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NPDA for accepting the language L = {anbm | n,m ≥ 1 and n ≠ m}

  • Difficulty Level : Medium
  • Last Updated : 08 Feb, 2019

Prerequisite – Pushdown automata, Pushdown automata acceptance by final state

Problem – Design a non deterministic PDA for accepting the language L = \{a^n b^m | n, m \geq 1 \text{and n!=m}\}, i.e.,

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L = {aab, abb, aaab, abbb, aaaab, aaabb, aabbb, abbbb ......} 

In each of the string, the number of a’s are followed by unequal 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 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:

\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. After that, when ‘b’ comes then pop one ‘a’ from the stack each time. So, at the end if the stack becomes empty and b is coming or string is end and stack is not 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, \epsilon )
\delta(q1, b, a) \vdash (q1, \epsilon )
\delta(q1, \epsilon, a) \vdash (qf, a)   
\delta(q1, b, z) \vdash (qf, z)   

Where, q0 = Initial state
qf = Final state
\epsilon = indicates pop operation

State Transition Diagram –

So, this is our required non deterministic PDA for accepting the language L = \{a^n b^m | n, m \geq 1 \text{and n!=m}\}.

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