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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

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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:

• 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, , , 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

• 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

• ^: 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

• 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

• 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 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’

• NFA

• DFA

0,1

1

q2

0

q1

q0

1

0

1

0

{q0,q1}

{q0,q2}

{q0 }

0

1

### 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-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

• 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

• 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

### 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

• 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

• UnionL1  L2

• ComplementationL1

• IntersectionL1  L2

• ConcatenationL1L2

• 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, 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

• 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

• 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, 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

• 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