Are decidable languages closed under difference?
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 it accepts then accept and vice versa.
Are decidable languages closed under intersection?
Theorem 6: The set of Turing-decidable languages is closed under union, intersection, and complement.
What operations are recognizable languages closed under?
Turing recognizable languages are closed under union and intersection. Explanation: A recognizer of a language is a machine that recognizes that language. A decider of a language is a machine that decides that language.
Is Turing recognizable languages are closed under union?
(b) Show that the class of Turing-recognizable languages is closed under union. Answer: For any two Turing-recognizable languages L1 and L2, let M1 and M2, respectively, be TMs that recognize them. Then w is in L1 or in L2 (or both). If w ∈ L1, then M1 accepts w, so M will eventually accept w.
Is Sigma * decidable?
But Sigma* is a regular, decidable and context free language.
Is the family of recursively enumerable languages closed under intersection?
Recursively enumerable languages are also closed under intersection, concatenation, and Kleene star.
Is Sigma * Decidable?
Is P closed under union?
P is closed under union. For any two P-language L1 and L2, let M1 and M2 be the TMs that decide them in polynomial time. We construct a TM M’ that decides the union of L1 and L2 in polynomial time: M’= “On input ”: Since both stages take polynomial time, the algorithm runs in polynomial time.
Is all DFA decidable?
E(dfa) is a decidable language. Proof: A DFA accepts some string iff reaching an accept state from the start state by >traveling along the arrows of the DFA is possible.
Is L G regular Undecidable?
Regularity is decidable for DCFL, but it is undecidable for general Context-Free Languages.
Is recursively enumerable closed under complement?
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.
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.
Which is the closure of a decidable language?
✦Decidable languages are closed under ∪, °, *, ∩, and complement. ✦Example: Closure under ∪. ✦Need to show that union of 2 decidable L’s is also decidable. Let M1 be a decider for L1 and M2 a decider for L2 A decider M for L1 ∪L2: On input w: 1. Simulate M1 on w. If M1 accepts, then ACCEPT w.
What makes a language a semi-decidable language?
Definition:A language is called semi-decidable(or recognizable) if there exists an algorithm that accepts a given string if and only if the string belongs to that language. In case the string does not belong to the language, the algorithm either rejects it or runs forever. Clearly, any decidable language is recognizable.
How are the decidable languages closed under Union and intersection?
Proposition: The decidable languages are closed under union and intersection. Proof: Let L and M be languages that are decided by algorithms A and B respectively. In order to decide their union (or intersection) simply run A and B in parallel on the same given input string until they either accept or reject.
How is a Union decided in a decidable language?
In order to decide their union (or intersection) simply run A and B in parallel on the same given input string until they either accept or reject. The input string is accepted iff either one (or both, respectively) accepts it, and rejected otherwise. Proposition:The decidable languages are closed under complementation.