Turing Machines And Undecidability Practice Quiz - 1


8 Questions MCQ Test RRB JE for Computer Science Engineering | Turing Machines And Undecidability Practice Quiz - 1


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QUESTION: 1

The number of states required to automate the last question i.e. {a,b}*{aba}{a,b}* using finite automata:

Solution:
QUESTION: 2

Which of the following problems are decidable?

Solution:
QUESTION: 3

Which of the following are decidable?

I. Whether the intersection of two regular languages is infinite
II. Whether a given context-free language is regular
III. Whether two push-down automata accept the same language
IV. Whether a given grammar is context-free

Solution:

(A) Intersection of two regular languages is regular and checking if a regular language is infinite is decidable.
(B) Deciding regularity of a context free language is undecidable. We check if L(CFG) contains any string with length between n and 2n−1 , where n is the pumping lemma constant. If so, L(CFG) is infinite otherwise it is finite.
(C) Equality problem is undecidable for all languages except in case of finite automata i.e. for regular languages.
(D) We have to check if the grammar obeys the rules of CFG. If, it obeys such rules then it is decidable.  Thus, option (B) is correct. Please comment below if you find anything wrong in the above post.

QUESTION: 4

Which of the following problems is undecidable?

Solution:

A set is closed under an operation means when we operate an element of that set with that operator we get an element from that set. 
Here, CFG generates a CFL and set of all CFLs is the set. But ambiguity is not an operation and hence we can never say that CFG is closed under such operation. 
Only ambiguity problem for CFGs are undecidable. 
 
Thus, option (B) is correct. 
 
Please comment below if you find anything wrong in the above post.

QUESTION: 5

Which of the following statements is/are FALSE?

1. For every non-deterministic Turing machine, there exists an equivalent deterministic Turing machine.
2. Turing recognizable languages are closed under union and complementation.
3. Turing decidable languages are closed under intersection and complementation.
4. Turing recognizable languages are closed under union and intersection.

Solution:

A recognizer of a language is a machine that recognizes that language. A decider of a language is a machine that decides that language. Both types of machine halt in the Accept state on strings that are in the language A Decider also halts if the string is not in the language A Recogizer MAY or MAY NOT halt on strings that are not in the language On all input: A Decider MUST halt (in Accept or Reject state) A Recogizer MAY or MAY NOT halt on some strings (Q: Which ones?) A language is Turing-decidable (or decidable) if some Turing machine decides it. Aka Recursive Language. A language is Turing-recognizable if some Turing machine recognizes it. Aka Recursively Enumerable Language.
Recursive (Turing Decidable) languages are closed under following Kleene star, concatenation, union, intersection, complement and set difference. Recursively enumerable language are closed under Kleene star, concatenation, union, intersection. They are NOT closed under complement or set difference.

QUESTION: 6

Let L1 be a recursive language. Let L2 and L3 be languages that are recursively enumerable but not recursive.
Which of the following statements is not necessarily true? 

Solution:

A) Always True (Recursively enumerable - Recursive) is Recursively enumerable
B) Not always true L1 - L3 = L1 intersection (Complement L3) L1 is recursive , L3 is recursively enumerable but not recursive Recursively enumerable languages are NOT closed under complement.
C) and D) Always true Recursively enumerable languages are closed under intersection and union.

QUESTION: 7

Which of the following is true for the language 

Solution:

Turing machine can be designed for ap using the concept of ‘Sieve of Eratosthenes’.
Suppose we are given an integer ‘n’ and we want to find out all the prime numbers less than or equal to ‘n’. We repeat the following steps : We find the smallest number in the list, declare it prime and eliminate all the multiples of that number from the list. We keep doing this until each element has been declared prime or eliminated from the list.
Now, if p = 0 or p = 1, we reject the input. Else, we replace the first and the last ‘a’ with symbol $.
In the above steps, what we do is we find the first non-black symbol from the left. Let this symbol occur at position ‘x’. Suppose ‘x’ is a prime number. If this non-blank symbol is $, input string will be accepted. But, if the symbol is ‘a’, we mark it as a* and replace all the multiples of ‘x’ with the symbol ‘blank’. If at the end, symbol $ is replaced with 'blank’, we reject the input string (because p will be multiple of some ‘x’ in that case). Else, we go back and repeat the steps.

Thus, the input is neither regular nor context-free, but is accepted by a Turing machine.

Please comment below if you find anything wrong in the above post.

QUESTION: 8

If L and L' are recursively enumerable, then L is

Solution:

If L is recursively enumerable, then L' is recursively enumerable if and only if L is also recursive.