computational models — lecture 2tau-cm2019.wdfiles.com/local--files/course-schedule/dfa2.pdf ·...

Computational Models — Lecture 2

Handout Mode

Roded Sharan.

Tel Aviv University.

March, 2019

Roded Sharan (TAU) Computational Models, Lecture 2 March, 2019 1 / 64

Computational Models - Lecture 2

I Non-Deterministic Finite Automata (NFA)

I Closure of Regular Languages Under⋃, ‖, ∗

I Regular expressions

I Equivalence with finite automata

I Sipser’s book, 1.1− 1.3


Part I

Non-Deterministic Finite Automata


DFA – formal definition, (reminder)

Definition 1 (DFA)

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

I Q is a finite set called the states,

I Σ is a finite set called the alphabet,

I δ : Q × Σ 7→ Q is the transition function,

I q0 ∈ Q is the start state, and

I F ⊆ Q is the set of accept states.


Formal model of computation, (reminder)

Definition 2

M = (Q,Σ, δ,q0,F ) accepts w ∈ Σ∗ if δ̂(q0,w) ∈ F .

Definition 3 (δ̂)

For DFA M = (Q,Σ, δ,q0,F ), define δ̂ : Q × Σ∗ 7→ Q by

δ̂(q,w) =

{δ(δ̂(q,w1,...,n−1),wn), n = |w | ≥ 1q, w = ε.

.


The language of a DFA, (reminder)

Definition 4The language of a DFA M, denoted L(M), is the set of strings that M accepts.

Definition 5A language is called regular, if some deterministic finite automaton accepts it.


NFA — non-deterministic Finite Automata

q4

q1

q2

q3

0

0,1 0,1

10

I May have more than one transition labeled with the same symbol,

I May have no transitions labeled with a certain symbol,

I The machine “splits” into multiple copies covering all possibilities

I Each branch follows one possibility

I If the input doesn’t appear, that branch “dies”.

I Automaton accepts if some branch accepts.


Computation on 1001

q4

q1

q2

q3

0

0,1 0,1

10

q1

q1

q2

q1

q2

q1

q3

q1

q4

1

0

0

1

symbol


NFA – Formal Definition

Let P(Q) denote the powerset of Q (i.e., all subsets of Q).

Definition 6 (NFA)

A non-deterministic finite automaton is a 5-tuple (Q,Σ, δ,S,F ), where

I Q is a finite set called the states

I Σ is a finite set called the alphabet

I δ : Q × Σ 7→ P(Q) is the transition function

I S ⊆ Q is the set of starting states

I F ⊆ Q are the set of accepting states

We sometimes consider an NFA (Q,Σ, δ,q0,F ).This is merely a “syntactic sugar” for the NFA (Q,Σ, δ, {q0},F )


Example

q4

q1

q2

q3

0

0,1 0,1

10

N1 = (Q = {q1,q2,q3,q4},Σ = {0,1}, δ,S = {q1},F = {q4})

for δ defined by

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

Note that ∅ is a valid output for δ


Formal model of computation

Definition 7

N = (Q,Σ, δ,S,F ) accepts w ∈ Σ∗, if δ̂N(S,w) ∩ F 6= ∅.

Definition 8 (δ̂)

For NFA N = (Q,Σ, δ,S,F ), define δ̂N : P(Q)× Σ∗ 7→ P(Q) by:

for Q′ ⊆ Q, δ̂N(Q′,w) =

{Q′, w = ε,⋃

q∈δ̂N (Q′,w1,...,n−1)δ(q,wn), n = |w | ≥ 1. .

When clear from the context we will write δ̂ (i.e., omitting the N ).


An equivalent definition

Definition 9 (Equivalent definition)

N = (Q,Σ, δ,S,F ) accepts w = w1, . . . ,wn ∈ Σn, if if ∃r0, . . . , rn ∈ Q s.t.

I r0 ∈ S

I rn ∈ F

I ri+1∈δ(ri ,wi+1), for all 0 ≤ i < n.


Equivalence of NFA’s and DFA’s

Easy: For any DFA M there exists a NFA N such that L(N) = L(M).

Other direction is also true.

Theorem 10For any NFA N there exists a DFA M such that L(N) = L(M).

I Given an NFA N, we construct a DFA M, that accepts the samelanguage.

I Make DFA emulates all possible NFA states.

I As consequence of the construction, if the NFA has k states, the DFAhas 2k states (an exponential blow up).


