CSC 473 Automata, Grammars & Languages 9/29/10 · CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 1 C SC 473 Automata, Grammars & Languages Automata, Grammars and Languages

CSC 473 Automata, Grammars & Languages 9/29/10

Lecture 03 1

C SC 473 Automata, Grammars & Languages

Automata, Grammars and Languages

Discourse 03

Finite Automata

C SC 473 Automata, Grammars & Languages 2

Finite Automata / Switching Theory(CS) / (CE)

• Boolean operators / Gates (Elem. Switching Ops)

01111

011001

11010

100000

¬ x x ⊕ y x ∨ y x ∧ yyx


Boolean Functions / Combinatorial Circuits• Circuit

∨

∧

∧

¬

1x

2x

1(sum)z

2(carry)z

1 2x x!

1 2x x!

1 2x x!

1 2x x!

H half adder

H H1x

2x

∨3x

1 2x x!

1(sum)z

1 2x x!

1 2 3( )x x x! "

2(new carry)z

F full adder

(old carry)


Lecture 03 2


Boolean Functions / Comb. Circuits (contʼd)• Table representing F

11 1 1 1

10 0 1 1

10 1 0 1

01 0 0 1

10 1 1 0

01 0 1 0

01 1 0 0

00 0 0 0

z2z1x3x2 x1


Boolean Functions / Comb. Circuits (contʼd)• Equations representing F

• General scheme (n inputs, m outputs)

1 1 1 2 3

2 2 1 2 3 1 2

( )

( ) (( ) ) ( )

z g x x x x

z g x x x x x x

= = ! !

= = ! " # "

1 2 1 2

1 1 2 1 2

( )

( , , , ) ( , , , )

( ( , , , ), , ( , , , ))

m n

n m n

z g x

z z z g x x x

g x x x g x x x

=

= =… …

… … …


Finite Automata / Sequential Circuits• Add “memory” elements = delay elements

• Finite # of delay elements possible ⇒ ∃ d

( )y i

( ) ( ( ), ( 1), ( 2), )z i f x i x i x i= ! ! …

( 1)y i +

x zfcombinatorialcircuit

( ) ( ( ), ( 1), ( ))z i f x i x i x i d= ! !…


Lecture 03 3


Finite Automata / Sequential Circuits• Ex: sequential adder: add 2 binary numbers; low order bits received first

• (a) sequential net (circuit):

100101011110

carry

1( )x i

2( )x i

1( )z i

fulladder F

1( 1)y i +

1( )y i

1

2

(0)

(0)

x

x


Finite Automata / Sequential Circuits• (b) Next-State & Output Equations:

• (c) Transition Table: state space

1 1 2 1 1 2

1 1 2 1

( 1) (( ( ) ( )) ( )) ( ( ) ( ))

( ) ( ) ( ) ( )

y i x i x i y i x i x i

z i x i x i y i

+ = ! " # "

= ! !

0 1{0,1} { , }q q= =

input alphabet ={(00),(01),(10),(11)}

output alphabet ={0,1}

!

"

(00) (01) (10) (11)

0q

1q

0/ 0q

next state/outputtable

q

0/ 1q

0/ 1q

0/ 1q

1/ 0q

1/ 0q

1/ 0q

1/ 1q

x


Finite Automata / Sequential Circuits• (d) State Diagram:

(00)/ 1

(00)/ 0

(01)/ 1

(10)/ 1

0q 1

q

(11)/ 0

(11)/ 1

(01)/ 0

(10)/ 0


Lecture 03 4


Finite Automata / Sequential Circuits• (e) Finite-State Transducer (Mealy Machine)

A 5-tuple where0

( , , , , )M Q q!= " #

0 1

0

0 0 0 0

0 0 0 1

1 1

{ , }

{(00),(01),(10),(11)}

{0,1}

:

( ,(00)) ( ,0) ( ,(01)) ( ,1)

( ,(10)) ( ,1) ( ,(11)) ( ,0)

( ,(01)) ( ,0) .

Q q q

q

Q Q

q q q q

q q q q

q q etc

!

! !

! !

!

=

" =

# =

$ " % $ #

= =

= =

=

finite set of states

start stateinput alphabet

output alphabet

transition/output function


General Sequential Network

1

n

x

x

!

1

m

z

z

!

1

s

y

y

!

…

:n s

m s

B

B

!+

+

"

{0,1}B =

state space =s

B

( ( ), ( )) ( ( 1), ( ))

( , ) ( , )

y i x i y i z i

q a q b

!

!

= +

"=

δ is abooleanfunction


Three Types of Automata

M

M

M

transducer

recognizer (acceptor)

Enumerator (generator)

3 2 1na a a a!

3 2 1nc c c c!

time

3 2 1na a a a!

3 2 1nc c c c!

ia ! "

Yes(1)No(0)


Lecture 03 5


Machines that Recognize• Detection of an “event”, i.e., a pattern in input• Recognition of just those words in some language L• Definition of a language• Ex: detect abab –all non-overlapping occurrences

ab

ab

b

a

a b a b


Ex: C Comments /* … */• Filter in the lexical scanner

• Recognizer

{,/}a !" # =

:aa:! !

/:!

/:!

:/a a

:!"

/:/

:!"

empty

/:!:a ! :!"

:a!

