computational models #1tau-cm2017b.wdfiles.com/.../course-schedule/dfa1_2017.pdf · 2017-03-13 ·...

Computational Models #1

Handout Mode

Nachum Dershowitz & Yishay Mansour

March 13-15, 2017

Nachum Dershowitz & Yishay Mansour Computational Models #1 March 13-15, 2017 1 / 41

Lecture Outline

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

I Sipser, Chapter 1


Time Magazine

• Put the right kind of software into a computer, and it will do whatever you want it to. There may be limits on what you can do with the machines themselves, but there are no limits on what you can do with software.

Chess

• Games: 101050

• Positions: 1040

• Positions: 20, 400, 5362, 71852, 809896(?), 9132484(?), ...

states initial

terminal

Part I

Languages, words and alphabets


Languages, words and alphabets

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

I ⌃ = {a, b, c, . . . , z} – the English alphabet.I ⌃ = {↵,�, �, . . . , ⇣} – the Greek alphabet.I ⌃ = {0, 1} – the binary alphabet.I ⌃ = {0, 1, . . . ,9} – the digital alphabet.

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

The collection of all strings over ⌃ is denoted by ⌃⇤.For {0, 1}, the binary alphabet, ", 1, 0, 000000000, 1111111000 are allmembers of ⌃⇤.Definition 3A language over ⌃ is a (possibly infinite) subset of ⌃⇤.


Language Examples

I Modern English.I Ancient Greek.I All prime numbers, written using digits.I A = {w 2 {0, 1}⇤ : w has at most seventeen 0’s}.I B = {0n1n : n � 0}.I C = {w 2 {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,⌃, �, q0,F ),where

IQ is a finite set called the states

I ⌃ is a finite set called the alphabetI � : Q ⇥ ⌃ ! Q is the transition functionI

q0 2 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,⌃, �, q1,F ) whereI

Q = {q1, q2, q3}, ⌃ = {0, 1},

I the transition function � is

0 1q1 q1 q2q2 q3 q2q3 q2 q2

Iq1 is the start state

IF = {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,⌃, �, q0,F ) accepts w 2 ⌃⇤ if b�M

(q0,w) 2 F .

Definition 6 (b�)b�

M

: Q ⇥ ⌃⇤ 7! Q is defined by

b�M

(q,w) =

⇢�(b�(q,w1,...,n�1),wn

), n = |w | � 1q, w = ".

.

Iw1,...,k stands for the word w1, . . . ,wk

— the k -letter prefix of w

(w1,...,0 = w; = ")

I Note that b�M

(q,�) = �(q,�) for � 2 ⌃.I We write b� when M is clear from the context.


An equivalent definition

Definition 7 (Equivalent definition)M = (Q,⌃, �, q0,F ) accepts w = w1w2 . . .wn

, if 9r0, . . . , rn

2 Q s.t.,I

r0 = q0.I �(r

i

,wi+1) = r

i+1, for all 0 i < n.I

r

n

2 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).

IM may accept many strings

IM 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

IQ = {q1, q2}, ⌃ = {0, 1}, F = {q2}, � =?

I What is L(M2) := {w 2 {0, 1}⇤ : b�(q1,w) = q2}?

Theorem 10L(M2) = {w 2 {0, 1}⇤ : #1(w) is odd}. #1(w) — number of ones in

w.

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

The j ’th assumption: w 2 {0, 1}j is in L(M) iff #1(w) is odd


Proving L(M2) = L0 := {w 2 ⌃⇤ : #1(w) is odd}

IBasis (length 0): " /2 L0 and b�(q1, ✏) = q1 =) ✏ /2 L(M2).

IStep: assume hypothesis holds for words of length j � 0.

I Let x = y� 2 {0, 1}j+1 for � 2 {0, 1} (hence, |y | = j).I If #1(y) is even. By assumption b�(q1, y) = q1.

I � = 1 =)1. #1(x) is odd =) x 2 L0

2. b�(q1, x) := �(b�(q1, y),�) = �(q1, 1) = q2 =) x 2 L(M2).I � = 0 =)

1. #1(x) is even =) x /2 L0

2. b�(q1, x) := �(b�(q1, y),�) = �(q1, 0) = q1 =) x /2 L(M2).

I If #1(y) is odd...I We proved for x 2 {0, 1}j+1: x 2 L(M2) () x 2 L0


The language of M1

q1

0

q2 3q

11 0

0,1

Theorem 11L(M1) = {w102k : k � 0,w 2 {0, 1}⇤}

Proof:Claim 12 (implies the theorem)

Let L0i

= {x 2 {0, 1}⇤ : b�(q1, x) = q

i

} and letI L1 = {0k : k � 0}I L2 = {w102k : k � 0,w 2 {0, 1}⇤}I L3 = {w102k+1 : k � 0,w 2 {0, 1}⇤}

Then, L0i

= Li

for every i 2 {1, 2, 3}


Proving Claim ??

