Turing Machine as Comparator in Automata Theory

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Read this chapter to learn how a Turing Machine can be used as a comparator. We will see how the input can be put inside the tape and check the corresponding inputs for comparison. We will see the steps in detail with examples for a clear understanding.

Turing Machine as a Comparator

A Turing Machine can perform various operations, including comparing two numbers to determine if they are equal, less than, or greater than each other.

This concept is similar to how Turing Machines handle addition and subtraction. Let us see how, we will use unary numbers to as input to a Turing Machine to work as a comparator. The basic logic and structure of the Turing Machine remain consistent with those used in other operations.

Unary Numbers

As we know, unary numbers represent quantities using repeated symbols, typically the digit "1." For instance, the number 3 is represented as "111" in unary, and 2 is represented as "11". We will use a "0" as a separator between the two numbers we want to compare.

Example: Comparing Two Numbers

Setting up the Problem − Let us start by comparing two numbers using a Turing Machine. We will work with the following example −

• Number A: "111" (which represents 3)
• Number B: "111" (which also represents 3)

In this case, we have a "0" separating the two numbers: "1110111".

The Process of Comparison − The Turing Machine begins by scanning the tape from the leftmost "1". Here is a step-by-step breakdown of the process −

• Convert the First "1" in A to "X" − The machine reads the first "1" in the left-hand side number (A) and converts it into an "X". This indicates that this digit has been processed.
• Move to the First "1" in B and Convert It to "X" − After processing the first "1" in A, the machine moves right, skipping any "1"s until it encounters the separator "0". It then continues moving right to find the first "1" in the right-hand side number (B) and converts it into an "X".
• Repeat the Process − The machine moves back to the left to find the next unprocessed "1" in A, converts it into an "X", and repeats the process for the next unprocessed "1" in B. This cycle continues until all "1"s in both numbers have been converted to "X".

Handling Different Outcomes

There are three final states: for equal case, for first one greater than the second, and for the second one greater than the first one.

Case 1: A Equal to B − If both numbers have the same number of "1"s, the machine will convert all "1"s in both A and B into "X". After processing, if the machine encounters only "X"s and no additional "1"s in either number, it concludes that A is equal to B.

Case 2: A Less Than B − If A has fewer "1"s than B, the machine will encounter an extra "1" in B after all "1"s in A have been processed. This extra "1" indicates that A is less than B.

Case 3: A Greater Than B − If A has more "1"s than B, the machine will continue processing "1"s in A even after all "1"s in B have been converted to "X". When the machine encounters a blank space (indicating the end of B) while A still has unprocessed "1"s, it concludes that A is greater than B.

Transition Diagram Construction

Once the logic is understood, constructing the transition diagram for the Turing Machine as a comparator becomes straightforward. The states and transitions follow the process described above, with each state representing a step in the comparison process.

State Transitions Diagram

The state transitions diagram looks like this −

Transition Table

Here is its Transition Table

• Q0 (Initial State) − On reading "1", convert it to "X", and move right. Continue moving right until encountering "0" (separator).
• Q1 − On reading "0", move right. On reading "1" in B, convert it to "X" and move left.
• Q2 − Move left to find the next unprocessed "1" in A. On finding "X" or "0", continue the cycle until all "1"s are processed.
• Q3 (Final Check) − After converting all "1"s to "X", move right to check if there are any remaining "1"s in B. If none are found, A equals B. If an extra "1" is found, A is less than B.

Example Execution

Let us apply this logic to our initial example:

Step 1 − Convert the first "1" in A ("1110111") to "X", resulting in "X110111".

Step 2 − Move right, skip the "0", convert the first "1" in B to "X", resulting in "X110X11".

Step 3 − Move left, find the next "1" in A, convert it to "X", and repeat the process.

$$\mathrm{1110111}$$

$$\mathrm{X110111}$$

$$\mathrm{X110X11}$$

$$\mathrm{XX10X11}$$

$$\mathrm{XX10XX1}$$

$$\mathrm{XXX0XX1}$$

$$\mathrm{XXX0XXX}$$

After completing these steps, both numbers will be fully converted to "X", confirming that A equals B.

Conclusion

In this chapter, we explained how Turing Machines can be applied for comparator. We have seen with examples along with state transition diagrams and state transition tables on how we can make the comparison operations.

There are three final states: for equal case, for first one greater than the second, and for the second one greater than the first one. This can handle all these three operations with the Turing Machine we designed.