Equivalence of NFA’s and DFA’s, the DFA

Let N = (Q,Σ, δ,S,F ).

Construction 11 (M = (QM ,Σ, δM ,d0,FM))

I QM = {[R] : R ⊆ Q}.I d0 = [S]

I FM = {[R] ∈ QM : R ∩ F 6= ∅}

I For [R] ∈ QM and σ ∈ Σ, let δM([R], σ) = [δ̂N(R, σ)] (=[⋃

r∈R δ(r , σ)]).

To prove equivalence, we need to prove that

δ̂N(S,w) ∩ F 6= ∅ ⇐⇒ δ̂M(d0,w) ∈ FM

The above is an immediate corollary of the following claim:

Claim 12

[δ̂N(S,w)] = δ̂M(d0,w) for every w ∈ Σ∗.


Proving [δ̂N(S,w)] = δ̂M(d0,w)

The proof is by induction on the length of w .

I |w | = 0, by definition.

I Assume for words of length (m− 1), and let x = yσ, where y is a word oflength (m − 1) and σ ∈ Σ.

I Let Qy = δ̂N(S, y) and dy = δ̂M(d0, y).

I Compute

δ̂M(d0, x) = δM(dy , σ) (By definition of δ̂M )= δM([Qy ], σ) (By i.h)

= [δ̂N(Qy , σ)] (By definition of δM )

=[ ⋃

q∈Qy

δ(q, σ)]

(By definition of δ̂N )

=[ ⋃

q∈δ̂N (S,y)

δ(q, σ)]

= [δ̂N(S, x)]. (By definition of δ̂N )


Example: NFA⇒ DFA

Non-Deterministic Automata:


Example: NFA⇒ DFA

Deterministic automata - set of states:


Example: NFA⇒ DFA

Transitions from [{q0}]:


Example: NFA⇒ DFA

Transitions from [{q0,q1}]:

Transitions from [∅] and [{q1}]?


NFA with ε-moves

q4

q1

q2

q3

0

0,1 0,1

10,ε

What is the interpretation of ε transitions ?What will happen with 101 ?

We will distinguish ε (that stands for the empty word) from the ε above (thatstands for “free transition”) by context


Example: NFA with ε-moves

L = {aibjck |i , j , k ≥ 0}


NFA — Formal definition with ε-moves

Transition function δ is going to be different.

I Let P(Q) denote the powerset of Q.

I Let Σε denote the set Σ ∪ {ε}.

Definition 13 (NFA, with ε-moves)

A non-deterministic finite automaton is a 5-tuple (Q,Σ, δ,S,F ):

I Q is a finite set called the states

I Σ is a finite set called the alphabet

I δ : Q × Σε 7→ P(Q) is the transition function

I S ⊆ Q is the set of starting state

I F ⊆ Q is the set of accepting states


Example

q4

q1

q2

q3

0

0,1 0,1

10,ε

N1 = (Q,Σ, δ,S,F ):

I Q = {q1,q2,q3,q4}, Σ = {0,1}, S = {q1} and F = {q4}.

I δ is

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


Formal model of computation, with ε-moves

Definition 14

N = (Q,Σ, δ,S,F ) accepts w ∈ Σ∗, if δ̂N(S,w) ∩ F 6= ∅.

Definition 15For NFA N = (Q,Σ, δ,S,F ), letE(q) = {q′ ∈ Q : q′ can be reached from q by 0 or more ε transitions}(i.e., {q′ : ∃q1, . . . , qk ∈ Q s.t.q1 = q ∧ qk = q′ ∧ ∀i ∈ [k − 1] qi+1 ∈ δ(qi , ε)})

E(Q′) =⋃

q∈Q′ E(q).

In particular, q ∈ E(q).

Definition 16 (δ̂)

For NFA N = (Q,Σ, δ,S,F ), define δ̂N : P(Q)× Σ∗ 7→ P(Q) by:

for Q′ ⊆ Q, δ̂N(Q′,w) =

{E(Q′), w = ε,

E(⋃

r∈δ̂(Q′,w1,...,wn−1)δ(r ,wn)

), n = |w | ≥ 1. .

When does N accept the empty string?Roded Sharan (TAU) Computational Models, Lecture 2 March, 2019 24 / 64

An equivalent definition

For a ∈ (Σε)∗, let d(a) ∈ Σ∗ be a without the ε symbols.

