My problem is how can I prove that a grammar is unambiguous? I have the following grammar: $$S → statement ∣ \mbox{if } expression \mbox{ then } S ∣ \mbox{if } expression \mbox{ then } S \mbox{ else } S$$

and make this to an unambiguous grammar, I think its correct:

  • $ S → S_1 ∣ S_2 $

  • $S_1 → \mbox{if } expression \mbox{ then } S ∣ \mbox{if } expression \mbox{ then } S_2 \mbox{ else } S_1$

  • $S_2 → \mbox{if } expression \mbox{ then } S_2 \mbox{ else } S_2 ∣ statement$

I know that a unambiguous grammar has one parse tree for every term.


4 Answers 4


There is (at least) one way to prove unambiguity of a grammar $G = (N,T,\delta,S)$ for language $L$. It consists of two steps:

  1. Prove $L \subseteq \mathcal{L}(G)$.
  2. Prove $[z^n]S_G(z) = |L_n|$.

The first step is pretty clear: show that the grammar generates (at least) the words you want, that is correctness.

The second step shows that $G$ has as many syntax trees for words of length $n$ as $L$ has words of length $n$ -- with 1. this implies unambiguity. It uses the structure function of $G$ which goes back to Chomsky and Schützenberger [1], namely

$\qquad \displaystyle S_G(z) = \sum_{n=0}^\infty t_nz^n$

with $t_n = [z^n]S_G(z)$ the number of syntax trees $G$ has for words of length $n$. Of course you need to have $|L_n|$ for this to work.

The nice thing is that $S_G$ is (usually) easy to obtain for context-free languages, although finding a closed form for $t_n$ can be difficult. Transform $G$ into an equation system of functions with one variable per nonterminal:

$\qquad \displaystyle \left[ A(z) = \sum\limits_{(A, a_0 \dots a_k) \in \delta} \ \prod\limits_{i=0}^{k} \ \tau(a_i)\ : A \in N \right] \text{ with } \tau(a) = \begin{cases} a(z) &, a \in N \\ z &, a \in T \\ \end{cases}.$

This may look daunting but is really only a syntactical transformation as will become clear in the example. The idea is that generated terminal symbols are counted in the exponent of $z$ and because the system has the same form as $G$, $z^n$ occurs as often in the sum as $n$ terminals can be generated by $G$. Check Kuich [2] for details.

Solving this equation system (computer algebra!) yields $S(z) = S_G(z)$; now you "only" have to pull the coefficient (in closed, general form). The TCS Cheat Sheet and computer algebra can often do so.


Consider the simple grammar $G$ with rules

$\qquad \displaystyle S \to aSa \mid bSb \mid \varepsilon$.

It is clear that $\mathcal{L}(G) = \{ww^R \mid w \in \{a,b\}^*\}$ (step 1, proof by induction). There are $2^{\frac{n}{2}}$ palindromes of length $n$ if $n$ is even, $0$ otherwise.

Setting up the equation system yields

$\qquad \displaystyle S(z) = 2z^2S(z) + 1$

whose solution is

$\qquad \displaystyle S_G(z) = \frac{1}{1-2z^2}$.

The coefficients of $S_G$ coincide with the numbers of palindromes, so $G$ is unambiguous.

  1. The Algebraic Theory of Context-Free Languages by Chomsky, Schützenberger (1963)
  2. On the entropy of context-free languages by Kuich (1970)
  • 3
    $\begingroup$ As you know @Raphael, ambiguity is not decidable, so at least one of your steps cannot be mechanised. Any idea which ones? Getting a closed form for $t_n$? $\endgroup$ Feb 10, 2013 at 7:32
  • 2
    $\begingroup$ The equation system may not be solvable algorithmically if the degree is too high, and pulling the exact coefficients out of the generating functions can be (too) hard. In "practice", though, one often deals with grammars of small "degree" -- note that, say, Chomsky normal form leads to equation systems of small degree -- and there are methods to get at least $\sim$-asymptotics for the coefficients; this may be sufficient to establish ambiguity. Note that in order to prove unambiguity, showing $S_L(z) = S_G(z)$ without pulling coefficients is enough; proving this identity may be hard, though. $\endgroup$
    – Raphael
    Feb 10, 2013 at 13:59
  • $\begingroup$ Thank you @Raphael. Do you know of any texts that develop in detail how undecidability comes into play even if one uses e.g. Chomsky normal form? (I can't get hold of Kuich.) $\endgroup$ Feb 11, 2013 at 10:32
  • $\begingroup$ @MartinBerger I just rediscovered your comment in my todo list; sorry for the long silence. There are three steps which (I think) are not computable in general: 1) Determine $S_G$. 2) Compute $|L_n|$. 3) Determine $[z^n]S_g(z)$. In particular, what representation of $L$ to use for 2)? $\endgroup$
    – Raphael
    Jan 30, 2014 at 11:24
  • $\begingroup$ Why is representation of $L$ a problem? We can use any of the multiple ways of representing CFGs for compilers for example. Maybe you mean how to represent $L_n$? $\endgroup$ Jan 30, 2014 at 16:00

