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Chapter 0: Q20E (page 1)

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

a.a*b*b.aba*bc.a*鈭�b*d.aaa*e.危*a危*b危*a危*f.aba鈭�babg.(蔚鈭�a)bh.(a鈭�ba鈭�bb)危*

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

Let危={a,b} . For each k猢�1, let Ckbe the language consisting of all strings that contain an a exactly K places from the right-hand end.

ThusCk=危*补危k-1 . Describe an NFA with k+1states that recognizes Ckin terms of both a state diagram and a formal description.

In the fixed-point version of the recursion theorem (Theorem 6.8), let the transformation t be a function that interchanges the states qacceptandqreject in Turing machine descriptions. Give an example of a fixed point for t.

a). Let C be a context-free language and R be a regular language. Prove that the languageC鈭�Ris context free.

b). Let A= { w|w鈭�{a,b,c}*andwcontains equal numbers of a鈥�s,b鈥�s,andc鈥�s}. Use part(a) to show that A is not a CFL

Modify the proof of Theorem 3.16 to obtain Corollary 3.19, showing that a language is decidable if some nondeterministic Turing machine decides it. (You may assume the following theorem about trees. If every node in a tree has finitely many children and every branch of the tree has finitely many nodes, the tree itself has finitely many nodes.)

Question: Describe the error in the following 鈥減roof鈥� that 0*1*is not a regular language. (An error must exist because 0*1*is regular.) The proof is by contradiction. Assume that 0*1*is regular. Let p be the pumping length for localid="1662103472623" 0*1*given by the pumping lemma. Choose s to be the string 0p1p . You know that s is a member of 0*1*, but Example 1.73 shows that s cannot be pumped. Thus you have a contradiction. So 0*1* is not regular.
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