Prerequisite – Pushdown automata, Pushdown automata acceptance by final state

**Problem –** Design a non deterministic PDA for accepting the language L = { : n>=1} U { : 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 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:

= { a, z }

Where,

= 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’. If ‘b’ comes then

- For : Whenever ‘a’ comes, push it in stack and if ‘a’ comes again then also push it in the stack.
- 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

qf1, qf2 = Final state

= indicates pop operation

So, this is our required non deterministic PDA for accepting the language L ={ : n>=1} U { : n>=1}.

Attention reader! Don’t stop learning now. Get hold of all the important CS Theory concepts for SDE interviews with the CS Theory Course at a student-friendly price and become industry ready.

## Recommended Posts:

- NPDA for accepting the language L = {a
^{n}b^{m}c^{n}| m,n>=1} - NPDA for accepting the language L = {a
^{2m}b^{3m}| m ≥ 1} - NPDA for accepting the language L = {a
^{m}b^{n}c^{(m+n)}| m,n ≥ 1} - NPDA for accepting the language L = {a
^{n}b^{n}c^{m}| m,n>=1} - NPDA for accepting the language L = {a
^{n}b^{n}| n>=1} - NPDA for accepting the language L = {a
^{m}b^{(2m)}| m>=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
^{m}b^{(m+n)}c^{n}| m,n ≥ 1} - NPDA for accepting the language L = {a
^{m}b^{(2m+1)}| m ≥ 1} - 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^{m}| n,m ≥ 1 and n ≠ m} - 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^{n}| m,n ≥ 1} - NPDA for accepting the language L = {wwR | w ∈ (a,b)*}
- NPDA for the language L ={w∈ {a,b}*| w contains equal no. of a's and b's}
- 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} - NPDA for L = {0
^{i}1^{j}2^{k}| i==j or j==k ; i , j , k >= 1} - DFA machines accepting odd number of 0’s or/and even number of 1’s
- NFA machines accepting all strings that ends or not ends with substring 'ab'

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.