learning regular - university of pennsylvania...(deterministic finite automaton) and the syntactic...

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LearningRegular

ω-Languages

The Goal

Device an algorithm for learning

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an unknown regularω-language L

The Goal

Device an algorithm for learning

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an unknown regularω-language L

set of infinite words (ω-words)

The Goal

Device an algorithm for learning

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an unknown regularω-language L

accepted by a finite automaton

The Goal

Device an algorithm for learning

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Learner

an unknown regularω-language L

The Goal

Device an algorithm for learning

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Learner

an unknown regularω-language L

L

Teacher

The Goal

Device an algorithm for learning

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Learner

an unknown regularω-language L

L

Teacher

membership queries

equivalencequeries

Coping with ω-words

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Learner

L

Teacher

Isabcccccabdebbbaaaabcdaa…

in L?

Coping with ω-words

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Learner

L

Teacher

Isabcccccabdebbbaaaabcdaa…

in L?

Coping with ω-words

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Learner

L

Teacher

Iswingardium laviosaω in L?

Coping with ω-words

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Learner

L

Teacher

Iswingardium laviosaω in L?

wingardium laviosa laviosa laviosa laviosa laviosa …

Coping with ω-words

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Learner

L

Teacher

Iswingardium laviosaω in L?

wingardium laviosa laviosa laviosa laviosa laviosa …

THM:

Two regular ω-languages are equivalent iff

they agree on the set of ultimately periodic words

The Goal

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Is uvω in L ?

Learner

L

Teacher

The Goal

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Is uvω in L ?

Yes / No

Learner

L

Teacher

The Goal

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Is uvω in L ?

Yes / No

Is H same as L ?

Learner

L

Teacher

The Goal

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Is uvω in L ?

Yes / No

Is H same as L ?

Yes / No, c.e: uvω

Learner

L

Teacher

The Goal

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Is uvω in L ?

Yes / No

Is H same as L ?

Yes / No, c.e: uvω

H s.t. H=L

Learner

L

Teacher

Motivation

Same problem for regular (finitary) languages is solved by the L* algorithm [Angluin]

L* has found applications in many areas including AI, neural networks, geometry, data mining, verificationand synthesis and many more.

Reactive systems concerns ω-languages, using L* for this limits application to safety (and excludes liveness, fairness)

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Previous Work on Learning ω-Langs.

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allregular

ω-languages

Safe

DP / DMDS / DR

DWP

DCDB

DWCDWB

strictly less expressive

Previous Work on Learning ω-Langs.

20

allregular

ω-languages

Safe

DP / DMDS / DR

DWP

DCDB

DWCDWB

strictly less expressive

• [de la Higuera & Janodet, 2004]

• [Jayasrirani et al, 2012]

• [Saoudi & Yokomori, 1993]

From Prefixes

From Lassos

Previous Work on Learning ω-Langs.

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allregular

ω-languages

Safe

DP / DMDS / DR

DWP

DCDB

DWCDWB

strictly less expressive

• [de la Higuera & Janodet, 2004]

• [Jayasrirani et al, 2012]

• [Saoudi & Yokomori, 1993]

• [Maler & Pnueli ,1995]

From Prefixes

From Lassos

From Lassos

Previous Work on Learning ω-Langs.

22

allregular

ω-languages

Safe

DP / DMDS / DR

DWP

DCDB

DWCDWB

strictly less expressive

• [de la Higuera & Janodet, 2004]

• [Jayasrirani et al, 2012]

• [Saoudi & Yokomori, 1993]

• [Maler & Pnueli ,1995]

From Prefixes

From Lassos

From Lassos

goal

Previous Work on Learning ω-Langs.

23

allregular

ω-languages

Safe

DP / DMDS / DR

DWP

DCDB

DWCDWB

strictly less expressive

• [de la Higuera & Janodet, 2004]

• [Jayasrirani et al, 2012]

• [Saoudi & Yokomori, 1993]

• [Maler & Pnueli ,1995]

From Prefixes

From Lassos

From Lassos

goal

Previous Work on Learning ω-Langs.

