Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The 10 -step halting problem is to decide, given any collection of Turing machine instructions, together with any initial tape contents, whether that Turing machine will halt within 10 steps when started on that tape. Explain why the 10 -step halting problem is computable.

Short Answer

Expert verified
The 10-step halting problem is computable because it involves a finite simulation.

Step by step solution

01

Understanding the Halting Problem

The general halting problem is to determine if a given Turing machine halts on a given input. Alan Turing proved that this problem is undecidable for an arbitrary Turing machine and input. However, in this case, we are tasked with determining if a Turing machine halts but limited to within 10 steps.
02

Restriction to 10 Steps

The 10-step halting problem is a restricted version of the halting problem, where we only need to determine halting within a finite number of steps—specifically, 10. This finite boundary allows us to list all possible configurations of the Turing machine and its tape after every step, up to 10 times.
03

Generating All Possible Configurations

Given a Turing machine and initial tape configuration, simulate each step of the machine. Since there are only 10 steps, the number of configurations (states and tape contents) is finite. You can systematically explore each step one by one up to the 10th step.
04

Simulation Process

Simulate the Turing machine: starting from the initial state and tape, observe changes to the state, tape, and head position for each instruction executed. Keep track of each state and tape content at each of the 10 steps.
05

Checking for Halting

During simulation, check after each step: if the Turing machine enters a halting state within those 10 steps, then it halts within 10 steps. If, after simulating all 10 steps, none lead to a halting state, then the machine does not halt within 10 steps.
06

Deciding Computability

Since we can explicitly simulate up to 10 steps, examining all possible states and tape scenarios in this finite set, we can conclusively determine whether a halting state is reached. Hence, it is computable.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Turing Machine
A Turing machine is an abstract computational model introduced by Alan Turing. It is used to formalize the concept of computation and algorithms. Imagine a machine with an infinite tape divided into cells where each cell can hold a symbol. There is a head that reads and writes symbols on the tape, and it can move left or right.
The machine follows a set of rules or instructions that tell it what to do based on the current symbol it reads and its current state. The Turing machine changes states, writes a new symbol, and moves the tape head during each step, depending on the specified instructions.
  • This concept is powerful because it can simulate any algorithmic process given enough time and space.
  • Turing machines are a foundational model for understanding computation.
While Turing machines are theoretical constructs, they help us define what can be calculated and what processes are computable.
Computability
Computability refers to the ability of a problem to be solved by a computational process. In other words, if we can devise a series of steps that a Turing machine can follow to arrive at a solution, the problem is computable.
The 10-step halting problem is an example of a computable problem because it has clear boundaries. We are only looking to determine whether the machine halts in a fixed, finite number of steps—10 in this case.
  • We can simulate each step sequentially, checking each possible state and corresponding tape configuration.
  • For each configuration, we determine if the machine reaches a halting state within this limited frame.
Since there is a definitive process, and we are dealing with finite steps, even if complex, computability of the 10-step halting problem is assured.
Simulation
Simulation involves imitating the process of a Turing machine by executing its instructions step-by-step. For the 10-step halting problem, simulation becomes practical because we narrow the process down to just ten steps.
To simulate, start by initializing the Turing machine with its initial state and tape content. Then, execute the following:
  • At each step, observe and note the changes in the state, tape content, and head position.
  • Follow the machine's instructions to simulate its behavior systematically.
  • Track whether a halting state is reached in any of these steps.
This simulation essentially means running the Turing machine programmatically and observing its behavior, which makes it easier to determine if it halts within the specified steps.
Decidability
Decidability refers to whether a problem has an algorithm that provides a yes or no answer in a finite amount of time. The general halting problem is undecidable because no algorithm can determine for every Turing machine and input pair whether the machine halts.
However, when we restrict this to a certain number of steps, as in the 10-step halting problem, it becomes decidable. Here's why:
  • There is a finite number of possible states and tape configurations when limited to only 10 steps.
  • An exhaustive check of all possible configurations can be performed within this limit.
  • If a halting state is encountered within these steps, we can firmly conclude that the machine halts.
Thus, the 10-step limit makes it possible to determine decidability for this restricted version of the halting problem.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What is the significance of the unsolvability of the halting problem?

A Turing machine contains only the following instructions: $$ \begin{aligned} &(1,1,1,1, R) \\ &(1, b, 1,2, R) \end{aligned} $$ Can this machine ever reach the following configuration? Explain your answer. $$ \ldots b 01 b \ldots $$

The following BNF grammar defines a set of binary strings. \(\langle\) string \(\rangle::=\langle\) zero \(\rangle\langle\mathrm{A}\rangle\langle\) zero \(\rangle\) \(\langle A\rangle::=\langle\) zero \(\rangle\langle B\rangle\langle\) zero \(\rangle\) \(\langle\mathrm{B}\rangle::=\langle\) one \(\rangle\langle\mathrm{B}\rangle \mid<\) one \(>\) \(<\) zero \(>::=0\) : := 1 a. Describe the language defined by this grammar. b. Write a Turing machine to decide whether any binary string is a string in this language by halting with a blank tape if the string is in the language and halting with a nonblank tape if the string is not in the language.

Write a Turing machine that takes as input the unary representation of any two different numbers, separated by a blank, and halts with the representation of the larger of the two numbers on the tape. (Hint: You may need to use a "marker" symbol such as \(X\) or \(Y\) to replace temporarily any input symbols you have already processed and do not want to process again; at the end, your program must "clean up" any marker symbols.)

Write a Turing machine that operates on any binary string and changes it to a string of the same length with all \(1 \mathrm{~s}\). It should, for example, change the tape to $$ \ldots b 011010 b \ldots $$ to \(\ldots b 111111 b \ldots\) However, you must write instructions that allow your Turing machine to work on any binary string, not just the one shown here.

See all solutions

Recommended explanations on Computer Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free