meljun cortes automata theory 6

8/13/2019 MELJUN CORTES Automata Theory 6

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CSC 3130: Automata theory and formal languages

DFA minimization algorithm

MELJUN P. CORTES MBA MPA BSCS ACS

MELJUN CORTES


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Example

Construct a DFA over alphabet {0, 1}thataccepts those strings that end in 111

This is big, isnt there a smallerDFA for this?

0

1

qe

q0

q1

q00

q10

q01

q11

q000

q001

q101

q111

0

1

0

1

0

1

1

1

1

1

0


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Smaller DFA

Yes, we can do it with 4 states:

The state remembers the number of consecutive 1s at

the end of the string (up to 3)

Can we do it with 3 states?

1q0

q1

q2

q3

1 11

0

0

0

0


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Even smaller DFA?

Suppose we had a 3 state DFAMfor L

We do not know what this M looks like

but lets imagine what happens when:

By the pigeonhole principle, on two of theseinputsMends in the same state

Minputs:

e, 1, 11, 111


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Pigeonhole principle

Here, balls are inputs, bins are states:

Suppose you are tossing mballs into nbins,

and m >n. Then two balls end up in the

same bin.

If you have a DFA with nstates and you run

it on minputs, and m >n, then two inputs end

up in same state.


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A smaller DFA?

Minputs:

e, 1, 11, 111

What goes wrong if

Mends up in same state on input 1and input 111?

Mends up in same state on input eand input 11?


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A smaller DFA

SupposeMends up in the same state afterreading inputs x= 1iandy= 1j, where 0 i


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No smaller DFA!

Conclusion

So, this DFA is minimal

In fact, it is the uniqueminimal DFA for L

There is no DFA with 3 states for L

1q0

q1

q2

q3

1 11

0

0

0

0


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DFA minimization

There is an algorithmto start with any DFA andreduce it to the smallest possible DFA

The algorithm attempts to identify classes ofequivalent states

These are states that can be merged togetherwithout affecting the answer of the computation


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Examples of equivalent states

q0, q1equivalent

q0

q1

q2

q3

0

1

0

1

0,1

0, 1

q0

q1

q2

q3

1

0, 1

0, 1

0, 1

0

q0, q1equivalent

q2, q3also equivalent


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Equivalent and distinguishable states

Two states q, qare equivalentif

Here, d(q, w)is the state that the machine is in if itstarts at qand reads the string w

q, qare distinguishableif they are not equivalent:

^

For every string w, the states d(q, w)and d(q,

w)are either both acceptingor both rejecting

^ ^

For some string w, one of the states d(q, w),

d(q, w)isacceptingand the other isrejecting


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Examples of distinguishable states

q0, q1equivalent

q3distinguishable from q0, q1, q2q3is accepting, others are rejecting

q0

q1

q2

q3

0

1

0

1

0,1

0, 1

(q0, q2) and(q1, q2) distinguishablethey behave differently on input 0

q01

q2

q3

0

10,1

0, 1


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DFA minimization algorithm

Find all pairs of distinguishablestates as follows:

For any pair of states q, q:

If qis accepting and qis rejecting

Mark(q, q)as distinguishable

Repeat until nothing is marked:For any pair of states (q, q):

For every alphabet symbol a:

If (d(q, a), d(q, a))are marked as distinguishable


For any pair of states (q, q):

If (q, q)is not marked as distinguishable

Mergeqand qinto a single state


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Example of DFA minimization

0

1

qe

q0

q1

q00

q10

q01

q11

0

1

0

1

1

0

01

1

0

0

1

q0

q1

q00

q01

q11

q10

q10

qe

q0

q1

q01

q00


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0

1

qe

q0

q1

q00

q10

q01

q11

0

1

0

1

1

0

01

1

0

0

1

x x x x x x

q0

q1

q00

q01

q11

q10

q10

qe

q0

q1

q01

q00

q11is distinguishable from all other states


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0

1

qe

q0

q1

q00

q10

q01

q11

0

1

0

1

1

0

01

1

0

0

1

x

x

x

x

x

xx x x x

q0

q1

q00

q01

q11

q10

q10

qe

q0

q1

q01

q00

q1is distinguishable from qe, q0, q00,q10On transition 1, they go to distinguishable states


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0

1

qe

q0

q1

q00

q10

q01

q11

0

1

0

1

1

0

01

1

0

0

1

x

x

x

x

x

x

x

xx

x

xxx x

q0

q1

q00

q01

q11

q10

q10

qe

q0

q1

q01

q00

q01is distinguishable from qe, q0, q00,q10On transition 1, they go to distinguishable states