24

allregular

ω-languages

Safe

DP / DMDS / DR

DWP

DCDB

DWCDWB

strictly less expressive

• [de la Higuera & Janodet, 2004]

• [Jayasrirani et al, 2012]

• [Saoudi & Yokomori, 1993]

• [Maler & Pnueli ,1995]

From Prefixes

From Lassos

From Lassos

goal

However …

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L$

SyntacticFORC

RecurrentFDFA

[Calbrix, Nivat & Podelski ‘93]

[Maler & Staiger ‘93]

However …

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L$

SyntacticFORC

RecurrentFDFA

May be exp. bigger

than

[Calbrix, Nivat & Podelski ‘93]

[Maler & Staiger ‘93]

However …

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L$

SyntacticFORC

RecurrentFDFA

May be exp. bigger

than

May be quadr.

bigger than

[Calbrix, Nivat & Podelski ‘93]

[Maler & Staiger ‘93]

Main Contribution

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Learner

Lω –

a learning algorithm for all 3 representations

(periodic, syntactic, recurrent)

Why is it difficult?

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ω-aut.

L*

1962

1987

2014

1993

1994

2005

2012

2008

Why is it difficult?

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states of the minimal DFA (deterministic finite automaton)

and the syntactic right congruence ~ of a language.

L* works due to the Myhill-NerodeTheorem, stating a 1-to-1 relationship between

1

2

53

4

a

b

c

a

bc

a

12

3 54

ω-aut.

L*

1962

1987

2014

1993

1994

2005

2012

2008

Why is it difficult?

31

states of the minimal DFA (deterministic finite automaton)

and the syntactic right congruence ~ of a language.

L* works due to the Myhill-NerodeTheorem, stating a 1-to-1 relationship between

1

2

53

4

a

b

c

a

bc

a

12

3 54

But for ω-langs, the syntactic right congruence is not informative enough!

ω-aut.

L*

1962

1987

2014

1993

1994

2005

2012

2008

For finite words

The Syntactic Right Congruence

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x »L y iff v2*. xv2L , yv2L

For finite words

Example:

L = { w w has an even number of a’s}

The Syntactic Right Congruence

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x »L y iff v2*. xv2L , yv2L

For finite words

Example:

L = { w w has an even number of a’s}

Then »L has two equivalence classes:

words with an odd number of a’s and

words with an even number of a’s

The Syntactic Right Congruence

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x »L y iff v2*. xv2L , yv2L

a

a

b b10

For finite words:

For ω-words:

The Syntactic Right Congruence

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x »L y iff v2*. xv2L , yv2L

x »L y iff w2!. xw2L, yw2L

For finite words:

For ω-words:

The Syntactic Right Congruence

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x »L y iff v2*. xv2L , yv2L

x »L y iff w2!. xw2L, yw2Lx »L y iff u,v2*. xuv!2L, yuv!2L

For ω-words:

The Syntactic Right Congruence

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x »L y iff u,v2*. xuv!2L, yuv!2L

Example:

L = { w | w has finitely many a’s}

For ω-words:

The Syntactic Right Congruence

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x »L y iff u,v2*. xuv!2L, yuv!2L

Example:

L = { w | w has finitely many a’s}

Then xuv!2L iff uv! has finitely may a’s.

For ω-words:

The Syntactic Right Congruence

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x »L y iff u,v2*. xuv!2L, yuv!2L

Example:

L = { w | w has finitely many a’s}

Then xuv!2L iff uv! has finitely may a’s.

Regardless of x. Thus »L has just one equivalence class.

For ω-words:

The Syntactic Right Congruence

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x »L y iff u,v2*. xuv!2L, yuv!2L

Example:

L = { w | w has finitely many a’s}

Then xuv!2L iff uv! has finitely may a’s.

Regardless of x. Thus »L has just one equivalence class.

For ω-words:

The Syntactic Right Congruence

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x »L y iff u,v2*. xuv!2L, yuv!2L

Example:

L = { w | w has finitely many a’s}

Then xuv!2L iff uv! has finitely may a’s.

Regardless of x. Thus »L has just one equivalence class.

But clearly an ω-automaton for L requires at least two states.