/a !

a

!/ ! /

(transducer) notationin : out


Finite Automaton (Finite State Machine, FSA)

• Defn 1.5: A (deterministic) finite automaton is a 5-tuple

• is a finite set, the states• is a finite set, the alphabet• is the transition function• is the start state• is the set of accepting (final) states

• Ex:

0( , , , ),M Q q F!= "

F Q!

Q

: Q Q! " # $

!

0q Q!

00( , , , ),M Q q F!= " 0 1 2

{ , , }Q q q q=

1{ , } { }a b F q! = =

0 1 0 2

1 2 1 0

2 2 2 2

( , ) ( , )

( , ) ( , )

( , ) ( , )

q a q q b q

q a q q b q

q a q q b q

! !

! !

! !

= =

= =

= =

0q 1

q

2q

a

b

b a

,a b

0M


Lecture 03 6


How FA Compute• FA is a finite structure—like

a program—fixed and static• Need to define the behavior of M on input w

Sequence of configurations Like trace of a program on given data Dynamic and input-dependent

• Ex: start on input Look at sequenceof “moves” determined by the transition function:

Since in accepting state when input exhausted, w isrecognized by

0( , , , ),M Q q F!= "

0M w ababa=

(q0,ababa)!(q

1,baba)!(q

0,aba)!

(q1,ba) ! (q

0,a)!(q

1,!)

" (q0,ababa)!

#(q

1,!)

0M

C SC 473 Automata, Grammars & Languages

www.jflap.org

JFLAP is a package of graphical tools which can be used as an aid inlearning the basic concepts of Formal Languages and Automata Theory.


Info on JFLAP• Website http://www.jflap.org/

Downloads Tutorial

• Lectura (linux) install cd /usr/local/jflap java -jar JFLAP.jar

• X11 forwarding (graphics) ssh -X lectura


Lecture 03 7


How FA Compute (contʼd)• Given a FA• Defn: configuration of M is an element of• Defn: yields in one step (or moves) relation

between configurations is defined by

where Notes: is a function, since δ is. Is undefined.

• Defn: yields is the relation Means “moves in zero or more steps to”

• Defn: A string w is recognized (accepted) by M ⇔

0( , , , ),M Q q F!= "

(q,aw)!( !q ,w)" #(q,a) = !q

Q!

" #

!

, , ,a w q q Q!

"# $ # $ #

!

(q,!)!

!!

(q,w)!

!(q,w)

(! f " F)(q

0,w)!

#(f,$)


How FA Compute (contʼd)• Defn: The language recognized (accepted) by M is

• Defn 1.16: A language S is regular iff there is some FAthat recognizes it, i.e.,

• Ex: In FA

( )[ is a FA ( )]M M S L M! " =

L(M) = {w : M accepts w}

= {w :(! f " F)(q0,w)!

#(f,$)}

(q0,a) !

!(q

1,")

(q0,aba)!

!(q

1,")

!

(q0,(ab)

ia)!

!(q

1,")

# L(M0) = {(ab)

ia : i $ 0}

0M


Example:• Coin checker for 30¢ coffee. Σ={n,d,q}

15

05

10

3025

20

35

40

45

50

i = make change for i – 30 ¢ & vend coffee

q

n

n

n

n

n

n

d

d

d

d

dd

q

q

q

q

q


Lecture 03 8


Regular Operations & Regular Expressions

• The regular operations on languages are: union (∪), concatenation (⋅ ) and Kleene star (*).

• So called because the class of regular languages areclosed under them—i.e., applying these operators toregular languages results in a regular language. (We willprove these closure results later.)

• In fact, these three operations (∪, ⋅ ,*) actually characterizewhat it means to be a regular language: any regularlanguage can be built up from alphabet symbols and afinite number of these regular operations.

• This motivates the notion of regular expression: asequence of symbols, like an arithmetic expression, thatdefines a regular language using regular ops.


Regular Expressions• A syntax for describing sets of strings (languages)

Terse Eliminates fussy “{ }” Reminiscent of arithmetic expressions Obeys some useful “algebra”, e.g., (E*F*)* = (E+F)*

• Syntax for regular expressions over Σ,+, ⋅ ,*,(,) E → (E+E) ( text uses ∪ not +; some authors use |) → (EE) (usually suppress the ⋅ in E ⋅ E) → (E*) → ε (some authors use λ ) → ∅ → a for each a in Σ

• suppress (,) where possible: (a+b)*a not ( ( (a + b)* ) ⋅ a )


Regular Expressions (contʼd)• Meaning rules for the syntax• The meaning (denotation) of an expression, L(E), is a set

of strings (a language)• Rules expression E language L(E) ∅ a {a} ε {ε} (E+F) L(E)∪L(F) (EF) L(E)⋅L(F) (E*) L(E)*

{}! =


Lecture 03 9


Reg. Expr.: Examples, Equivalence(=)• (a+b)*a• (a*b*)*• =(a+b)*• (a+b)*a(a+b)*a(a+b)*• (b*ab*ab*ab*)*PASCAL unsigned numbers. d={0,1,…,9}• dd*(ε +.dd*)(ε+E(+ + – + ε)dd*)• aΣ*a+b Σ*b+a+bDefn: E=F ⇔L(E)= L(F)

∅*= ε (E*F*)*=(E+F)* E∅=∅E=∅ E(FG)=(EF)G E(F+G)=EF+EG Eε=εE=E

L { , }{ }a b a!

({ }{ })a b! ! !

{ , }a b!

