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For each of the following languages, give two strings that are members and two strings that are not members—a total of four strings for each part. Assume the Σ=a,balpha-alphabet in all parts.

a.a*b*b.aba*bc.a*b*d.aaa*e.Σ*aΣ*bΣ*aΣ*f.abababg.(εa)bh.(ababb)Σ*

Short Answer

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Step by step solution

01

To Concern the Two Strings

There are two strings which are concerned with members, and some of the two strings are concerned with non-members. In the question, there are some calculations we have to solve, and they're related to the step-2 answer also.

02

To Explain the Given Expression

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Most popular questions from this chapter

Question: Let S={{M}|MisaTMandLM={M'}.Show that S nor S' neither is Turing recognizable.

Let X be the set {1,2,3,4,5}and Y be the set {6,7,8,9,10}.The unary function f:XYand the binary function g:X×YYare described in the following tables.

g12345f(n)67676 g123456789101010101010789106789106789106789106

a. What is the value of f(2)?
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c. What is the value of g (2, 10) ?
d. What are the range and domain ofg?
e. What is the value ofg(4, f (4))?

Show that the set of incompressible strings contains no infinite subset that is Turing-recognizable.

Give informal English descriptions of PDAs for the languages in Exercise 2.6

Give context-free grammars generating the following languages.

a. The set of strings over the alphabet a,bwith more a's than b's

b. The complement of the language anbnn0.

c. w#xwRis a substring of x for w,x 0,1*

d. localid="1662105288591" x1#x2#...#xkk1,each xilocalid="1662105304877" a,b*,and for some i and j ,localid="1662105320570" xi=xjR

In the following solitaire game, you are given anm×m board. On each of itsm2 positions lies either a blue stone, a red stone, or nothing at all. You play by removing stones from the board until each column contains only stones of a single color and each row contains at least one stone. You win if you achieve this objective. Winning may or may not be possible, depending upon the initial configuration. LetSOLITAIRE={<G>|G is a winnable game configuration}. Prove that SOLITAIREis NP-complete.

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