Nondeterministic finite automata
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Nondeterministic Finite Automata. CS 130: Theory of Computation HMU textbook, Chapter 2 (Sec 2.3 & 2.5). NFAs: Nondeterministic Finite Automata. Same as a DFA, except:

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Nondeterministic finite automata

Nondeterministic Finite Automata

CS 130: Theory of Computation

HMU textbook, Chapter 2(Sec 2.3 & 2.5)


Nfas nondeterministic finite automata

NFAs:Nondeterministic Finite Automata

  • Same as a DFA, except:

    • On input a, state q may have more than one transition out, implying the possibility of multiple choices when processing an input symbol

    • On input a, state q may have no transition out, implying the possibility of “being stuck”

  • A string w is acceptable as long as there exists an admissible state sequence for w


Nondeterministic finite automata

NFAs

  • A nondeterministic finite automaton M is a five-tuple M = (Q, , , q0, F), where:

    • Q is a finite set of states of M

    •  is the finite input alphabet of M

    • : Q    power set of Q, is the state transition function mapping a state-symbol pair to a subset of Q

    • q0 is the start state of M

    • F  Q is the set of accepting states or final states of M


Example nfa

Example NFA

  • NFA that recognizes the language of strings that end in 01

0,1

Exercise:draw the complete transition table for this NFA

1

q2

0

q1

q0

note:(q0,0) = {q0,q1}(q1,0) = {}


Definition for an nfa

^ definition for an NFA

  • ^: Q X *  power set of Q

  • ^(q, ) = {q}

  • ^(q, w), w = xa(where x is a string and a is a symbol)is defined as follows:

    • Let ^(q, x) = {p1,p2,…pk}

    • Then, ^(q, w) =  (pi, a)


Language recognized by an nfa

Language recognized by an NFA

  • A string w is accepted by an NFA M if^(q0, w)  F is non-empty

    • Note that ^(q0, w) represents a subset of states since the automaton is nondeterministic

    • Equivalent definition: there exists an admissible state sequence for w in M

  • The language L(M) recognized by an NFA is the set of strings accepted by M

    • L(M) ={ w | ^(q0, w)  F is non-empty }


Converting nfas to dfas

Converting NFAs to DFAs

  • Given a NFA, M = (Q, , , q0, F), build a DFA, M’ = (Q’, , ’, {q0}, F’) as follows.

    • Q’ contains all subsets S of states in Q.

    • The initial state of M’ is the set containing q0

    • F’ is the set of all subsets of Q that contain at least one element in F (equivalently, the subset contains at least one final state)


Converting nfas to dfas1

Converting NFAs to DFAs

  • ’ is determined by putting together, for each state in the subset and each symbol, all states that may result from a transition:’(S, a) =  (q, a)qS

  • May remove “unreachable” states in Q’


Example conversion

Example conversion

  • NFA

  • DFA

0,1

1

q2

0

q1

q0

1

0

1

0

{q0,q1}

{q0,q2}

{q0 }

0

1


Nfa with transitions

NFA with -transitions

  • NFA that allows the transition of an empty string from a state

    • Jumping to a state is possible even without input

  • Revision on NFA definition simply allows the  “symbol” for 


Nfa with transitions1

NFA with -transitions

  • A nondeterministic finite automaton with -transitions (or -NFA) is a five-tupleM = (Q, , , q0, F), where:

    • Q is a finite set of states of M

    •  is the finite input alphabet of M

    • : Q  ( + )  power set of Q, is the state transition function mapping a state-symbol pair to a subset of Q

    • q0 is the start state of M

    • F  Q is the set of accepting states or final states of M


Converting nfas to nfas

Converting -NFAs to NFAs

  • Task: Given an -NFA M = (Q, , , q0, F),build a NFA M’ = (Q, , ’, q0, F’)

