NPDA for accepting the language L = {a^2mb^3m|m>=1}

Before learning this, you should know about pushdown automata and how they accept inputs using final states.

Problem

Design a non deterministic PDA for accepting the language L = {a^2mb^3m| m ≥ 1}, i.e.,

L = {aabbb, aaaabbbbbb, aaaaaabbbbbbbbb, aaaaaaaabbbbbbbbbbbb, ......}

In each of the string, for every 2 'a's there is 3 'b'. For every 2 a's we have to pop 3 b's in the stack.

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. Here, we have 3 'b's for every 2 'a's. 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

As we want to design a NPDA, thus every time ‘a’ comes before ‘b’. We will push three ‘a’s into the stack for every 'a's out of the two and again for the next two ‘a’s, we will push three 'a' into the stack. That is, for the first 'a' we will do nothing only state will change and for the next 'a' we will do the pushing operation push 3 a's and similarly we perform this alternatively, i.e., For two a’s we push three ‘a’. After that, when ‘b’ 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:

δ(q₀ , a, z) ⊢ (q1, z) [ Indicates no operation only state change ]

δ(q₁ , a, z) ⊢ (q2, aaaz) [ Indicates push operation for alternate 'a']

δ(q₂ , a, aaaz) ⊢ (q₁, aaaz) [ Indicates no operation only state change]

δ(q₁ ,a, aaaz) ⊢ (q₂, aaaa) [ Indicates push operation for alternate 'a']

δ(q₂ ,b, a) ⊢ (q₃ ,Є) [Indicates pop operation ]

δ(q₃ ,b, a) ⊢ (q_{3 ,}Є) [Indicates pop operation ]

δ(q₃ ,Є, z) ⊢ (q_{f ,}Є) [Indicates final state ]

Where, q₀ = Initial state q_f = Final state

є = indicates pop operation

Correct transition diagram-