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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. Same as a DFA, except:

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Nondeterministic Finite Automata

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  1. Nondeterministic Finite Automata CS 130: Theory of Computation HMU textbook, Chapter 2(Sec 2.3 & 2.5)

  2. 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

  3. 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

  4. 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) = {}

  5. ^ 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)

  6. 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 }

  7. 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)

  8. 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’

  9. Example conversion • NFA • DFA 0,1 1 q2 0 q1 q0 1 0 1 0 {q0,q1} {q0,q2} {q0 } 0 1

  10. 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 

  11. 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

  12. 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

  13. 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

  14. 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

  15. 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

  16. 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

  17. Closure of Regular Languages under certain operations • Union L1  L2 • Complementation L1 • Intersection L1  L2 • Concatenation L1L2 • Goal: ensure a FA can be produced from the FAs of the “operand” languages

  18. 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

  19. 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?

  20. Intersection • By De Morgan’s Law,L1  L2 = ( L1L2 ) • Applications of the constructions provided for and complementation provide a construction for 

  21. 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

  22. 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

  23. 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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