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a. Write a Turing machine that, when run on the tape \(\ldots\) b 11111 b... produces an output tape of \(\ldots\) b 11110 b... b. Write a Turing machine that, when run on any tape containing a unary string, changes the rightmost 1 to 0 and then halts. (If your solution to Exercise 15 a was sufficiently general, you will not have to change it here.)

Short Answer

Expert verified
The Turing machine changes the last '1' to '0', resulting in b 11110 b.

Step by step solution

01

Define States and Symbols

For the Turing machine, we need to define the states and symbols. Let's use two states: q0 (initial state) and q_accept (halting state). We have symbols: 'b' (blank), '1', and '0'.
02

Transition for Replacing the Last '1' with '0'

From state q0, scan the tape until the first 'b' moving right. Then transition to the left when '1' is encountered and change the last '1' to '0'. Use the following transitions: 1. (q0, 1) -> (q0, 1, R) [stay in q0 moving right on '1'] 2. (q0, b) -> (q1, b, L) [move back to the rightmost '1'] 3. (q1, 1) -> (q_accept, 0, R) [replace '1' with '0' and halt]
03

Apply to a Specific Input Sequence

Place the tape head at the beginning of the non-blank sequence \(...\)b 11111 b... Moving right, q0 traverses '1's until it reaches the blank 'b'. It then shifts to the previous position in state q1 and changes the last '1' to '0', resulting in \(...\)b 11110 b..., and moves to q_accept state.

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Key Concepts

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

Halting State
In the world of Turing machines, a halting state is a crucial concept. It's the state that signifies when the machine should stop processing the input and halt its operations. Think of it as the "off switch" for the machine.

The purpose of having a halting state, such as 'q_accept' in our exercise, is to provide clarity that the machine has completed its task. When reaching this state, no more transitions occur, and it effectively ends the computation. This allows users to know when the desired output has been produced.

Without a halting state, a Turing machine might continue operating indefinitely, leaving us with incomplete results or an endless loop. That's why designing the transition to a halting state carefully is essential. It ensures that the machine finishes its task precisely at the expected output. In our exercise, reaching 'q_accept' after altering the unary string is the indication that the Turing machine has successfully completed its operation.
State Transitions
State transitions are the backbone of a Turing machine. They dictate how the machine moves between different states based on the input it reads.

For our exercise, these transitions are well-defined:
  • From 'q0' reading a '1', the machine stays in 'q0' and continues moving right.
  • Once a blank 'b' is encountered, it transitions to 'q1' and moves left.
  • If a '1' is found in 'q1', it transitions to 'q_accept', changes the '1' to '0', and moves right, marking the task's completion.
These transitions are pivotal because they map out the machine's journey from starting to ending operations. They provide the instruction manual that tells the machine exactly what to do step by step.

Understanding these rules is like understanding the gears in a clock. Each rule is a small but crucial piece of how the Turing machine functions. The goal is always to reach that halting state with the desired change made on the tape.
Unary String Manipulation
Unary string manipulation involves operations on strings composed entirely of the same symbol, typically '1's, such as the sequence 11111. In Turing machines, manipulating such unary strings involves adjusting these '1's precisely according to given rules.

In our exercise, the task is to replace the rightmost '1' with a '0' and halt. This form of manipulation is simple yet demonstrates the Turing machine's ability to perform specific tasks on string algorithms. Here is the breakdown:

  • The machine reads the tape from left to right until the space, or blank 'b', is reached.
  • At the first blank space, it shifts one position left.
  • The rightmost '1' encountered is then changed to a '0'.
This type of string operation, though basic, underscores the role of a Turing machine in processing and transforming data systematically. Unary string manipulations serve as great introductory exercises to grasp the fundamental operation of Turing machines, highlighting both their power and their structured simplicity.

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Most popular questions from this chapter

The following BNF grammar defines a set of binary strings. \(\langle\) string \(\rangle::=\langle\) zero \(\rangle\langle\) one \(\rangle \mid\langle\) zero \(\rangle\langle\) string \(\rangle\langle\) one \(\rangle\) ::= 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 to perform a unary decrement (the opposite of an increment). Assume that \(n>0\).

Say an automobile manufacturer designs a new car using a sophisticated and detailed computer simulation, but no prototype vehicles, and the automobile is later found to have a defect. Do you think the manufacturer is accountable? Is the manufacturer accountable if it builds prototypes that do not reveal the defect, but it does not do a simulation?

Write a Turing machine that operates on any string of \(1 \mathrm{~s}\) and changes it to a string of alternating \(1 \mathrm{~s}\) and \(0 \mathrm{~s}\).

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.)

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