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Nondeterministic Finite Automata

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Nondeterministic Finite Automata

CS 130: Theory of Computation

HMU textbook, Chapter 2(Sec 2.3 & 2.5)

- 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

- 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

- 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) = {}

- ^: 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)

- 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 }

- 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)

- ’ 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)qS
- May remove “unreachable” states in Q’

- NFA
- DFA

0,1

1

q2

0

q1

q0

1

0

1

0

{q0,q1}

{q0,q2}

{q0 }

0

1

- 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

- 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

- 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

- 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

- 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

- 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

- 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

- UnionL1 L2
- ComplementationL1
- IntersectionL1 L2
- ConcatenationL1L2
- Goal: ensure a FA can be produced from the FAs of the “operand” languages

- 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}Q1Q2,12, 3, q30,F1F2)
- where 3 is just 12 plus the following epsilon transitions: q30 q10 and q30 q20

- M3 recognizes L3

- 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?

- By De Morgan’s Law,L1 L2 = ( L1L2 )
- Applications of the constructions provided for and complementation provide a construction for

- 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 = (Q1Q2, 12, 3, q10, F2)
- where 3 is just 12 plus the following epsilon transitions: q1i q20 for all q1i in F1

- M3 recognizes L3

- Moore Machines
- Output symbol for each state encountered

- Mealy Machines
- Output symbol for each transition encountered

- Exercise: formally define Moore and Mealy machines

- 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