# NPDA for accepting the language L = {a2mb3m | m ≥ 1}

Last Updated : 28 Aug, 2019
Prerequisite – Pushdown automata, Pushdown automata acceptance by final state Problem â€“ Design a non deterministic PDA for accepting the language L = {a2mb3m | m ≥ 1}, i.e.,
L = {aabbb, aaaabbbbbb, aaaaaabbbbbbbbb, aaaaaaaabbbbbbbbbbbb, ......}
In each of the string, for every 2 ‘a’s there is 3 ‘b’. Explanation â€“ Here, we need to maintain the order of aâ€™s and bâ€™s.That is, all the aâ€™s are coming first and then all the bâ€™s are coming. Thus, we need a stack along with the state diagram. The count of aâ€™s and bâ€™s is maintained by the stack. Here, we have 3 ‘b’s for every 2 ‘a’s. We will take 2 stack alphabets:
 = {a, z}
Where,  = set of all the stack alphabet
z = stack start symbol 
Approach used in the construction of PDA â€“ As we want to design a NPDA, thus every time â€˜aâ€™ comes before â€˜bâ€™. We will push three â€˜aâ€™s into the stack for two consecutive two ‘a’s and again for the next two â€˜aâ€™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 and similarly we perform this alternatively, i.e., For two aâ€™s we push three â€˜aâ€™ For four bâ€™s we push six â€˜aâ€™ After that, when â€˜bâ€™ comes then pop one â€˜aâ€™ 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 â€“
(q0, a, z) (q1, z)     [ Indicates no operation only state change ]
(q1, a, z) (q2, aaaz)  [ Indicates push operation for alternate 'a']
(q2, a, aaaz) (q1, aaaz) [ Indicates no operation only state change ]
(q1, a, aaaz) (q2, aaaa) [ Indicates push operation for alternate 'a']
(q2, b, a) (q3,  )  [Indicates pop operation ]
(q3, b, a) (q3,  )  [Indicates pop operation ]
(q3, , z) (qf, z )

Where, q0 = Initial state qf = Final state = indicates pop operation Correct transition diagram-

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