=

{ : has 2 a's}w w! " #

{w ! " : # a's divisable by 3}

{ : ??}w ! "

{ : begins & ends same}w w! "


Nondeterminism• Real computing devices are deterministic: the current

configuration and instruction determines the nextconfiguration. The relation is a function.

• Why the concept of nondeterminism? Provides powerful, economical descriptive ability Provides a way to specify languages without over-specifying and

complex handling of cases Can be algorithmically converted to a deterministic description (at

the sacrifice of some economy and with added complexity) Generalization of determinism

• Ex: abab occurs somewhere in w: …abab…

!

a b a b

a,ba,b a,b

ab a


Nondeterminism (contʼd)• Ex: w∈Σ* has penultimate symbol b: w = …b?

• Ex: w∈Σ* has ≥ 2 a’s: w = … a … a …

b b

a,b a

b

a,ba

a a

a,b a,ba,b


Lecture 03 10


ε-Moves Can Be Useful• SNOBOL arithmetic constants (no floating E)

Use to specify “optional characters” like Unix command line [opt]

ε

−

d

d=digit

d d

ε

••

d


Nondeterministic Finite Automaton• Defn 1.5: A nondeterministic finite automaton is a 5-tuple

is a finite set, the states is a finite set, the alphabet transition function is the start state is the set of accepting (final) states

• Ex:

0( , , , ),M Q q F!= "

F Q!

Q

: ( { }) ( )Q Q! "# $ % & P

!

0q Q!

01( , , , ),M Q q F!= "

0 1{ , }Q q q=

1{ , } { }a b F q! = =

0 0

0 0 1

1 1 1 1

( , ) { }

( , ) { , }

( , ) { } ( , ) { }

q a q

q b q q

q a q q b q

!

!

! !

=

=

= =

1Mb

a,b a,b

0q

1q


DFA vs NFA• DFA δ

• For each state q and input symbola, there is exactly one choice ofnew state (or no transition isdefined at all). Each transition“consumes” an input symbol

• Special case of NFA!

• NFA δ

• There may be multiple choices forthe same input symbol

• There may be ε-moves that do not“consume” an input character

• There can be “chains” of ε-moves• ε-moves can create even more

choice for the next input character

qa

b

qa

a

b

b

ε

ε


Lecture 03 11


How NFA Compute• Given a NFA• Defn: configuration –• Defn: yields in one step (or moves) relation

between configurations

• Defn: yields = Means “COULD move in zero or more steps to”

• Defn: w is recognized (accepted) by M ⇔

Same as before, but has the meaning “if there exists somesequence of moves from the start config to some accepting config”

0( , , , ),M Q q F!= "

(q,aw)!( !q ,w)" !q # $(q,a)

( , )q w Q!

" # $

!

!!

(! f " F)(q

0,w)!

#(f,$)

(q,w)!( !q ,w)" !q # $(q,%) (ε-move)


How NFA Compute (contʼd)• Defn: The language recognized (accepted) by M is

• Ex: In NFA This provides no “evidence” that aabbba is accepted (or not) However, also via a separate computation sequence:

And so aabbba is recognized!

L(M) = {w : M accepts w}

= {w :(! f " F)(q0,w)!

#(f,$)}

(q

0,aabbba) !

!(q

0,")

1M

(q

0,aabbba) !

!(q

1,")


“Tree” of Computations• Ex: NFA M1 0

( , )q aabbba

0( , )q abbba

0( , )q bbba

1( , )q bba

0( , )q bba

0( , )q ba 1

( , )q ba

0( , )q a

1( , )q a

0( , )q !

1( , )q !X null

“evidence”

∃ acceptingComputation⇒ w∈L(M1)

1q F!

! =


Lecture 03 12


Computation Tree: Example• Ex: L = {w: w begins & ends same } 2 a

3

4

a,b

1

a,b

bb

a

( , )2 ! ( , )3 !

( , )1 ababa

( , )2 baba

( , )2 aba

( , )2 ba ( , )3 ba

( , )2 a

X3∈F

( , )1 abab

( , )2 bab

( , )2 ab

( , )2 b ( , )3 b

( , )2 !

XX

reject abab¬∃ path to Faccept ababa

∃ some path to F


Example with ε-Moves• String length a multiple of 2 or 3

0

2

4

3ε

ε

a

5

6a

aaa( , )0 aaa

( , )2 aaa ( , )4 aaa

( , )3 aa

( , )2 a

( , )5 aa

( , )6 a

( , )3 ! ( , )4 !

!

X 4∈F ⇒Accept aaa

← ε-moves


Example with ε-Moves• a*b*

( , )0 aab

( , )1 aab ( , )0 ab

( , )1 ab ( , )0 b! X

1∈F ⇒Accept aab

X ( , )1 b

( , )1 !

ε

a

0 1

b

ε-moves “consume” no input symbols


Lecture 03 13


Equivalence of NFA to DFA• There is an algorithm to convert any NFA into a DFA

We show basic idea assuming NFA has no ε-moves Then (later) modify the construction for NFAs with ε-moves

• Ex: L = {x : last symbol of x appeared previously } Σ={a,b}

• Idea: given input string, keep track of all possible reachedstates after reading each letter. At end of input, see if afinal state is among those reached

p

q

r

s

a

b

a,ba,b

a,b

a

b

N0:


Equivalence of NFA and DFA (contʼd)• Computation paths through NFA N0 on w = abba

p p p p p

r r r

q q q q

r r

s

s

q

a

a b b a

a

b a

b b a

b

b b a

as F!


