Exercise 9 1 (P) (a) Show that P is closed under union, complement, and concatenation (b) The complexity class coP contains all languages L whose
exercise
Prove that the class P is closed under intersection, complement and concatenation Solution: • Intersection Let L1,L2 ∈ P We want to show that L1 ∩ L2 ∈ P
poly
P is closed under union, intersection, complement, and concatenation Additionally, polynomials are closed under addition and multiplication Closure under
p and np
only if there exist positive constants c and n0 such that n2 ≤ cn log2 n for all n ≥ n0, which holds if (b) Show that P is closed under concatenation Answer:
hwsoln
(a) (1 point ) Prove that the complexity class P is closed under complement (That is, show that if L ∈ P then Lc ∈ P, where Lc
homework
Prove that the class NP is closed under union, intersection, concatenation and Kleene star Is the class NP closed also under complement? Solution:
t sol
(a) Demonstrate that the class P is closed under union, intersection, complement, concatenation and Kleene star (b) Prove that the class NP is closed under
chap
Problem 2: Closure of P Part 1: Prove that P is closed under union, intersection, and concatenation That is, if L1,L2 ∈ P, prove that each of the following are
Ex
This means that P is closed under complement Based on the previous proof, we also have ¯L ∈ NP According to the definition of coNP, L ∈ coNP Thus,
ps sol
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
Prove that the class P is closed under intersection complement and concatenation. Solution: • Intersection. Let L1
Since P is closed under complement (Le)e is in P and we can conclude that L ∈ P. Exercise 9.2 (Reduction). Given an undirected graph G := 〈G
NP is closed under union intersection
Prove that the class NP is closed under union intersection
closed under complement. Symmetric difference. Let LI Lz be two sets in CK Three relativizations have been given
23 февр. 2001 г. If P=NP then since P is closed under complement
5 нояб. 2013 г. If P = NP then NP is also closed under complementation. In other words
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
Exercise 9.1 (P). (a) Show that P is closed under union complement
Prove that the class P is closed under intersection complement and concatenation. Solution: • Intersection. Let L1
(c) Show that P is closed under complementation. Answer: Suppose that language L1 ? P so there is a polynomial-time TM M1 that decides L1. A Turing machine M2
This means that P is closed under complement. Based on the previous proof we also have ¯L ? NP. According to the definition of coNP
P is closed under union intersection
Prove that the class NP is closed under union intersection
4 oct. 2017 The converse problem is in NP and PSPACE is stable by complement ... only if M does not accept w (PSPACE is closed under complementation).
the regular languages are closed under complementation. However we ... In particular
Prove that the class P is closed under intersection complement and concatenation Solution: Intersection Let L 1;L 2 2P We want to show that L 1 L 2 2P Because L 1 2P then there exists a TM M 1 with time complexity O(nk 1) for some constant k 1 Because L 2 2P then there exists a TM M 2 with time complexity O(nk 2) for some constant k 2
P vs NP Not much is known unfortunately Can think of NP as the ability to appreciate a solution P as the ability to produce one P NP Dont even know if NP closed under complement i e NP = co-NP? Does L NP imply ? NP?
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 : 1 Run M on w 2 If M accepts reject If it rejects accept ” M’ decides the complement of L Since M runs in polynomial time M’ also
Polynomials are closed under many oper-ations (e g addition multiplication) hence P is closed under many operations (e g concatention) Classes likeDT IM E(n)and evenDT IM E(O(n)) are thought to not be closed under concatenation and many other operations (We do not know ifthey are ) Theorem 2 LetL2P ThenL 2P Proof
24 2 1 5 P is closed under complementation Proposition 24 2 3 Decision problem X is in P if and only if X is in P Proof: (A) If X is in P let A be a polynomial time algorithm for X (B) Construct polynomial time algorithm A? for X as follows: given input x A? runs A on x and if A accepts x A? rejects x and if A rejects x then A
of P from PSPACE for which polynomial-time reductions are appropriate We now consider po-tential separations of classes within P and from NP For these complexity classes the notion of polynomial-time mapping reductions is too coarse since it does not distinguish L from P We need a ?ner notion of reduction De?nition 1 1
Is NP closed under complement?
Also, if NP is not closed under complement, then P != NP. The converse of "If P=NP, then NP is closed under complement" would be "If NP is closed under complement, then P=NP". However, this is not known to be true: it is possible that NP is closed under complement but is still different from P. Thanks for the correction @DavidRicherby.
How to prove that class P is closed under intersection complement and concatenation?
Prove that the class P is closed under intersection, complement and concatenation. Solution: Intersection. Let L 1;L 22P. We want to show that L 1L 22P. Because L 12P then there exists a TM M 1with time complexity O(nk 1) for some constant k 1. Because L 22P then there exists a TM M 2with time complexity O(nk 2) for some constant k 2.
What is the complementary event of P?
Complementary event of P will be (1 - Probability of event occurred). Login Study Materials BYJU'S Answer NCERT Solutions NCERT Solutions For Class 12 NCERT Solutions For Class 12 Physics
How do you know if a class is closed under complement?
A class is said to be closed under complement if the complement of any problem in the class is still in the class. Because there are Turing reductions from every problem to its complement, any class which is closed under Turing reductions is closed under complement. Any class which is closed under complement is equal to its complement class.
CSE105 Homework 3