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A Turing machine with left reset is similar to an ordinary Turing machine, but the transition function has the form

未 : Q 脳 螕鈭掆啋Q 脳 螕 脳 {R, RESET}.

If 未(q, a) = (r, b, RESET), when the machine is in state q reading an a, the machine鈥檚 head jumps to the left-hand end of the tape after it writes b on the tape 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 left reset recognize the class of Turing-recognizable languages.

Short Answer

Expert verified

Answer :

This statement is proved.

Step by step solution

01

Turing machine.

A Turing Machine (TM) 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

Left reset in a Turing machine.

A Turing machine with left reset is similar to an ordinary Turing machine, but the transition function has the form:

:QQ{leftarrow}.

The difference is that 鈥渓eft arrow鈥 takes the head back to the first square, in one move. So, the head can either move one square to the right, with 鈥溾啋鈥, or return all the way to the beginning of the tape with 鈥渓eft arrow鈥. That is,

(q,a)=(p,b,leftarrow)

Means that from state q and symbol a, we go to state is overwritten with b, and the head moves all the way back to the first square of the tape.A Turing machine with left reset is similar to an ordinary Turing machine, but the transition function has the form

:Q{R,RESET}

If (q,a)=(r,b,RESET),when the machine is in state q reading an the machine鈥檚 head jumps to the left-hand end of the tape after it writes on the tape and enters state Note that these machines do not have the usual ability to move the head one symbol left

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Most popular questions from this chapter

In the following solitaire game, you are given anmm board. On each of itsm2 positions lies either a blue stone, a red stone, or nothing at all. You play by removing stones from the board until each column contains only stones of a single color and each row contains at least one stone. You win if you achieve this objective. Winning may or may not be possible, depending upon the initial configuration. LetSOLITAIRE={<G>|G is a winnable game configuration}. Prove that SOLITAIREis NP-complete.

For each part, give a relation that satisfies the condition.

  1. Reflexive and symmetric but not transitive
  2. Reflexive and transitive but not symmetric
  3. Symmetric and transitive but not reflexive

Examine the following formal descriptions of sets so that you understand which members they contain. Write a short informal English description of each set.

  1. {1,3,5,7,...}
  2. {...,-4,-2,0,2,4,...}
  3. {n|n=2mfor someminN}
  4. {n|n=2mfor someminN, andn=3kfor somekinN}
  5. {w|wis a string of0sand1sandwequals the reverse ofw}
  6. {n|nis an integer andn=n+1}

This problem is inspired by the single-player game Minesweeper, generalized to an arbitrary graph. Let Gbe an undirected graph, where each node either contains a single, hidden mine or is empty. The player chooses nodes, one by one. If the player chooses a node containing a mine, the player loses. If the player chooses an empty node, the player learns the number of neighboring nodes containing mines. (A neighboring node is one connected to the chosen node by an edge.) The player wins if and when all empty nodes have been so chosen.

In the mine consistency problem, you are given a graphG along with numbers labeling some of G鈥檚 nodes. You must determine whether a placement of mines on the remaining nodes is possible, so that any node v that is labeled m has exactly m neighboring nodes containing mines. Formulate this problem as a language and show that it isNPcomplete.

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 .

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