Formal Definition of a Nondeterministic Finite...

Formal Definition of a Nondeterministic FiniteAutomaton

September 7, 2014

Formal Definition of a Nondeterministic Finite Automaton

A comment first

I The formal definition of an NFA is similar to that of a DFA.Both have states, an alphabet, transition function, one startstate, and a collection of accept states

I They differ in one essential way: the transition functions

I In a DFA transition function is δ : Q × Σ→ Q;In an NFA it is δ : Q × (Σ ∪ {ε})→ P(Q)where P(Q) is the collection of subsets of Q called the powerset of Q


A comment first





A comment first





Notation

For any alphabet Σ, Σε = Σ ∪ {ε}


NFA formal definition (Def. 1.37)

A nondeterministic finite automaton is a 5-tuple(Q,Σ, δ, q0,F ) where

1. Q is a finite set of states

2. Σ is a finite alphabet

3. δ : Q × Σε → P(Q) is the transition function

4. q0 ∈ Q is the start state

5. F ⊆ Q is the set of accept states


Example NFA (Example 1.38)

Recall the NFA N1, Figure 1:

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q2 -ε,0 ��

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q4

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Figure : Nondeterministic Finite Automaton N1

N1 = (Q,Σ, δ, q1,F ) where:1. Q = {q1, q2, q3, q4}2. Σ = {0, 1}3. δ is given in the table:

δ 0 1 εq1 {q1} {q1, q2} ∅q2 {q3} ∅ {q3}q3 ∅ {q4} ∅q4 {q4} {q4} ∅

4. q1 is the start set; 5. F = {q4}




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N1 = (Q,Σ, δ, q1,F ) where:

1. Q = {q1, q2, q3, q4}2. Σ = {0, 1}3. δ is given in the table:

δ 0 1 εq1 {q1} {q1, q2} ∅q2 {q3} ∅ {q3}q3 ∅ {q4} ∅q4 {q4} {q4} ∅





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q2 -ε,0 ��

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q4

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N1 = (Q,Σ, δ, q1,F ) where:1. Q = {q1, q2, q3, q4}

2. Σ = {0, 1}3. δ is given in the table:

δ 0 1 εq1 {q1} {q1, q2} ∅q2 {q3} ∅ {q3}q3 ∅ {q4} ∅q4 {q4} {q4} ∅





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q2 -ε,0 ��

q3 -1 ��

q4

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N1 = (Q,Σ, δ, q1,F ) where:1. Q = {q1, q2, q3, q4}2. Σ = {0, 1}

3. δ is given in the table:δ 0 1 εq1 {q1} {q1, q2} ∅q2 {q3} ∅ {q3}q3 ∅ {q4} ∅q4 {q4} {q4} ∅





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q2 -ε,0 ��

q3 -1 ��

q4

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δ 0 1 εq1 {q1} {q1, q2} ∅q2 {q3} ∅ {q3}q3 ∅ {q4} ∅q4 {q4} {q4} ∅





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q2 -ε,0 ��

q3 -1 ��

q4

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δ 0 1 εq1 {q1} {q1, q2} ∅q2 {q3} ∅ {q3}q3 ∅ {q4} ∅q4 {q4} {q4} ∅



Computation performed by an NFA

Let N = (Q,Σ, δ, q0,F ) be an NFA and w a string over Σ. We saythat N accepts w if we can write w as w = w1w2 . . .wm, yi ∈ Σε,1 ≤ i ≤ m, and a sequence of states r0, r1, . . . , rm exists in Q suchthat:

1. r0 = q0

2. ri+1 ∈ δ(ri ,wi+1), for i = 0, 1, . . . ,m − 1

3. rm ∈ F

I Condition 1 says that the machine starts its computation inthe start state

I Condition 2 says that state ri+1 is one of the allowable newstates when N is in state ri and reads wi+1. Note thatδ(ri ,wi+1) is a set

I Condition 3 says that the machine accepts the input if the laststate is in the accept state set.




1. r0 = q0

2. ri+1 ∈ δ(ri ,wi+1), for i = 0, 1, . . . ,m − 1

3. rm ∈ F







1. r0 = q0

2. ri+1 ∈ δ(ri ,wi+1), for i = 0, 1, . . . ,m − 1

3. rm ∈ F







1. r0 = q0

2. ri+1 ∈ δ(ri ,wi+1), for i = 0, 1, . . . ,m − 1

3. rm ∈ F





Equivalence of NFA and DFA

I DFAs and NFAs recognize the same class of languages

I This equivalence is both surprising and useful

1. It is surprising because NFAs appears to have more power thanDFA, so we might expect that NFA recognizes more languages

2. It is useful because describing an NFA for a given languagesometimes is much easier than describing a DFA



I DFAs and NFAs recognize the same class of languagesI This equivalence is both surprising and useful














Theorem 1.39

Every nondeterministic finite automation (NFA) has an equivalentdeterministic finite automaton (DFA).Proof idea: Convert the NFA into an equivalemt DFA thatsimulates the NFA

I The approach is to pretend to be an DFA simulating an NFA.I How would you simulate an NFA if you were pretending to be

a DFA?I What does one need to keep track of as the input is processed?

