Questions tagged [semi-decidability]

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

28 questions with no upvoted or accepted answers
Filter by
Sorted by
Tagged with
3
votes
1answer
146 views

Efficient algorithm to determine if a lambda calculus term is equivalent to one without a given free variable

Consider the following problem: given a lambda calculus term $t$ and free variable $v$ determine whether $\phi(t,v)$, where $\phi(t,v) := \exists t'. t' \equiv t \land v \notin FV(t')$. This problem ...
2
votes
1answer
66 views

How To Show That B is Semi-Decidable Given A

I am preparing for my Computational Theory final and ran into this exact problem : B={ x | there exists a prefix of x that is in A}. Show that B is semi-decidable. In other words, you need to ...
2
votes
0answers
164 views

How to prove a Language is neither a Computably enumerable nor Co-Computably enumerable?

What would be the general approach for that? And what are the things that generally overlooked while proving such things? For example, I have a Language, L ={e:$L(M_e)$ such that it accepts only 'a ...
1
vote
0answers
34 views

k-limited solution for PCP

So there's following problem, that has been bugging me for the last few days: A solution of a PCP $ i_{1},\dots,i_{n}$ with the cards $(x_{1} ,y_{1}),\dots,(x_{m}, y_{m})$ is considered as $k$-...
1
vote
0answers
36 views

How to show that these two disjoint sets $A$ and $B$ exist

I came across this problem which asks to show the existence of two disjoint Turing-recognizable sets $A$ and $B$ such that no decidable set $C$ can separate them... In this case, a set $C$ is said to ...
1
vote
2answers
91 views

Need help understanding what co-recursively enumerable means

Lets say I have a set: $ L = \{\langle G \rangle | L(G) = \Sigma^{\star}\}$ and the question asks if it is co-RE. I know that if something is co-RE, it halts on every input not in L but may or may not ...
1
vote
0answers
49 views

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 ...
1
vote
0answers
603 views

Showing that $H'$ is not semi-decidable

I have an introductory class in computability theory and I'm currently working on my first exercises. I'm wondering if I'm on the right track with proving undecidable languages. Could you please have ...
1
vote
1answer
98 views

Check if language is decidable

I would like to determine if the following language is decidable or not. L = { w $\in$ $\Sigma^*$ | $T(M_w)$ is recognized by a Turing machine with at most 42 states}. I know that every finite ...
0
votes
0answers
28 views

Decidability for $ \exists w´, w´´\in L:$ so that |w´´| - |w´| is prime

I tried to decide wheter the given Language $ L = \{ \langle M \rangle | M \space is \space TM \space and \space \exists \space w´,w´´\in L(M):|w´´|-|w´| \space is \space prime \} $ is recursive or ...
0
votes
1answer
41 views

Prove that a language is decidable

I need some help to prove that the language is decidable. $K$ = {$N$ : $N$ is a DFA (Sigma = {a, b, c}) and $L$($N$) contains at least one word in which there is no a}. It tried to make an algorithm ...
0
votes
0answers
32 views

A TM that doesn't decide Σ*, and a TM that doesn't decide the empty set?

I was wondering if it was possible to create a TM that semi-decides (but doesn't decide) either of the following two languages: $\emptyset$ $\Sigma^{*}$ I assume for 2, a one-state TM that just ...
0
votes
1answer
98 views

Proving a language is not Semidecidable

I have the language $L = \{ \langle M_1, M_2 \rangle : L(M_1) \subset L(M_2)\}$ and I'd like to prove that it is not Semidecidable. To do so, I need to use a reduction from $\neg H$. I cannot use Rice'...
0
votes
0answers
28 views

A set that is not recursively enumerable and not (K'≤ A)

Is there a set A such that it's not recursively enumerable and not(K'≤ A) ? where K' is complement of K= {n| φ n (n) halts} Thanks!
0
votes
0answers
34 views

How to prove that for a decidable problem the problem and the compliment of the problem are semi-decidable?

Given a decidable problem, how would I go about proofing that the problem and the complement of the problem have to be semi-decidable?
0
votes
0answers
341 views

Decidability of intersection of two languages of same type

Given two context-sensitive languages, $L_1$ and $L_2$ is the problem of "whether $L_1 \cap L_2$ also belongs to CSL" decidable? I have the same question for the case when $L_1$ and $L_2$ belongs to ...
0
votes
0answers
41 views

Is the language below not decidable , if yes , it is then R.E ?

I am given a Turing machine M as input and i have to find out if this language below decidable and if not is it then in this case recursively enumerable . $$L=\\\{<M> \mid \text{ is ...
0
votes
0answers
216 views

Turing machine accepts two different strings

I am having hard time to proving this problem $C=\{\langle M \rangle \mid M \text{ is a Turing Machine , } L(M) \text { only contains two different strings}\}$ some ideas that i have tried are : i ...
0
votes
0answers
107 views

Decidability of {(M,w); M terminates on input w and tape of M is empty after computation}

I am currently trying to prove whether the above language is decidable, partially decidable or fully undecidable. I am certain that this language is partially decidable and reducible to the halting ...
0
votes
0answers
113 views

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 $\overline{E_{TM}}$ ...
0
votes
0answers
55 views

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-...
0
votes
0answers
68 views

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 ...
0
votes
0answers
49 views

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 ...
0
votes
0answers
112 views

Are there any RE-complete languages w.r.t. polynomial reduction?

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 ...
-1
votes
1answer
13 views

Is the Languague which contains all TMs which write the blank symbol at firs by the given input w decidable?

Consider the problem of determining whether a Turing machine M on an input w writes the blank symbol at first. Is this decidable ?
-1
votes
2answers
50 views

Is determining if a Turing Machine stops for at least one entry decidable?

I can't find how to prove the decibility with a reduction. EDIT: I've tried the reduction from the halting problem and the aceptance problem. Stopping for at least one entry has infinite inputs (you ...
-1
votes
2answers
82 views

Relation between sets and partially computable functions

I encountered this problem. Let $A$ , $B$ , $C$ be disjoint sets $(A\cap B = B\cap C = A\cap C = \emptyset)$. The $f_1, f_2$ and $f_3$ are partially computable functions that are defined as follow:...
-2
votes
1answer
35 views

Show that $L = L_\phi \cup L_{\{\sum^*\}} \notin RE$ with Rice theorem

Show that $L = L_\phi \cup L_{\{\sum^*\}} \notin RE$ with Rice theorem. Well I did show that with reduction, by using $HP'$. Simply by creating a function from $f(\langle M \rangle, x) = (M')$ Thus, ...