  • Need to eliminate -transitions

  • Need epsilon closure concept

  • Add transitions to enable transitions previously allowed by the -transitions

  • Note: the conversion process in the textbook instead builds a DFA from an -NFA

    • The conversion described in these slides is simpler


Epsilon closure

Epsilon closure

  • In an NFA M, let q  Q

  • ECLOSE(q) represents all states r that can be reached from q using only -transitions

  • Recursive definition for ECLOSE

    • If (q, ) is empty, ECLOSE(q) = {q}

    • Else, Let (q, ) = {r1, r2,…, rn}.ECLOSE(q) = ECLOSE(ri)  {q}

  • Note: check out constructive definition of ECLOSE in the textbook


Additional transitions

Additional transitions

  • NFA M’ = (Q, , ’, q0, F’) such that ’ is described as follows

  • Suppose ECLOSE(q) = {r1, r2,…, rn }.

  • For each transition from state ri tostate sj on (non-epsilon) symbol a,add a transition from qto sj on symbol a

  • (For each transition from state sj tostate qon (non-epsilon) symbol a,add a transition from sj to rj on symbol a, for each rj)

  • ’ =  minus the epsilon transitions plus the additional transitions mentioned above


Final states

Final states

  • NFA M’ = (Q, , ’, q0, F’) such that F’ is described as follows

  • F’ = F plus all states q such that ECLOSE(q0) contains a state in F


Equivalence of finite automata

Equivalence of Finite Automata

  • Conversion processes betweenDFAs, NFAs, and -NFAs show that no additional expressive capacity (except convenience) is introduced by non-determinism or -transitions

  • All models represent regular languages

  • Note: possible exponential explosion of states when converting from NFA to DFA


Closure of regular languages under certain operations

Closure of Regular Languages under certain operations

  • UnionL1  L2

  • ComplementationL1

  • IntersectionL1  L2

  • ConcatenationL1L2

  • Goal: ensure a FA can be produced from the FAs of the “operand” languages


Union

Union

  • Given that L1 and L2 are regular, then there exists FAs M1 = (Q1, 1, 1, q10, F1) andM2 = (Q2, 2, 2, q20, F2) that recognizeL1 and L2 respectively

  • Let L3 = L1  L2. Define M3 as follows:M3 = ({q30}Q1Q2,12, 3, q30,F1F2)

    • where 3 is just 12 plus the following epsilon transitions: q30  q10 and q30  q20

  • M3 recognizes L3


Complementation

Complementation

  • Given that L1 is regular, then there exists DFA M1 = (Q, , , q0, F) that recognizes L1

  • Define M2 as follows:M2 = (Q, , , q0, Q - F)

  • M2 recognizes L1

    • Strings that used to be accepted are not, strings not accepted are now accepted

  • Note: it is important that M1 is a DFA; starting with an NFA poses problems. Why?


Intersection

Intersection

  • By De Morgan’s Law,L1  L2 = ( L1L2 )

  • Applications of the constructions provided for and complementation provide a construction for 


Concatenation

Concatenation

  • Given that L1 and L2 are regular, then there exists FAs M1 = (Q1, 1, 1, q10, F1) andM2 = (Q2, 2, 2, q20, F2) that recognizeL1 and L2 respectively

  • Let L3 = L1L2. Define M3 as follows:M3 = (Q1Q2, 12, 3, q10, F2)

    • where 3 is just 12 plus the following epsilon transitions: q1i  q20 for all q1i in F1

  • M3 recognizes L3


Finite automata with output

Finite Automata with Output

  • Moore Machines

    • Output symbol for each state encountered

  • Mealy Machines

    • Output symbol for each transition encountered

  • Exercise: formally define Moore and Mealy machines


Next regular expressions

Next: Regular Expressions

  • Defines languages in terms of symbols and operations

  • Example

    • (01)* +(10)* defines all even-length strings of alternating 0s and 1s

  • Regular expressions also model regular languages and we will demonstrate equivalence with finite automata


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