I In the examples of NFA we kept track of various branches ofthe computation by placing tokens on each state that couldbe activated at given points in the input

I Tokens were updated by moving, adding, and removing themaccording to the way NFA operates

Hence, all we needed to keep track of is the set of states withtokens on them


Theorem 1.39








Theorem 1.39








Theorem 1.39








Theorem 1.39








Conclusion

I An NFA with k states has 2k subsets of states

I Each subset correspond to one of the possibilities that DFAmust remember. So, the DFA simulating an NFA with kstates has 2k states

I The only thing we need now is to figure out which are thestart state, the accept states, and transition function


The formal proof

Let N = (Q,Σ, δ, q0,F ) be the NFA recognizing the language A.We construct the DFA M recognizing A. Before doing the fullconstruction, consider first the easier case when N has no εtransitions.Constructing M = (Q′,Σ, δ′, q′0,F

′)

1. Q′ = P(Q); every state of M is a set of states of N

2. For R ∈ Q′ and a ∈ Σ let δ′(R, a) = {q ∈ Q|q ∈ δ(r , a) for some r ∈ R}Note: δ′(R, a) = ∪r∈Rδ(r , a)

3. q′0 = {q0}; M starts computing in the same state as N

4. F ′ = {R ∈ Q′|R contains an accept state of N}that is, the machine M accepts if one of the possible states that N could be inat this point is an accept state


The formal proof


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The formal proof


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The formal proof


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The formal proof


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Consider the ε transitions

Notation: for any R ∈ Q ′ define E (R) to be the collection ofstates that can be reached from R by going only along εtransitions, including the members of R themselves. Formally:E(R) = R ∪ {q ∈ Q|∃r ∈ R ∧ δ(r , ε) = q}Note: any q ∈ E (R) can be reached from R by traveling along 0or more ε arrows.


Modifications

I Modify δ′ to place additional tokens on all states that can bereached by going along ε arrows after every step. This effectis achieved by replacing δ(r , a) by E (δ(r , a)) in definition ofδ′. That isδ′(R, a) = {q ∈ Q|q ∈ E(δ(r , a)) for some r ∈ R}

I Modify the start state of M to put tokens initially on all statesthat can be reached from q0 along ε transitions, i.e.,q′0 = E ({q0})


Modifications

I Modify δ′ to place additional tokens on all states that can bereached by going along ε arrows after every step. This effectis achieved by replacing δ(r , a) by E (δ(r , a)) in definition ofδ′. That isδ′(R, a) = {q ∈ Q|q ∈ E(δ(r , a)) for some r ∈ R}

I Modify the start state of M to put tokens initially on all statesthat can be reached from q0 along ε transitions, i.e.,q′0 = E ({q0})


Note

I M thus constructed obviously simulates N

I At every step in the computation of M it clearly enters a statethat corresponds to the subset of states that N could be in atthat point


Corollary 1.40

A language is regular iff some NFA recognizes it

Proof:

I If a language A is recognized by an NFA then A is recognizedby the DFA equivalent, hence, A is regular

I If a language A is regular, it means that it is recognized by aDFA. But any DFA is also an NFA hence, the language isrecognized by an NFA


Example conversion NFA to DFA

Consider the NFA N4 described in Figure 2:

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Figure : The NFA N4

Formal description of N4

N4 = ({1, 2, 3}, {a, b}, δ, 1, {1}) where δ is given in the table:

δ a b ε1 ∅ {2} {3}2 {2, 3} {3} ∅3 {1} ∅ ∅


Example conversion NFA to DFA

Consider the NFA N4 described in Figure 2:

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Figure : The NFA N4

Formal description of N4

N4 = ({1, 2, 3}, {a, b}, δ, 1, {1}) where δ is given in the table:

δ a b ε1 ∅ {2} {3}2 {2, 3} {3} ∅3 {1} ∅ ∅


Construction of D4 = (Q′, {a, b}, δ′, q′0,F ′)

I Since N4’s set of states is Q = {1, 2, 3}Q′ = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}

I The start state of Q ′: the set of states reachable from 1 (thestart state of N4) traveling by ε: E ({1}) = {1, 3}

I The accept state of Q ′: are those states of Q ′ that contain theaccept states of Q: i.e., F ′ = {{1}, {1, 2}, {1, 3}, {1, 2, 3}}


Transition function

δ′ : P(Q)× Σ→ P(Q), where for each q′ ∈ P(Q) and x ∈ Σ wehave δ′(q′, x) ∈ P(Q)


Constructing D4’s transitions

I State ∅: goes to ∅ on both a, b, i.e., δ′(∅, a) = δ′(∅, b) = ∅I State {1}: goes to ∅ on a and to {2} on b

I State {2}: goes to {2, 3} on a and to {3} on b

I State {1, 2}: goes to {2, 3} on a and to {2, 3} on b

I And so on, for each state


Putting all together

Figure 3 shows the resulting DFA D4 equivalent to NFA N4

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Figure : DFA D4 equivalent to NFA N4


Simplifying D4

The automaton D4 may be simplified by observing that no arrowspoint at the states {1} and {1, 2}. Removing these states weobtain the automaton in Figure 4

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Figure : The simplified version of D4


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