schützenberger's star-free theorem and what followed · regular language = language than can...
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Schutzenberger’s Star-Free Theorem and whatFollowed
Thomas Place
LaBRI, Bordeaux University
March 22, 2016
Schutzenberger’s Theorem
Regular Languages
Important Reminder: Regular Languages
Finite words over a finite alphabet A:
Word = finite concatenations of letters in A.(Example: “abba” and “bbaabba” for A = {a, b}).Language = set of words. The language of all words is A⇤.
Regular language = language than can be defined bya finite automaton or a regular expression.
q0 q1
a
b
Language (ab)⇤
r0 r1
a, b
a, b
Language of words of even length
Regular Languages
Important Reminder: Regular Languages
Finite words over a finite alphabet A:
Word = finite concatenations of letters in A.(Example: “abba” and “bbaabba” for A = {a, b}).Language = set of words. The language of all words is A⇤.
Regular language = language than can be defined bya finite automaton or a regular expression.
q0 q1
a
b
Language (ab)⇤
r0 r1
a, b
a, b
Language of words of even length
Star-free Languages: What are they ?
Definition by Induction
all finite and co-finite languages are star-free({a} is star-free, {b, bab, aa} is star-free, A⇤ is star-free,...)
if L1, L2 are star-free then,the union L1 [ L2 is star-free,the intersection L1 \ L2 is star-free,the complement L1 is star-free.
if L1, L2 are star-free then the concatenation, L1 · L2 isstar-free as well.
The language of words that do not contain a “a” is star-free:
A
⇤ · a · A⇤
The star-free languages form a strict subclass of the regularlanguages.
Star-free Languages: What are they ?
Definition by Induction
all finite and co-finite languages are star-free({a} is star-free, {b, bab, aa} is star-free, A⇤ is star-free,...)
if L1, L2 are star-free then,the union L1 [ L2 is star-free,the intersection L1 \ L2 is star-free,the complement L1 is star-free.
if L1, L2 are star-free then the concatenation, L1 · L2 isstar-free as well.
The language of words that do not contain a “a” is star-free:
A
⇤ · a · A⇤
The star-free languages form a strict subclass of the regularlanguages.
Star-free Languages: What are they ?
Definition by Induction
all finite and co-finite languages are star-free({a} is star-free, {b, bab, aa} is star-free, A⇤ is star-free,...)
if L1, L2 are star-free then,the union L1 [ L2 is star-free,the intersection L1 \ L2 is star-free,the complement L1 is star-free.
if L1, L2 are star-free then the concatenation, L1 · L2 isstar-free as well.
The language of words that do not contain a “a” is star-free:
A
⇤ · a · A⇤
The star-free languages form a strict subclass of the regularlanguages.
Star-free Languages: What are they ?
Definition by Induction
all finite and co-finite languages are star-free({a} is star-free, {b, bab, aa} is star-free, A⇤ is star-free,...)
if L1, L2 are star-free then,the union L1 [ L2 is star-free,the intersection L1 \ L2 is star-free,the complement L1 is star-free.
if L1, L2 are star-free then the concatenation, L1 · L2 isstar-free as well.
The language of words that do not contain a “a” is star-free:
A
⇤ · a · A⇤
The star-free languages form a strict subclass of the regularlanguages.
Star-free Languages: What are they ?
Definition by Induction
all finite and co-finite languages are star-free({a} is star-free, {b, bab, aa} is star-free, A⇤ is star-free,...)
if L1, L2 are star-free then,the union L1 [ L2 is star-free,the intersection L1 \ L2 is star-free,the complement L1 is star-free.
if L1, L2 are star-free then the concatenation, L1 · L2 isstar-free as well.
The language of words that do not contain a “a” is star-free:
A
⇤ · a · A⇤
The star-free languages form a strict subclass of the regularlanguages.
Star-free Languages: Examples
q0 q1
a
b
Language (ab)⇤
This language is star-freeA word is in the language i↵
• It does not begin with a “b”
• It does not contain two consecutive “a”
• It does not contain two consecutive “b”
• It does not end with a “a”
A word is in the language i↵
• It does not belong to b · A⇤
• It does not belong to A
⇤ · aa · A⇤.
• It does not belong to A
⇤ · bb · A⇤.
