NPDA for accepting the language L = {aⁿb^m cⁿ | m,n>=1}

Before attempting this problem, you should have a working knowledge of Pushdown Automata concepts.

Problem

The problem can be solved if you have prior knowledge about NPDA.

Design a non deterministic PDA for accepting the language L = {aⁿ b^m cⁿ | m, n >= 1}, i.e.,

L = { abc, abbc, abbbc, aabbcc, aaabccc, aaaabbbcccc, ...... }

In each of the string, the number of a's is equal to number of c's and the number of b's is independent of the number of a's and c's.  

Explanation

Here, we need to maintain the order of a’s, b's and c’s. That is, all the a's  are coming first and then all the b's and then c's are coming. Thus, we need a stack along with the state diagram. The count of a’s and c’s is maintained by the stack. The number of a's is exactly equal to the number of c's We will take 2 stack alphabets:

\Gamma = { a, z }

Where, \Gamma  = set of all the stack alphabet z = stack start symbol 

Approach used in the construction of PDA

As we want to design a NPDA, thus every time 'a' comes before 'b'. When ‘a’ comes then push it in stack and if again ‘a’ comes then also push it.

For 'b', we will do nothing in the stack only change the state in state diagram and keep skipping b's.

When ‘c’ comes then pop one 'a' from the stack each time .So, at the end if the stack becomes empty then we can say that the string is accepted by the PDA.

Stack Transition Functions

\delta(q0, a, z) \vdash (q0, az)
\delta(q0, a, a) \vdash (q0, aa)
\delta(q0, b, a) \vdash (q1, a)
\delta(q1, b, a) \vdash (q1, a)
\delta(q1, c, a) \vdash (q2, \epsilon )
\delta(q2, c, a) \vdash (q2, \epsilon )
\delta(q2, \epsilon, z) \vdash (qf, z )

Where, q0 = Initial state qf = Final state \epsilon  = indicates pop operation  So, this is our required non deterministic PDA for accepting the language L = {aⁿ b^m cⁿ | m, n >= 1}.