I was referring to slide 4 of this, which states following:

It is decidable whether a given CFG accepts a non-empty language?

Then I was reading Intro to Automata theory book by Ullman et al. It states following:

Let $L_{ne}$ be the language of all codes for Turing machines that accept atleast one input string. $$L_{ne}=\{M | L(M)\neq \phi\}$$ $L_{ne}$ is not recursive.

Knowing that "not recursive" is same as "undecidable" and the functionality of Turing machine is expressed as a CFG, I feel that the problems in above two statements sound same and the statements contradicts each other in terms of their decidability claims. Thus, one of them must be wrong. However I also feel that I still miss something basic and both of them are non contradictory and equally true. But then what I miss?


The thing you missed is that Turing Machines are strictly more expressive than Context-Free Grammars. CFGs are as expressive as push-down automata (PDA, i.e. automata with one stack).

E.g. the language $L = \{a^nb^nc^n | n > 0\}$ cannot be expressed by CFG (or decided by a PDA), but can be decided by a Turing Machine.

So, for CFG, the emptiness problem is decidable, but somewhere on the way to Turing completeness, this property is lost.

  • $\begingroup$ Ahhhh I mistook CFG for unrestricted grammar...The functionality of TM can be expressed with the help of unrestricted grammar (of form $(V\cup T)^+\rightarrow(V\cup T)^*$), whereas PDA functinality is expressed as CFG (taking form $V\rightarrow(V\cup T)^*$). So I think I can say that "its decidable whether language accepted by a PDA is empty or not". Right? $\endgroup$ – Maha Dec 20 '16 at 12:11

The solution is a bit subtle. Wikipedia says the following:


That means you can reduce the halting problem to the problem, "Does this CFG generate all strings?", but not necessarily to the problem "Does this CFG generate the empty language?".


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