For some grammars, a proof by induction (over word length) is possible.

Consider for example a grammar $G$ over $\Sigma = \{a,b\}$ given by the following rules:

$\qquad \displaystyle S \to aSa \mid bSb \mid \varepsilon$

All words of length $\leq 1$ in $L(G)$ -- there's only $\varepsilon$ -- have only one left-derivation.

Assume that all words of length $\leq n$ for some $n \in \mathbb{N}$ have only one left-derivation.

Now consider arbitrary $w = w_1 w' w_n \in L(G) \cap \Sigma^n$ for some $n > 0$. Clearly, $w_1 \in \Sigma$. If $w_1 = a$, we know that the first rule in every left-derivation has to be $S \to aSa$; if $w_1 = b$, it has to be $S \to bSb$. This covers all cases. By induction hypothesis, we know that there is exactly one left-derivation for $w'$. In combination, we conclude that there is exactly one left-derivation for $w$ as well.

This becomes harder if

  • there are multiple non-terminals,
  • the grammar is not linear, and/or
  • the grammar is left-recursive.

It may help to strengthen the claim to all sentential forms (if the grammar has no unproductive non-terminals) and "root" non-terminals.

I think the conversion to Greibach normal form maintains (un)ambiguity, to applying this step first may take care of left-recursion nicely.

The key is to identify one feature of every word that fixes (at least) one derivation step. The rest follows inductively.


This is a good question, but some Googling would have told you that there is no general method for deciding ambiguity, so you need to make your question more specific.

  • 4
    $\begingroup$ The OP asks for proof techniques, not algorithms. $\endgroup$
    – Raphael
    Jun 11, 2012 at 10:46
  • $\begingroup$ I think so, too; it might be mentioned in the question. $\endgroup$ Jun 11, 2012 at 15:37
  • 1
    $\begingroup$ Google is not an oracle of truth, because knowlede is not democratic, and Google results are. I wouldn't count on Google in this case, because people often copy-cat one from another without checking the correctness of what they copy. Without showing a proof, they might be wrong. $\endgroup$
    – SasQ
    Aug 6, 2012 at 15:18
  • 8
    $\begingroup$ @SasQ: You read my words too literally. What Google gives me is the URLs to aticles that explain things. $\endgroup$ Oct 5, 2012 at 11:35
  • $\begingroup$ The result is for general case, as much as $a^nb^n, n\in \mathbb {Z^+}$ cannot be generated by a regular grammar, for all (infinite) values of $n$. But, for finite values of $n$, it can be generated by a regular grammar. Similarly, is there a condition on the size of grammar, or what parameters decide the complexity of proving the grammar is unambiguous? $\endgroup$
    – jiten
    Aug 7, 2023 at 6:24

Basically, it's a child generation problem. Start with the first expression, and generate it's children .... Keep doing it recursively (DFS), and after quite a few iterations, see if you can generate the same expanded expression from two different children. If you are able to do that, it's ambiguous. There is no way to determine the running time of this algorithm though. Assume it's safe, after maybe generating 30 levels of children :) (Of course it could bomb on the 31st)

  • 3
    $\begingroup$ The OP asks for proof techniques, not algorithms. $\endgroup$
    – Raphael
    Jun 11, 2012 at 10:46
  • 2
    $\begingroup$ that can't possibly be a way to prove if a grammar is ambiguous or not. As a matter of fact when that bombing happens is undecidable. $\endgroup$ Apr 6, 2015 at 12:35

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