we already have a grammar for the whole tree; we just have to make sure all sentential forms are words (change the alphabets), filter away those that do not contain $\beta$ (that is a regular property as $\beta$ is fixed) and cut away everything after (before) $\beta$, including $\beta$. This cutting should also be possible.

Now to a formal proof. We will transform the grammar as outlined and use closure properties of $\mathrm{CFL}$ to do the filtering and cutting, i.e. we perform a non-constructive proof.

Let $G = (N, T, \delta, S)$ a context-free grammar. It is easy to see that $\operatorname{SF}(G)$ is context-free; construct $G'=(N',T',\delta',N_S)$ like this:

  • $N' = \{N_A \mid A \in N\}$
  • $T' = N \cup T$
  • $\delta' = \{\alpha(A) \to \alpha(\beta)\mid A\to\beta \in \delta \} \cup \{N_A \to A \mid A\in N\}$

with $\alpha(t)=t$ for all $t \in T$ and $\alpha(A)=N_A$ for all $a\in N$. It is clear that $\mathcal{L}(G')=\operatorname{SF}(G)$; therefore the corresponding prefix closure $\operatorname{Pref}(\operatorname{SF}(G))$ and suffix closure $\operatorname{Suff}(\operatorname{SF}(G))$ are context-free, too¹.

Now, for any $\beta \in (N\cup T)^*$ are $\mathcal{L}(\beta(N\cup T)^*)$ and $\mathcal{L}((N\cup T)^*\beta)$ regular languages. As $\mathrm{CFL}$ is closed under intersection and right/left quotient with regular languages, we get

$\qquad \displaystyle \operatorname{Before}(\beta) = (\operatorname{Pref}(\operatorname{SF}(G))\ \cap\ \mathcal{L}((N\cup T)^*\beta))\,/\,\beta \in \mathrm{CFL}$


$\qquad \displaystyle \operatorname{After}(\beta) = (\operatorname{Suff}(\operatorname{SF}(G))\ \cap\ \mathcal{L}(\beta(N\cup T)^*))\,\backslash\, \beta \in \mathrm{CFL}$.

¹ $\mathrm{CFL}$ is closed under right (and left) quotient; $\operatorname{Pref}(L) = L / \Sigma^*$ and similar for $\operatorname{Suff}$ yield prefix resp. suffix closure.

  • $\begingroup$ I started to write an answer then realized my proof was the same as yours. I'd have put it this way (compressed to fit here): form a grammar $G'$ by adding a new terminal $\hat A$ (a metavariable) for each non-terminal $A$ and a production $A\to\hat A$. Then sentential forms of $G$ are the words recognized by $G$ that consist of metavariables. This is the intersection of a CFG with a regular language and thus is regular. The prefix set of a CFG is a CFG (take a PDA and make every state final). $\mathrm{Before}(\gamma) = \{\gamma \mid \gamma\beta\in L(\mathrm{Prefix}(\hat G))\}$ is again a CFG. $\endgroup$ Commented Mar 26, 2012 at 21:21
  • 1
    $\begingroup$ @Gilles, three comments on that: 1) the sentential forms typically (properly) contain the language. 2) "make every state final" -- that won't work; you'll accept prefixes of non-words, too. 3) The last step of "cutting off" a suffix seems to be tricky to get rigorous. :/ Do you have a rigorous but more compact proof than mine? $\endgroup$
    – Raphael
    Commented Mar 26, 2012 at 21:58
  • $\begingroup$ 1) doesn't matter (change $G$ to have or not have a nonterminal for each terminal). 2) Oops, I cut off too much: make every state that can reach a final state final. 3) Do it one terminal $b$ at a time; in the PDA, mark the states from which one can reach a final state by consuming $b$ as final instead. Yes, it would take more expansion to make this rigorous. $\endgroup$ Commented Mar 26, 2012 at 22:16

Yes, $\mbox{Before}(\beta)$ and $\mbox{After}(\beta)$ are context-free languages. Here's how I would prove it. First, a lemma (which is the crux). If $L$ is CF then:

$\mbox{Before}(L,\beta) = \{ \gamma \ |\ \exists \delta . \gamma \beta \delta \in L \}$


$\mbox{After}(L,\beta) = \{ \gamma \ |\ \exists \delta . \delta \beta \gamma \in L \}$

are CF.

Proof? For $\mbox{Before}(L,\beta)$ construct a non-deterministic finite-state transducer $T_{\beta}$ that scans a string, outputting every input symbol it sees and simultaneously searches non-deterministically for $\beta$. Whenever $T_{\beta}$ sees the first symbol of $\beta$ it forks non-deterministically and ceases outputting symbols until either it finishes seeing $\beta$ or it sees sees a symbol that deviates from $\beta$, stopping in either case. If $T_{\beta}$ sees $\beta$ in full, it accepts upon stopping, which is the only way it accepts. If it sees a deviation from $\beta$, it rejects.

The lemma can be jiggered to handle cases where $\beta$ could overlap with itself (like $abab$ -- keep looking for $\beta$ even while in the midst of scanning for a prior $\beta$) or appears multiple times (actually, the original non-determinisic forking already handles that).

It's fairly clear that $T_\beta(L) = \mbox{Before}(L,\beta)$, and since the CFLs are closed under finite-state transduction, $\mbox{Before}(L,\beta)$ is therefore CF.

A similar argument goes for $\mbox{After}(L,\beta)$, or it could be done with string reversals from $\mbox{Before}(L,\beta)$ , CFLs also being closed under reversal:

$\mbox{After}(L,\beta) = \mbox{rev}(\mbox{Before}(\mbox{rev}(L),\mbox{rev}(\beta)))$

Actually, now that I see the reversal argument, it would be even easier to start with $\mbox{After}(L,\beta)$, since the transducer for that is simpler to describe and verify -- it outputs the empty string while looking for a $\beta$. When it finds $\beta$ it forks non-deterministically, one fork continuing to look for further copies of $\beta$, the other fork copying all subsequent characters verbatim from input to output, accepting all the while.

What remains is to make this work for sentential forms as well as CFLs. But that is pretty straightforward, since the language of sentential forms of a CFG is itself a CFL. You can show that by replacing every non-terminal $X$ throughout $G$ by say $X^\prime$, declaring $X$ to be a terminal, and adding all productions $X^\prime \rightarrow X$ to the grammar.

I'll have to think about your question on unambiguity.


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