Let $L$ be a language over an alphabet $\Sigma$ and let

$$ \text{MIN}(L) = \{ w \in L \mid \forall x,y \in \Sigma^* : (w = xy \land x \in L) \implies y = \varepsilon \} $$ $$ \text{MAX}(L) = \{ w \in L \mid \forall x \in \Sigma^+ : wx \not\in L \} $$

Simply put, $\text{MIN}(L)$ consists of words from $L$ such that none of their proper prefixes is in $L$, and $\text{MAX}(L)$ consists of words from $L$ which are not a proper prefix of any word in $L$.

Now the question is: For given context-free grammar $G$, is it decidable whether language $\text{MIN}(L(G))$ is context-free or not, similarly with language $\text{MAX}(L(G))$ (of course we could define the problem with pushdown automaton instead of context-free grammar, or with any other finite description of context-free language). The problem is not trivial since the class of context-free languages is not closed under these operations:

$$\text{MIN}(\{ a^i b^j c^k \mid i, j, k > 0 \land (k \geq i \lor k \geq j) \}) = \{ a^i b^j c^k \mid i, j, k > 0 \land k = \min(i,j) \}$$

is not CFL, similarly

$$\text{MAX}(\{ a^i b^j c^k \mid i, j, k > 0 \land (k \leq i \lor k \leq j) \}) = \{ a^i b^j c^k \mid i, j, k > 0 \land k = \max(i,j) \}$$

is also not CFL (both can be shown using Ogden's lemma).

I would guess that the problem is undecidable, and I tried (somehow) proving it by contradiction with reduction of the undecidable Post correspondence problem, but got nowhere. Any ideas proving undecidability or creating an algorithm?

  • $\begingroup$ I believe if $L(G)$ is a DCFL, then so is $\text{MIN}(L(G))$ (build a DPDA, and keep track of whether any prefix would have been accepted). This doesn't answer your question, of course. $\endgroup$
    – D.W.
    Commented Apr 23 at 21:18
  • $\begingroup$ Thank you for the advice, I'm relatively new here. Is it okay now? $\endgroup$
    – JimmyB
    Commented Apr 24 at 12:06
  • $\begingroup$ Looks great! Thank you! I appreciate your contribution, I hope to see you back on the site. $\endgroup$
    – D.W.
    Commented Apr 24 at 16:16

1 Answer 1


Ok so I think I figured it myself at last.

In case of $\text{MIN}$ operation, for the sake of contradiction, let's assume it is decidable. Now let $(X = (x_1,...,x_n), Y = (y_1,...,y_n))$ be an instance of PKP over an alphabet $\Sigma$ such that $\# \not\in \Sigma$. We could construct CFG for language

$$ L_{XY}^C = \{ x_{i_1} ... x_{i_k} \# i_k...i_1 \# j_1 ... j_l \# y_{j_l} ... y_{j_1} \mid i_1,...,i_k,j_1,...,j_l \in \{ 1,...,n \} \}^C $$

(I leave it for the reader to realize why this language is context-free) as well as CFG for language

$$ L_{pal} = \{ w \in \Gamma^* \mid w = w^R \} $$

where $\Gamma = \Sigma \cup \{ 1,...,n, \# \}$. Now it should not be hard to see, that $L_{XY} \cap L_{pal}$ is empty language (and therefore CFL) iff the instance of PKP has no solution, and is not CFL iff the instance of PKP has a solution (because then it has infinitely many solutions and it can be simply shown using pumping lemma for context-free languages that it is not CFL).

Now WLOG assume that $ {\\\$} \not\in \Gamma $. We will construct language

$$ L = L_{pal} {\\\$} L_{XY}^C {\\\$} \cup L_{XY}^C {\\\$} $$

Language $L$ is obviously also context-free, because CFLs are closed under concatenation and union. It's easy to see that the only words from $L$ that have also some proper prefix from $L$ are precisely words from $L_{XY}^C {\\\$} L_{XY}^C {\\\$} \cap L$. Therefore

$$ \text{MIN}(L) = (L_{pal} {\\\$} L_{XY}^C {\\\$} \cup L_{XY}^C {\\\$}) - L_{XY}^C {\\\$} L_{XY}^C {\\\$} = (L_{pal} - L_{XY}^C) {\\\$} L_{XY}^C {\\\$} \cup L_{XY}^C {\\\$} = (L_{pal} \cap L_{XY}) {\\\$} L_{XY}^C {\\\$} \cup L_{XY}^C {\\\$} $$

Finally, language $\text{MIN}(L)$ is CFL iff language $L_{pal} \cap L_{XY}$ is CFL, because of closure of CFLs under right quotient with regular languages (language $ {\\\$} \Gamma^* {\\\$} $), and also because of closure under concatenation and union. So if we could decide whether $\text{MIN}(L)$ is CFL or not, we could also decide if an instance of PKP has solution, which derives a contradiction.

In case of $\text{MAX}$, it can be also proved very similarly that the problem is undecidable - realize what is the result of $ \text{MAX} (L_{XY}^C {\\\$} L_{pal} {\\\$} \cup L_{pal} {\\\$}) $ and utilize that CFLs are closed under intersection with regular languages.


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