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I've read somewhere that a Turing machine cannot compute this and it's therefore undecidable but why? Why is it computationally impossible for a machine to generate the parse tree's and make a decision? Perhaps I'm wrong and it can be done?

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    $\begingroup$ Yes you are correct, a Turing machine cannot decide whether a context-free language is ambiguous or not, and this can be reduced from the post correspondence problem, which is undecidable. Note that a parse tree can be infinitely large, and we cannot decide when we stop the computation. $\endgroup$ Commented Jan 17, 2011 at 1:35
  • $\begingroup$ Hsien-Chih, are you referring to "parse trees" for words not in the language (i.e. unsuccessful parses), or are you trying to say that parse-trees can become arbitrarily large? $\endgroup$
    – Raphael
    Commented Jan 17, 2011 at 8:31

3 Answers 3

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We reduce from Post's Correspondence Problem. Suppose we can, in fact, decide the language $\{\langle G\rangle\mid G\textrm{ is an ambiguous CFG}\}$.

Given $\alpha_1, \ldots, \alpha_m, \beta_1, \ldots, \beta_m$: Construct the following CFG $G = (V,\Sigma,R,S)$: $V = \{S, S_1, S_2\}$, $$\begin{align} R = \{S_{\phantom0}&\rightarrow S_1\mid S_2,\\ S_1&\rightarrow \alpha_1 S_1 \sigma_1 \mid \cdots \mid \alpha_m S_1 \sigma_m \mid \alpha_1 \sigma_1 \mid \cdots \mid \alpha_m \sigma_m,\\ S_2&\rightarrow \beta_1 S_2 \sigma_1\mid \cdots \mid \beta_m S_2 \sigma_m \mid \beta_1 \sigma_1\mid \cdots \mid \beta_m \sigma_m\} \end{align}$$ (where $\sigma_i$ are new characters added to the alphabet, e.g., $\sigma_i = \underline{i}$).

If the grammar is ambiguous, then there is a derivation of some string $w$ in two different ways. Supposing, wlog, that the derivations both start with the rule $S\rightarrow S_1$, reading the new characters backwards until they end makes sure there can only be one derivation, so that's not possible. Hence, we see that the only ambiguity can come from one $S_1$ and one $S_2$ 'start'. But then, taking the substring of $w$ up to the beginning of the new characters, we have a solution to the PCP (since the strings of indices used after those points match).

Similarly, if there is no ambiguity, then the PCP cannot be solved, since a solution would imply an ambiguity that just follows $S\Rightarrow S_1\Rightarrow^* \alpha\tilde{\sigma}$ and $S\Rightarrow S_2\Rightarrow^* \beta\tilde{\sigma}$, where $\alpha = \beta$ are strings of matching $\alpha$'s and $\beta$'s (since the $\tilde{\sigma}$'s match).

Hence, we've reduced from PCP, and since that's undecidable, we're done.

(Let me know if I've done anything boneheaded!)

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  • $\begingroup$ I wonder if there is any another way to prove this without using PCP. Also what is the difference when we talk about an algorithm that tests ambiguity of an arbitrary CFG vs an algorithm that tests ambiguity of a particular CFG. $\endgroup$
    – rsonx
    Commented Aug 19, 2020 at 7:36
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    $\begingroup$ For anyone looking for the original source of this undecidability argument, I believe it is here: www-igm.univ-mlv.fr/~berstel/Mps/Travaux/A/… The reference is apparently: N.ChomskyM.P.Schützenberger, The Algebraic Theory of Context-Free Languages, Studies in Logic and the Foundations of Mathematics, volume 26, pages 118-161, 1959. Look for Ambiguity Theorem 2. $\endgroup$
    – a3nm
    Commented Dec 7, 2021 at 17:16
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The answer by apolge presents the proof that it is undecidable whether an arbitrary context free grammar is ambiguous. The question of whether a context free grammar defines an inherently ambiguous language is a separate one. The undecidability of inherent ambiguity of a CFG was proved by Ginsburg and Ullian (JACM, January 1966). https://dl.acm.org/doi/10.1145/321312.321318

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    $\begingroup$ This does not make much sense to me. A language is an infinite object, hence it cannot form an input to an algorithm in the first place. The question is only meaningful for languages given in some finite presentation. In fact, while not indicated very clearly, the Ginsburg and Ullian paper uses “language” as a shorthand for “context-free language” (this is stated in the Preliminaries), and they assume the languages are presented by grammars. Hence the result they actually prove is that it is undecidable whether a given context-free grammar generates an inherently ambiguous language. $\endgroup$ Commented Apr 24, 2020 at 9:55
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    $\begingroup$ @Emil the input is still a context free grammar but we ask if there exists a context free grammar that is unambiguous and language equivalent to the input grammar. $\endgroup$
    – Faustus
    Commented Apr 24, 2020 at 14:38
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    $\begingroup$ @Faustus That’s exactly what I’m saying: we ask if a given context-free grammar (not language) generates an inherently unambiguous language. $\endgroup$ Commented Apr 24, 2020 at 15:59
  • $\begingroup$ @Emil: since the usual representations of context-free languages can be easily transformed into each other (grammars, PDAs, you name it), it makes perfect sense to say that the question at hand is undecidable for context-free languages. Especially, in an informal description. $\endgroup$
    – Thomas S
    Commented May 16, 2023 at 19:46
  • $\begingroup$ @ThomasS In another context, this could work. But not here, as the problem is explicitly contrasted with alpoge's answer. The answer here falsely and misleadingly suggests, with emphasis, that the two problems take input of a different kind (a CFG in alpoge's answer, and some direct input of a CFL here, whatever it might mean). In fact, the inputs of both problems are identical. $\endgroup$ Commented Dec 11, 2023 at 8:28
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A language is a set of strings drawn from an alphabet. Then, paradoxically, the language per se, should be defined unambiguously by enumerating all strings of it. This is true and can be proved for a finite language. But for infinite languages, it is impossible to give an explicit representation of it. Thus, a "finite" way of describing (defining or representing the language) should be used. English prose (or say, natural language) is amgibuous. Now, in case of a CFL, we use a grammar to define it (or eventually the equivalent, the PDA), and that grammar is context free (CFG).

In my opinion, then, the ambiguitiy is on the grammar, and not on the language. That is why they use the term "inherently" ambiguous; to refer to the CFG but not the language. Indeed the CFG refers to the CFL, whose definition is tied to the CFG.

But in the end, we are proving that the grammar is ambiguous, not the language.

I am wondering if I am missing something here.

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    $\begingroup$ There context-free languages for which no unambiguous context-free grammar exist. Such languages are called inherently ambiguous. According to Wikipedia the union of $\{a^{n}b^{n}c^{m}d^{m}\mid n,m>0\}$ with $\{a^{n}b^{m}c^{m}d^{n}\mid n,m>0\}$ is context-free and inherently ambiguous. $\endgroup$ Commented Dec 10, 2023 at 21:25

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