Finite Automata without output - Review

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Finite Automata Without Output

NFA with and without Epsilon Transition

Constructing DFA from NFA with Epsilon Transition

Steps to convert NFA with Epsilon move to DFA: Step 1 : Take ∈ closure for the beginning state of NFA as beginning state of DFA. Step 2 : Find the states that can be traversed from the present for each input symbol (union of transition value and their closures for each states of NFA present in current state of DFA). Step 3 : If any new state is found take it as current state and repeat step 2. Step 4 : Do repeat Step 2 and Step 3 until no new state present in DFA transition table. Step 5 : Mark the states of DFA which contains final state of NFA as final states of DFA.

NFA with Epsilon Transition

If any finite automata contains ε (null) move or transition, then that finite automata is called NFA with ∈ moves

Example of NFA with Epsilon Move:

Equivalence and Minimization of Finite Automata

Main topic

Introduction to Finite Automata

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DFA Designing

DFA for Languages

Example: Asked in Previous year University Exam: Design a deterministic finite automata(DFA) for accepting the language L = {a^n b^{m} | m+n=even}

DFA for strings

Example: Asked in Previous year University Exam: Construct a DFA machine over input alphabet \sum_= {0, 1}, that accept even number of 1's

Subtopic

Example: Asked in Previous year University Exam: Construct a DFA machine over input alphabet Sigma= {0, 1}, that accepts odd number of 0's and even number of 1's

Difference between NFA and DFA

In DFA no transition for empty input i.e for epsilon In NFA transition for empty input i.e. for epsilon

Difference in Transition function DFA transition function: Q x Sigma->Q NFA transition Function: Q x Sigma->2^Q

String Acceptance by FA

String Acceptance by NFA

Testing Lets take one input string abab First input is a, so from state A we will go to state B Second input is b, so from state B we will go to state C Third input is a, so from state C we will go to state C itself Fourth input is b, so from state C we will go to state C itself(final state)

String Acceptance by DFA

Testing Lets take one input string abab First input is a, so from state A we will go to state B Second input is b, so from state B we will go to state C Third input is a, so from state C we will go to state C itself Fourth input is b, so from state C we will go to state C itself(final state)

A string is said to be accepted by a Finite Automata when it is accepted by final state starting from initial state while consuming one input during each transition

Conversion from NFA to DFA

While converting NFA to DFA, if NFA has n states than the corresponding DFA will have 2^n states. DFA is not designed for all 2^n states but only for those states reachable from initial state.

Example: Convert the following NFA into DFA Solution: Transition table is

Example: Previous year University Exam Question Convert the following NFA into DFA

Representation of Finite Automata

It can be represented in two ways: 1. Transition/State diagram 2. Transition/State table

Transition Table:

Non-Deterministic Finite Automata (NFA/NDFA)

NFA is also of 5-tuples (Q, Sigma, Del, q0, qf) Where, 4-tuples are same as DFA but the difference is in Del only. Del- Transition function which maps Q x Sigma-> 2^Q, For example- Del (q0, a) = {q0,q1}

NFA Example:

Deterministic Finite Automata (DFA)

Tuples

Q-f

qf- Final state or set of final states enclosed in circle or marked by a special symbol in Transition table and enclosed in double circle in Transition diagram

Q-0

q0- Initial/starting state, denoted by Arrow in Transition table and circle with arrow in transition diagram

Delta

Del- Transition function which maps Q x Sigma->Q i.e. there is one transition for each input from one state to another. For exam Del(q0,a)= q1

Sigma

Sigma- Finite set of nonempty inputs denoted by a....z/0.....9

Q

Where. Q- Finite nonempty set of states denoted by q0,q1,q2....../A,B........Z,