Computational Models - Lecture 11tau-cm2017a.wdfiles.com/local--files/course-schedule/DFA1_h.pdf ·...

Computational Models - Lecture 11

Handout Mode

Iftach Haitner.

Tel Aviv University.

October 31, 2016

1Based on frames by Benny Chor, Tel Aviv University, modifying frames by Maurice Herlihy, Brown University.

Iftach Haitner (TAU) Computational Models, Lecture 1 October 31, 2016 1 / 41

Talk Outline

I Languages, words and alphabetsI Finite automata and regular languagesI Regular operations

I Sipser’s book, chapter 1.1


Part I

Languages, words and alphabets


Languages, words and alphabets

Definition 1An alphabet Σ is a finite set of letters.

I Σ = {a,b, c, . . . , z} – the English alphabet.I Σ = {α, β, γ, . . . , ζ} – the Greek alphabet.I Σ = {0,1} – the binary alphabet.I Σ = {0,1, . . . ,9} – the digital alphabet.

Definition 2A word (i.e., string) over Σ, is a finite sequence of letters from Σ.

The collection of all strings over Σ is denoted by Σ∗.

For {0,1}, the binary alphabet, ε, 1, 0, 000000000, 1111111000 are allmembers of Σ∗.

Definition 3A language over Σ is a (possibly infinite) subset of Σ∗.


Language Examples

I Modern English.I Ancient Greek.I All prime numbers, written using digits.I A = {w ∈ {0,1}∗ : w has at most seventeen 0’s}.I B = {0n1n : n ≥ 0}.I C = {w ∈ {0,1}∗ : w has an equal number of 0’s and 1’s}.

Make sure you understand what the above notions stand for...


Part II

Finite Automata


Example: A One-Way Automatic Door

frontpad

rearpad

door

I open when person approachesI hold open until person clearsI don’t open when someone standing behind door


The Automatic Door as DFA

closed open

FRONT

NEITHER

FRONTREARBOTH

REARBOTH

NEITHER

I States:I OPENI CLOSED

I Sensor:I FRONT: someone on front padI REAR: someone on rear padI BOTH: someone(s) on both padsI NEITHER no one on either pad.


The Automatic Door as DFA

A DFA is Deterministic Finite Automata

closed open

FRONT

NEITHER

FRONTREARBOTH

REARBOTH

NEITHER

neither front rear bothclosed closed open closed closedopen closed open open open


DFA: Informal definition

The machine M1:

q1

0

q2 3q

11 0

0,1

I States: q1,q2, and q3.I Start state: q1 (arrow from “outside”).I Accept state: q2 (double circle).I State transitions: arrows tagged with letters.


DFA: Informal definition (cont.)

q1

0

q2 3q

11 0

0,1

I On an input stringI DFA begins in start state q1I after reading each symbol, DFA makes

state transition with matching label.I After reading last symbol, DFA “produces" output:

I accept if DFA is an accepting state.I reject otherwise.


DFA: Informal definition (cont..)

q1

0

q2 3q

11 0

0,1

What happens on the following input strings:I 1101I 0010I 01100I In general?!


DFA: Informal definition (cont...)

q1

0

q2 3q

11 0

0,1

This DFA acceptsI All input strings that end with a 1I All input strings that contain at least one 1, and end with an even

number of 0’sI No other strings

Proof: ?


DFA - Formal Definition

Definition 4A deterministic finite automaton (DFA) is a 5-tuple (Q,Σ, δ,q0,F ),where

I Q is a finite set called the statesI Σ is a finite set called the alphabetI δ : Q × Σ→ Q is the transition functionI q0 ∈ Q is the start stateI F ⊆ Q is the set of accept states


Back to M1

q1

0

q2 3q

11 0

0,1

M1 = (Q,Σ, δ,q1,F ) whereI Q = {q1,q2,q3}, Σ = {0,1},

I the transition function δ is

0 1q1 q1 q2q2 q3 q2q3 q2 q2

I q1 is the start stateI F = {q2}.


Another Example

q1

q2

a b

a

a a

a

b

b

b

b

r1

s

r2


Formal Model of Computation

Definition 5

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

Definition 6 (δ̂)

δ̂M : Q × Σ∗ 7→ Q is defined by

δ̂M(q,w) =

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

.

I w1,...,k stands for the word w1, . . . ,wk — the k -letter prefix of w(w1,...,0 = w∅ = ε)

I Note that δ̂M(q, σ) = δ(q, σ) for σ ∈ Σ.I We write δ̂ when M is clear from the context.


An equivalent definition

Definition 7 (Equivalent definition)M = (Q,Σ, δ,q0,F ) accepts w = w1w2 . . .wn, if ∃r0, . . . , rn ∈ Q s.t.,

I r0 = q0.I δ(ri ,wi+1) = ri+1, for all 0 ≤ i < n.I rn ∈ F .


The language of a DFA

Definition 8L(M), the language of a DFA M, is the set of strings that M accepts.

We sometime say that M accepts the language L(M).

