# Questions tagged [semi-decidability]

Questions about which problems are semi-decidable, also known as recognizable or recursively enumerable.

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### Showing that the language $L = \{\langle M, w \rangle\ |\ M$ moves left at least three times while computing $w \}$ is decidable or undecidable

How would you go about showing that the language $L = \{\langle M, w \rangle\ |\ M$ moves left at least three times while computing $w \}$ is decidable or undecidable? Intuitively my thoughts are ...
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### The Church-Turing-Thesis in proofs

Currently I'm trying to understand a proof of the statement: "A language is semi-decidable if and only if some enumerator enumerates it." that we did in my lecture. One direction of the proof goes ...
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### The set of words accepted by TMs simultaneously is finite is non semi-decidable

Consider $L = \{(M_1,M_2):\text{the set of words accepted by both TM at the same time is finite}\}$. I want to determine if this language is decidable, semi-decidable or not semi-decidable. My ...
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### which of the following languages are Recursively Enumerable?

Which of the following languages are recursively enumerable? A={⟨M⟩∣ TM M accepts at most 2 distinct inputs} B={⟨M⟩∣ TM M accepts more than 2 distinct inputs} For first language I think that we can ...
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### Why is $L=\{\langle M \rangle \mid |L(M)| \geq k\}$ not recognizable?

Here $M$ denotes a turing machine. By set theory, $L = \overline{E_{TM}} \cap \overline{L_0} \cap \overline{L_1}$ where $L_i=\{\langle M \rangle \mid |L(M)|=i\}$. And I know that $\overline{E_{TM}}$ ...
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### cardinality of recursive/r.e/not r.e languages? [duplicate]

I was just looking into properties of languages and wondered about the cardinality of them are all recursive languages countable or can they also be uncountable (can u have a recursive language which ...
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### How can I prove that there is a decidable language which is not in P?

Generally, I want to use the diagonal argument to prove it. I tried to define a language $A$ which is constructed by a Turing machine $D$: It will only take a input which has a form of a ...
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### Can a semi-decidable problem be also decidable?

As far as I understand, a semi-decidable (recursively enumerable) problem could be: decidable (recursive) or undecidable (nonrecursively enumerable) ...
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### Semidecidability/Decidability for strings seperated by a non alphabet symbol

I am trying to prove the following: Let $\Sigma$ be an alphabet not containing the symbol "$;$", and suppose that $L \subseteq \Sigma^*$; $\Sigma^*$ is recursively enumerable. If this is the case ...
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### Which Turing machine problems are Decidable?

Let $M_0$, $M_1$, $M_2$,..., be an effective enumeration of all Turing machines. Which of the following problems is (are) decidable ? Given a natural number $N$, does $M_N$ starting with an empty ...
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### Is the Rice Theorem applicable for these problems?

I have 1 problem :--> L = { < M > | TM halts on no inputs } I have solved the above problems by reductions given in the book and even there are many links ...
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### Is the set of programs that compute some function other than $h$ recursively enumerable?

Let $h$ be a total computable function. Is $S = \{x \mid f_x \neq h\}$ recursively enumerable? Originally this was an exercise that restricted $h$ to: $h(x) = x + 1$ . However, it can be formulated ...
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### Show that the set of programs whose Kolmorgorov complexity is smaller than their length is recursively enumerable

Define the language $\qquad R = \{x \in \{0,1\}^\ast \mid C(x) \ge |x| \}$ where $C(x)$ is the Kolmorgorov Complexity of $x$ and $|x|$ denotes the length of $x$. Prove that $R$ is co-recursively ...
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### Suppose $L_1, L_2, ..., L_k$ are recursively enumerable languages forming a partition of $\Sigma^*$. How do I show that each $L_i$ are recursive?

Suppose $L_1, L_2, ..., L_k$ are recursively enumerable languages forming a partition of $\Sigma^*$. How do I show that each $L_i$ are recursive ? I see that for $x \in \Sigma^*$, $x$ belongs to ...
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### Function is recursive iff its graph is recursively enumerable

So I understand that a function is recursive if there exist a Turing Machine that accepts it and halts on every input, since function is defined everywhere. But how to prove that function is recursive ...
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### Is undecidability of TMs' properties a statistical statement?

