Existence of non-context free but decidable languages

Question

I've been reading the decidablity and undecidability chapters in Sipser's "Intro to Theory of Computation" however I could not find an explanation on the existence of a language that is both non-context free and decidable.

The only reference to this was a simple language hierarchy diagram showing where the decidable/recognisable bounds were in relation to language types.

I'm unsure as to how I should approach this but I've thought about proving this by diagonlisation:

• Let $M$ be the set of all decideable Turing Machines, and $L$ the set of all languages that are context-free. (Assume finite alphabet)
• By drawing up and filling the table where each language corresponds to a Turing Machine, I was hoping that I could find a contradiction in some $m \in M$ where there is no corresponding language $l \in L$.

I know that this will not work as both $M$ and $L$ are countable.

Any ideas on how I should approach this?

Answer 1

You could simply consider the language $L = \{a^k | k \text{ is a prime number }\}$. This is not a context free language, but surely this is decidable by a Turing Machine that checks if the length N of the input is not divisible by any number between 2 and N-1 .

Answer 2

I had not the enough reputation in this community to leave a comment on @Umamg's answer; so, I try to complete Umang's answer in mine.

One way to show that the language $L=\{a^p: \text{p is a prime number}\}$ is not context-free is to use pumping lemma for CFLs in the following way:

If $L$ was a CFL, then given an arbitrary long string in this language, say $a^p$ with $p$ being greater than the pumping length of $L$, this string would be decomposed into five parts $uvxyz$ with $|vy|\geq 1$ and so that for any $i\geq 0$ the string $uv^ixy^iz$ would belong to $L$.

Let $l:=|vy|$, then the above discussion shows that all the following strings are part of the language $L$: $$a^p, a^{p+l}, a^{p+2l}, \ldots, a^{p+nl}, \ldots$$

That is, the set of prime numbers contains an infinite arithmetic progression which is impossible since the number $p+pl$ is divisible by p.

Answer 3

A direct diagonalization argument is very much usable here (even though a somewhat unorthodox choice, as we have "more natural" separating examples available to us).

We can construct in a straight-forward way a notation system for context-free grammars, which then lifts to a notation system $(L_w)_{w \in \Sigma^*}$ for the context-free languages. We can then define the diagonal language $L_\Delta = \{w \in \Sigma^* \mid w \notin L_w\}$. The usual proofs that each context-free language is decidable are all uniform in the grammar, and thus give us that $L_\Delta$ is decidable. By construction, $L_\Delta$ is not context-free.

Everything in the preceeding paragraph remains true if we replace "context-free" by "context-sensitive".

Answer 4

There are two quick answers within easy reach of undergrad courses:

1. CFL ⊊ CSL ⊊ R
2. CFL ⊊ P ⊊ R

Basically, "all" of complexity theory happens "between" CFL and R (with some notable exceptions). Just have a look at the complexity zoo and note how far down CFL is, and how far up R.