Example: d(ε01εε3ε) = 013

Definition 17 (Equivalent definition)

N = (Q,Σ, δ,S,F ) accepts w ∈ Σ∗, if exist a = (a1a2 . . . ak ) ∈ (Σε)k and

r0, . . . , rk ∈ Q s.t.

I w = d(a).

I r0 ∈ S

I rk ∈ F

I ri+1 ∈ δ(ri ,ai+1), for all 0 ≤ i < k .


Removing ε-transitions

Given NFA N = (Q,Σ, δ,S,F ) with ε-transitions, we create an equivalent NFAN ′ = (Q,Σ, δ′,S′,F ) with no ε-transitions.

I S′ = E(S)

I δ′(q,a) = E(δ(q,a))

It is not hard to prove that δ̂N(S,w) = δ̂N′(S′,w) for any w ∈ Σ∗.

Thus, L(N) = L(N ′).


Example: Removing ε-transitions

Non-Deterministic Automata with ε-transitions

The non-Deterministic automata without ε-transitions

I S′ = {q0,q1,q2,q3}

I δ′ is

a b cq0 ∅ ∅ ∅q1 {q1,q2,q3} ∅ ∅q2 ∅ {q2,q3} ∅q3 ∅ ∅ {q3}


Part II

Closure of Regular Languages, Revisited


Regular languages, revisited

By definition, a language is regular if it is accepted by some DFA.

Corollary 18

A language is regular if and only if it is accepted by some NFA.

This is an alternative way of characterizing regular languages.

We will now use the equivalence to show that regular languages are closedunder the regular operations (union, concatenation, star).


Closure under union (alternative proof)

N1

N2


Closure under union, cont.

N1

N2

ε

ε


Closure under union cont..

I NFA N1 = (Q1,Σ, δ1,S1,F1) accept L1, and

I NFA N2 = (Q2,Σ, δ2,S2,F2) accept L2.

wlg. Q1 ∩Q2 = ∅.(?)

Define NFA N = (Q = {q0} ∪Q1 ∪Q2,Σ, δ,S = {q0},F = F1 ∪ F2),

for δ(q,a) =

δ1(q,a) q ∈ Q1δ2(q,a) q ∈ Q2S1 ∪ S2 q = q0 and a = ε

Alternatively, let S = S1 ∪ S2 and omit the last line of δ.

Claim 19L(N) = L(N1) ∪ L(N2).

Proof: Easily follows by the the alternative definition of acceptance by NFA,DIY...


Closure under concatenation

N2

N1


Closure under concatenation, cont.

N2N

1

ε

ε

Remark: Accepting states are exactly those of N2.


Closure under concatenation, cont..

I NFA N1 = (Q1,Σ, δ1,S1,F1) accept L1

I NFA N2 = (Q2,Σ, δ2,S2,F2) accept L2

Define NFA N = (Q1 ∪Q2,Σ, δ,S1,F2):

δ(q,a) =

δ1(q,a) q ∈ Q1 ∧ a 6= εδ1(q,a) (q ∈ Q1 \ F1) ∧ a = εδ1(q,a) ∪ S2 q ∈ F1 ∧ a = εδ2(q,a) q ∈ Q2

Claim 20L(N) = L1‖L2.

Proof: Need to prove w ∈ L(N)⇐⇒ w ∈ L1‖L2.

We assume wlg. that N1 and N2 have no ε moves.


Proving w ∈ L(N)⇐⇒ L1‖L2

Let w ∈ L1‖L2:

=⇒ ∃w1 ∈ L1 and w2 ∈ L2, s.t. w = w1w2

=⇒ ∃ r10 , . . . r

1|w1| and r2

0 , . . . r2|w2|, s.t.

I r10 ∈ S1 (2) r1

|w1| ∈ F1 (3) ∀i : r1i+1 ∈ δ1(r1

i ,w1i+1).

I r20 ∈ S2 (2) r2

|w2| ∈ F2 (3) ∀i : r2i+1 ∈ δ2(r2

i ,w2i+1).

=⇒ r10 , . . . r

1|w1|, r

20 , . . . r

2|w2| proves that w ∈ L(N)

Let w ∈ L(N) be of length k :

=⇒ ∃j and r0 . . . rk+1 s.t. for (a1, . . . ,ak+1) = (w1, . . . ,wj , ε,wj+1, . . .wk ):

I r0 ∈ S1 and rk+1 ∈ F2I ∀i : ri+1 ∈ δ(ri ,ai+1).