We need to prove that 8i 2 {1, 2, 3}:

x 2 Li

() x 2 L0i

.

Recall that

x 2 L0i

() b�(x) = q

i

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


L1 = {0k : k � 0}

Recall x = y� 2 {0, 1}j+1 for � 2 {0, 1}.

Proving: x 2 L1 =) b�(q1, x) = q1 ( =) x 2 L01).

Ix = 0j+1, y = 0j and � = 0.

I Since y 2 L1, by i.h. b�(q1, y) = q1

I Therefore, b�(q1, x) = �(b�(q, y),�) = �(q1, 0) = q1.

Proving:

b�(q1, x) = q1 =) x 2 L1.

I Let q

y

= b�(q1, y) (hence, b�(q1, x) = �(qy

,�) = q1)I

q

y

= q1 and � = 0. (?)I By i.h. y = 0j .I Hence, x = y� = 0j0 = 0j+1 2 L1.


L2 = {w102k : k � 0,w 2 {0, 1}⇤}

Recall x = y� 2 {0, 1}j+1 for � 2 {0, 1}.

Proving x 2 L2 =) b�(q1, x) = q2.

I Assume � = 1I Since �(q

i

, 1) = q2 for any i =) b�(q1, x) = q2.

I Assume x = w102k for k > 0 ( =) y = w102k�1 and � = 0)I Hence, y 2 L3.I By i.h. b�(q1, y) = q3I Thus, b�(q1, x) = �(q3, 0) = q2.

Proving

b�(q1, x) = q2 =) x 2 L2.

I Assume � = 1 =) x 2 L2. (?)I Assume � = 0

Iq

y

:= b�(q1, y) = q3I By i.h. y = w102k+1 for some k � 0I Therefore x = y� = w102k+10 2 L2.


L3 = {w102k+1 : k � 0,w 2 {0, 1}⇤}

Recall x = y� 2 {0, 1}j+1 for � 2 {0, 1}.

Proving x 2 L3 =) b�(q1, x) = q3.

Ix = w102k+1, y = w102k and � = 0

Iy 2 L2

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

Proving

b�(q1, x) = q3 =) x 2 L3.

I Let q

y

= b�(q1, y)

I Hence, q

y

= q2 and � = 0 (?)I By i.h. y = w102k

I Therefore, x = y� = w102k0 2 L3.


Part III

Regular Operations


Additional examples of regular languages

Let ⌃ = {0, 1}.I {w 2 {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 2 A _ x 2 B}

The concatenation operation:

AkB = {xy : x 2 A ^ y 2 B}

The star operation:

A⇤ = {x1x2 . . . xk

: k � 0 and each x

i

2 A}


The regular operations – Examples

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

UnionA [ B = {good, bad, boy, girl}

Concatenation

AkB = {goodboy, goodgirl, badboy, badgirl}

StarA⇤ = {", 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,⌃, �1, q1,F1) accepts L1,I

M2 = (Q2,⌃, �2, q2,F2) accepts L2.Define M as follows (M will accept L1 [ L2):

IQ = Q1 ⇥ Q2.

I ⌃ is the same.I For each (r1, r2) 2 Q and a 2 ⌃,

�((r1, r2), a) = (�1(r1, a), �2(r2, a))

Iq0 = (q1, q2)

IF = {(r1, r2) : r1 2 F1 or r2 2 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

b�M

((q1, q2), x) = (b�1(q1, x), b�2(q2, x)).

Proof: By induction on word length. DIY ... |Proving Claim ??:

Ix 2 L(M1) =) b�1(q1, x) = r1 2 F1. (similar if x 2 L(M2).)

I By Claim ??, b�M

((q1, q2), x) = (r1, ·) 2 F =) x 2 L(M).

Ix 2 L(M) =) b�

M

((q1, q2), x) = (r1, r2) 2 F .I By Claim ??, (r1, r2) = (b�1(q1, x), b�2(q2, x)).I Hence, either r1 2 F1 or r2 2 F2 =) x 2 L(M1) [ L(M2).


What about concatenation?

Theorem 16If L1, L2 are regular languages, then so is L1kL2.

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

L1kL2 = {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

q1q2

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 ", the symbol of the empty

string. Will deal with this latter

Every DFA “is” also an NFA.


Non-deterministic computation

q4

q1q2

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

q1q2

q3

0

0,1 0,1

10

q1

q1

q2

q1

q2

q1 q3

q1q4

1

0

0

1

symbol


Why non-determinism?

Theorem 17 (Informal, to be proved soon)Deterministic and non-deterministic finite automata, accept exactly the

same 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

q1q2

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

q111q001

q010

q011

q100

q101

q110

0

0 0

0

000

1 1

1

1 1 1

DFA has few bugs...Nachum Dershowitz & Yishay Mansour Computational Models #1 March 13-15, 2017 41 / 41

computational models #1tau-cm2017b.wdfiles.com/.../course-schedule/dfa1_2017.pdf · 2017-03-13 ·...

Documents