​Analytical Engine: 

• An analytical engine is a machine which is considered to be the concept for the first general mechanical computer.
• It was programmed using punch cards and also have an integrated memory.
• It was first proposed by Charles Babbage in 1837 and it was considered as the first design concept of a general-purpose computer.

EDSAC:

• EDSAC stands for Electronic Delay Storage Automatic Calculator
• It is considered to be the second stored-program electronic computer.
• It is also considered as the first computer to run a graphical computer game.

ENIAC:

• ENIAC stands for Electronic Numerical Integrator And Computer, was the first general-purpose electronic computer. It was the first Turing-complete, digital computer capable of being reprogrammed to solve a full range of computing problems.

The correct answer is option 1 i.e Analytical Engine

The correct answer is Ambiguity problem for CFGs.

Key Points

• The Ambiguity problem for CFGs (Context-Free Grammars) is undecidable.
  • A context-free grammar is ambiguous if there exists a string that can be generated by the grammar in more than one way (i.e., it has more than one distinct parse tree).
  • The ambiguity problem asks whether a given CFG is ambiguous.
  • This problem is undecidable, meaning there is no algorithm that can determine the ambiguity of any CFG in general.

Additional Information

• Other problems listed in the options are decidable:
  • The membership problem for CFGs is decidable. This problem asks whether a given string can be generated by a CFG.
  • The finiteness problem for FSAs (Finite State Automata) is decidable. This problem asks whether the language recognized by the FSA is finite.
  • The equivalence problem for FSAs is decidable. This problem asks whether two FSAs recognize the same language.

The correct answer is (A), (B), (C) Only

Key Points

To evaluate the statements about Turing Machines, let's analyze each one:

• (A) A Turing machine can simulate any algorithmic computation.
  • True. Turing machines are powerful computational models that can simulate any computation that can be described algorithmically.
• (B) Turing machines can have a finite number of states and an infinite tape.
  • True. A Turing machine has a finite number of states but an infinite tape, allowing it to perform computations beyond finite memory limits.
• (C) The transition function of a Turing machine can depend on the current state and the symbol being read.
  • True. The transition function of a Turing machine is defined based on both the current state and the symbol currently being read from the tape.
• (D) A Turing machine cannot recognize context-free languages.
  • False. Turing machines can recognize a broader class of languages, including context-free languages; in fact, every context-free language can be recognized by a Turing machine.

Based on this analysis, the true statements are (A), (B), and (C).

Therefore, the correct answer is: 1) (A), (B), (C) Only.

The correct answer is (A), (B), (D) Only

Key Points

To evaluate the statements about Finite Automata (FA), let's analyze each one:

• (A) A finite automaton can recognize regular languages.
  • True. Finite automata are specifically designed to recognize regular languages.
• (B) Deterministic Finite Automata (DFA) and Nondeterministic Finite Automata (NFA) are equivalent in terms of the languages they can recognize.
  • True. Both DFAs and NFAs can recognize the same set of regular languages, although they may differ in the number of states used.
• (C) An FA can have an infinite number of states.
  • False. A finite automaton, by definition, has a finite number of states.
• (D) The transition function of a DFA is defined for every state and input symbol.
  • True. In a DFA, the transition function must be defined for every combination of state and input symbol, ensuring that it always has a valid transition.

Based on this analysis, the true statements are (A), (B), and (D).

Therefore, the correct answer is: 2) (A), (B), (D) Only.

The correct answer is Output tape.

Key Points

• A Turing machine is a mathematical model of computation that defines an abstract machine.
• The standard components of a Turing machine include:
  • Input tape: A tape divided into cells, each containing a symbol from a finite alphabet. The tape is infinitely long and serves as the input and working memory.
  • Control unit: A finite set of states and a transition function that dictates the machine's operations based on the current state and the symbol being read.
  • Read/write head: A head that reads and writes symbols on the tape and can move left or right.
• Output tape is not a standard component of a Turing machine.

Additional Information

• A Turing machine operates based on a set of rules (transition function) that describe how the machine's state changes based on the current state and tape symbol.
• Turing machines are used to model the logic of any computer algorithm and are essential in the theory of computation.
• They help in understanding the limits of what can be computed.
• The concept was introduced by Alan Turing in 1936, and it has become a fundamental model in computer science.

Problem mentioned in option (2) Ambiguity problem for context-free grammar (CFG_S) is undecidable.

