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## Is the class P closed under complement?

**P is closed under complement**. For any P-language L, let M be the TM that decides it in polynomial time. We construct a TM M’ that decides the complement of L in polynomial time: M’= “On input <w>: 1.

## What operations is P closed under?

We show that P is closed under the **star operation** by dynamic programming. Let A be any language in P, and let M be the TM deciding A in polynomial time.

### Regular Languages Closed Under Complement Proof

### Images related to the topicRegular Languages Closed Under Complement Proof

## Is the complement of a language in P also in P?

**Every language in P has its complement also in P**, and therefore in NP.

## Is P closed under intersection?

If x /∈ A then there is NO proof that x ∈ A. **The class P is closed under union, intersection, concatenation, and ∗**.

## Is the class of recognizable languages closed under complement?

– Turing recognizable languages are **not closed under complement**.

## Is Turing recognizable closed under concatenation?

– 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. – **Turing recognizable languages are not closed under complement**.

## What is NP problem?

That is, co-NP is the set of decision problems where there exists a polynomial p(n) and a polynomial-time bounded Turing machine M such that for every instance x, x is a no-instance if and only if: for some possible certificate c of length bounded by p(n), the Turing machine M accepts the pair (x, c).

## Is linear on?

**An algorithm is said to take linear time, or O(n) time, if its time complexity is O(n)**. Informally, this means that the running time increases at most linearly with the size of the input.

## What is Kleene star automata?

Definition − The Kleene star, ∑*, is **a unary operator on a set of symbols or strings, ∑, that gives the infinite set of all possible strings of all possible lengths over ∑ including λ**.

## How do you show a language in P?

To prove that a given language is in P: **Construct an algorithm that decides the language**. This algorithm may “call” any other algorithms from the textbook, lectures, class handouts, or homework assignments (but you should cite the appropriate reference).

## Is a set of languages whose complement is in NP?

Since language L is NP-Complete all NP and NP Complete problems can be reduced to L in polynomial time. And it is given that complement of language L is in NP. Hence **complement of all NP problems is in NP**.

## What is the complement of NP?

The complement of an NP-complete problem is **co-NP-complete**; if any co-NP-complete problem is in a class C that is closed under polynomial-time reductions, then every co-NP⊆C; NP is closed under polynomial-time reductions.

### Regular Languages Closed Under Complement Proof

### Images related to the topicRegular Languages Closed Under Complement Proof

## Is NP closed under concatenation?

**NP is closed under concatenation**. For any two NP languages L1 and L2, let M1,M2 be the NTMs that decide them in polynomial time. We construct a NTM N1 that decides L1L2 in polynomial time: N1 = “On input string w: 1.

## Is NP closed under star?

**NP is closed under Kleene star**. Given a NTM N which decides L 2 NP in nondeterministic polynomial time, the following NTM N0 decides L⇤ in nondeterministic polynomial time.

## Is NP closed under union?

**NP is closed under union, intersection, and concatenation**; but is not known to be closed under complement.

## Is complement of recognizable language recognizable?

Theorem: **A language is decidable iff both it and its complement are recognizable**. Proof: Surely, a decidable language is recognizable. Moreover, if a language is decidable, then so is its complement, and hence that complement is recognizable.

## Is the complement of a recognizable language also recognizable?

Flipping the accept and reject states generates a TM to decide the complement of this language. Not all Recognizable languages are closed under complement. **If the complement of a recognizable language is also recognizable, the language is, in fact, decidable.**

## Is the complement of a Turing recognizable language Turing recognizable?

That is, a TM that accepts any input . **A Language L is co-Turing recognizable if the complement of L, , is Turing recognizable**. A language L Is decidable, if and only if it is Turing recognizable and co-Turing recognizable. If L is decidable so is its complement, .

## What are Turing 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.

## Are recursive languages closed under complement?

**Recursively enumerable languages are closed under complement**. Proof. Same as previous machine. This fails because M only needs to halt if w in L(M) – doesn’t have to say “no”.

## Is Re closed under union?

A language is recursive if it is the set of strings accepted by some TM that halts on every input. For example, any regular language is recursive. Fact. (a) **The set of r.e. languages is closed un- der union and intersection**.

## What is DSA Class P?

The class P **consists of those problems that are solvable in polynomial time**, i.e. these problems can be solved in time O(n^{k}) in worst-case, where k is constant. These problems are called tractable, while others are called intractable or superpolynomial.

### 16. Complexity: P, NP, NP-completeness, Reductions

### Images related to the topic16. Complexity: P, NP, NP-completeness, Reductions

## What is P and NP class problems?

NP is set of problems that can be solved by a Non-deterministic Turing Machine in Polynomial time. P is subset of NP (any problem that can be solved by deterministic machine in polynomial time can also be solved by non-deterministic machine in polynomial time) but P≠NP.

## What is P and NP class in automata?

Step 1 − If a problem is in class P, it is nothing but we can find a solution to that type of problem in polynomial time. Step 2 − If a problem is in class NP, it is nothing but that we can verify a possible solution in polynomial time.

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