Prove complement a^nb^nc^n is contextfree

Question

So the complement of L1 = $\{a^{n}b^{n}c^{n} \mid n \geq 1\}$ would be L2 = $\{a,b,c\}^* \setminus \{a^{n}b^{n}c^{n} \mid n \geq 1\}$.

In other words, any combinations of a,b and c where we dont have an equal number of all three letters and w = $\varepsilon$ is also legit.

However, while I'm certain that there should be a contextfree grammar for L2, I can't seem to find a grammer that allows you to generate the terminals freely without allowing $a^{n}b^{n}c^{n}$ with n $\geq$ 1.

My attempt was to make a starting Rule S -> $Q_{_{a}}$ ; $Q_{_{b}}$ ; $Q_{_{c}}$ ; $\varepsilon$ so that the individual Q rules would make it possible for two of the terminals a, b and c to have an equal number, but not for the third. (in $Q_{_{a}}$, a is restricted by max(b,c), in $Q_{_{b}}$, b is restricted by max(a,c), in $Q_{_{c}}$, c is restricted by max(a,b) so that the restricted terminal can never show up as often as the unrestricted terminal with the highest count)

The rules I set up though, only allow to have unlimited numbers of unrestricted terminals in any order. I'm not sure how to implement a rule for the restricted terminal, without allowing it to have as high a count as the unrestricted ones, if the unrestricted ones have the same count.

here's my P for G$_{_{L2}}$ so far

S $\rightarrow$ $Q_{_{a}}$ ; $Q_{_{b}}$ ; $Q_{_{c}}$ ; $\varepsilon$

$Q_{_{a}}$ $\rightarrow$ $Q_{_{a}}$b $Q_{_{a}}$ ; $Q_{_{a}}$c $Q_{_{a}}$ ; $\varepsilon$

$Q_{_{b}}$ $\rightarrow$ $Q_{_{b}}$a $Q_{_{b}}$ ; $Q_{_{b}}$c $Q_{_{b}}$ ; $\varepsilon$

$Q_{_{c}}$ $\rightarrow$ $Q_{_{c}}$a $Q_{_{c}}$ ; $Q_{_{c}}$b $Q_{_{c}}$ ; $\varepsilon$

These are still missing the rules for the individual restricted variable. In what fashion can I add those, without breaking the [ $a^{n}b^{n}c^{n}$ | n = 0 ] rule?

Edit: it occured to me that I could refine the restricting rules, such that the restricted terminal can be derived from a rule if, and only if one (and only one) of the unrestricted ones is always created with it. For example:

S $\rightarrow$ $Q_{_{a}}$ ; $Q_{_{b}}$ ; $Q_{_{c}}$ ; $\varepsilon$

$Q_{_{a}}$ $\rightarrow$ $Q_{_{ab}}$b $Q_{_{ab}}$ ; $Q_{_{ac}}$c $Q_{_{ac}}$

$Q_{_{ab}}$ $\rightarrow$ $Q_{_{ab}}$b $Q_{_{ab}}$ ; $Q_{_{ab}}$c $Q_{_{ab}}$ ; $Q_{_{ab}}$ a $Q_{_{ab}}$ b $Q_{_{ab}}$ ; $Q_{_{ab}}$ b $Q_{_{ab}}$ a $Q_{_{ab}}$ ; $\varepsilon$

$Q_{_{ac}}$ $\rightarrow$ $Q_{_{ac}}$b $Q_{_{ac}}$ ; $Q_{_{ac}}$c $Q_{_{ac}}$ ; $Q_{_{ac}}$ a $Q_{_{ac}}$ c $Q_{_{ac}}$ ; $Q_{_{ac}}$ c $Q_{_{ac}}$ a $Q_{_{ac}}$ ; $\varepsilon$

At this point my brain starts running in circles. Would this set me on the right path?

Answer 1

You're overcomplicating. There is no need for the grammar to be deterministic, so it's OK for cases to overlap.

Say $\omega \in \overline{L_1}$. At least one of the following statements must be true:

• $\omega = a^ib^j\nu \text{ where } i \ge 1, j\ge 0 \text{ and } i \ne j, \text{ and } \nu \text{ does not start with } b$
• $\omega = a^ib^jc^k \text{ where } i, j, k \ge 1, \text{ and }j \ne k$
  (I don't require $i = j$, although if $i \ne j$ then the first condition also applies.)
• $\omega \text{ does not start with } a$
• $\omega \text{ contains } ba, ca \text{ or }cb$

These all have reasonably simple grammars.

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Just in case it's not obvious, there are a few ways to do $a^jb^k | j\ne k$. One simple one:

$$\begin{align} S &\to A \mid B \\ A &\to aC \mid aA \\ B &\to Cb \mid Bb \\ C &\to \epsilon \mid aCb \end{align} $$

$C$ is a balanced sequence of $a$ and $b$, $A$ has more $a$s, $B$ has more $b$s, and $S$ has either more $a$s or more $b$s.

Answer 2

Informally let us first see the structure of the language.

$$L_2= \Sigma^* \text{\\} \{a^nb^nc^n | n \geq 1\}$$

This language has all strings which :

1. starts with $b$.
2. starts with $c$.
3. starts with $a$ and have the form $a^i b^j c^k . (a+b+c)^*$ where $\lnot((i=j) \land (i=k) \land (j=k)) \equiv ((i\neq j) \lor (i\neq k) \lor (j\neq k))$. Now strings of this form can be represented as follows: $$(a^ib^jc^* + a^i b^* c^j + a^+b^ic^j). (a+b+c)^*, \text{where $(i\neq j)$}$$
4. starts with $a$ and have the form (to handle strings such as $abcabc$ for instance): $$(a^+b^+c^+)((a^++b^+)^+c^*)(a+b+c)^*$$

Cases 1,2,4 are regular and case 3 is the one which requires stack. So we can easily create a PDA out of this. It shall be NPDA I guess, because on seeing we should work in parallel in two possible path ways, (3 and 4).