NPDA for accepting the language L = {a^mbⁿc^m+n | m,n ≥ 1}

The problem below require basic knowledge of Pushdown Automata.

Problem

Design a non deterministic PDA for accepting the language L = {a^m bⁿ c^m+n | m,n ≥ 1} for eg. ,

L = {abcc, aabccc, abbbcccc, aaabbccccc, ......}

In each of the string, the total sum of the number of 'a’ and 'b' is equal to the number of c’s. And all c's are come after 'a' and 'b'. 

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 are coming after that all the c's. Thus, we need a stack along with the state diagram. The count of a’s, b's and c’s is maintained by the stack. We will take 2 stack alphabets:

Г= { a, z }

Where, Г= 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’ and 'b' comes before 'c'. When ‘a’ comes then push it in stack and if again ‘a’ comes then also push it.

After that, when ‘b’ comes then push ‘a’ into the stack and if again ‘b’ comes then also push it.

And then when 'c' comes, 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, aa)
\delta(q1, b, a) \vdash(q1, aa)
\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