But can app ear on arcs, and means the empt y string - - PDF document

SLIDE 1 Finite Automata With

• T

ransition s Allo w

• to

b e a lab el

• n

arcs.

• Nothing

else c hanges: acceptance

• f

w is still the existence

• f

a path from the start state to an accepting state with lab el w .

✦

But

• can

app ear

• n

arcs, and means the empt y string (i.e., no visible con tribution to w ). Example Start q r s 1 1

• 001

is accepted b y the path q ; s; r ; q ; r ; s, with lab el 001 = 001. Eliminati

• n
• f
• T

ransitions

• transitions

are a con v enience, but do not increase the p

• w

er

• f

F A's. T

• eliminate
• transitions:

1. Compute the transitiv e closure

• f

the

• arcs
• nly

.

✦

Example: q r s

• q

! fq g; r ! fr ; sg; s ! fr ; sg. 2. If a state p can reac h state q b y

• arcs,

and there is a transition from q to r

• n

input a (not ), then add a transition from p to r

• n

input a. 3. Mak e state p an accepting state if p can reac h some accepting state q b y

• arcs.

4. Remo v e all

• transitions.

1

SLIDE 2 Example Start q r s 1 0,1 0,1 Regular Expressions An algebraic equiv alen t to nite automata.

• Used

in man y places as a language for describing simple but useful patterns in text. Op erators and Op erands If E is a regular expression, then L(E ) denotes the language that E stands for. Expressions are built as follo ws:

• An
• p

erand can b e: 1. A v ariable, standing for a language. 2. A sym b

• l,

standing for itself as a set

• f

strings, i.e., a stands for the language fag (formally , L(a) = fag). 3. , standing for fg (a language). 4. ;, standing for ; (the empt y language).

• The
• p

erators are: 1. +, standing for union. L(E + F ) = L(E ) [ L(F ). 2. Juxtap

• sition

(i.e., no

• p

erator sym b

• l,

as in xy to mean x

• y

) to stand for c

• nc

atenation. L(E F ) = L(E )L(F ), where the concatenation

• f

languages L and M is fxy j x is in L and y is in M g. 3.

• to

represen t closur e. L(E

• )

=

• L(E

)

• ,

where L

• =

fg [ L [ LL [ LLL [

• .
• P

aren theses ma y b e used to alter grouping, whic h b y default is

• (highest

precedence), then concatenation, then union (lo w est precedence). 2

SLIDE 3 Examples

• L(001)

= f001g.

• L
• +

10

• )

= f0; 1; 10; 100; 100 0; : : : g.

• L
• 0(0

+ 1)

• =

the set

• f

strings

• f

0's and 1's,

• f

ev en length, suc h that ev ery

• dd

p

• sition

has a 0. Equiv alence

• f

F A Languages and RE Languages

• W

e'll sho w an NF A with

• transitions

can accept the language for a RE.

• Then,

w e sho w a RE can describ e the language

• f

a DF A (same construction w

• rks

for an NF A).

• The

languages accepted b y DF A, NF A,

• NF

A, RE are called the r e gular languages. RE to

• NF

A

• Key

idea: construction

• f

an

• NF

A with

• ne

accepting state is b y induction

• n

the heigh t

• f

the expression tree for the RE.

• Pictures
• f

the basis and inductiv e constructions are in the course reader. Example W e'll go

• v

er the general construction in class and w

• rk

the example

• f
• 0(0

+ 1 )

• .

F A-to-RE Construction Tw

• algorithms:

1. State elimination : giv es smaller expression, in general, and easier to apply . Co v ered in course reader. 2. A simple, inductiv e construction, whic h w e'll do here (also in reader).

• Let

A b e a F A with states 1; 2; : : : ; n.

• Let

R (k ) ij b e a RE whose language is the set

• f

lab els

• f

paths that go from state i to state j without p assing thr

• ugh

an y state n um b ered ab

• v

e k .

• Construction,

and the pro

• f

that the expressions for these RE's are correct, are inductions

• n

k . Basis : k = 0. P ath can't go through any states. 3

SLIDE 4

• Th

us, path is either an arc

• r

the n ull path (a single no de).

• If

i 6= j , then R (0) ij is the sum

• f

all sym b

• ls

a suc h that A has a transition from i to j

• n

sym b

• l

a (; if none).

• If

i = j , then add

• to

ab

• v

e. Induction : Assume w e ha v e correctly dev elop ed expressions for the R (k 1) 's. Then for the R (k ) 's:

• R

(k ) ij = R (k 1) ij + R (k 1) ik (R (k 1) k k )

• R

(k 1) k j Pr

• f

it works : A path from i to j that go es through no state higher than k either: 1. Nev er go es through k , in whic h case the path's lab el is (b y the IH) in the language

• f

R (k 1) ij ,

• r

2. Go es through k

• ne
• r

more times. In this case:

✦

R (k 1) ik con tains the p

• rtion
• f

the path that go es from i to k for the rst time.

✦

(R (k 1) k k )

• con

tains the p

• rtion
• f

the path (p

• ssibly

empt y) from the rst k visit to the last.

✦

R (k 1) k j con tains the p

• rtion
• f

the path from the last k visit to j . Final step : The RE for the en tire F A is the sum (union)

• f

the RE's R (n) ij , where i is the start state and j is

• ne
• f

the accepting states.

• Note

that sup erscript (n) represen ts no restriction

• n

the path at all, since n is the highest-n um b ered state. Example The follo wing is the \clamping" automaton, with states named b y in tegers: 1 1 0,1 Start 3 1 2 Some basis expressions:

• R

(0) 11 = . 4

SLIDE 5

• R

(0) 12 = 1.

• R

(0) 22 =

• +

+ 1.

• R

(0) 31 = 1.

• R

(0) 32 = R (0) 21 = ;. Tw

• inductiv

e examples:

• R

(1) 32 = R (0) 32 + R (0) 31 (R (0) 11 )

• R

(0) 12 = ; + 1

• 1

= 11.

✦

Uses algebr aic laws:

• =

; R = R = R ( is the iden tit y for con tatenation); ; + R = R + ; = R (; is the iden tit y for union).

• R

(1) 22 = R (0) 22 + R (0) 21 (R (0) 11 )

• R

(0) 12 =

• +

+ 1 + ;

• 1

=

• +

+ 1.

✦

Additional algebraic la w used: ;R = R; = ; (; is the annihilator for concatenation). 5