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Chapter 0: Q22P (page 1)

Show that A is Turing-recognizable Am鈮�ATM

Short Answer

Expert verified

It is proved that A is Turing Recognizable.

Step by step solution

01

Introduction of Turing Recognizable

A language L is said to be Turing Recognizable if and only if there exist any Turing Machine (TM) which recognize it i.e., TM halt and accept strings belong to language L and will reject or not halt on the input strings that doesn鈥檛 belong to language L .

02

To show A is Turing recognizable

Before we proceed, we know that ATM is Turing Recognizable.

Let us assume that A is Turing Recognizable, this means we can construct a Turing Machine M for A , such that LM=A.

Assume f is Mapping Reduction Function which is defined as:
fw=M,w

So,w鈭�Aif and only ifM,w鈭�ATM.

So, if, Am鈮�ATMthen A must be Turing Recognizable as ATM is Turing Recognizable

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Most popular questions from this chapter

Show that EQTM is recognizable by a Turing machine with an oracle for ATM.

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

Let X=M,wM is a single-tape TM that never modifies the portion of the tape that contains the input w. Is X decidable? Prove your answer.

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.

Show that EQTM'mEQTM'
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