# Construct a Turing Machine for a language L = {aibjck | i<j<k or i>j>k} ∩ {aibjck | i>j>k or i>j>k}

Prerequisite – Turing Machine
The language L = {aibjck | i < j < k or i > j > k} ∩ {aibjck | i > j > k or i > j > k} is same as the languages L={aibjck | j > max(i, k)}
In this language, every string of ‘a’, ‘b’ and ‘c’ have certain number of a’s, then certain number of b’s and then certain number of c’s.

The condition is as follows

• Count of 1st symbols should be atleast 1. ‘b’ and ‘c’ can have thereafter be as many but count of a is less than count of ‘b’ and count of ‘b’ is less than count of ‘c’.
• And, The count of 3rd symbols should be atleast 1. ‘a’ and ‘b’ can have thereafter be as many but count of c is less than count of ‘b’ and count of ‘b’ is less than count of ‘a’
• On Conclusion: Count of 2nd symbol must be greater than max of count of 1st and 3rd symbol and count of 1st and 3rd symbol may or may not be equal
• Assume that string is ending with ‘\$’.

Examples:

```Input: a a a b b c
Here a = 3, b = 2, c = 1
Output: NOT ACCEPTED
// |a| < |b| > |c|

Input: a b b b c c
Here a = 1, b = 3, c = 2
Output: ACCEPTED

Input: a a a b b b b c c c
Here a = 3, b = 4, c = 3
Output: ACCEPTED ```

Tape Representation: Approach:

1. Camparing two elements by making two element as a single element.
2. If |Second| is greater than max(|First|, |Third|), then it is accepted.
3. Else it is not accepted.

Steps:

• Step-1: Convert A into X and move right and goto step 2. If Y is found ignore it and move right to step-5.
• Step-2: Keep ignoring A and Y and move towards right. Convert D into Y and move right and goto step-3.
• Step-3: Keep ignoring D and Z and move towards right. If C is found make it Z and move left to step 4.If B is found ignore it and move left and goto step-8.
• Step-4: Keep ignoring Z, A, Y and D and move towards left. If X is found ignore it and move right and goto step-1.
• Step-5: Keep ignoring Y and move towards right. If D is found make it Y and move right to step-6.
• Step-6: Keep ignoring D and Z and move towards right. Convert C into Z and move left and goto step-7. If B Ignore it move left and goto step-11.
• Step-7: Keep ignoring D and Z and move towards left. If Y is found ignore it and move right and goto step-5.
• Step-8: Keep ignoring D, Y and A and move towards left. Ignore X move right and goto step-9. If D ignore and move right and goto step-11
• Step-9: Convert A into X and move right and goto step-10.
• Step-10: Keep ignoring Y and A and move towards right. If D make it Y and move right and goto step-8.
• Step-11: Stop the Machine (String is accepted).

State transition diagram : Here, Q0 shows the initial state and Q1, Q2, Q3, Q4, Q5, Q6, Q8, Q9, Q10 shows the transition state and Q7 and Q11 shows the final state. A, C, D are the variables used and R, L shows right and left.

Explanation:

• Using Q0, when A is found make it X and go to right and to state Q1.And, when Y is found ignore it and go to right and to state Q4
• On the state Q1, ignore all A and Y and goto right. If D found make it Y and goto right into next state Q2.
• In Q2, ignore all D, Z and move right. If B found ignore it, move left and goto the state Q4, If C found make it Z move left and to Q3.
• In Q3 state, ignore all Z, D, Y, A and move left. If X found ignore it move right to Q0.
• In Q4, ignore all Y and move right. If D is found make it Y and move to right to Q5.
• In Q5 state, ignore all D, Z and move right. If C found make it Z move left to state Q6.If B is found ignore move left to state Q7
• In Q6, ignore all D, Z and move left. If Y found ignore it and move right to state Q4.
• If Q7 state is reached it will produced the result of acceptance of string.
• In Q8, ignore all A, Y, D and move left. If X found ignore it move right to state Q9. If B is found ignore it and move right to Q11.
• In Q9 state, if A found make it X move right to state Q10
• In Q10, ignore all A, Y and move right. If D found make it Y and move right to state Q8.
• If Q11 state is reached it will produced the result of acceptance of string.

Note: For comparison of |A|, |D|, |C|, the concept of Turing Machine as Comparator is used

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