Recursively enumerable languages are also
closed
under intersection, concatenation, and Kleene star.
Are recursively enumerable languages closed under difference?
The class of recursively enumerable languages
is not closed under complementation
, because there are examples of recursively enumerable languages whose complement is not recursively enumerable.
Is the set of non re languages closed under union or intersection?
Intersection Both decidable and Turing recognizable languages are closed under intersection
. – Decidable languages are closed under complementation. To design a machine for the complement of a language L, we can simulate the machine for L on an input. … If L is decidable then it is Turing recognizable.
Is Re closed under intersection?
This can make things seem very nice and symmetric:
r.e. sets are closed under both union and intersection
. However, this is not the case once we start talking about infinite unions and intersections. Obviously the r.e. sets are not closed under arbitrary infinite unions/intersections.
Are recursively enumerable languages closed under complementation?
Recursive enumerable languages
are not closed under complementation
.It signifies that Y′ may/may not be recursive enumerable. But the answer will be Y′ is not recursive Enumerable. Why? If a language and its complement are both recursively enumerable, then both are recursive.
Are recursively enumerable languages infinite?
Proof: The set of strings is an infinite countable set. The set of languages is not countable because it is the powerset of the set of strings. Recursively enumerable languages are countable because TMs are countable. Therefore, recursively enumerable languages ⊂
all languages
.
Why is it called recursively enumerable?
Recursive Enumerable (RE) or Type -0 Language
It means
TM can loop forever for the strings which are not a part of the language
. RE languages are also called as Turing recognizable languages.
What is a language L called if L is a recursively enumerable language?
In mathematics, logic and computer science,
a formal language
is called recursively enumerable (also recognizable, partially decidable, semidecidable, Turing-acceptable or Turing-recognizable) if it is a recursively enumerable subset in the set of all possible words over the alphabet of the language, i.e., if there …
Are all enumerable languages decidable?
Yes
. In particular, recursive (decidable) languages are a subset of the recursively enumerable languages, so anything that’s not recursively enumerable isn’t recursive (decidable).
What is the difference between recursive and recursively enumerable?
The main difference is that in recursively enumerable language
the machine halts for input strings which are in language L
. but for input strings which are not in L, it may halt or may not halt. When we come to recursive language it always halt whether it is accepted by the machine or not.
Is undecidable closed under complement?
The
complement of an undecidable language is also undecidable
. Suppose a Turing machine can decide a language . Then we can construct a machine that decides its complement by running and then switching from “accept” to “not accept” and vice versa.
Is Sigma * decidable?
But Sigma
*
is
a regular, decidable and context free language
.
Is the union of two undecidable languages undecidable?
If L is the union of two undecidable languages, then L
is undecidable
. L is accepted by some NFA with states if and only if L is accepted by some DFA with states. If L ∈ P, then L is not regular.
Which among the following is not true for 2 way infinite TM?
6. Which among the following is not true for 2-way infinte TM? c)
Any computation that can be performed by 2-way infinite tape can also be performed by standard TM
. Explanation: All of the mentioned are correct statements for a two way infinite tape turing machine.
What is the intersection of CFL and regular language?
It is well known that the intersection of a context free language and a regular language is
context free
. This theorem is used in several proofs that certain languages are not context free. The usual proof of this theorem is a cross product construction of a PDA and a DFA.
What is co Turing recognizable?
Intuitively, if a language is co-Turing-recognizable, it means that
there is a computer program that, given a string not in the language, will eventually confirm that the string is not in the language
. It might loop infinitely if the string is indeed within the language, though.
