1 FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY 15-453 For next time: Read 2.1 & 2.2.

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FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY

15-453

For next time: Read 2.1 & 2.2

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MINIMIZING DFAs

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IS THIS MINIMAL?

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1

0

0

00

NO

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IS THIS MINIMAL?

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THEOREMFor every regular language L, there exists a unique (up to re-labeling of the states)

minimal DFA M such that L = L(M)

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NOT TRUE FOR NFAs

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0

0

0

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EXTENDING Given DFA M = (Q, Σ, , q0, F) extend

to : Q Σ* → Q as follows:

(q, ε) =

(q, ) =

(q, w1 …wk+1 ) = ( (q, w1 …wk ), wk+1 )

^

^

^

^ ^

A string w Σ* distinguishes states q1 from q2 if

(q1, w) F (q2, w) F ^ ^

q

(q, )

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00,1

00

1

1

1

q0

q1

q2

q3

ε distinguishes accept from non-accept states

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Fix M = (Q, Σ, , q0, F) and let p, q, r Q

Definition:

p is distinguishable from q iff there is a w Σ* that distinguishes p from q

p is indistinguishable from q iff p is not distinguishable from q

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Define relation “~”:

p ~ q iff p is indistinguishable from q

p ~ q iff p is distinguishable from q /

Proposition: “~” is an equivalence relation

p ~ p (reflexive)

p ~ q q ~ p (symmetric)

p ~ q and q ~ r p ~ r (transitive)

p q r

w wSuppose not:

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so “~” partitions the set of states of M into disjoint equivalence classes


q0

Qq

[q] = { p | p ~ q }

Proposition: “~” is an equivalence relation

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Algorithm MINIMIZE

Input: DFA M

Output: DFA MMIN such that:

M MMIN

MMIN has no inaccessible states

MMIN is irreducible

states of MMIN are pairwise distinguishable||

Theorem: MMIN is the unique minimum

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TABLE-FILLING ALGORITHMInput: DFA M = (Q, Σ, , q0, F)

(2) EM = { [q] | q Q }

(1) DM = { (p,q) | p,q Q and p ~ q }/

•Make best effort to find pairs of states that are distinguishable.

•Pairs left over will be indistinguishable.

IDEA!

Output:

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Output:

(2) EM = { [q] | q Q }

(1) DM = { (p,q) | p,q Q and p ~ q }/

q0

q1

qi

qn

q0 q1 qi qn

Base Case: p accepts and q rejects p ~ q/

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Output:

(2) EM = { [q] | q Q }

(1) DM = { (p,q) | p,q Q and p ~ q }/

q0

q1

qi

qn

q0 q1 qi qn

Base Case: p accepts and q rejects p ~ q/

Recursion:d d d

d

d

p p

q q~/

p ~ q/

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1 0

1

0 1

0,1

0

q0

q0 q1

q1

q2

q2 q3

q3

q0 q1 q2 q3

d d d

d

d

d

21

11

1

1

0

0

00

q0

q0 q1

q1

q2

q2 q3

q3 d d

d

d

q0 q1

q2q3

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If 2 states are not distinguished by table-filling algorithm, then they are equivalent

Proof (by contradiction):

Suppose there exist states p and q that aren’t distinguished by the T-F algorithm, but p ~ q /

Then there exists w such that:

(p, w) F and (q, w) F ^ ^

Of all such bad pairs, let (p, q) be a pair with the smallest |w| So, w = w, where ΣLet p = (p,) and q = (q,)

Then (p, q) is also a bad pair

(Why is |w| >0 ?)

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Algorithm MINIMIZE

Input: DFA M

Output: DFA MMIN

(1) Remove all inaccessible states from M

(2) Apply Table-Filling algorithm to get EM = { [q] | q is an accessible state of M }

MMIN = (QMIN, Σ, MIN, q0 MIN, FMIN)

QMIN = EM, q0 MIN = [q0], FMIN = { [q] | q F }

MIN( [q], ) = [ ( q, ) ]

Must show MIN is well defined!

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Algorithm MINIMIZE

Input: DFA M

Output: DFA MMIN

(1) Remove all inaccessible states from M

(2) Apply Table-Filling algorithm to get EM = { [q] | q is an accessible state of M }

MMIN = (QMIN, Σ, MIN, q0 MIN, FMIN)

QMIN = EM, q0 MIN = [q0], FMIN = { [q] | q F }

MIN( [q], ) = [ ( q, ) ]

Claim: MMIN M

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1

0

1010

1

0

1

0q0 q1

q2

MINIMIZE

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1 0

1

0

1

0,1

0

q0 q1

q3

q5

q4

0

1

q2

0,1q0

q0 q1

q1

q3

q3 q4

q4

d

dd

d

d d

q5

q5

d

d

d

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1 0

1

00,1

0

q0 q1

q3

q5

q4

0

1

q0

q0 q1

q1

q3

q3 q4

q4

d

dd

d

d d

q5

q5

d

d

d

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*MMIN is the unique minimal DFA equivalent to M

Suppose MMMIN, and M has no inaccessible states and is irreducible

Claim: There exists a 1-1 onto correspondence (preserving transitions) between M and MMIN

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NB: If M is minimal, then M has no inaccessible states and is irreducible. (So claim implies * )

This Claim is also implying the converse, ie if M has no inaccessible states and is irreducible, then M is minimal.

Proof. Let Mmin ( M MMIN) be minimal. Then Mmin MMIN

So, by Claim, both Mmin and M are isomorphic to MMIN

NOT TRUE for NFAs !

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Proof: We will construct a map recursively

Base Case: q0 MIN → q0

Recursive Step: If p → p

q

qThen q → q




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q

qThen q → q

We need to prove:

The map is defined everywhere

The map is well defined

The map is a bijection

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That is, for all q MMIN there is a q M such that q → q

If q MMIN, there is w such that MIN(q0 MIN,w) = q



q

qThen q → q

The map is defined everywhere

^

Let q = (q0,w) ^

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q

qThen q → q

The map is well defined

Suppose there exist q and q such that q → q and q → q

We show that q and q are indistinguishable, so it must be that q = q

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MMIN M

Suppose there exist q and q such that q → q and q → q

q

u

q0

q

v

q0

q

u

q0 MIN

q

v

q0 MIN

Suppose q and q are not indistinguishable

w

Accep

t

w Reject

w Reject

w

Accep

t

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q

qThen q → q

The map is onto

For all q M there is a q MMIN such that q → q

If q M, there is w such that (q0,w) = q^

Let q = MIN(q0 MIN,w) ^

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MMIN M

Suppose there exist p and q such that p → q and q → q

q

u

q0

q

v

q0

p

u

q0 MIN

q

v

q0 MIN

Suppose p and q are not indistinguishable

w

Accep

t

w Reject

w Reject

w

Accep

t

The map is 1-1

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How do we prove two regular expressions are equivalent?

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Finish Reviewing Chapter 1 of the book andRead 2.1 & 2.2 for next time

1 FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY 15-453 For next time: Read 2.1 & 2.2.

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