Equivalence of NFA and DFA (contʼd)• Idea: keep a list of all possible states reachable by each

prefix of w (“parallel worlds”). For NFA N0:

0

{ } { , { , { , { ,

} , , ,

} , ,

} }

{ } { , , , }

( ) since { , , , }

a b b a

abba

p p p p p

q q q q

r r r

s s

p p q r s

abba L N p q r s F

! ! ! !

" !

# $ % &


Lecture 03 14


Equivalence of NFA and DFA (contʼd)• Equivalent DFA M will have:

State set P (Q) Alphabet Σ Start state “set” {q0} Accepting states {X ⊆ Q : X ∩ F ≠ ∅} Deterministic transition function δ′ : P (Q) × Σ → P (Q)

• Ex: For NFA N0:({ , }, ) { , , }

({ , }, ) { , , }

({ }, ) { , }

({ }, ) { , }

p q a p q s

p q b p q r

p a p q

p b p r

!

!

!

!

" =

" =

" =

" =

…


Equivalence of NFA and DFA (contʼd)• Thm: [Rabin-Scott Construction]. Let L = L(N) for some NFA

N with no ε-moves. There is an algorithm to constuct a DFAM equivalent to N, i.e. with L(M) = L(N).

Pf: Given N we construct a DFA M and then verify that itrecognizes the same set as N.

Construction: Given NFA construct where

( , , , , )N Q s F!= "

( , , , , )M Q s F!" " " "= #

!Q = P(Q), !s = {s}, !F = {X " !Q : X # F $ %}

and !& : !Q ' ( ) !Q is defined as:

(*S + Q,a " () !&(S,a)

= {q " Q :(,p " S)q " &(p,a)}


Equivalence of NFA and DFA (contʼd)• Picture of δ′

!"(S,a) = !S

a

aa

b

a

b

S!

S

S Q S Q!" "


Lecture 03 15


Equivalence of NFA and DFA (contʼd)• Verification: Show (1) M is a DFA and (2) L(M)=L(N) (1) δ′ is a function by the construction, and Q′ is finite: |Q′ | = 2|Q| . So M is a DFA. (2) To show equivalence we prove the• Lemma:

Pf: By induction on the length of the input string w. Base |w|=0.

Step Suppose (IH) the lemma is true Let To show:

(p,w) !

N

!(q,") # $Q.q % Q & ({p},w)!

M

!(Q,")

(p,!) !

N

"(q,!) # p = q # ({p},!)!

M

"({q},!)

.| | .w w k! "

| w |= k + 1,w = ua,|u|= k.

(p,ua) !

N

!(q,") # $Q.q % Q & ({p},ua)!

M

!(Q,")


Equivalence of NFA and DFA (contʼd) ⇒. Assume Then ∃ state r with and Then By (IH) (*)By construction of M Let ThenUsing this with (*) results in:

So

(p,ua) !

N

!(q,").

(p,ua) !

N

!(r,a) !

N(q,")

(p,u) !

N

!(r,").

!R.r " R # ({p},u)!

M

$(R,%)

( , ).q r a!"

( , ).q R a! "#

( , ).Q R a! "= !Q.q " Q # (R,a)!

M

$(Q,%).

!Q.q " Q # ({p},ua)!

M

$(R,a)!

M

$(Q,%).

!Q.q " Q # ({p},ua)!

M

$(Q,%).


Equivalence of NFA and DFA (contʼd) ⇐. Assume Then ∃ state R with So (1)By construction and (2)Since we have from (IH) (3)Combining (2) & (3):

So (p,ua) !

N

!(q,"). !

(p,ua) !

N

!(r,a) !

N(q,")

({p},u)!

M

!(R,")

. ( , )r R q r a!" # #

( , ) .R a Q! " =

({p},ua)!

M

!(R,a)!

M

!(Q,").

!Q.q " Q # ({p},ua)!

M

$(Q,%).

(r,a) !

N

!(q,").

(p,u) !

N

!(r,")


Lecture 03 16


Equivalence of NFA and DFA (contʼd) We now finish the verification proof. Let From the Lemma

That is, for some for some

( !s,w)!

M

"(Q,#)

!Q.f " Q # ({s},w)!

M

$(Q,%) &

.f F!

(s,w) !

N

!(f,") #

( !s,w)!

M

"(Q,#) $ Q % F & '.

(s,w) !

N

!(f,") f F! "

.Q F !"

( ) ( ).L M L N! = !


Example: ε-Free NFA → DFA• Consider the previous NFA

0( , , , ),{ }N Q p s!= "

0( ( ), , ,{ )},M Q p F! "= # "P

{p}

{p,q}

{p,r}

{p,q,s}

{p,q,r}

{p,r,s}

{p,q,r,s}

a

a

b

b

a

b

a

ba

b

a,b

b

a


• ε-closure(R) = E(R) for a set of

states R

a

1

2

34

5

6

7

8

9 10

11

12

1314

15

ε

ε

a

ε

ε

ε

a

ε

b

ε

ε

ε

c

b

d

!-closure({4, 5})=E({4, 5})={1, 2, 3, 4, 5, 6,7, 8}

!"closure(R) = {q # Q :($r # R)(r,!) !

