Common

When a string is accepted by a PDA?

When a string is accepted by a PDA?

In final state acceptability, a PDA accepts a string when, after reading the entire string, the PDA is in a final state. From the starting state, we can make moves that end up in a final state with any stack values. The stack values are irrelevant as long as we end up in a final state.

How many states are there in the PDA which recognizes the language containing 0’s followed by equal no of 1’s?

a*b* examples

state input stack
2 bb$ c
3 b$ ε
3 $ ε
3 ε ε

What are the languages accepted by PDA?

The languages which can be accepted by PDA are called context-free languages (CFL), denoted by LCF. Diagrammatically, a PDA is a finite state automaton (see Fig. 5.1), with memories (push-down stacks). In this chapter, we shall study PDA and their associated languages, context-free languages LCF.

Which of the following regex allows strings?

Discussion Forum

Que. Which of the following regular expression allows strings on {a,b}* with length n where n is a multiple of 4.
b. (bbbb+aaaa)*
c. ((a+b)(a+b)(a+b)(a+b))*
d. None of the mentioned
Answer:((a+b)(a+b)(a+b)(a+b))*
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What is the difference between acceptance of string in PDA with null stack or final state?

Push down automaton, In a final state PDA accepts a string, when and after reading the entire string, the PDA is in a final state. The stack values in PDA can be irrelevant as long as the values ends up in final state. Empty stack accepts a string when after reading the entire string, the PDA has emptied its stack.

What is addition in pushdown automata as compare to finite state automata?

Pushdown automata has the additional stack for storing long sequence of alphabets. Finite Automata doesn’t has any space to store input alphabets.

Which data structure is used by pushdown automata?

Explanation: A push down automata uses a stack to carry out its operations.

How do I know what language my PDA is?

The language acceptable by empty stack can be defined as: N(PDA) = {w | (q0, w, Z) ⊢* (p, ε, ε), q ∈ Q}…The δ will be:

  1. δ(q0, 0, Z) = (q1, 0Z)
  2. δ(q1, 0, 0) = (q1, 0)
  3. δ(q1, 0, Z) = (q0, ε) (indicate that one 0 and one 1 is already read, so simply read the second 0)
  4. δ(q1, 1, 0) = (q1, ε)