Undecidability MCQ Quiz in தமிழ் - Objective Question with Answer for Undecidability - இலவச PDF ஐப் பதிவிறக்கவும்

Concepts:

L(M) is accepted by Turing Machine therefore the language is recursively enumerable language.

Recursively Enumerable language is closed under intersection and L(M₁) ∩ L(M₂) is accepted by M

Is decidable

∴ option 1 is correct

Important points:

Recursively Enumerable language is not closed under complementation and subset.

A recursive model is one which has all causal effects unidirectional and disturbance terms uncorrelated.

Key Points

• A recursive model is a special case of an equation system where the endogenous variables are determined one at a time in sequence.
• Thus the right-hand side of the equation for the first endogenous variable includes no endogenous variables, only exogenous variables.
• Causal effects are unidirectional and disturbance terms are uncorrelated.

The correct answer is option 4.

Key Points

S1: Regular languages are a subset of the set of languages accepted by TMs which do not write anything on the tape.

True, Regular languages are a subset of the set of languages accepted by TMs Which do not write anything on the tape. The subset of regular languages is not closed under regular grammar. 

S2: The set of languages accepted by halting TMs with a bidirectional infinite tape is a proper superset of decidable language.

False, The set of languages accepted by halting TMs with a bidirectional infinite tape is a proper superset of decidable language. Halting Turing machine accepts recursive languages. Proper superset of recursive languages is an undecidable language.

S3: Every decidable language can be accepted by a DFA with a priority queue.

True, Every decidable language can be accepted by a DFA with a priority queue. 

Answer: 2

Explanation: 

1: L = { | D is a DFA and w is not in L(D)}

This language is decidable since L is just a complement of the DFA, and the complement of DFA can be obtained by simply making final states to Non-final states and non-final states to the final states.

2: L = { | D and Q are DFAs and L(D) ∩ L(Q) = ϕ }

This language is decidable since the intersection of the language of two DFAs will be regular.

3: L = {1, M2> | M1 , M2 are TMs and L(M1) = L(M2) }

This language is undecidable since Equivalence of TWO TMs not decidable.

There exists no algorithm that can find if two Turing machines accept the same language or not.

There exists a machine M2 that accepts the same language as accepted by Turing machine M1 but, if two Turing machines are equal or not, is not decidable. This is a non-trivial property in Turing machines.

4: L = {1> | M1 is a TM and does M1 accepts regular language}

This language is undecidable since the Regularity problem for TMs is not decidable. It is not necessary for a TM to accept a Regular language It may or may not accept. 

The correct answer is option 4

CONCEPT:

Decidable language: A language is decidable if there exists a Turing machine which halts or accepts it.

EXPLANATION:

1: Undecidable

There is no TM to determine whether a given program will produce an output.

2: Undecidable

Context-free languages are not closed under complementation.

3: Decidable

Regular languages are closed under complementation.

4: Decidable

Recursive languages are closed under complementation.

So option 4 is correct

Key Points

Note that the following statements are true

1. Every infinite regular language L has a subset S which is undecidable. 
2. Every infinite regular language L has a subset S which is unrecognizable.
3. Every infinite language L has a subset S which is undecidable.
4. Every infinite language L has a subset S which is unrecognizable.

So, S₁ and S₂ both are correct.  

the correct answer is option 4

Explanation: for statements S₂ consider the following explanation

every infinite language (regular or not) has an undecidable subset.

With the particular focus on regular languages, the intended solution might have been something like using the pumping lemma to find x,y,z such that xyⁿz is in your language for every n,  then consider the subset A = {xyⁿz | n ∈ N}, where A is your favorite undecidable set of natural numbers.

it proves s1 is correct.

Option 1: Decidable

Every NFA is a PDA with finite memory.

Option 2: Undecidable

Membership algorithm does not exist for unrestricted grammars

Option 3: Undecidable

Regularity problem for TM is undecidable.

Option 4: Undecidable

Equivalence of Two grammar is undecidable.

This is an example of a State Entry Problem. This problem can be reduced to a halting problem.

Taking two Turing machines into consideration.

1. A Turing machine M with final state ‘q’
2. A Turing machine R with inputs – M, q, w.

Giving word ‘w’ as input to machine M, there are two possibilities.

• If M halts in its final state q, then R accepts the input. This makes the problem partially decidable.
• If M enters in an infinite loop, then M cannot produce any output. R rejects the input, so this problem becomes undecidable.

Hence, Option_1 is correct.

Concept:

If a NP complete problem is reducible to another problem in polynomial time than that other problem becomes NP hard.

If X ϵ NP complete and X reduces to Y in polynomial time than Y is NP hard.

3 - SAT is NP complete problem

Explanation:

Q₁ reduces in polynomial time to 3- SAT. As 3 - SAT is NP complete problem, Q₁ is easier than NPC. So, Q₁ can’t be NP hard. It is in NP.

3- SAT reduces in polynomial time to Q₂. As, 3- SAT is NP complete so, Q₂ will be NP hard.

Therefore, Q₁ is NP and Q₂ is NP hard.

Properties:

Recursive language is closed under complementation and it is Turing decidable

Recursively enumerable language is not closed under complementation

Explanation:

L̅ is recursive language and every recursive language is Turing decidable

E is recursively enumerable language and every recursively enumerable language is recognizable

but not Turing decidable.