# NPDA for accepting the language L = {a^{2m}b^{3m} | m ≥ 1}

Prerequisite – Pushdown automata, Pushdown automata acceptance by final state

**Problem –** Design a non deterministic PDA for accepting the language L = {: m ≥ 1}, i.e.,

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

In each of the string, for every 2 ‘a’s there is 3 ‘b’ after it.

**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. 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 –**

As we want to design a NPDA, thus every time ‘a’ comes before ‘b’. We will push three ‘a’s into the stack for two consecutive two ‘a’s 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 and similarly we perform this alternatively, i.e.,

For two a’s we push three ‘a’

For four b’s we push six ‘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 –**

(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

## Recommended Posts:

- NPDA for accepting the language L = {a
^{i}b^{j}c^{k}d^{l}| i==k or j==l,i>=1,j>=1} - NPDA for accepting the language L = {a
^{n}b^{n}| n>=1} - NPDA for accepting the language L = {a
^{n}b^{(2n)}| n>=1} U {a^{n}b^{n}| n>=1} - NPDA for accepting the language L = {a
^{n}b^{n}c^{m}| m,n>=1} - NPDA for accepting the language L = {a
^{m}b^{(2m)}| m>=1} - NPDA for accepting the language L = {a
^{m}b^{(2m+1)}| m ≥ 1} - NPDA for accepting the language L = {a
^{m}b^{(m+n)}c^{n}| m,n ≥ 1} - NPDA for accepting the language L = {a
^{m}b^{n}c^{p}d^{q}| m+n=p+q ; m,n,p,q>=1} - NPDA for accepting the language L = {a
^{n}b^{m}c^{n}| m,n>=1} - NPDA for accepting the language L = {a
^{n}b^{m}| n,m ≥ 1 and n ≠ m} - NPDA for accepting the language L = {a
^{m}b^{n}c^{n}| m,n ≥ 1} - NPDA for accepting the language L = {a
^{(m+n)}b^{m}c^{n}| m,n ≥ 1} - NPDA for accepting the language L = {a
^{m}b^{n}c^{(m+n)}| m,n ≥ 1} - DFA for accepting the language L = { a
^{n}b^{m}| n+m=even } - DFA for accepting the language L = {a
^{n}b^{m}| n+m=odd}

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.