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0

1

qe

q0

q1

q00

q10

q01

q11

0

1

0

1

1

0

01

1

0

0

1

A

x

A

x

Ax

x

A

x

Ax

x

B

xx

x

Ax

xx x

q0

q1

q00

q01

q11

q10

q10

qe

q0

q1

q01

q00

Merge states not marked distinguishable

qe, q0, q00, q10are equivalent group A

q1, q01are equivalent group Bq11cannot be merged group C

A

B

C


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0

1

qe

q0

q1

q00

q10

q01

q11

0

1

0

1

1

0

01

1

0

0

1

A

x

A

x

Ax

x

A

x

Ax

x

B

xx

x

Ax

xx x

q0

q1

q00

q01

q11

q10

q10

qe

q0

q1

q01

q00

1qA

qB

qC1

1

0

0

0

minimized DFA:

A

B

C


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Why does DFA minimization work?

We need to convince ourselves of threeproperties:

Consistency

The new DFA Mis well-defined

Correctness

The new DFA Mis equivalentto the original DFA

Minimality

The new DFA Mis the smallestDFA ossiblefor L

M M


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Main claim

Proof outline:

First, we assume q, qare marked distinguishable, and

show they must be distinguishable

Then, we assume q, qare not marked distinguishable,and show they must be equivalent

Any two states qand qin Maredistinguishable if and only ifthe algorithm

marks them as distinguishable


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Proof of part 1

First, suppose q, qare marked as distinguishable Could it be that they are equivalent?

No, because recall how the algorithm works:

q

q

w1

w1

w2

w2

wk

wk

wk-1

wk-1

xxxx


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Proof idea for part 2

Now suppose q, qare not marked asdistinguishable

Could it be that they are distinguishable?

Suppose so

Then for some string w, d(q, w)accepts, but d(q, w)

rejects

Working backwards, the algorithm will mark qand

qas distinguishable at some point

^ ^


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Proof of part 2

q

q

w1

w1

w1

w1

wk

wk

wk-1

wk-1

For any pair of states q, q:If qis accepting and qis rejecting



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Proof of part 2

q

q

w1

w1

w2

w2

wk

wk

wk-1

wk-1

x

Repeat until nothing is marked:For any pair of states (q, q):

For every alphabet symbol a:

If (d(q, a), d(q, a))are marked as distinguishable


xxx


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Why does DFA minimization work?

We need to convince ourselves of threeproperties:

Consistency

The new DFA Mis well-defined

Correctness

The new DFA Mis equivalentto the original DFA

Minimality

The new DFA Mis the smallestDFA ossiblefor L

M M


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Consistency of transitions

Why are the transitions between the merged

states consistent?

0

1

qe

q0

q1

q00

q10

q01

q11

0

1

0

1

1

0

01

1

0

0

1

A

B

C


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Proof of consistency

Suppose there were two inconsistent transitionsin Mlabeled aout of merged stateqA

Aq3

q7q

1

Bq5

q2

Cq6

a

a


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States q5and q6must be distinguishable because

they were not merged together

Aq3

q7q

1

Bq5

q2

q6

a

a


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Then, q3and q7must also be distinguishable, and

this is impossible

Aq3

q7q

1

Bq5

q2

Cq6

a

a

w

w


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Consistency of states

Why are the merged states either all acceptingor

all rejecting?

0

1

qe

q0

q1

q00

q10

q01

q11

0

1

0

1

1

0

01

1

0

0

1

A

B

C


32/36

Consistency of states

Because merged states are not distinguishable,

so they are mutually equivalent

0

1

qe

q0

q1

q00

q10

q01

q11

0

1

01

1

0

01

1

0

0

1

A

B

C


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Correctness

Each state of Mcorresponds to a class of

mutually equivalent statesin M

0

1qe

q0

q1

q00

q10

q01

q11

0

1

0

1

1

0

01

1

0

0

1

A

B

C


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Proof of correctness

Claim:After reading input w,

Proof: True in start state, and stays true after,

because transitions are consistent

At the end, Maccepts win state qiif and only if Mon input wlands in the state qAthat represents qi qAmust be accepting by consistency of states

Mis in state qAif and only ifMis in one of

the states represented byA


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Proof of minimality

Now suppose there is some smallerMfor L

By the pigeonhole principle, there is a state q ofMsuch that

All pairs of states in Mare distinguishable

M M

q

q

v

v

q

v,v


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Proof of minimality

Since qand qare distinguishable, for some w

But in M,d(q, w)cannot both accept and reject!

So, Mcannot be smallerthan M

All pairs of states in Mare distinguishable

M M

q

q

v

v

q

v,v

w

w

?w

meljun cortes automata theory 6

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