Solution Foundation

Families of DFAs (FDFA)

A non-traditional representation of ω-automata

inspired by Maler & Staiger’s ’97 families of right congruences (FORC)

and their canonical representation of a Syntactic FORC

for which they have shown a Myhill-Nerode Theorem

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Family of Right Congruences [MS97]

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Leading Right Congruence

2

2

1

1

4

4

5

5

Plus some restriction (details ommited)

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(»,¼1,¼2,¼3,¼4,¼5)

ProgressLeading

Family of DFAs (FDFA)

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Leading DFAM

(M, P1, P2, P3, P4, P5 )

ProgressLeading

Family of DFAs (FDFA)

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Leading DFAM

11P1

(M, P1, P2, P3, P4, P5 )

ProgressLeading

Family of DFAs (FDFA)

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Leading DFAM

11P1

2

2P2

(M, P1, P2, P3, P4, P5 )

ProgressLeading

Family of DFAs (FDFA)

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Leading DFAM

11P1

2

2P2

33P3

(M, P1, P2, P3, P4, P5 )

ProgressLeading

Family of DFAs (FDFA)

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Leading DFAM

11P1

2

2P2

33P3

4

4P4

(M, P1, P2, P3, P4, P5 )

ProgressLeading

Family of DFAs (FDFA)

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Leading DFAM

11P1

2

2P2

33P3

4

4P4

5

P5

(M, P1, P2, P3, P4, P5 )

ProgressLeading

Family of DFAs (FDFA)

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Leading DFAM

11P1

2

2P2

33P3

4

4P4

5

P5

(M, P1, P2, P3, P4, P5 )

ProgressLeading

That restriction removed.

FDFA Exact Acceptance

(u,v)2 M, P1, P2, P3, P4, P5

P1P1P1

22

44

33 55

MM

2211

44

33 552

4

3 5

M

21

4

3 5

P4P4P4 P5P5P5

P3P3P3

P1P1P1

?uv!

FDFA Exact Acceptance

(u,v)2 M, P1, P2, P3, P4, P5

P1P1P1

22

44

33 55

MM

2211

44

33 552

4

3 5

M

21

4

3 5

P4P4P4 P5P5P5

P3P3P3

P1P1P1

u

?uv!

FDFA Exact Acceptance

(u,v)2 M, P1, P2, P3, P4, P5

P1P1P1

22

44

33 55

MM

2211

44

33 552

4

3 5

M

21

4

3 5

P4P4P4 P5P5P5

P3P3P3

P1P1P1

u

v

?uv!

FDFA Exact Acceptance

(u,v)2 M, P1, P2, P3, P4, P5

P1P1P1

22

44

33 55

MM

2211

44

33 552

4

3 5

M

21

4

3 5

P4P4P4 P5P5P5

P3P3P3

P1P1P1

u

v

?uv!

FORC and FDFA – Example

L = { w | w has finitely many a’s}

1

a,b

1

No a’s

Some a’s

FORC FDFA

Leading

Progress

All prefixes are equally good

FORC and FDFA – Example

L = { w | w has finitely many a’s}

1

a,b

a,ba

P1

b

1

1

No a’s

Some a’s

FORC FDFA

Leading

Progress

All prefixes are equally good

FORC and FDFA – Example

L = { w | w has finitely many a’s}

1

a,b

a,ba

P1

b

1

1

No a’s

Some a’s

FORC FDFA

(abba,bbb)

(abba,bab)

Leading

Progress

All prefixes are equally good

4-state automaton

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0

1

2

2

1

2

1

00,1,2

4-state automaton

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0

1

2

2

1

2

1

00,1,2

Some periods of are easy to find:

11

22

100201

10020122

4-state automaton

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0

1

2

2

1

2

1

00,1,2

Some periods of are easy to find:

11

22

100201

10020122

Some are hard:

1012

4-state automaton

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0

1

2

2

1

2

1

00,1,2

Some periods of are easy to find:

11

22

100201

10020122

Some are hard:

10121012 1012

1012 1012

4-state automaton

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0

1

2

2

1

2

1

00,1,2

Some periods of are easy to find:

11

22

100201

10020122

Some are hard:

10121012 1012

1012 1012

The DFA for periods of has 23 states

=> L$ > 23

4-state automaton

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0

1

2

2

1

2

1

00,1,2

All periods that loop back are easy to find:

11

22

100201

10020122

4-state automaton

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0

1

2

2

1

2

1

00,1,2

All periods that loop back are easy to find:

11

22

100201

10020122

The DFA that accepts only periods of that loop back has only 5 states !

Saturation (Consistency)

How can we use this without losing saturation?

