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:  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

NFAs

 A nondeterministic finite automaton M is a five-tuple M = (Q,  ,  , q 0 , F), where:  Q is a finite set of states of M    is the finite input alphabet of M  : Q of Q    power set of Q, is the state transition function mapping a state-symbol pair to a subset   q 0 F  is the start state of M of M Q is the set of accepting states or final states

Example NFA

 NFA that recognizes the language of strings that end in 01 0,1 q 0 0 q 1 1 q 2 Exercise: draw the complete transition table for this NFA note:  (q 0 ,0) = {q 0 ,q 1 }  (q 1 ,0) = {}



^ definition for an NFA

    ^: Q X  *   ^(q,  ) = {q} power set of Q  ^(q, w), w = xa (where x is a string and a is a symbol) is defined as follows:   Let  ^(q, x) = {p 1 ,p 2 ,…p k } Then,  ^(q, w) =   (p i , a)

Language recognized by an NFA

  A string w is accepted by an NFA M if  ^(q 0 , w)  F is non-empty  Note that  ^(q 0 , 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 |  ^(q 0 , w)  F is non-empty }

Converting NFAs to DFAs

 Given a NFA, M = (Q,  ,  , q 0 , F), build a DFA, M’ = (Q’,  ,  ’, {q 0 }, F’) as follows.

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

The initial state of M’ is the set containing q 0 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 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

 NFA 0,1 0 1 q 0 q 1  DFA 1 {q 0 } 0 0 {q 0 ,q 1 } 1 0 1 q 2 {q 0 ,q 2 }

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



-transitions

 A nondeterministic finite automaton with  transitions (or  -NFA) is a five-tuple M = (Q,  ,  , q 0 , F), where:      Q is a finite set of states of M   is the finite input alphabet of M : Q 

(



+

 a subset of Q

)

 power set of Q, is the state transition function mapping a state-symbol pair to q 0 F  is the start state of M of M Q is the set of accepting states or final states

Converting



-NFAs to NFAs

     Task: Given an  -NFA M = (Q,  ,  , q 0 , F), build a NFA M’ = (Q,  ,  ’, q 0 , 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

    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,  ) = {r 1 , r 2 ,…, r n }.

ECLOSE(q) =  ECLOSE(r i )  {q} Note: check out constructive definition of ECLOSE in the textbook

Additional transitions

     NFA M’ = (Q,  ,  ’, q 0 , F’) such that  ’ is described as follows Suppose ECLOSE(q) = {r 1 , r 2 ,…, r n }.

For each transition from state r i to state s j on (non-epsilon) symbol a, add a transition from q to s j on symbol a (For each transition from state s add a transition from s j to r j j to state q on (non-epsilon) symbol a, on symbol a, for each r j )  ’ =  minus the epsilon transitions plus the additional transitions mentioned above

Final states

  NFA M’ = (Q,  ,  ’, q 0 , F’) such that F’ is described as follows F’ = F plus all states q such that ECLOSE(q 0 ) contains a state in F

Equivalence of Finite Automata

   Conversion processes between DFAs, NFAs, and determinism or   -NFAs show that no additional expressive capacity (except convenience) is introduced by non -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

    Union Complementation Intersection Concatenation L1  L1 L1  L1L2 L2 L2  Goal: ensure a FA can be produced from the FAs of the “operand” languages

Union

   Given that L1 and L2 are regular, then there exists FAs M1 = (Q1,  1,  1, q1 0 , F1) and M2 = (Q2,  2,  2, q2 0 , F2) that recognize L1 and L2 respectively Let L3 = L1  L2. Define M3 as follows: M3 = ({q3 0 }  Q1  Q2,  1  2,  3, q3 0 ,F1  F2)  where  3 is just  1  2 plus the following epsilon transitions: q3 0   q1 0 and q3 0   q2 0 M3 recognizes L3

Complementation

    Given that L1 is regular, then there exists DFA M1 = (Q,  ,  , q 0 , F) that recognizes L1 Define M2 as follows: M2 = (Q,  ,  , q 0 , 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

 By De Morgan’s Law,

L1



L2 = (

L1  L2

)

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

Concatenation

   Given that L1 and L2 are regular, then there exists FAs M1 = (Q1,  1,  1, q1 0 , F1) and M2 = (Q2,  2,  2, q2 0 , F2) that recognize L1 and L2 respectively Let L3 = L1L2. Define M3 as follows: M3 = (Q1  Q2,  1  2,  3, q1 0 , F2)  where  3 is just  1  2 plus the following epsilon transitions: q1 i   q2 0 for all q1 i in F1 M3 recognizes L3

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

   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