%(q,!)}

For R ⊆ Q the ε-closure, E(R) of R is:

R

NFA with ε-Moves


Lecture 03 17


ε-closure of a set of states• Coalesce all nodes reachable from {4,5} by ε-moves:

a

910

11

12

131415

a

ab

c

bd{1,2,3,4,5,6,7,8}

Note: still anNFA

E({4,5})

E({9})

E({13})

a

a

Etc.a

!-closure({4, 5})=E({4, 5})={1, 2, 3, 4, 5, 6,7, 8}


Conversion: NFA → DFA• Thm: There is an algorithm to convert any NFA to an

equivalent DFA .• Pf: Construction: Given NFA

construct new NFA where

Verification.

Pf: By induction on |w|

M = (2Q,!, "# , "s, "F )

( , , , , )N Q s F!= "

!F = {S " Q | S # F $ %}

!" : 2Q# $ % 2

Q

!s = E(s)

!"(S,a) = !{E(p)| p # "(q,a) for some q # S}

(q,w) !N

!(p,") # (E(q),w) !

M

!(P,")

for some set P containing p


Conversion: NFA → DFA• Thm 1.39: [Rabin-Scott Theorem]: There is an algorithm

to convert any NFA into an equivalent DFA.• Corollary 1.40: A language is regular ⇔ some NFA

recognizes it.• Ex: Start with an NFA N1 as follows:

ε

−

1 2 3 4b

b b

dε

1N

! = {b,",d}


Lecture 03 18


Conversion: NFA → DFA• Ex:

ε

−

1 2 3 4b

b b

dε

1N

E(1) = {1} E(2) = {1,2,3}

E(3) = {1,3} E(4) = {4}

1!

b

2 2!

b

2 3!

b

4 4!

b

"

1!

#

" 2!

#

3 3!

#

" 4!

#

"

1!

d

" 2!

d

" 3!

d

" 4!

d

3

Useful summary


Conversion: NFA → DFA• Ex:

ε

−

1 2 3 4b

b b

dε

1N

!s = E(1) = {1}

!"( !s,b) = E(2) = {1,2,3}

!"( !s,#) = $

!"( !s,d) = $

!"({1,2,3},b) = E(1)%E(2)%E(3)%E(4) = {1,2,3,4}

!"({1,2,3},#) = E(3) = {1,3}

!"({1,2,3},d) = $


Ex: NFA → DFA (contʼd)• Ex:

ε

−

1 2 3 4b

b b

dε

1N

!"({1,3},b) = E(2)#E(4) = {1,2,3,4}

!"({1,3},$) = %

!"({1,3},d) = %

!"({1,2,3,4},b) = E(2)#E(4) = {1,2,3,4}

!"({1,2,3,4},$) = E(3) = {1,3}

!"({1,2,3,4},d) = E(3) = {1,3}


Lecture 03 19


Conversion: NFA → DFA (contʼd)

1M

−1 123 13 1234

b

b

b

d

− b


Regular Expression → NFA• Thm 1.55: There is an algorithm that, given a regular

expression E, constructs a NFA N such that L(E) = L(M). Pf: Induction on the # of operator symbols in E. Base: E = ε ∅ a∈Σ

Step: Assume (IH) the result is true of all expressionswith ≤ operator symbols (+,⋅,*). Let E have k+1 ops.

Three cases: Case E = (E1+E2). By IH, ∃ FA M1 , M2 with L(E1) = L(M1)

and L(E2) = L(M2). Construct the following NFA M.

a


1M

2M

ε

ε1F

2F

M

1 2F F F= !

1 2( ) ( ) ( )L M L M L M= !

Case +


Lecture 03 20


Regular Expression → NFA (cont”d)• Case E = (E1⋅E2). By IH, ∃ FA M1 , M2 with L(E1) = L(M1)

and L(E2) = L(M2). Construct the following NFA M.


2M

ε

M

2F F=

1 2( ) ( ) ( )L M L M L M= i

Case •

2F

1F

1M

ε

Unmark final states in M1


Regular Expression → NFA (cont”d)• Case E = (E1)*. By IH, ∃ FA M1 with L(E1) = L(M1).

Construct the following NFA M.


Lecture 03 21


s

M

1{ }F F s= !

1( ) ( )L M L M

!=

Case *

1F

1M

ε

εε

QED


ε

b

εε

ε a ε a

ε

b

a

a

Example: Reg. Exp.→NFA• (b+aa)*

Not very economical


Regular Expressions—Applications• Regexp used in various development tools

qed – interactive text editor. 1st version Lampson & Deutsch 1967◆ Regexp added by Ken Thompson, Bell Labs, ca. 1968

Regexp compiled into NFA in machine code Rabin-Scott idea used to scan “on the fly” One of the first software patents

◆ Offspring ed by Ken for Unix◆ Many others followed: em, vi / ex, sam, qedx, …

grep, egrep - pattern search in a file shell – command line interpreter lex – lexical analyzer generator sed – non-interactive stream editor awk – pattern scanning and processing language perl – pattern-driven programming language


Lecture 03 22


Applications (contʼd)• Regular expressions = “patterns” meaning awk regexp matches >=1 r r+ matches >=0 r r* matches 0 or 1 r r? matches r then s rs

matches r or s r|s match literal “c” \c match begin/end line ^ $

match any char . group exprs (s)

character list [abc…] negated char list [^abc…]


Applications--Examples -?[0-9]+ nonempty digit strings, optional sign[^0-9] any char except digit\[.*\] reference citations in a paperg/^[ ]*$/d delete blank linesg/[ ]+/d delete lines with a blankEx: match is always (1) leftmost and (2) longest file: abcddddef vi: s/d*/x/ → xabcddddef s/d+/x/ → abcxefEx: csh: sort roll[1-5] | egrep “C SC|MATH” | pr


Applications--ExamplesEx: traditional spelling mnemonic• “i before e, except after c, or when pronounced ̒ aʼ, as in ʻneighborʼ and ʻweighʼ --except for ʻweirdʼ examples.”• grep “[^c]ei” /usr/share/dict/words > foo cat foo abseil Aeneid ageing Alamein albeit atheistBoeing Budweiser caffein canoeist deice deicticdilettanteism dreidl ...

