# Can there be a context-sensitive pumping lemma?

A "pumping" property (words of a certain length imply the existence of loops in the language-defining mechanism) are known to exist for regular and context-free languages and a few more (usually used to disprove a language's membership to a certain class).

Within the discussion around this question, Daisy's answer suggests that there can't be a pumping lemma for context-sensitive languages - since they're so complex.

Is that true - can it be shown that there can't be some type of pumping property - and is there a good reference for that (or against that)?

• Can you give a formal definition of "pumping lemma"? If not, there can be no such proof by principle. – Raphael Jul 18 '16 at 20:35
• Perhaps one can refute Parikh's theorem for context-sensitive languages, and this leads us to expect that no pumping lemma resembling the ones we know exists. – Yuval Filmus Jul 18 '16 at 21:27
• @YuvalFilmus What do you mean? Clearly the prime numbers are context sensitive, and are not semilinear. So Parikh does not hold for context sensitive. That means that "linear pumping" does not apply. Like Raphael I am curious what other methods would be considered pumping. – Hendrik Jan Jul 18 '16 at 22:11
• You're right, it comes down to knowing what exactly "pumping" is... I was hoping for some suggestions... What is Parikh's theorem for context-sensitive languages? I only found one for context-free languages. – lukas.coenig Jul 19 '16 at 6:39
• @lukas.coenig There is no Parikh's theorem for context-sensitive languages, but there might have been one if there were a simple pumping lemma for context-sensitive languages. – Yuval Filmus Jul 19 '16 at 6:42

## 1 Answer

Here is some evidence that there is no pumping lemma for the context-sensitive languages.

Of course, an answer hinges on the question what constitutes a pumping lemma. The weakest reasonable definition I could think of is this: A language class $\mathcal{C}$ has a pumping lemma if there is a decidable ternary predicate $P(\cdot,\cdot,\cdot)$ where $P(g,w,d)$ means:

• $g$ is a word encoding a language $L(g)$ from $\mathcal{C}$ (think: grammar),
• $w$ is a word in the language encoded by $g$
• $d$ is a word encoding a pumpable computation/derivation for $w$ (think: NFA computation with repeated state or CFG derivation tree with repeated nonterminal). Here, pumpable means: there exist infinitely many words in $L(g)$.

Moreover, we want that given a language $L$ in $\mathcal{C}$ encoded by $g$, for every sufficiently long word $w\in L$, there exists a word $d$ such that $P(g,w,d)$.

For example, the pumping lemma for regular languages would give rise to the predicate "$g$ encodes an $\varepsilon$-free NFA and $d$ encodes a run that repeats a state and reads $w$". For suitable encodings, this clearly satisfies the above conditions.

Now let us show that such a predicate does not exist for the context-sensitive languages.

Observe that if a language class has a pumping lemma, then the infinity problem (Given a grammar, does it generate an infinite language?) is recursively enumerable: Given an encoding $g$, we can enumerate words $w$ and $d$ and check whether $P(g,w,d)$. If we found such $w,d$, we answer 'yes', otherwise, we continue the enumeration.

However, we show that the infinity problem for the context-sensitive languages is not recursively enumerable. Recall that $\Pi_2^0$ is a level of the arithmetic hierarchy that strictly includes the recursively enumerable languages. Hence, it suffices to prove:

Claim: The infinity problem for the context-sensitive languages is $\Pi_2^0$-complete.

It is well-known that the infinity problem for recursively enumerable languages is $\Pi_2^0$-complete (more often, one finds the formulation that the finiteness problem is $\Sigma_2^0$-complete). Hence, it suffices to reduce the latter problem to the infinity problem for the context-sensitive languages.

Given a TM $M$, we construct an LBA $A$ for the language $$\{u\#v \mid \text{v is a shortlex-minimal accepting computation of M on input u}\}.$$ Then, $L(A)$ is infinite iff $L(M)$ is infinite, which completes our proof.

Update: Tried to be clearer. Update: Added example.

• – Raphael Jul 20 '16 at 20:33