# NPDA for accepting the language L = {a^{(m+n)}b^{m}c^{n} | m,n ≥ 1}

Prerequisite – Pushdown automata, Pushdown automata acceptance by final state

**Problem –** Design a non deterministic PDA for accepting the language

The strings of given language will be:

L = {aabc, aaabcc, aaabbc, aaaabbcc, ......}

In each of the string, the total sum of the number of ‘b’ and ‘c’ is equal to the number of ‘a’s. And all c’s are come after ‘a’ and ‘b’.

**Explanation –**

Here, we need to maintain the order of a’s, b’s and c’s. That is, all the a’s are are coming first and then all the b’s are coming after that all the c’s. Thus, we need a stack along with the state diagram. The count of a’s, b’s and c’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 –**

As we want to design a NPDA, thus every time ‘a’ comes before ‘b’and ‘b’ comes before ‘c’.First we have to count number of a’s and that number should be equal to number of b’s. When all b’s are finished, then count number of a’s and that should be equal to number of c’s.

For all the ‘a’ we will push ‘a’ into the stack each time and then start popping them when ‘b’s are coming. After finishing the popping by ‘b’s and after ‘c’s are coming we will pop these ‘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) (q0, az) (q0, a, a) (q0, aa) (q0, b, a) (q1, ) (q1, b, a) (q1, ) (q1, c, a) (q2, ) (q2, c, a) (q2, ) (q2,, z) (qf, z)

Where, q0 = Initial state

qf = Final state

= indicates pop operation.

**State transition diagram**

**Note:** This language is similar to the language , but we use instead of .

## Recommended Posts:

- NPDA for accepting the language L = {a
^{n}b^{m}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
^{n}b^{n}c^{m}| m,n>=1} - NPDA for accepting the language L = {a
^{m}b^{(m+n)}c^{n}| m,n ≥ 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
^{m}b^{n}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^{(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
^{n}b^{(2n)}| n>=1} U {a^{n}b^{n}| n>=1} - NPDA for accepting the language L = {a
^{2m}b^{3m}| m ≥ 1} - NPDA for accepting the language L = {a
^{n}b^{n}| n>=1} - NPDA for accepting the language L = {a
^{m}b^{n}c^{(m+n)}| m,n ≥ 1} - NPDA for accepting the language L = {wwR | w ∈ (a,b)*}
- 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.