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The most obvious one that I found was, $$S \rightarrow SSS | A | B | C$$ $$A \rightarrow Aa | \epsilon$$ $$B \rightarrow Bb | \epsilon$$ $$C \rightarrow Cc | \epsilon$$

However, I realize this CFG is ambious because we can generate different parse trees for the same string due to the rule $S \rightarrow SSS$.
So I revised my language to avoid ambiguity as follows:

$$S \rightarrow SABC | ASBC | ABSC | ABCS $$ $$A \rightarrow Aa | \epsilon$$ $$B \rightarrow Bb | \epsilon$$ $$C \rightarrow Cc | \epsilon$$

where $S$ is the starting variable in both grammars.

I wonder if this CFG is correct? What I wasn't sure is the first rule $S$, it looks a bit redundant but I don't know how to fix it so that it can generate all strings over the alphabet $\Sigma = \{a, b, c\}$. Any idea?

Update
The question is motivated from this problem:

Show that the complement of the language $L = \{a^nb^mc^k, k = n + m\}$ is context-free.

Since I already found the CFG for the language $L_1 = \{a^nb^mc^k, k \neq n + m\}$, what I'm trying to do is finding the CFG for its complement which includes two parts:

  1. $L_1$
  2. $L_2$ = Any string over alphabet $\Sigma = \{a, b, c\}$ that is not in the form $a^mb^nc^k$.

So if I can find the CFG for 2, the union of these two languages is also CFG by adding an extra rule: $$S \rightarrow S_1 | S_2$$ where $S_1$ is starting variable of $L_1$, and $S_2$ is starting variable for $L_2$.

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3 Answers

up vote 6 down vote accepted

Why don't you use the following easy grammar, where S is the starting variable?

S $\rightarrow$ BS | $\varepsilon$

B $\rightarrow$ a | b | c

Edit for your update:

You want to prove that $L_2$ is context free.
However, as Yuval said $L_3 = \{a^mb^nc^k \mid m,n,k \in \mathbb{N}\}$ is regular.
Now, hence $\Sigma^*$ is regular and $\Sigma^*$\ $L_3$ is your wanted language, we know that $L_2$ is regular since regular languages are closed under difference, which implies that $L_2$ context free.

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Thanks a lot. I must be blind ^_^! In fact, I was obsessed by the problem statement because what I was trying to achieve is a CFG generates all strings over alphabet $\Sigma = \{a, b, c\}$ excluding the string of the form $a^mb^nc^k$. –  Chan Aug 21 '11 at 21:41
    
Oh! Well, you won't succeed by trying to find a difference of two context free languages, as they aren't closed under difference. You will have to try to do it another way:) –  sxd Aug 21 '11 at 21:48
    
Interesting! I didn't notice that they are not closed under difference although I knew that they're not under intersection. See my edit for the complete problem. –  Chan Aug 21 '11 at 21:50
    
Not necessarily, after all $a^*b^*c^*$ is regular. –  Yuval Filmus Aug 21 '11 at 21:51
    
@Yuval, yes you are right, the intersection between a CFL and a regular language is again a CFL. However, in general CFL's are not closed under intersection. –  sxd Aug 21 '11 at 21:54
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For the motivating problem,

Show that the complement of the language $L = \{a^nb^mc^k, k = n + m\}$ is context-free.

A string is in the complement of L if and only if it contains (at least) one of $ba$, $ca$ or $cb$ as a substring, or a letter other than $a,b$ or $c$ in case of a larger alphabet. This is easily expressed as a union of several languages, or as an automaton that scans the string in search of any of the forbidden substrings.

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The answer by @zyx goes a long way, but if the string is in $a^* b^* c^*$ you have to ensure the condition on the number of symbols is violated. This condition can go wrong because (1) there are too many $c$, or (2) there are too few. Written in terms of a grammar, with $L$ for less $a$ and $b$ than $c$, and $G$ for more: $$ \begin{align*} S &\rightarrow L \mid G \\ L &\rightarrow A c \\ A &\rightarrow A c \mid B \\ B &\rightarrow a B c \mid C \\ C &\rightarrow b C c \mid \epsilon \\ G &\rightarrow a D \mid E \\ D &\rightarrow a D \mid B \\ E &\rightarrow a E c \mid b F \\ F &\rightarrow b F \mid C \end{align*} $$ The languages mentioned by zyx are regular, so context free, and context free languages are closed respect to union.

Another take is to obtain the language $\{ a^n b^m c^k \mid n + m \ne k \}$ by operations that preserve context freeness. First consider $L_1 = \{0^n 1^m \mid n \ne m \}$, generated by: $$ \begin{align*} S &\rightarrow 0 A \mid B 1 \\ A &\rightarrow 0 A \mid E \\ B &\rightarrow B 1 \mid E \\ E &\rightarrow 0 E 1 \mid \epsilon \end{align*} $$ The $A$ branch ensures too many 0, the $B$ branch too many 1; $E$ gives as many 0 as 1, and is context free by closure properties.

Take now the homomorphism defined by $h(a) = h(b) = 0$, $h(c) = 1$. Then $h^{-1}(L_1) \cap a^* b^* c^*$ is the language we are looking for.

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