MELJUN CORTES -- AUTOMATA THEORY LECTURE - 6

7/31/2019 MELJUN CORTES -- AUTOMATA THEORY LECTURE - 6

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CSC 3130: Automata theory and formal languages

Andrej Bogdanov

http://www.cse.cuhk.edu.hk/~andrejb/csc3130

DFA minimization algorithm


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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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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 1 and input 111?

Mends up in same state on input e and 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 unique minimal DFA forL

There is no DFA with 3 states forL

1q0

q1

q2

q3

1 11

0

0

0

0


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DFA minimization

There is an algorithm to 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, q1 equivalent

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, q1 equivalent

q2, q3 also equivalent


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Equivalent and distinguishable states

Two states q, q are equivalent if

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

q, q are distinguishable if they are not equivalent:

^

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

w) are eitherboth accepting orboth rejecting

^ ^

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

d(q, w) is accepting and the other is rejecting


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Examples of distinguishable states

q0, q1 equivalent

q3 distinguishable from q0, q1, q2q3 is 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 ofdistinguishable states as follows:

For any pair of states q, q:

Ifqis accepting and q is 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 q into 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

q11 is 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

q1 is 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

q01 is 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, q10 are equivalent group A

q1, q01 are equivalent group Bq11 cannot 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 M is well-defined

Correctness

The new DFA M is equivalent to the original DFA

Minimality

The new DFA M is the smallestDFA ossible forL

M M


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Main claim

Proof outline:

First, we assume q, q are marked distinguishable, and

show they must be distinguishable

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

Any two states qand q in M aredistinguishable if and only ifthe algorithm

marks them as distinguishable


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Proof of part 1

First, suppose q, q are 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, q are 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

q as 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:Ifqis accepting and q is 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 M is well-defined

Correctness

The new DFA M is equivalent to the original DFA

Minimality

The new DFA M is the smallestDFA ossible forL

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 M labeled aout ofmerged stateqA

Aq3

q7q

1

Bq5

q2

Cq6

a

a


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States q5 and q6 must be distinguishable because

they were not merged together

Aq3

q7q

1

Bq5

q2

q6

a

a


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Then, q3 and q7 must 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 eitherall accepting or

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


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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 ofM corresponds to a class of

mutually equivalent states in 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, M accepts win state qi if and only ifMon input wlands in the state qA that represents qi qA must be accepting by consistency of states

M is in state qAif and only ifM is in one of

the states represented byA


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Proof of minimality

Now suppose there is some smallerM forL

By the pigeonhole principle, there is a state q ofM such that

All pairs of states in M are distinguishable

M M

q

q

v

v

q

v, v


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Proof of minimality

Since qand q are 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 M are distinguishable

M M

q

q

v

v

q

v, v

w

w

?w

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MELJUN CORTES -- AUTOMATA THEORY LECTURE - 6