I r0 . . . rj proves that (w1, . . . ,wj ) ∈ L1

I rj+1 . . . rk proves that (wj+1, . . . ,wk ) ∈ L2

=⇒ w ∈ L1‖L2


Closure under star


Closure under star, cont.

Let N = (Q,Σ, δ,S,F ) accepting L, assuming wlg. that q0 /∈ Q.

Define N ′ = (Q′ = Q ∪ {q0},Σ, δ′,S′ = {q0},F ′ = {q0}):

δ′(q,a) =

δ(q,a) q ∈ Q ∧ a 6= εδ(q, ε) q /∈ F ∧ a = εδ(q, ε) ∪ {q0} q ∈ F ∧ a = εS q = q0 ∧ a = ε

Claim 21L(N ′) = L(N)∗.

Proof: DIY


Summary

I Regular languages are closed under

I unionI concatenationI star

I Non-deterministic finite automata

I are equivalent to deterministic finite automataI but much easier to use in some proofs and constructions.


Part III

Regular Expressions


Regular expressions

Notation for building up languages by describing them as expressions, e.g.,(0 ∪ 1)0∗.

I 0 and 1 are shorthand for the set (languages) {0} and {1}I so 0 ∪ 1 = {0,1}.I 0∗ is shorthand for {0}∗.I Concatenation, is implicit. So 0∗10∗ stands for{w ∈ {0,1} : w has exactly a single 1}.

I Just like in arithmetic, operations have precedence:I star firstI concatenation nextI union lastI parentheses used to change default order i.e., ab∗ 6= (ab)∗

Q.: What does (0 ∪ 1)0∗ stand for?

Remark: Regular expressions are often used in text editors or shell scripts.Roded Sharan (TAU) Computational Models, Lecture 2 March, 2019 41 / 64

Regular expressions – formal definition

Definition 22A string R is a regular expression over alphabet Σ, if R is of form

I a for some a ∈ Σ

I ε

I ∅I (R1 ∪ R2) for regular expressions R1 and R2

I (R1‖R2) for regular expressions R1 and R2

I (R∗1 ) for regular expression R1

R(Σ) denotes all (finite) regular expression over Σ.

Parenthesis and ‖ are omitted when their role is clear from the context.


Formal definition, cont.

Definition 23The language L(R) of regular expression R, is defined by

R L(R)a {a}ε {ε}∅ ∅(R1 ∪ R2) L(R1) ∪ L(R2)(R1R2) L(R1)‖L(R2)(R∗1 ) L(R1)∗

Isn’t this definition circular?


Examples of regular expressions

For Σ = {0,1}, write regular expression for the following languages:

I The third letter from the end is 1

(0 ∪ 1)∗1(0 ∪ 1)2

I The number of 1’s is even

(0 ∪ 10∗1)∗

I The number of 1’s is odd

(0 ∪ 10∗1)∗10∗


Part IV

Regular Expressions and Regular Languages


A remarkable fact

Theorem 24A language is described by a regular expression iff it is regular.

⇐=: Given a regular language, construct a regular expression describing it.

=⇒: Given a regular expression, construct an NFA accepting its language.


Given RE R, build NFA accepting it (=⇒)

1. R = a , for some a ∈ Σ

a

2. R = ε

3. R = ∅


Given R, Build NFA Accepting It (=⇒), cont.

N1

N2

ε

ε

N2N

1

ε

ε

R = (R1 ∪ R2) R = (R1‖R2)

R = (R1)∗

♣


Examplesa

a

b

b

ab ε

ba

ab ε

a

ε

ε

ba U a

Formal proof by induction on the length of the regular expression


Regular expression from a DFA (⇐=)

Proof relies on a generalization of NFAs.

NFA:

I Each transition is labeled with a symbol or ε.

I Reads zero or one symbols.

I Takes matching transition, if any.

Generalized non-deterministic finite automata (GNFA):

I Each transition is labeled with a regular expression.

I Reads zero or more symbols.

I Takes matching regular expression, if any.


GNFA – Formal Definition

Let R(Σ) be the set of regular expressions over Σ.

Definition 25A generalized deterministic finite automaton (GNFA) is (Q,Σ, δ,qs,qa)

I Q is a finite set of states

I Σ is the alphabet

I δ : (Q \ {qa})× (Q \ {qs}) 7→ R(Σ) is the transition function.

