# Pushdown Automata Acceptance by Final State

We have discussed Pushdown Automata (PDA) and its acceptance by empty stack article. Now, in this article, we will discuss how PDA can accept a CFL based on final state. Given a PDA P as:

P = (Q, Σ, Γ, δ, q0, Z, F)

The language accepted by P is the set of all strings consuming which PDA can move from initial state to final state irrespective to any symbol left on the stack which can be depicted as:

L(P) = {w |(q0, w, Z) =>(qf, ɛ, s)}

Here, from start state q0 and stack symbol Z, final state qf ɛ F is reached when input w is consumed. The stack can contain string s which is irrelevant as final state is reached and w will be accepted.

**Example:** Define the pushdown automata for language {a^nb^n | n > 0} using final state.

**Solution:** M = where Q = {q0, q1, q2, q3} and ∑ = {a, b} and Γ = { A, Z } and F={q3} and δ is given by:

δ( q0, a, Z ) = { ( q1, AZ ) } δ( q1, a, A) = { ( q1, AA ) } δ( q1, b, A) = { ( q2, ɛ) } δ( q2, b, A) = { ( q2, ɛ) } δ( q2, ɛ, Z) = { ( q3, Z) }

Let us see how this automata works for aaabbb:

**Explanation:** Initially, the state of automata is q0 and symbol on stack is Z and the input is aaabbb as shown in row 0. On reading a (shown in bold in row 1), the state will be changed to q1 and it will push symbol A on stack. On next a (shown in row 2), it will push another symbol A on stack and remain in state q1. After reading 3 a’s, the stack will be AAAZ with A on the top.

After reading b (as shown in row 4), it will pop A and move to state q2 and stack will be AAZ. When all b’s are read, the state will be q2 and stack will be Z. In row 7, on input symbol ɛ and Z on stack, it will move to q3. As final state q3 has been reached after processing input, the string will be accepted.

This type of acceptance is known as *acceptance by final state*.

Next we will see how this automata works for aab:

As we can see in row 4, the input has been processed and PDA is in state q2 which is non-final state, the string aab will not be accepted.

Let us discuss question based on this:

**Que-1.** Consider the transition diagram of a PDA given below with input alphabet ∑ = {a, b}and stack alphabet Γ = {X, Z}. Z is the initial stack symbol. Let L denote the language accepted by the PDA. (GATE-CS-2016)

**Solution:** We first label the state of given PDA as:

Next, the given PDA P can be written as:

Q = {q0, q1, q2} and ∑ = {a, b} And Γ = {A, Z} and F={q0,q2} and δ is given by : δ( q0, a, Z ) = {( q0, XZ)} δ( q0, a, X) = {( q0, XX )} δ( q0, b, X) = {( q1, ɛ)} δ( q1, b, X) = {( q1, ɛ)} δ( q1, ɛ, Z) = {( q2, Z)}

As we can see, q0 is initial as well as final state, ɛ will be accepted. For every a, X is pushed onto stack and PDA remains in final state. Therefore, any number of a’s can be accepted by PDA.

If input contains b, X is popped from stack for every b. Then PDA is moved to final state if stack becomes empty after processing input (δ( q1, ɛ, Z) = {( q2, Z)}). Therefore, number of b must be equal to number of a if they exist.

As there is only one move for a given state and input, the PDA is deterministic. So, correct option is (D).

GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details

## Recommended Posts:

- Difference between Pushdown Automata and Finite Automata
- Introduction of Pushdown Automata
- Construct Pushdown automata for L = {a
^{(2*m)}c^{(4*n)}d^{n}b^{m}| m,n ≥ 0} - Construct Pushdown automata for L = {0
^{n}1^{m}2^{(n+m)}| m,n ≥ 0} - Construct Pushdown automata for L = {0
^{(n+m)}1^{m}2^{n}| m, n ≥ 0} - Construct Pushdown automata for L = {0
^{n}1^{m}2^{m}3^{n}| m,n ≥ 0} - Construct Pushdown automata for L = {0
^{m}1^{(n+m)}2^{n}| m,n ≥ 0} - Construct Pushdown Automata for given languages
- Detailed Study of PushDown Automata
- Construct Pushdown Automata for all length palindrome
- Automata Theory | Set 8
- Automata Theory | Set 9
- Automata Theory | Set 10
- Applications of various Automata
- Automata Theory | Set 5

This article is contributed by **Sonal Tuteja**. 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.