• It does not belong to A
⇤ · a
A word is in the language i↵
• It belongs to b · A⇤
• It belongs to A
⇤ · aa · A⇤
• It belongs to A
⇤ · bb · A⇤
• It belongs to A
⇤ · a
b · A⇤ TA
⇤ · aa · A⇤T
A
⇤ · aT
A
⇤ · bb · A⇤
r0 r1
a, b
a, b
Words of even length
This language is not star-free
Being star-free is a semantic property of a language.Not obvious from the syntax that defines it.
Star-free Languages: Examples
q0 q1
a
b
Language (ab)⇤
This language is star-freeA word is in the language i↵
• It does not begin with a “b”
• It does not contain two consecutive “a”
• It does not contain two consecutive “b”
• It does not end with a “a”
A word is in the language i↵
• It does not belong to b · A⇤
• It does not belong to A
⇤ · aa · A⇤.
• It does not belong to A
⇤ · bb · A⇤.
• It does not belong to A
⇤ · a
A word is in the language i↵
• It belongs to b · A⇤
• It belongs to A
⇤ · aa · A⇤
• It belongs to A
⇤ · bb · A⇤
• It belongs to A
⇤ · a
b · A⇤ TA
⇤ · aa · A⇤T
A
⇤ · aT
A
⇤ · bb · A⇤
r0 r1
a, b
a, b
Words of even length
This language is not star-free
Being star-free is a semantic property of a language.Not obvious from the syntax that defines it.
Star-free Languages: Examples
q0 q1
a
b
Language (ab)⇤
This language is star-free
A word is in the language i↵
• It does not begin with a “b”
• It does not contain two consecutive “a”
• It does not contain two consecutive “b”
• It does not end with a “a”
A word is in the language i↵
• It does not belong to b · A⇤
• It does not belong to A
⇤ · aa · A⇤.
• It does not belong to A
⇤ · bb · A⇤.
• It does not belong to A
⇤ · a
A word is in the language i↵
• It belongs to b · A⇤
• It belongs to A
⇤ · aa · A⇤
• It belongs to A
⇤ · bb · A⇤
• It belongs to A
⇤ · a
b · A⇤ TA
⇤ · aa · A⇤T
A
⇤ · aT
A
⇤ · bb · A⇤
r0 r1
a, b
a, b
Words of even length
This language is not star-free
Being star-free is a semantic property of a language.Not obvious from the syntax that defines it.
Star-free Languages: Examples
q0 q1
a
b
Language (ab)⇤
This language is star-free
A word is in the language i↵
• It does not begin with a “b”
• It does not contain two consecutive “a”
• It does not contain two consecutive “b”
• It does not end with a “a”
A word is in the language i↵
• It does not belong to b · A⇤
• It does not belong to A
⇤ · aa · A⇤.
• It does not belong to A
⇤ · bb · A⇤.
• It does not belong to A
⇤ · a
A word is in the language i↵
• It belongs to b · A⇤
• It belongs to A
⇤ · aa · A⇤
• It belongs to A
⇤ · bb · A⇤
• It belongs to A
⇤ · a
b · A⇤ TA
⇤ · aa · A⇤T
A
⇤ · aT
A
⇤ · bb · A⇤
r0 r1
a, b
a, b
Words of even length
This language is not star-free
Being star-free is a semantic property of a language.Not obvious from the syntax that defines it.
Star-free Languages: Examples
q0 q1
a
b
Language (ab)⇤
This language is star-free
A word is in the language i↵
• It does not begin with a “b”
• It does not contain two consecutive “a”
• It does not contain two consecutive “b”
• It does not end with a “a”
A word is in the language i↵
• It does not belong to b · A⇤
• It does not belong to A
⇤ · aa · A⇤.
• It does not belong to A
⇤ · bb · A⇤.
• It does not belong to A
⇤ · a
A word is in the language i↵
• It belongs to b · A⇤
• It belongs to A
⇤ · aa · A⇤
• It belongs to A
⇤ · bb · A⇤
• It belongs to A
⇤ · a
b · A⇤ TA
⇤ · aa · A⇤T
A
⇤ · aT
A
⇤ · bb · A⇤
r0 r1
a, b
a, b
Words of even length
This language is not star-free
Being star-free is a semantic property of a language.Not obvious from the syntax that defines it.