• if you think this spelling rule is sufficient, you will be deficient,inefficient, unscientific and far from omniscient


Lecture 03 23


Applications--ExamplesEx: lex – generates a lexical analyzer yylex(). Example: wordcount (wc)%{int nchar, nword, nline;%}%%\n { nline++; nchar++; }[^ \t\n]+ { nword++, nchar += yyleng;}

// yyleng = length of matched string. { nchar++;}%%int main(void) { yylex(); // invoke generated lexer printf("%d\t%d\t%d\n", nchar, nword, nline); return 0;}


L Regular ⇔ L Denoted by a Reg. Expr.• Weʼve defined “regular” as meaning: recognized by a DFA

(equiv. to rec. by an NFA)• This equivalence result is known as “Kleeneʼs Theorem”• Weʼve already shown the ⇐ direction—we constructed an

NFA from a regular expression (Using Rabin-Scott wecould convert this NFA to a DFA.)

• Now we show the ⇒ direction: given a DFA M constructa regular expr. E with L(M) = L(E).

• Thm (Kleene): There is an algorithm that, given a DFA M ,computes a regular expression E such that L(M) = L(E).

• Pf: Given the graph of the DFA, use the “node eliminationalgorithm” to gradually eliminate all nodes in favor ofexpressions on the edges of the graph.


B

C

1

0

01

A 1 0

Sa

ε ε

B

C

1

0

01

A 1 0

B1

00

0

1

ε ε

Kleeneʼs Thm: use “Generalized NFA”

A

Sa

01

A

Sa

1(1+01)*00+0

εε

S

a

ε(1(1+01)*00+0)*ε

E =(1(1+01)*00+0)*

1. Add init. S, &final a

2. Elim. C

3. Elim. B 4. Elim. A

Order: CBA


Lecture 03 24


Ex: Node Elimination AlgorithmB

A

b

b

ab

a

C

B C

A

b

b

ab

a

Add ε-moves:

S εa

ε


B Cb

b

ba*a

S εa

ε

B

bb

bba*a

S εa

b

(bb+bba*a)*bS a

Elim. A:

Elim. C:

Elim. B:

ACB

Ex: Node Elimination Algorithm (ACB)

A

AC


B C

A

b

b

ab

a

S εa

ε

Ex: Node Elimination: other orders


Lecture 03 25


Bb

S εa

B

bb

bba*a

S εa

b

(bb+bba*a)*bS a

Elim. C:

Elim. A:

Elim. B:

CAB=ACB

Ex: Other elimination orders (CAB)

bb

A

bba

a

CA= AC (above)

C


Bb

S εa

A

S a

(bb)*b

(bb)*b+(bb)*bba*a(bb)*bS a

Elim. C:

Elim. B:

Elim. A:

CBA

Ex: Other elimination orders (CBA)

bb

A

bba

a

CB

C

(bb)*bb

a

a(bb)*b


A

S a

b(bb+ba*ab)*S a

Elim. B:

Elim. A:

BAC=ABC

Ex: Other elimination orders (BAC)

BA =AB

B

a

Cb

bbε

bab

S aCb bb

ba*ab

ε

Elim. C:


Lecture 03 26


Ex: All elimination orders equiv (BAC = ACB)

BAC b(bb+ba*ab)*

ACB (bb+bba*a)*b

Easy to prove by induction: for any expression E, b(Eb)*=(bE)*b.Using this identity:

b(bb+ba*ab)* = b[ (b+ba*a) b]* = [b (b+ba*a) ]*b

= (bb+bba*a)*b

Further regular expression simplication is possible:

b(bb+ba*ab)*=b[b(b+a*ab)]*=b[b(ε+a*a)b]* =b[ba*b]*

Good exercise: show results of all other elimination orders areequivalent to these, using regular expression algebra


Algebra of Regular Expressions• ∃ an algebra for symplifying regular expressions• Can use this algebra to construct RegExps from FSA

r+s = s+r (r+s)+t=r+(s+t) r+r=r r+∅=r

(rs)t = r(st) rε = εr = r r∅ = ∅r = ∅

r(s+t) = rs + rt (r+s)t = rt + st

∅* = ε (r*)* = r* r* = r + r*

r* = ε + rr* (r*s*)* = (r+s)*


Solving Regular Expr Equations• Can solve “linear equations” with regexp variables

X = aX + b

=a(aX + b) + b = a2X + ab + b

= a2 (aX + b) + ab + b = a3X + a2b + ab + b … X = a*bCheck: a[a*b] + b = aa*b + b = (aa*+ε)b = a*bEx:

Xb

a

YX = aX + bYY = ε X = a*b


Lecture 03 27


Solving RegExp Equations (contʼd)• Ex: NFA RegExp B

C

1

0

01

A 1 00 1

1 00 1

___ elim. ____ 10

0 110

0 110

___ elim. ____ (110)0

0 110(110)0

[0 110(110)0]

[0 110(110)0]Simplify using reg. algebra:

0 110(110)0 [ 110(11

A A B

B B C

C A B

B B C

A A C

C A C

C C A

A A A

A

!