I M may accept many stringsI M accepts only one language.

What language does M accept if it accepts no strings?

Definition 9A language is called regular, if some deterministic finite automatonaccepts it.


Example: DFA M2

I Q = {q1,q2}, Σ = {0,1}, F = {q2}, δ =?

I What is L(M2) := {w ∈ {0,1}∗ : δ̂(q1,w) = q2}?

Theorem 10L(M2) = {w ∈ {0,1}∗ : #1(w) is odd}. #1(w) — number of ones inw.

I Proof by induction on the word lengthI What does it mean?

The j ’th assumption: w ∈ {0,1}j is in L(M) iff #1(w) is odd


Proving L(M2) = L′ := {w ∈ Σ∗ : #1(w) is odd}

I Basis (length 0): ε /∈ L′ and δ̂(q1, ε) = q1 =⇒ ε /∈ L(M2).I Step: assume hypothesis holds for words of length j ≥ 0.I Let x = yσ ∈ {0,1}j+1 for σ ∈ {0,1} (hence, |y | = j).I If #1(y) is even. By assumption δ̂(q1, y) = q1.

I σ = 1 =⇒1. #1(x) is odd =⇒ x ∈ L′

2. δ̂(q1, x) := δ(δ̂(q1, y), σ) = δ(q1, 1) = q2 =⇒ x ∈ L(M2).I σ = 0 =⇒

1. #1(x) is even =⇒ x /∈ L′

2. δ̂(q1, x) := δ(δ̂(q1, y), σ) = δ(q1, 0) = q1 =⇒ x /∈ L(M2).

I If #1(y) is odd...I We proved for x ∈ {0,1}j+1: x ∈ L(M2)⇐⇒ x ∈ L′


The language of M1

q1

0

q2 3q

11 0

0,1

Theorem 11

L(M1) = {w102k : k ≥ 0,w ∈ {0,1}∗}

Proof:

Claim 12 (implies the theorem)

Let L′i = {x ∈ {0,1}∗ : δ̂(q1, x) = qi} and letI L1 = {0k : k ≥ 0}I L2 = {w102k : k ≥ 0,w ∈ {0,1}∗}I L3 = {w102k+1 : k ≥ 0,w ∈ {0,1}∗}

Then, L′i = Li for every i ∈ {1,2,3}


Proving Claim 12

We need to prove that ∀i ∈ {1,2,3}:

x ∈ Li ⇐⇒ x ∈ L′i .

Recall that

x ∈ L′i ⇐⇒ δ̂(x) = qi

I Proof by induction on word length.I Induction basis: Easy to see that hypothesis holds for ε.I Induction step: Assume hypothesis holds for words of length j ≥ 0.I Let x = yσ ∈ {0,1}j+1 for σ ∈ {0,1}.I We prove the hypothesis for x , separately for each i ∈ {1,2,3}


L1 = {0k : k ≥ 0}

Recall x = yσ ∈ {0,1}j+1 for σ ∈ {0,1}.

Proving: x ∈ L1 =⇒ δ̂(q1, x) = q1 ( =⇒ x ∈ L′1).

I x = 0j+1, y = 0j and σ = 0.I Since y ∈ L1, by i.h. δ̂(q1, y) = q1

I Therefore, δ̂(q1, x) = δ(δ̂(q, y), σ) = δ(q1,0) = q1.

Proving: δ̂(q1, x) = q1 =⇒ x ∈ L1.

I Let qy = δ̂(q1, y) (hence, δ̂(q1, x) = δ(qy , σ) = q1)I qy = q1 and σ = 0. (?)I By i.h. y = 0j .I Hence, x = yσ = 0j0 = 0j+1 ∈ L1.


L2 = {w102k : k ≥ 0,w ∈ {0,1}∗}

Recall x = yσ ∈ {0,1}j+1 for σ ∈ {0,1}.

Proving x ∈ L2 =⇒ δ̂(q1, x) = q2.

I Assume σ = 1I Since δ(qi ,1) = q2 for any i =⇒ δ̂(q1, x) = q2.

I Assume x = w102k for k > 0 ( =⇒ y = w102k−1 and σ = 0)I Hence, y ∈ L3.I By i.h. δ̂(q1, y) = q3I Thus, δ̂(q1, x) = δ(q3,0) = q2.

Proving δ̂(q1, x) = q2 =⇒ x ∈ L2.

I Assume σ = 1 =⇒ x ∈ L2. (?)I Assume σ = 0

I qy := δ̂(q1, y) = q3I By i.h. y = w102k+1 for some k ≥ 0I Therefore x = yσ = w102k+10 ∈ L2.


L3 = {w102k+1 : k ≥ 0,w ∈ {0,1}∗}

Recall x = yσ ∈ {0,1}j+1 for σ ∈ {0,1}.

Proving x ∈ L3 =⇒ δ̂(q1, x) = q3.

I x = w102k+1, y = w102k and σ = 0I y ∈ L2

I By i.h. δ̂(q1, y) = q2.I Therefore, δ̂(q1, x) = δ(q2,0) = q3.