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Saturation (Consistency)

How can we use this without losing saturation?

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We say that a language is saturated if

uv!=xy! implies (u,v)2 L, (x,y)2 L

Saturation (Consistency)

How can we use this without losing saturation?

To achieve saturation, we change the definition of acceptance.

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We say that a language is saturated if

uv!=xy! implies (u,v)2 L, (x,y)2 L

FDFA Normalized Acceptance

P1P1P1 22

44

33 55

MM

2211

44

33 552

4

3 5

M

21

4

3 5

P4 P5P5P5

P3P3P3

P1P1P1

(u,v)2 M, P1, P2, P3, P4, P5 ?

FDFA Normalized Acceptance

P1P1P1 22

44

33 55

MM

2211

44

33 552

4

3 5

M

21

4

3 5

P4 P5P5P5

P3P3P3

P1P1P1

Normalization seeks for the smallest repetition

of the period that loops back

(u,v)2 M, P1, P2, P3, P4, P5 ?

FDFA Normalized Acceptance

P1P1P1 22

44

33 55

MM

2211

44

33 552

4

3 5

M

21

4

3 5

P4 P5P5P5

P3P3P3

P1P1P1

Normalization seeks for the smallest repetition

of the period that loops back

(u,v)2 M, P1, P2, P3, P4, P5 ?

FDFA Normalized Acceptance

P1P1P1 22

44

33 55

MM

2211

44

33 552

4

3 5

M

21

4

3 5

P4 P5P5P5

P3P3P3

P1P1P1

Normalization seeks for the smallest repetition

of the period that loops back

(u,v)2 M, P1, P2, P3, P4, P5 ?

FDFA Normalized Acceptance

P1P1P1 22

44

33 55

MM

2211

44

33 552

4

3 5

M

21

4

3 5

P4 P5P5P5

P3P3P3

P1P1P1

Normalization seeks for the smallest repetition

of the period that loops back

(u,v)2 M, P1, P2, P3, P4, P5 ?

FDFA Normalized Acceptance

P1P1P1 22

44

33 55

MM

2211

44

33 552

4

3 5

M

21

4

3 5

P4 P5P5P5

P3P3P3

P1P1P1

Normalization seeks for the smallest repetition

of the period that loops back

We term Recurrent FDFA the FDFA where progress DFA recognize only periods that loop back. It is saturated under normalized acceptance!

(u,v)2 M, P1, P2, P3, P4, P5 ?

More generally

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SyntacticFDFA

RecurrentFDFA

May be exp. bigger

than

Periodic FDFA (L$)

Generalizing to arbitrary n we get the following lower bound

The Learner Lω

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Same scheme for all 3

canonical FDFAs

Minimality?

The Recurrent FDFA for a given L is not necessarilythe minimal FDFA for L.

According to the normalized acceptance criterion a progress DFA Pu should give correct results only to extensions that close a cycle to u. On other extensions it has freedom.

This has the flavor of minimization with unknowns, which is NP complete.

The Recurrent FDFA chooses to treat all don’t cares as rejecting.

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Time Complexity

Interestingly, [Klarlund, 1994] has shown that choosing a leading automaton which is more refined (bigger) than the syntactic right congruence, may yield an overall smaller FDFA.

If we are given such a leading automaton, we can feedit to the learning algorithm, in which case it will yield the smaller FDFA.

However, the same phenomenon can cause the learning algorithm for the syntactic/recurrent FDFAs to work as hard as the one for the periodic FDFA

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Time Complexity

The worst case time complexity for all 3 families is thus polynomial in the size of the periodic FDFA.

Some positive results on sizes obtained via recurrent FDFA learner vs. periodic FDFA learner on randomly generated Muller automata.

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Future Direction

Find smaller canonical representations ?

Find smallest FDFA ?

Learning the leading first ?

L*-learning for (traditional) ω-automata ?

Other types of learning for ω-languages ?

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THE ENDThank you for your

attention!

Comments or questions?

Learning

Regular

Languagesvia

Alternating

Automata

IJCAI’15

Time Complexity

The worst case time complexity for all 3 families is thus polynomial in the size of the periodic FDFA.

Some positive results on sizes obtained via recurrent FDFA learner vs. periodic FDFA learner on randomly generated Muller automata.

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