Star-free Languages: Examples
q0 q1
a
b
Language (ab)⇤
This language is star-free
A word is in the language i↵
• It does not begin with a “b”
• It does not contain two consecutive “a”
• It does not contain two consecutive “b”
• It does not end with a “a”
A word is in the language i↵
• It does not belong to b · A⇤
• It does not belong to A
⇤ · aa · A⇤.
• It does not belong to A
⇤ · bb · A⇤.
• It does not belong to A
⇤ · a
A word is in the language i↵
• It belongs to b · A⇤
• It belongs to A
⇤ · aa · A⇤
• It belongs to A
⇤ · bb · A⇤
• It belongs to A
⇤ · a
b · A⇤ TA
⇤ · aa · A⇤T
A
⇤ · aT
A
⇤ · bb · A⇤
r0 r1
a, b
a, b
Words of even length
This language is not star-free
Being star-free is a semantic property of a language.Not obvious from the syntax that defines it.
Star-free Languages: Examples
q0 q1
a
b
Language (ab)⇤
This language is star-free
A word is in the language i↵
• It does not begin with a “b”
• It does not contain two consecutive “a”
• It does not contain two consecutive “b”
• It does not end with a “a”
A word is in the language i↵
• It does not belong to b · A⇤
• It does not belong to A
⇤ · aa · A⇤.
• It does not belong to A
⇤ · bb · A⇤.
• It does not belong to A
⇤ · a
A word is in the language i↵
• It belongs to b · A⇤
• It belongs to A
⇤ · aa · A⇤
• It belongs to A
⇤ · bb · A⇤
• It belongs to A
⇤ · a
b · A⇤ TA
⇤ · aa · A⇤T
A
⇤ · aT
A
⇤ · bb · A⇤
r0 r1
a, b
a, b
Words of even lengthThis language is not star-free
Being star-free is a semantic property of a language.Not obvious from the syntax that defines it.
Star-free Languages: Examples
q0 q1
a
b
Language (ab)⇤
This language is star-free
A word is in the language i↵
• It does not begin with a “b”
• It does not contain two consecutive “a”
• It does not contain two consecutive “b”
• It does not end with a “a”
A word is in the language i↵
• It does not belong to b · A⇤
• It does not belong to A
⇤ · aa · A⇤.
• It does not belong to A
⇤ · bb · A⇤.
• It does not belong to A
⇤ · a
A word is in the language i↵
• It belongs to b · A⇤
• It belongs to A
⇤ · aa · A⇤
• It belongs to A
⇤ · bb · A⇤
• It belongs to A
⇤ · a
b · A⇤ TA
⇤ · aa · A⇤T
A
⇤ · aT
A
⇤ · bb · A⇤
r0 r1
a, b
a, b
Words of even lengthThis language is not star-free
Being star-free is a semantic property of a language.Not obvious from the syntax that defines it.
Schutzenberger’s Theorem (Schutzenberger’65)
Given a regular language L, the following are equivalent:
1
2
L is star-free
the syntactic monoid of L is aperiodic
3 the minimal automaton of L is counter-free(McNaughton-Papert’71).
semantic
syntactic
Why is it interesting ? - A First Argument
• The syntactic monoid is a finite computable canonicalrepresentation of L
• Aperiodicity is a syntactic and easy to decide propertyof this algebraic object
) e↵ective characterization of the star-free languages.
Schutzenberger’s Theorem (Schutzenberger’65)
Given a regular language L, the following are equivalent:
1
2
L is star-free
the syntactic monoid of L is aperiodic
3 the minimal automaton of L is counter-free(McNaughton-Papert’71).
semantic
syntactic
Why is it interesting ? - A First Argument
• The syntactic monoid is a finite computable canonicalrepresentation of L
• Aperiodicity is a syntactic and easy to decide propertyof this algebraic object
) e↵ective characterization of the star-free languages.
Schutzenberger’s Theorem (Schutzenberger’65)
Given a regular language L, the following are equivalent:
1
2
L is star-free
the syntactic monoid of L is aperiodic
3 the minimal automaton of L is counter-free(McNaughton-Papert’71).
semantic
syntactic
Why is it interesting ? - A First Argument
• The syntactic monoid is a finite computable canonicalrepresentation of L
• Aperiodicity is a syntactic and easy to decide propertyof this algebraic object
) e↵ective characterization of the star-free languages.
