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Chapter 0: Q8P (page 1)

Show that EQTM'mEQTM'

Short Answer

Expert verified

Answer:

The proof is given below.

Step by step solution

01

Turing Machine

A Turing Machine is computational model concept that runs on the unrestricted grammar of Type-zero. It accepts recursive enumerable language. It comprises of an infinite tape length where read and write operation can be perform accordingly.

02

EQTM 'mEQTM

Let us understand about TM equality. Here,

EQTM={(<M>,<N>):M,Nare Turing Machine and language LM=LN}

EQTM={(<M)>,<N>):M,Nare Turing Machine and language LM=LN},

Now, EQTM'MEQTmmeansthatEQTMisnotmapreducibletoEQTM

We will first prove that is not Turing Recognizable.

  1. We know that A' mB: if and only if both A and B are Turing Recognizable or Not-Turing Recognizable.

  2. Now, in our prolem EQTM, is complement to EQTM.

    Thus, if is EQTMTuring Recognizable then EQTMwill not be Turing Recognizable and vice versa.

From (1) and (2), we can conclude that is not map reducible to EQTM, i.e,

EQTM'MEQTM

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Most popular questions from this chapter

Question:Consider the algorithm MINIMIZE, which takes a DFA as input and outputs DFA .

MINIMIZE = 鈥淥n input , where M=(Q,危,未,q0,A) is a DFA:

1.Remove all states of G that are unreachable from the start state.

2. Construct the following undirected graph G whose nodes are the states of .

3. Place an edge in G connecting every accept state with every non accept state. Add additional edges as follows.

4. Repeat until no new edges are added to G :

5. For every pair of distinct states q and r of and every a鈭�危 :

6. Add the edge (q,r) to G if 未q,a,未r,a is an edge of G .

7. For each state q,let[q] be the collection of statesq={r鈭�Q|noedge joins q and r in G }.

8.Form a new DFA M'=Q',危,未',q'0,A'where

Q'={[q]|q鈭�Q}(ifq=r,onlyoneofthemisinQ'),未'(q,a)=[未q,a]foreveryq鈭�Qanda鈭�危,q00=[q0],andA0={[q]|q鈭�A}

9. Output ( M')鈥�

a. Show that M and M' are equivalent.

b. Show that M0 is minimal鈥攖hat is, no DFA with fewer states recognizes the same language. You may use the result of Problem 1.52 without proof.

c. Show that MINIMIZE operates in polynomial time.

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*

Consider the undirected graph G=(V,E)whereV, the set of nodes, is{1,2,3,4}
andE, the set of edges, is{{1,2},{2,3},{1,3},{2,4},{1,4}}. Draw the graphG. What are the degrees of each node? Indicate a path from node 3 to node 4 on your drawing ofG.

Recall, in our discussion of the Church鈥揟uring thesis, that we introduced the language is a polynomial in several variables having an integral root}. We stated, but didn鈥檛 prove, thatis undecidable. In this problem, you are to prove a different property of鈥攏amely, thatDis -hard. A problem is -hard if all problems in are polynomial time reducible to it, even though it may not be inNPitself. So you must show that all problems in NPare polynomial time reducible to D .

Show that for any two languages AandB , a language J exists, where A鈮�TJandB鈮�TJ.

See all solutions

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