Construct a Turing machine for L = {aibjck | i < j < k or i > j > k}

26 Aug, 2021
Prerequisite – Turing MachineThe language L = {aibjck | i < j < k or i > j > k} is same as the union of two languages L1={aibjck | i < j < k } and L2={aibjck | 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.
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 ‘$’.
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.
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
anikaseth98
GATE CS
Theory of Computation & Automata
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