Introduction to Algorithms
3rd Edition
ISBN: 9780262033848
Author: Thomas H. Cormen, Ronald L. Rivest, Charles E. Leiserson, Clifford Stein
Publisher: MIT Press
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Chapter 34.2, Problem 4E
Program Plan Intro
To prove that the class NPof languages is closed under union, intersection, concatenation and Kleene star and also discuss the closure of NP under complement.
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It is well-known that regular languages are closed under the following operations: union, complement, and intersection. It is also well-known that all finite languages are regular. For each of the operations, union, intersection and complement, are finite languages closed under them? If yes, prove it. Else provide couter examples.
Question 34
The correct statements are:
Decidable languages are closed under union.
Decidable languages are closed under intersection.
Decidable languages are closed under complement.
Semi-decidable languages are closed under complement.
Show using a cross-product construction that the class of regular languages is closed under set difference. You do not need an inductive proof, but you should convincingly explain why your construction works.
Chapter 34 Solutions
Introduction to Algorithms
Ch. 34.1 - Prob. 1ECh. 34.1 - Prob. 2ECh. 34.1 - Prob. 3ECh. 34.1 - Prob. 4ECh. 34.1 - Prob. 5ECh. 34.1 - Prob. 6ECh. 34.2 - Prob. 1ECh. 34.2 - Prob. 2ECh. 34.2 - Prob. 3ECh. 34.2 - Prob. 4E
Ch. 34.2 - Prob. 5ECh. 34.2 - Prob. 6ECh. 34.2 - Prob. 7ECh. 34.2 - Prob. 8ECh. 34.2 - Prob. 9ECh. 34.2 - Prob. 10ECh. 34.2 - Prob. 11ECh. 34.3 - Prob. 1ECh. 34.3 - Prob. 2ECh. 34.3 - Prob. 3ECh. 34.3 - Prob. 4ECh. 34.3 - Prob. 5ECh. 34.3 - Prob. 6ECh. 34.3 - Prob. 7ECh. 34.3 - Prob. 8ECh. 34.4 - Prob. 1ECh. 34.4 - Prob. 2ECh. 34.4 - Prob. 3ECh. 34.4 - Prob. 4ECh. 34.4 - Prob. 5ECh. 34.4 - Prob. 6ECh. 34.4 - Prob. 7ECh. 34.5 - Prob. 1ECh. 34.5 - Prob. 2ECh. 34.5 - Prob. 3ECh. 34.5 - Prob. 4ECh. 34.5 - Prob. 5ECh. 34.5 - Prob. 6ECh. 34.5 - Prob. 7ECh. 34.5 - Prob. 8ECh. 34 - Prob. 1PCh. 34 - Prob. 2PCh. 34 - Prob. 3PCh. 34 - Prob. 4P
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Similar questions
- Show using a cross-product construction that the class of regular languages is closed under intersection. You do not need an inductive proof, but you should convincingly explain why your construction works.arrow_forwardShow that language is not regular using pumping lemma. Proofarrow_forward1. Given L₁ = {a, abc, ac} and L₂ = {ab, c ba} are the languages over Σ = {a,b,c}, determine L₁ L₂ Obtain the regular expressions for the set (b, b, b,..), using the appropriate Kleene's closure.arrow_forward
- Present a language B such that B is complete for the class of Turing-decidable languges under ≤m reductions.arrow_forwardProve that upside-down regular expressions without complement operation cannot describe all regular languages. **You can prove this by finding a regular language and showing that an upside-down RE cannot describe it. You can pick a finite language, then show that if there is an upside-down RE that describes it then the language must contain some extra elements. For the last part, you can make a proof by induction over the RE.arrow_forwardProve that the following languages are not regular. You may use the pumping lemma and the closure of the class of regular languages under union, intersection, and complement. a. {0"1"0"| m, n2 0} Ab. (0"1"| m + n}arrow_forward
- . Prove that xn A yn is a non-regular language.Hint: as you know that according to pumping lemma version-I an bn is a non-regular.arrow_forward2. Consider the language DOUBLE-SAT, all of the boolean expressions with at least two satisfying assignments. (a) What would a possible DOULBE-SAT solution look like for an expression? (b) Is DOUBLE-SAT in NP?arrow_forwardProve that the language L1={} is not regular language with the pumping lemma.arrow_forward
- The correct statements are: Group of answer choices Regular languages are closed under union. Regular languages are closed under intersection. Regular languages are not closed under complement. Regular languages are not closed under set difference.arrow_forwardMake a conjecture whether or not it is regular and prove your assertion. When you prove a language is not regular, use the game with demon. Do not apply Pumping Lemma directly. Given hint: F is not regular but applying pumping lemma, i.e., the game with demon, will not be easy. You should use the closure property of regular languages along with the game with demon.arrow_forwardDefine the following operation, called SUB, on languages SUB (A,B) = {w∣w∈A and w∉B}. Prove that SUB is closed for regular languages.arrow_forward
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