# Construct a Turing machine for L = {a^{i}b^{j}c^{k} | i < j < k or i > j > k}

Prerequisite – Turing Machine

The language L = {a^{i}b^{j}c^{k} | i < j < k or i > j > k} is same as the union of two languages L1={a^{i}b^{j}c^{k} | i < j < k } and L2={a^{i}b^{j}c^{k} | i > j > 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 either
- 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’.
- 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’

Assume that string is ending with ‘$’.

**Examples:**

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

**Tape Representation:**

**Approach:**

- Comparing two elements by making two element as a single element.
- After that the elements which are treated as single element are compared again .
- If |First| is greater than |(Second, Third)| and |Second| greater than |Third|, then it is accepted.
- If |Third| is greater than |(First, Second)| and |First| greater than |Second|, then it is accepted.
- 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. Ignore Z move left and goto step-11.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.**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.**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 B is found ignore it and move left and goto step-11.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 Z found ignore it move left to state Q6.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
- 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.
- 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 B is found ignore it and move left to Q11
- 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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