!

!

!

!

"

"

"

" "

" " "

" " "

" " " "

" " " " "

= + +

= +

= +

=

= + +

= +

=

= + +

= + +

= +

+ = + 0)]0

(110)0

((110) 0)A

"

" "

" " "

=

# = Gauss-Jordan elimination& back-substitution


Ex: Node Elimination Example via Algebra

• Want B. C is accept state.

• Elim. A:

• Elim B:• Elim C:• ∴

B

A

b

b

ab

a

C

A aA aB

B bC

C bA bB !

= +

=

= + +

*A a aB=

*

B bC

C ba aB bB !

=

= + +

B bC= *( * )

C ba abC bbC

ba ab bb C

!

!

= + +

= + +( * ) *C ba ab bb= +

( * ) *B bC b ba ab bb= = + Simplifies to = ( * ) *B b ba b


Closure Properties• A class of languages is said to be closed under an

operation if applying that operation to members of theclass results in a language that is again a member of theclass. Example: the regular languages are closed underthe operations of union, concatenation and Kleene star.

• Thm: The regular languages are closed under intersectionand complementation.

Pf: Complementation. Let L = L(M) where is a DFA. Then the FA is also deterministic, and So w leads to a non-

accepting state in ⇔ w leads to an accepting state in So

( , , , , )M Q s F!= "

( , , , , )M Q s F!= "

(s,w) !

M

!(q,") # (s,w) !

M

!(q,").

.MM

( ) ( ).L M L M!

= " #


Lecture 03 28


Closure Properties (contʼd) Intersection. Let be regular. By DeMorganʼs law

Since the regular languages are closed undercomplementation and union, the result follows.

1 2,L L

1 2 1 2( ).L L L L! = "

!


Closure Properties (contʼd)• Another proof of closure under ∩ illustrates the technique

called “cross-product construction”. See Sipser text,Theorem 1.25.

• Thm: The class of regular languages is closed under theintersection operation.

Pf: Assume and , where the automata

are deterministic. Construction. Construct a “cross-product machine” M as

follows: where the transition function is defined by:

Machine M simulates the two given machines “in parallel”, keeping each machine state in one component of

1 1 1 1 1 1 1 1( ), ( , , , , )L L M M Q s F!= = "

2 2 2 2 2 2 2 2( ), ( , , , , )L L M M Q s F!= = "

1 2 1 2 1 2 1, 2 1 2( , , ,( ), )M Q Q s s F F! != " # $ # " "

1 2 1 2 1 1 2 2(( , ), ) ( ( , ), ( , )).q q a q a q a! ! ! !" =

1 2( , ).q q


Closure Properties (contʼd) Verification By an easy induction on |x|, can show that

Therefore, for a pair of final states

This says that

i.e., that

((q1,q

2),x) !

M

!((p

1,p

2),") #

(q1,x)!

M1

!(p

1,") $ (q

2,x)!

M2

!(p

2,") !

((s1,s

2),x)!

M

!((f

1,f

2),") #

(s1,x)!

M1

!(f

1,") $ (s

2,x)!

M2

!(f

2,")

1 1 2 2,f F f F! !

1 2( ) ( ) ( )x L M x L M x L M! " ! # !

1 2( ) ( ) ( ).L M L M L M= ! !


Lecture 03 29


Closure Properties (contʼd)• Defn: A homomorphism h is a function that maps each

symbol of to a string over somealphabet Δ, i.e.,

• The homomorphism is extended to operate on stringscharacter-by-character, i.e.,

• It is further extended to languages element-wise, i.e.,

• Thm: If L is regular and h is a homomorphism, then h(L) isregular.

Pf: Assume L is recognized by a DFA

1 2{ , , , }

na a a! = …

1 1 2 2( ) , ( ) , , ( )

n nh a w ha w ha w= = =…

1 2 1 2( ) ( )( ) ( ).

n nhc c c hc hc hc=… …

( ) {( ): }.h L hw w L= !

( , , , , ).M Q s F!= "


Closure Properties (contʼd)• Construction: Construct the machine where for each transition in

M:

put into Mh the transition An easy induction establishes that

from which it follows that

( , , , , )h h

M Q s F!= "

ap q h(a)

p q

(s,w) !

M

!(q,") # (s,h(w)) !

Mh

!(q,"),

( ) ( ( )).h

L M h L M= !

Note: GNFA


What is Not Regular?• FA have a very limited computing ability. They cannot, for

example, recognized strings of well-nested parentheses,or well-formed arithmetic expressions, or even thelanguage of strings of the form w#w, having two copies ofthe same substring.

• How can we show some languages are not regular? Wewill give a property that all regular languages must have(called the pumping property). Then, to show that alanguage L is not regular, we argue that it lacks thispumping property.


Lecture 03 30


Pumping Lemma• Thm [Pumping Lemma for Regular Languages]. Suppose

that L is an infinite¹ regular language. Then

¹All finite languages are regular, so only infinite languages areof interest.

