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I have a Turing Machine M. Let L be the set of all strings representing the encoding of M that has input alphabet {1,2}, where M accepts infinitely many strings that start with 1 and finitely many strings that start with 2. I'm trying to determine if L is decidable, semi-decidable (computably enumerable) or undecidable.

Using Rice's Theorem, I believe I can prove that L is not decidable:

P is the property of a language where the language has infinitely many strings that start with one symbol and finitely many strings that start with another symbol. P is non-trivial and is a property of languages of TMs. Therefore, according to Rice's Theorem, M = {⟨M⟩∣ L includes infinitely many strings that start with 1 and finitely many strings that start with 2 and Σ = {1,2}} is undecidable.

But I'm struggling to prove whether it's semi-decidable or completely undecidable. This post mentions using an Generalized Rice's Theorem to prove that a language is recursively enumerable (aka semi-decidable) but I'm having trouble understanding their approach.

Can anyone shed some light on this approach and/or if I'm heading down the right path? Many thanks in advance.

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Amal Kahalee
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  • Rice's theorem tells you that the set of Turing machines $M$ that accept your language $L(M)$ is not recursively enumerable. $L(M)$ certainly is recursively enumerable. You need to say more about $M$ or $L(M)$. – Rob Arthan Jan 05 '15 at 00:28

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Hint: It may be clearer to state that the set of internal states of the Turing machine $T$ is either $\{0,1\}$ for the infinitely (or finitely many strings that it may produce on its tape), especially if the language $\Sigma$ is recursively enumerable (or equivalently, in the range of a partial recursive function $\psi$).

I assume, then, that the language $\Sigma$ is recursively enumerable. We say that $\Sigma$ is decidable if $\Sigma$ lays in the range of a total recursive function $\psi_e(n)$ where $e$ is the code number of the machine $T^e$ that halts on input $n$, and where $n \in \mathbb N$. Intuitively, total recursion holds if $\psi_e$ is defined for every input.

Rice's theorem states that any nontrivial property $P$ of a partial recursive function $\psi_e$ is undecidable (computably unsolvable by $T^e$). We take $P$ to be decidable (either $\psi_e(n)=0$ $\vee$ $\psi_e(n)=1$ if $P$ holds or does not hold, respectively). More directly, let $0'$ be the degree of unsolvability of the problem for deciding if $T^e$ halts on arbitrary input $n$. If the solution set for deciding if $P$ holds is unsolvable, then $P$ is undecidable and the solution set is $0'$.

Therefore, Enumerating the pairs $(n,i)$ if $P$ holds for a string $s_n$ in $\Sigma$ is simply taking $P$ as computable and showing that any $P^i, i \in \mathbb N$ holds or does not hold for a finite (or infinite) $s_n$. Supposing that it does hold, then the solution set it produces should be decidable, as the characteristic $\chi_\Sigma(x) = 0$ if $P(x)$ holds for $x \in \Sigma$ or $\chi_\Sigma(x) = 1$ if $P(x)$ does not hold for $x \in \Sigma$. If this pair is decidable, then the solution set is decidable, since a set is computably enumerable iff its characteristic function is, and not if otherwise. A property of $\psi$ could taken as any property that the sequence of partial functions $\psi_1(n),...,\psi_e(n)$ terminates on input $n$, or if each member of the solution set is a proof that $P(x)$ holds.

Amal Kahalee
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cmn1
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  • @AmalKahalee No problem! I hope it was clear. I think the main point is that any nontrivial property $P$ you wish to prove is essentially undecidable, given the fact that your recursive enumerable language $\Sigma$ is indeed only recursively enumerable. The essential essence of it all is that the sequence $\psi_i, i \in \mathbb N$ is enumerable itself, and therefore in the range of a partial recursive (rather than total recursive) function, since if it were total, then it all properties of $\psi_i$ would be decidable, which is contradictory. Another way of putting this is $0' \subset \Re$. – cmn1 Jan 05 '15 at 02:14
  • @AmalKahalee Yes that is correct! The sharp point to keep in mind is that, given a set $K \subseteq \mathbb N$, we say $K$ is decidable (computable) iff its characteristic function is. This function, if decidable, is total, as it is defined for every input in $K$. Essentially, every total function is partial, but not every partial function is total, since partial functions may enumerate sets such as $K$ which have no computable characteristic function. – cmn1 Jan 05 '15 at 04:38