; (Basis) is; and a are not. (Induction) E + F is i - - PDF document

SLIDE 1 Decision Prop erties

• f

Regular Languages Giv en a (represen tation, e.g., RE, F A,

• f

a) regular language L, what can w e tell ab

• ut

L?

• Since

there are algorithms to con v ert b et w een an y t w

• represen

tations, w e can c ho

• se

the rep that mak es the test easiest. Mem b ership Is string w in regular language L?

• Cho
• se

DF A represen tation for L.

• Sim

ulate the DF A

• n

input w . Emptiness Is L = ;?

• Use

DF A represen tation.

• Use

a graph-reac habilit y algorithm to test if at least

• ne

accepting state is reac hable from the start state. Finiteness Is L a nite language?

• Note

ev ery nite language is regular (wh y?), but a regular language is not necessarily nite. DF A metho d:

• Giv

en a DF A for L, eliminate all states that are not reac hable from the start state and all states that do not reac h an accepting state.

• T

est if there are an y cycles in the remaining DF A; if so, L is innite, if not, then L is nite. RE metho d: Almost, w e can lo

• k

for a

• in

the RE and sa y its language is innite if there is

• ne,

nite if not. Ho w ev er, there are exceptions, e.g.

• 1
• r
• ;.

Th us: 1. Find sub expressions equiv alen t to ; b y:

✦

(Basis) ; is;

• and

a are not.

✦

(Induction) E + F is i b

• th

E and F are; E F is if either E

• r

F are; E

• nev

er is. 2. Eliminate sub expressions equiv alen t to ; b y:

✦

Replace E + F

• r

F + E b y F whenev er E is and F isn't.

✦

Replace E

• b

y

• whenev

er E is equiv alen t to ;. 1

SLIDE 2 3. No w, nd sub expressions that are equiv alen t to

• b

y:

✦

(Basis)

• is;

a isn't.

✦

(Induction) E + F is i b

• th

E and F are; ditto E F ; E

• is

i E is. 4. No w, w e can tell if L(R) is innite b y lo

• king

for a sub expression E

• suc

h that E is not equiv alen t to . Example Consider (0 + 1;)

• +

1;

• .
• Step

1: ; (t wice) and 1; are sub expressions equiv alen t to ;.

• Step

2:

• +

1 remains.

• Step

3:

• nly

sub expression

• is

equiv alen t to .

• Since

is starred, language is innite. Minimization

• f

States

• Real

goal is testing equiv alence

• f

(reps

• f

) t w

• regular

languages.

• In

teresting fact: DF A's ha v e unique (up to state names) minim um

• state

equiv alen ts.

✦

But pro

• f

in course reader do esn't quite get to that p

• in

t. Distingu ish abl e States Key idea: nd states p and q that are distinguishable b ecause there is some input w that tak es exactly

• ne
• f

p and q to an accepting state.

• Basis:

an y nonaccepting state is distinguishable from an y accepting state (w = ).

• Induction:

p and q are distinguishable if there is some input sym b

• l

a suc h that

• (p;

a) is distinguishable from

• (q

; a).

✦

All

• ther

pairs

• f

states are indistinguishable, and can b e merged in to

• ne

state. Example (V ery Simple) Consider: 2

SLIDE 3 1 1 1 Start p q r

• p

is distinguishable from q and r b y basis. Can w e distinguish q from r ?

• No

string b eginning with w

• rks,

b ecause b

• th

states go to p, and therefore an y string

• f

the form 0x tak es q and r to the same state.

• No

string b eginning with 1 w

• rks.

✦

T ec hnically ,

• (q

; 1) = r and

• (r

; 1) = q are not distinguishable. Th us, induction do es not tell us q and r are distinguishable.

✦

What happ ens is that, starting in either q

• r

r , as long as w e ha v e inputs 1, w e are in

• ne
• f

the accepting states, and when a is read, w e go to the same state forev er after. Constructing the Minim um-S tat e DF A

• F
• r

eac h group

• f

indistinguishable states, pic k a \represen tativ e."

✦

Note a group can b e large, e.g., q 1 ; q 2 ; : : : ; q k , if all pairs are indistinguishable.

✦

Indistinguishabilit y is transitiv e (wh y?) so indistinguishabilit y partitions states.

• If

p is a represen tativ e, and

• (p;

a) = q , in minim um

• state

DF A the transition from p

• n

a is to the represen tativ e

• f

q 's group (to q itself if q is either alone in a group

• r

a represen tativ e).

• State

state is represen tativ e

• f

the

• riginal

start state.

• Accepting

states are represen tativ es

• f

groups

• f

accepting states.

✦

Notice w e could not ha v e a \mixed" (accepting + nonaccepting) group (wh y?). 3

SLIDE 4

• Delete

an y state that is not reac hable from the start state. Example F

• r

the DF A ab

• v

e, p is in a group b y itself; fq ; r g is the

• ther

group. Start p 0,1 q r 1 Wh y Ab

• v

e Minimizati

• n

Can't b e Beaten Supp

• se

w e ha v e a DF A A, and w e minim ize it to construct a DF A M . Y et there is another DF A N that accepts the same language as A and M , y et has few er states than M . Pro

• f

con tradiction that this can't happ en:

• Run

the state-distinguishabilit y pro cess

• n

the states

• f

M and N together.

• Start

states

• f

M and N are indistinguishable b ecause L(M ) = L(N ).

• If

fp; q g are indistinguishable, then their successors

• n

an y

• ne

input sym b

• l

are also indistinguishable.

• Th

us, since neither M not N could ha v e an inaccessible state, ev ery state

• f

M is indistinguishable from at least

• ne

state

• f

N .

• Since

N has few er states than M , there are t w

• states
• f

M that are indistinguishable from the same state

• f

N , and therefore indistinguishable from eac h

• ther.
• But

M w as designed so that all its states ar e distinguishable from eac h

• ther.
• W

e ha v e a con tradiction, so the assumption that N exists is wrong, and M in fact has as few states as an y equiv alen t DF A for A.

• In

fact (stronger), there m ust b e a 1-1 corresp

• ndence

b et w een the states

• f

an y

• ther

minim um

• state

N and the DF A M , sho wing that the minim um

• state

DF A for A is unique up to renaming

• f

the states. 4