Question

A Turing Machine with Left Reset is similar to an ordinary Turing Machine, but the transition function has the form:

δ: Q x τ x {R,RESET}

If δ(q,a) = {r,b,RESET} when the turing machine is in state q reading an a, the machine's head jumps to the left-hand end of the tape after it writes b on thetape and enters state r.Note that these machines do not have the usual ability to move the head one symbol left.Show that turing machines with leftreset with left reset recognize the class of Turing-recognizable languages.

Answer 1

d: Q x t x {R,RESET}

Answer 2

Given a Turing-recognizable language, by definition there exists a standard Turing

Machine M1 that recognizes it. We need to convert M1 into a TM with left reset; call it M2. Since

it can’t move left and often gets confused, M2 will keep track of where its head is supposed to be

by marking that square with a dot. After each transition is complete, there is exactly one dot on

the tape, but during the transition procedure, there can be more than one dot on the tape.

For rules in M1 of the form x → y, R, M2 performs the following procedure:

Step 1. Read an x, overwrite it with a dotted y (this dot will help it find the correct final position

and will be erased later), and move right.

Step 2. Mark the current square (one square to the right of the original square) with a dot, but

then it must move (because it can only move right or reset left, not stay in one position), so reset

left.

Step 3. To get its head back to the correct place, move right until you read a dotted square

(this will be the square with the dotted y). Erase the dot and move right. Now the head is over the

correct square, and this square already has a dot on it (it was marked in Step 2).

For rules in M1 of the form x → y, L, M2 performs the following procedure:

Stage 1 (during which M2 marks the eventual destination square, i.e., the square to the left of

the original square, with a dot)

Step 1. Read an x, overwrite it with a dotted y, then reset left. If this square (the square at the

left end of the tape) is dotted, then it must be the dotted y, so quit (because M1 would also have

failed to move left, so we have successfully simulated the behavior of M1). If this square is not

dotted, mark this square with a dot; the move right that must accompany this marking is part of

Step 2.

Step 2. Move right. Read the current square; if it is dotted, it must be the dotted y, so erase

the dot over the y, reset left, and go to Stage 2, Step 1 (because the destination square has been

successfully marked). If this square is not dotted, mark it with a dot (it is our new candidate for

the destination square) and reset left.

Step 3. Move right until you read a dotted square. (Note that this first dotted square keeps track

of the extent of your search thus far.) Erase the dot and move right (onto the current candidate

square). Go to step 2.

Stage 2 (during which M2 positions its head on the destination square by first dotting the square

to the left of the destination square)

Step 1. (Note that we have arrived here from Stage 1, Step 2, so the current head position is

the square at the left end of the tape.) Read the current square; if it is dotted, it is the destination

square, so reset left and quit. If it is not dotted, mark it with a dot (this is our first candidate for the

square-to-the-left-of-the-destination-square, or STTLOTDS for short); again, the move right that

must accompany this marking is part of Step 2.

Step 2. Move right. Read the current square; if it is dotted, it is the destination square, so

reset left and go to step 4. If it is not dotted, mark it with a dot (this is the new candidate for the

STTLOTDS) and reset left.

Step 3. Move right until you read a dotted square (this is the square marking the extent of our

search so far). Erase the dot and move right. Go to step 2.

Step 4. Move right until you read a dotted square (this is the STTLOTDS). Erase the dot, move

right, and quit.