( )( )[

( , , )(

( 0) )]i

p w w L w p

x y z w xyz y xy p

i xy z L

!

" # $ % & '

" = % ( % )

% # & $


Pumping Lemma (English)• Thm [Pumping Lemma for Regular Languages]. Suppose

that L is an infinite¹ regular language. Then there is somenumber p (called the pumping length) such that:

if w is any string in L with |w| ≥ p, then w can be factored into 3 substrings, w = xyz, that satisfy the following 3 conditions: (i) y ≠ ε [y is not empty] (ii) |xy| ≤ p [the prefix and pumped part are short] (iii) for every i ≥ 0, xyiz ∈ L [“pumped up” and

“pumped down” (i = 0) versions of the string must also be in L]

¹All finite languages are regular, so only infinite languages areof interest.


Pumping Lemma (contʼd)• Pf: Let be a DFA recognizing L and

let p be the number of its states. Let be an input string of length n where n

≥ p. Let be the sequence of states that M

enters while processing w so that for This state sequence has length n+1 ≥ p+1. Among the first p+1 states of this sequence, at least 2

must be the same state [“pigeonhole principle”]. Call thefirst of these 2 and the second . Because

occurs among the first p+1 places in thesequence we have that k ≤ p+1.Define the following substrings of w:

( , , , , )M Q s F!= "

1 2 nw a a a= !

1 2 1, , ,

n nr r r r

+!

1( , )

i i ir r a!

+=

1 .i n! !

jr

k jr r=

1 2 1, , ,

n nr r r r

+!

1 1 1, ,

j j k k nx a a y a a z a a

! != = =! ! !

k jr r=


Lecture 03 31


a

rk

rn+1r1

1 1jx a a

!= !

1j ky a a

!= !

=rjk n

z a a= !

Pumping Lemma (contʼd)• Picture:

• From the picture, we see that there is an accepting pathfrom to a final state for all the strings ofthe form . Also, since it mustbe that Furthermore, so

.

1s r= 1n

r+

, 0i

xy z i ! j k!

.y !" 1k p! +

.xy p! !


Non-regular Examples• Ex: is not regular. Pf: By contradiction. Suppose L is regular. Then by the

Pumping Lemma,

Then it follows that is a perfect square. This is impossible. For suppose for a so large that Then

Hence falls “in between” perfect squares—a contradiction.

2

{ : 0}k

L a k= !

( , , ) , ( 0),

[( 0) ( ) .

p q r

p q n r

x y z x a y a q z a

n a a a L

! = = > =

" # $ $ %

( 0)n p q n r! " + # +

2

0 0p r q n k+ + ! =

0k

02 1 .k q+ >

0( 1)p r q n+ + ! +

2 2 2

0 0 0 02 1 ( 1).k q k k k= + < + + = +

0( 1)p r q n+ + ! +

!


Non-regular Examples (contʼd)• Ex: is not regular. Pf: By contradiction. Suppose L is regular. Then by

closure properties of the regular languages is regular. Now We show cannot

be regular, which provides a contradiction. If is regular, then there are substrings with such that Case 1. y is entirely in the aʼs. Assume it is in the aʼs

before the 2 bʼs (The other subcase is symmetric). Then

and But then where This is a contradiction.

{ : { , }}R

L w w w a b!

= " #

* *

1L L a bba= !

1{ : 0}.

n n

L a bba n= !1L

1L

y !"

, ,x y z

10 .

nn xy z L! " #

, , ( 0)p q r s

x a y a z a bba q= = = >

.p q r s+ + =p r s

a a bba L!

.p r s+ <


Lecture 03 32


Non-regular Examples (contʼd)• Case 2. y contains a b. Then has more than 2

bʼs, and so This is a contradiction.

Contradictions in all cases ⇒ contradiction to theassumption that is regular. So is not regular.

• Ex: is not regular. See Text,Example 1.73.

• Ex:

is not regular. Pf: Suppose B is regular. Then so is as is its homomorphic image Contradiction.

1L

3xy z

3

1.xy z L!

1L

{ : 0}n n

L a b n= !

{ : is a well-nested

string of parentheses}

B w w=

{( ) : 0}n n

n= !

()B B! !

" = #

{ : 0}.n na b n !

!

!


Decision Problems• For a property/predicate P the decision problem for P is:

Given: x Question: Is P(x) true?

• Ex: Given DFA M, is it true that L(M) = ∅?• Thm: For DFA M all the following decision problems are

solvable, i.e. there exists an algorithm to decide thequestion for any input:

Given Question1. M,w w ∈ L(M) ?2. M L(M) = ∅ ?3. M L(M) = Σ* ?4. M, M′ L(M) ⊆ L(M′ ) ?5. M, M′ L(M) = L(M′ ) ?


Decision Problems (contʼd)

Pf: Assume given DFAs for inputs. (1) Trace w through M. “Yes” if leads from s to some q ∈ F (2) “Yes” if there is some q ∈ F reachable from s (3) Convert (4)

(5) Use (4) twice

. ( ) ( )M M L M L M!

" = # $ = %

1 2 2 1( ) ( ) ( ) ( )L M L M L M L M! " # = $

2( )L M

1( )L M!

CSC 473 Automata, Grammars & Languages 9/29/10 · CSC 473 Automata, Grammars & Languages 9/29/10 Lecture 03 1 C SC 473 Automata, Grammars & Languages Automata, Grammars and Languages

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