NPDA for accepting the language L = {aⁿb²ⁿ| n>=1} U {aⁿbⁿ| n>=1}

To understand this question, you should first be familiar with pushdown automata and their final state acceptance mechanism.

Problem

Design a non deterministic PDA for accepting the language L = {aⁿ b²ⁿ : n>=1} U {aⁿ bⁿ : n>=1}, i.e.,

L = {abb, aabbbb, aaabbbbbb, aaaabbbbbbbb, ......} U {ab, aabb, aaabbb, aaaabbbb, ......}

In each string, the number of a's are followed by double number of b's or the number of a’s are followed by equal number of b’s. 

Explanation

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

Γ= { a, z }

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

Approach used in the construction of PDA

In designing a NPDA, for every 'a' comes before 'b'.

• For : Whenever ‘a’ comes, push it in stack and if ‘a’ comes again then also push it in the stack. Then for every b pop a.
• For : Whenever ‘a’ comes, push 'a' two time in stack and if ‘a’ comes again then do the same. When ‘b’ comes (remember b comes after 'a') then pop one ‘a’ from the stack each time.

So that the stack becomes empty. If stack is empty then we can say that the string is accepted by the PDA. 

Stack transition functions

(q0, a, z) (q1, az)
(q0, a, z) (q3, aaz)
(q1, a, a) (q1, aa)
(q1, b, a) (q2, )
(q2, b, a) (q2, )
(q2, , z) (qf1, z)
(q3 a, a) (q3, aaa)
(q3, b, a) (q4, )
(q4, b, a) (q4, )
(q4, , z) (qf2, z)

Where, q₀ = Initial state , q_f1, q_f2 = Final state, ϵ = indicates pop operation. 👁 Image
 So, this is our required non deterministic PDA for accepting the language L = {aⁿ b²ⁿ : n>=1} U {aⁿ bⁿ : n>=1}.