Explanation:- 

The ambiguity of grammar is undecidable, i.e there is no particular algorithm for removing the ambiguity of grammar. Here are some examples of ambiguous grammar:

• S → aS | Sa | €
• E → E + E | E*E | id
• A → AA | (A) | a

 

Additional InformationEquivalence problem for FSA_S:- An automation is a machine that has a finite number of states. Any two automation is said to be equivalent if both accept exactly the same set of strings.

Finiteness problem for FSA_S:- The finiteness problem is decidable for FSAs ( also for CFG_S) as we just need to check for a loop DFA.

Membership problem for CFG_S:- The membership problem for a set is the problem of deciding which of two possible elements is logically more likely to belong to that set. In other words, given two elements of which at least one is in the set, to distinguish the member from the non-member.

The correct answer is option 1.

Key Points

• The problem "Is L1 ∧ L2 = ϕ?" is undecidable for context-sensitive languages L1 and L2. Here L₁∧ L₂ is a disjoint operator it's undecidable problem.

Hence statement I is correct.

• The problem "Is WϵL?" is decidable for context-sensitive language L, (where W is a string). Here W is a string that is accepted by the particular language is a membership problem and its decidable problem.

Hence statement II is correct.

Additional Information

Context-sensitive language is accepted by linear bounded automata and only membership(WϵL) problems are decidable.

Context-free language is accepted by pushdown automata and only membership problems(WϵL), finiteness L=finite, emptiness L=Φ are decidable.

C is false as the set of all recursively enumerable languages (semi-decidable) is a STRICT super set of the set of all recursive languages (decidable).

D is false as the set of all recursively enumerable languages (set of all Turing machines) is an infinite but countable set. 

The decidability problem of Turing Machines can directly be determined using Rice’s theorem.

L1 is undecidable. According to Rice’s theorem, emptiness problem of Turing machine is undecidable.

L2 is decidable. This is because here we have to check whether the Turing machine reaches a particular step ‘q’ on a given input in finite steps or not. This is a decidable problem.

L3 is undecidable. There is no algorithm to computationally determine whether a Turing machine accepts a recursive language or not. Some Turing machines may accept recursive languages, while other may not.

L4 is also undecidable. According to Rice’s theorem, membership problem of Turing machine is undecidable.

Language A = {0ⁿ 1ⁿ 2ⁿ | n = 1, 2, 3, .........}

It is context sensitive language.

• This language cannot be a context free grammar. Since all 0, 1 and 2’s is equal in number.

Reason:

If we take the case of push down automata. Then first push all the 0’s into the stack then with each 1 pop all 0’s. But at the end, there is no symbol at the stack and input is remaining. So, it is not accepted by push down automata and cannot be context free grammar.

• It is accepted by linear bounded automata with the help of forward and backward movement of tape.

Machine that accepts Language A:

Statement S₁: There exists no algorithm for deciding if any two Turing machines M₁ and M₂ accept the same language.

Statement S₁ is correct.

Reason:

• The equivalence of two Turing machines is undecidable. 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.

Statement S₂: The problem of determining whether a Turing machine halts on any input is undecidable.

Statement S₂ is correct.

Reason:

It is related to the halting problem of Turing machine. A recursive language is decidable if there exists a Turing machine that either halts or rejects an input. But it cannot be decided whether it halts on any input. According to the concept of rice theorem, it is undecidable. 

Concept:

Post correspondence problem (PCP): It defines the un-decidability of strings. For the solution of PCP , we have to check whether there is a finite sequence (i₁,….i_p) with i_j belongs to {1,….,m} for j = 1,….p so that u_i1u_i2…u_ip = v_i1v_i2………v_ip.

Explanation:

CASE: for pair (A, B)

A = {001, 0011, 11, 101}

B = {01, 111, 111, 010}.

Consider it as: A₁ = 001, A₂ = 0011, A₃ = 11, A₄ = 101

B₁ = 01, B₂ = 111, B₃ = 111, B₄ = 010

Here A₃A₄ = B₂B₁ or A₃A₄ = B₃B₁

i.e. 11101= 11101

So, pair (A, B) has a PCP solution.

CASE 2: Consider pair (C, D)

C = {00, 001, 1000}

D = {0, 11, 011}.

There does not exist any such combination of strings which satisfy the property for PCP solution. So, Pair (C, D) does not have PCP solution.

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.

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.