Proving δ̂(q1, x) = q3 =⇒ x ∈ L3.

I Let qy = δ̂(q1, y)

I Hence, qy = q2 and σ = 0 (?)I By i.h. y = w102k

I Therefore, x = yσ = w102k0 ∈ L3.


Part III

Regular Operations


Additional examples of regular languages

Let Σ = {0,1}.I {w ∈ {0,1}∗ : #1(w) ≡ 0 mod 7}.I Sequence of 0 followed by sequence of 1, i.e., {0m1n : m,n ≥ 0}.I Any finite language.

All the above languages are regular

Is there a simple proof?


The regular operations

Let A and B be languages.

The union operation:

A∪B = {x : x ∈ A ∨ x ∈ B}

The concatenation operation:

A‖B = {xy : x ∈ A ∧ y ∈ B}

The star operation:

A∗ = {x1x2 . . . xk : k ≥ 0 and each xi ∈ A}


The regular operations – Examples

Let A= {good, bad} and B = {boy, girl}.

UnionA ∪ B = {good, bad, boy, girl}

Concatenation

A‖B = {goodboy, goodgirl, badboy, badgirl}

StarA∗ = {ε, good, bad, goodgood, goodbad, badbad, badgood, . . .}


Closure under union

Theorem 13If L1 and L2 are regular languages, then so is L1 ∪ L2.

Approach to Proof:I Some DFA M1 accepts L1

I Some DFA M2 accepts L2

I Construct DFA M that accepts L1 ∪ L2.

Attempted Proof Idea:I first emulate M1, andI if M1 doesn’t accept, then emulate M2.

What’s wrong with this?Fix: Emulate both machines simultaneously.


Closure Under Union: Correct Proof

SupposeI M1 = (Q1,Σ, δ1,q1,F1) accepts L1,I M2 = (Q2,Σ, δ2,q2,F2) accepts L2.

Define M as follows (M will accept L1 ∪ L2):I Q = Q1 ×Q2.I Σ is the same.I For each (r1, r2) ∈ Q and a ∈ Σ,δ((r1, r2),a) = (δ1(r1,a), δ2(r2,a))

I q0 = (q1,q2)

I F = {(r1, r2) : r1 ∈ F1 or r2 ∈ F2}.I Formal proof (next slide) ♣

(hey, why not choose F = F1 × F2?)


Correctness of the construction

Claim 14

L(M) = L(M1) ∪ L(M2).

Follows by the next claim.

Claim 15

δ̂M((q1,q2), x) = (δ̂1(q1, x), δ̂2(q2, x)).

Proof: By induction on word length. DIY ... ♣Proving Claim 14:

I x ∈ L(M1) =⇒ δ̂1(q1, x) = r1 ∈ F1. (similar if x ∈ L(M2).)I By Claim 15, δ̂M((q1,q2), x) = (r1, ·) ∈ F =⇒ x ∈ L(M).

I x ∈ L(M) =⇒ δ̂M((q1,q2), x) = (r1, r2) ∈ F .I By Claim 15, (r1, r2) = (δ̂1(q1, x), δ̂2(q2, x)).I Hence, either r1 ∈ F1 or r2 ∈ F2 =⇒ x ∈ L(M1) ∪ L(M2).


What about concatenation?

Theorem 16If L1, L2 are regular languages, then so is L1‖L2.

Example: L1= {good, bad} and L2 = {boy, girl}.

L1‖L2 = {goodboy, goodgirl, badboy, badgirl}

This is much harder to prove.

Idea: Simulate M1 for a while, then switch to M2.

Problem: But when do you switch?

This leads us into non-determinism, wait for next class...


Part IV

Non-deterministic Finite Automata


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 May have transitions labeled with ε, the symbol of the empty

string. Will deal with this latter

Every DFA “is” also an NFA.


Non-deterministic computation

q4

q1

q2

q3

0

0,1 0,1

10

What happens when more than one transition is possible?I The machine “splits” into multiple copiesI Each branch follows one possibilityI Together, branches follow all possibilities.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


Why non-determinism?

Theorem 17 (Informal, to be proved soon)Deterministic and non-deterministic finite automata, accept exactly thesame set of languages.

Q.: So why do we need NFA’s?

Design a finite automaton for the language L— all binary strings witha 1 in their third-to-the-last position?


NFA for L

q4

q1

q2

q3

1

0,1

0,10,1

I “Guesses” which symbol is third from the last, andI checks that indeed it is a 1.I If guess is premature, that branch “dies”, and no harm occurs.


DFA for LI Have 8 states, encoding the last three observed letters.

I A state for each string in {0,1}3.

I Add transitions on modifying the suffix, give the new letter.

I Mark as accepting, the strings 1 ∗ ∗

1

q000

q111

q001

q010

q011

q100

q101

q110

0

0 0

0

000

1 1

1

1 1 1

DFA has few bugs...Iftach Haitner (TAU) Computational Models, Lecture 1 October 31, 2016 41 / 41

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