We know (by Rice's theorem) that is it not possible to decide a non-trivial property of a given TM. We could say therefore that we cannot be sure at 100 percent that a given TM has a certain non-...
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### Given language consisting Turing machines is decidable or not?

$$L=\{M\mid \text{there exist }x,y\in\Sigma^* \text{ s.t. }x \in L(M)\text{ and } y \notin L(M)\}\,.$$ I think it's not recursively enumerable because this language reduces to complement of the ...
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### Showing an equivalent definition of $CO-RE$

Let $L_1$ be a language. I would like t prove that $L_1\in CO-RE$ if and only if exists a language $L_2$ s.t. $L_1=\{ u\,\, |\,\, \forall_{v\in\Sigma^*} : <u,v>\in L_2 \}$ and $L_2\in R$. I'm ...
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### Partial recursive function with no total recursive extension

We define a partial recursive function $f:\{0,1\}^* \longrightarrow \{0,1\}^*$ to be semi-good if we can define a total recursive function $g:\{0,1\}^* \longrightarrow \{0,1\}^*$ from $f$, such for ...
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### Language of TMs that accept some x in less than 50 steps. Is it in co-RE?

L = {M | M is a TM and there exists an input that the TM M accepts in less than 50 steps} I need to find a minimal class it belongs to between R/ RE/ co-RE/ not in RE∪co-RE. I managed to show that ...
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### how to prove that the property 'doesn't halt for some input' is not semidecidable

I am taking a computer theory class and one of the exercises is to prove that the property "doesn't halt for some input" is not semidecidable. This property is the negation of the property "halts for ...
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### Decidability of $\{p|∃y : \operatorname{Dom}(φ p ) ⊇ \operatorname{Dom}(φ y )\}$

I need to classify the set $$\{p|∃y : \operatorname{Dom}(φ p ) ⊇ \operatorname{Dom}(φ y )\}$$ as decidable, semidecidable or not semidecidable. I don't know how to start. Any ideas? Dom (φp) it's ...
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### Is the given language decidable for turing machine?

I am going through undecidability of TM and found this question $L=\left \{ \left \langle M \right \rangle |M\ is\ TM \ and \ number\ of\ strings\ in\ the\ language\ \ is\ prime\right \}$ I think it ...
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### Does it matter for this function if the set we check membership of is finite?

I have the following problem. Let $\Phi$ be an admissible numbering of the single-parameter partially-recursive functions. That is, $\Phi(i, x) = f_i(x)$ with $f_i$ the $i$th partially-recusive ...
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### Dovetailing in Turing Machines?

I was going through TM here and encountered a term "Dovetailing". What is exactly the dovetailing in Turing Machines? Can anyone mind to provide a good explaination with some examples?
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### Is the given language decidable?

L = { < M > | M is a turing machine and } Obviously, the language which L(M) is polynomially reducible to, is context free and hence recursive, so it is a decidable language . Now, L(M) is ...
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### Is the set of pairs of TMs at least one of which accepts the empty word semi-deciable?

L = { < M1,M2 > | M1,M2 are TM's and Ɛ ∈ L(M1) ∪ L(M2) } Where Ɛ = Epsilon I know that this language is undecidable, but why it is semidecidable too. What i have tried is => Using Rice's ...
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### Is this language semi-decidable?

Let $M_w$ be the DTM encoded by the binary string $w$ and let $$L=\{w\#x\,|\,\text{all states are reached when running }M_w\text{ on }x\}.$$ I've already proved that this language is undecidable (the ...
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### The Hindley-Milner type system plus polymorphic recursion is undecidable or semidecidable?

I have often read that Hindley-Milner extended to allow polymorphic recursion is undecidable. However is the term used what is actually meant? Or do people actually mean semidecidable when they ...
I need to decide if there exists $L\in RE$ so that for every $L'\in RE$ we have $L' \leqslant_p L$, meaning a polynomial-time reduction. I've tried to use $L=A_{TM}$ (the accepting problem), but got ...