I qs ∈ Q is the start state

I qa ∈ Q is the unique accept state

It is a special type of GNFA, but still it is easy to transform any DFA/NFA intothis form.


GNFA – Model of Computation

Definition 26A GNFA G = (Q,Σ, δ,qs,qa) accepts a string w ∈ Σ∗, if there exists

I parsing w = a1a2 · · · ak ∈ (Σ∗)k , and

I r0, . . . , rk ∈ Q,

such that

I r0 = qs

I rk = qa

I ai ∈ L(δ(ri−1, ri )), for every 0 < i ≤ k .


The Transformation: DFA→ Regular Expression

Strategy – sequence of equivalent transformations

I Given a k -state DFA

I Transform into (k + 2)-state GNFA (how?)

I While GNFA has more than 2 states, transform it into equivalent GNFAwith one fewer state

I Eventually reach 2-state GNFA (states are just start and accept).

I Label of single transition is the desired regular expression. ♠


Removing a state

We remove one state qr , and then repair the machine by altering regularexpression of other transitions.

qi jq

qr

R1 3R

4R

qi jq

2R

2R*R1 3R U 4R


Conversion - Example


The StateReduce and Convert algorithms

Algorithm 27 (StateReduce)

Input: a (k > 2)-state GNFA G = (Q,Σ, δ,qs,qa).

I Select any state qr ∈ Q \ {qs,qa}.I Let Q′ = Q \ {qr}.I For any qi ∈ Q′ \ {qa} and qj ∈ Q′ \ {qs}, let

I R1 = δ(qi ,qr ), R2 = δ(qr ,qr ), R3 = δ(qr ,qj ) and R4 = δ(qi ,qj ).

I Define δ′(qi ,qj ) = (R1)(R2)∗(R3) ∪ (R4).

I Return the resulting (k − 1)-state GNFA G′ = (Q′,Σ, δ′,qs,qa).

Algorithm 28 (Convert)

Input: a (k ≥ 2)-state GNFA G.

I If k = 2, return the regular expression labeling the only arrow of G.

I Otherwise, return Convert(StateReduce(G)).


Correctness proof

Claim 29G and Convert(G) accept the same language.

Proof: By induction on k – the number of states of G.

Basis. k = 2: Immediate by the definition of GFNA.

Induction step: Assume claim for (k − 1)-state GNFA, where k > 2, prove fork -state GNFA.

Let G′ = StateReduce(G) (note that G′ has k − 1 states), and let qr be theremoved state.We prove (in a very high level) that L(G) = L(G′).


w ∈ L(G) =⇒ w ∈ L(G′)

Let w ∈ L(G) and let p = qs, . . . ,qa be (a possible) “path of states” traversedby G on w .

I If qr /∈ p, then G′ accepts w (the new regular expression on each edgeof G′ contains the old regular expression in the “union part”.)

I If p = qS, . . . ,qi ,qr ,qj , . . . ,qa, the regular expression (Ri,r )(Rr ,r )∗(Rr ,j )linking qi and qj in G′, causes G′ to accept w .

Hence, w ∈ L(G′).


w ∈ L(G′) =⇒ w ∈ L(G)

I Let qs = qj0 ,qj1 , . . . ,qjn = qa be (a possible) “path of states” traversed byG′ on w .

∃ parsing w = w1, . . . ,wn s.t. wi ∈ L(δG′(qji−1 ,qji )).

I For every i ∈ [n]:δG′(qji−1 ,qji ) =

(δG(qji−1 ,qr )δG(qr ,qr )∗δG(qr ,qji )

)∪ (δG(qji−1 ,qJi ).

I Hence for all i ∈ [n], the word wi corresponds to a possible traverse fromqji−1 to qji in G.

I Hence, w ∈ L(G′) =⇒ w ∈ L(G).


Summing it up

I We proved L(G) = L(G′).

I Hence, G and (the regular expression) Convert(G) accept the samelanguage.

I Thus, we proved: Every regular language can be described by a regularexpression.


Summary

I Non-Deterministic Automata (with ε-moves)

I Equivalence to DFA

I Closure properties

I unionI concatenationI star

I Regular expressions.

I Equivalence to DFA


computational models — lecture 2tau-cm2019.wdfiles.com/local--files/course-schedule/dfa2.pdf ·...

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