Question

Write a Turing machine that takes any unary string of an even number of \(1 \mathrm{~s}\) and halts with the first half of the string changed to 0 s. (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.)

Short answer

Execute the Turing machine to change the first half of an even length unary string from '1's to '0's, cleaning markers in the process.

Step-by-step solution

Understand the Problem

You need to write a Turing machine for a unary string input. If the string has an even number of '1's, change the first half into '0's and ensure the output does not contain marker symbols.

Define the Turing Machine States and Alphabet

Let the states be \( Q = \{q_0, q_1, q_2, q_3, q_{accept}, q_{reject}\} \). The alphabet contains the symbols \( \Sigma = \{1, 0, X, Y, \_\} \), where \( \_ \) represents the blank symbol.

Initialize and Mark Processing

Start at the leftmost '1'. Switch state from \( q_0 \) to \( q_1 \) as you replace the first '1' with 'X'. Move right to mark it processed and find the next unmatched '1'.

Check for Even Parity

Set state \( q_1 \) to search for the second '1'. Convert this '1' to state 'Y' and move to the right, transitioning to \( q_2 \). If there is no next '1', that means odd parity, move to \( q_{reject} \).

Continue Traversing

In state \( q_2 \), head back left to locate the first 'Y'. Upon finding it, move towards the next unprocessed '1'. Transition back to \( q_0 \) from \( q_2 \) and iterate marking 'X' and 'Y'.

Clean Up the Markers

While in state \( q_2 \) and no unprocessed '1's are left, convert 'Y' with '0' and return markers 'X' back to '0'. When only 'X's or '0's remain, finalize the transition to \( q_{accept} \).

Transition Table and Execution

Based on above steps, implement a transition table specifying rules for each state-action pair. Execute the process ensuring even '1's convert accordingly and halt in the accept state.

Key concepts

Unary String

A unary string is a simple form of numeric representation using only one type of symbol; in this case, the symbol '1'. Imagine you want to represent the number 5. In unary, this is simply consecutive instances of '1', i.e., '11111'. Unary strings are uncomplicated, making them suitable when simplicity is desired or for certain theoretical computer science explorations.

In the given problem, the Turing machine processes unary strings that have an even number of '1's. This particular requirement ensures that when these '1's are split into two equal parts, they can be processed symmetrically, such as converting the first half to another symbol '0'. Understanding and working effectively with unary strings often serves as an introduction to more complex number systems and data representations.

Marker Symbols

Marker symbols in a Turing machine serve as temporary placeholders. They indicate that a particular part of the input has been processed without altering the initial input entirely. In the context of this problem, two marker symbols, say 'X' and 'Y', are introduced.

• 'X' is used to mark the first '1' converted into a temporary state.
• 'Y' marks the second '1' in pair processing, signifying this portion is temporarily processed, ensuring no repetition occurs.

Using markers allows the Turing machine to remember where it has been, without requiring complex memory setups. They are crucial in intermediate steps, as they help transition the machine's state between finding '1's and ultimately replacing them with '0's only if the parity condition of even '1's is met. A "clean up" step is necessary, which replaces these markers back to the final desired symbol, reinforcing the importance of efficient marker management.

Transition Table

The transition table is the heart of a Turing machine's operations. It's effectively a set of instructions guiding the machine on what actions to perform in different states. Each entry in the table corresponds to a rule dictating:

• the current state the machine is in,
• the symbol currently being read,
• the symbol that should replace it,
• the next state to transition into, and finally,
• which direction to move (left or right).

In our exercise, the transition table will include rules for:

• marking the initial '1's as 'X' or 'Y',
• checking parity by testing for more '1's in alternate passes,
• converting processed marker symbols back to '0's after all necessary changes are done,
• fully directing the machine to halt in the accept state if the string meets the even criteria, or reject otherwise.

The intricacy of crafting a transition table lies in ensuring each interaction handles all possible contingencies and transitions smoothly until the task's logic concludes. This ensures the Turing machine not only performs the desired action but does so in a logically sound manner.