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If I have a Context-Free Grammar G such that the language of G is nil, is G decidable?

I know the answer is yes, but I am having trouble proving this. My first thought is to say there is only one state, q1, which is the start state and accept state for a Turing Machine that is the equivalent of G. This machine will accept no input and immediately halt and accept because it has reached an accept state. Is this an acceptable answer, or am I way off here?

EDIT:

As Joel said below, the language I described accepts all strings. To counter this, I propose a second machine, G'. G' has 3 states, the start state q1, an accept state q2, and a reject state q3. q1 transitions to q3 on all symbols in the alphabet of G', and so does q2. q1 has an epsilon transition to q2. Therefore, if any symbols exist in the string being fed to G', G' will reject. If there are no symbols, the only option is to take the epsilon transition into the accept state. How does that sound?

EDIT:

The above solution was proven to accept the language L(G') = {""}.

Darkhydro
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  • I think we may be using different terminology. By "language of G is nil" do you mean L(G) = { }, i.e. the empty set? Or do you mean L(G) = { "" }, i.e. the language consisting of exactly one string, namely the empty string? It is often assumed without loss of generality that there are no transitions from accepting states, so that when you get to an accept state you halt. With that assumption, your original TM accepts everything. The TM you describe in your Edit accepts L = { "" }, which I think is what you intend. The TM described in my answer accepts nothing. That is, it accepts L = { }. – Joel Lee Mar 22 '11 at 07:08
  • @Joel I mean L(G) = {nil}, which i'm sure is equivalent to your statement L(G) = {}. Can explain a little more how your answer accepts this language if it never reaches an accept state (because there are none?) – Darkhydro Mar 22 '11 at 17:18
  • By definition, machine _M_ accepts a string _s_ if it reaches an accepting state when given input _s_. Also by definition, _L(M)_, referred to as the "Language accepted by M", is the set of all strings accepted by _M_. _L(M)_ is a **set** of strings. But that set could, in fact, be empty. If the machine has no accepting states, then it cannot possibly accept any string given to it. Therefore _L(M)_ is the empty set. Your G' machine accepts exactly one string, namely, the zero-length string. Therefore, _L(G')_ = { "" }. – Joel Lee Mar 22 '11 at 22:53
  • @Joel I'm having a really hard time wrapping my head around this. If the language of M is the set of all strings accepted by M, then I can understand how L(M) = {nil} (the empty set), because it accepts no strings. But to be decidable you must accept, right? So regardless of whether the language of M is nil or not, I don't see a situation in which M accepts, and therefore it can never be 'decided'. – Darkhydro Mar 22 '11 at 23:44
  • @Joel I just read your comment below and it cleared things up a bit for me. Thanks. – Darkhydro Mar 23 '11 at 00:12

1 Answers1

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As you said, the answer is yes. A general proof is the fact that given a CFG G you can easily (well sort of) construct a TM simulates derivations using that grammar. However, you are looking for a short proof for the empty language. (The fact that you have a CFG in this case is irrelevant.)

You are on the right track in that if you can construct a TM that always halts for a given language L, then L is decidable. However, the machine that you describe will actually accept every string, i.e. the language consisting of every possible string over the alphabet. That's because if the start state is also an accept state, then the Turing machine will accept immediately when it starts. It does not have to read the entire input string (or any part of it) to accept.

To define a TM that accepts nothing, let the set of accepting states be empty. To guarantee that your machine always halts, your transition function can be empty as well.

Joel Lee
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  • I'm not sure I understand. I was under the impression that the machine must not only halt, but also accept. The machine can halt and then reject the language, as (I would think) would be the case in your example. – Darkhydro Mar 22 '11 at 06:25
  • Note on terminology / concepts. We talk about a machine accepting or rejecting a given string. From that we define the notion of the "Language accepted by M", and we say that "machine M accepts language L". However, we do not say that "machine M rejects language L". Instead, you would say "the language accepted by M is not equal to L". – Joel Lee Mar 22 '11 at 23:41