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Question:A Turing machine with stay put instead of left is similar to an ordinary Turing machine, but the transition function has the form

δ:Q×Γ-Q×Γ×{R,S}.

At each point, the machine can move its head right or let it stay in the same position. Show that this Turing machine variant is not equivalent to the usual version. What class of languages do these machines recognize?

Short Answer

Expert verified

Answer:

The statement is proved below.

Step by step solution

01

Step 1: Turing machine.

A Turing Machine is a mathematical model which consists of an infinite length tape divided into cells on which input is given. It consists of a head which reads the input tape. A state register stores the state of the Turing machine.

02

Language accept by the machine.

a).

A Turing machine with stay put instead of left is similar to an ordinary Turing machine, but the transition function has the form of,

δ:Q×Γ-Q×Γ×{R,S}.

At each point, the machine can move its head right or let it stay in the same position.

Because head of TM can't move left, it will behave like a finite automata.

Motion of head in right is equivalent to consuming next symbol, and staying at a place is equivalent to consuming an epsilon symbol. When we hit blank symbol check if last state was accept state of TM, if yes then our finite automata accepts. Otherwise automata will reject the input string.
Not giving a formal definition of this TM and finite automata, because it is quite easy to visualization . Hence, the statement is proved.

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Most popular questions from this chapter

In Theorem 3.21 we showed that a language is Turing-recognizable iff some enumerator enumerates it. Why didn’t we use the following simpler algorithm for the forward direction of the proof? As before, s1,s2,... is a list of all strings in *

E = “Ignore the input.

1. Repeat the following for i=1,2,3,...

2. Run M on si.

3. If it accepts, print out si.”

Show that a language is decidable if some enumerator enumerates the language in the standard string order.

Explain why the following is not a description of a legitimate Turing machine. Mbad= “On input (p), a polynomial over variablesx1,...,xk:

1. Try all possible settings of x1,...,xk:to integer values.

2. Evaluate p on all of these settings.

3. If any of these settings evaluates to 0, accept; otherwise, reject.”

Let a k - PDA be a pushdown automaton that has k stacks. Thus a 0 - PDA is an NFA and a 1 - PDA is a conventional PDA. You already know that 1 - PDAs are more powerful (recognize a larger class of languages) than 0 - PDAs.

a. Show that 2 - PDAs are more powerful than 1 - PDAs.

b. Show that 3 - PDAs are not more powerful than2 - PDAs. (Hint: Simulate a Turing machine tape with two stacks.

A Turing machine with doubly infinite tape is similar to an ordinary Turing machine, but its tape is infinite to the left as well as to the right. The tape is initially filled with blanks except for the portion that contains the input. Computation is defined as usual except that the head never encounters an end to the tape as it moves leftward. Show that this type of Turing machine recognizes the class of Turing-recognizable languages.

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