Schutzenberger’s Theorem (Schutzenberger’65)
Given a regular language L, the following are equivalent:
1
2
L is star-free
the syntactic monoid of L is aperiodic
3 the minimal automaton of L is counter-free(McNaughton-Papert’71).
semantic
syntactic
Why is it interesting ? - A First Argument
• The syntactic monoid is a finite computable canonicalrepresentation of L
• Aperiodicity is a syntactic and easy to decide propertyof this algebraic object
) e↵ective characterization of the star-free languages.
Schutzenberger’s Theorem (Schutzenberger’65)
Given a regular language L, the following are equivalent:
1
2
L is star-free
the syntactic monoid of L is aperiodic
3 the minimal automaton of L is counter-free(McNaughton-Papert’71).
semantic
syntactic
Why is it interesting ? - A First Argument
• The syntactic monoid is a finite computable canonicalrepresentation of L
• Aperiodicity is a syntactic and easy to decide propertyof this algebraic object
) e↵ective characterization of the star-free languages.
The counter-free condition
McNaughton-Papert-Schutzenberger
Given L regular, the following are equivalent:
1L is star-free.
2 the minimal automaton of L is counter-free.
Counters
A counter is a non-trivial sequence of states q0, . . . , qn (n � 1)along with a word w such that,
q0w�! q1
w�! q2w�! · · · w�! qn
w�! q0
Easy to bound the length of the sequences and wordswe have to look for, which yields decidability.
The counter-free condition
McNaughton-Papert-Schutzenberger
Given L regular, the following are equivalent:
1L is star-free.
2 the minimal automaton of L is counter-free.
Counters
A counter is a non-trivial sequence of states q0, . . . , qn (n � 1)along with a word w such that,
q0w�! q1
w�! q2w�! · · · w�! qn
w�! q0
Easy to bound the length of the sequences and wordswe have to look for, which yields decidability.
Examples (1)
r0 r1
a, b
a, b
Words of even length
q0 q1
a
b
Language (ab)⇤
Has no counter: star-freeHas a counter: not star-free
r0a�! r1
a�! r0
Examples (1)
r0 r1
a, b
a, b
Words of even length
q0 q1
a
b
Language (ab)⇤
Has no counter: star-free
Has a counter: not star-freer0
a�! r1a�! r0
Examples (1)
r0 r1
a, b
a, b
Words of even length
q0 q1
a
b
Language (ab)⇤
Has no counter: star-freeHas a counter: not star-free
r0a�! r1
a�! r0
Examples (2)
q0
q1
q2
q3
q4
a
b
a
b
a
b
Has a counter: not star-freeq0
ab�! q4ab�! q2
ab�! q0
Examples (2)
q0
q1
q2
q3
q4
a
b
a
b
a
b
Has a counter: not star-freeq0
ab�! q4ab�! q2
ab�! q0
Examples (3)
q0 q1 q2 q3
q4
a a a
bbb
a
a, b
No counter: star-freeGood luck finding a star-free description
Examples (3)
q0 q1 q2 q3
q4
a a a
bbb
a
a, bNo counter: star-free
Good luck finding a star-free description
Schutzenberger’s Theorem (Schutzenberger’65)
Given a regular language L, the following are equivalent:
1
2
L is star-free
the syntactic monoid of L is aperiodic
3 the minimal automaton of L is counter-free(McNaughton-Papert’71).
semantic
syntactic
Why is it interesting ? - Second Argument
• The proofs of the hard direction (2, 3 ) 1) is constructive.
• Assuming aperiodicity or counter-free, we have a generic wayto construct a star-free description of L by induction
) We get normal forms for star-free descriptions
A
B
E↵ective characterization of the star-free languages.
Generic construction of star-free descriptions.
) Altogether, we learn a lot about the star-free languages.
Schutzenberger’s Theorem (Schutzenberger’65)
Given a regular language L, the following are equivalent:
1
2
L is star-free
the syntactic monoid of L is aperiodic
3 the minimal automaton of L is counter-free(McNaughton-Papert’71).
semantic
syntactic
Why is it interesting ? - Second Argument
• The proofs of the hard direction (2, 3 ) 1) is constructive.
• Assuming aperiodicity or counter-free, we have a generic wayto construct a star-free description of L by induction
) We get normal forms for star-free descriptions
A
B
E↵ective characterization of the star-free languages.
Generic construction of star-free descriptions.
) Altogether, we learn a lot about the star-free languages.
Schutzenberger’s Theorem (Schutzenberger’65)
Given a regular language L, the following are equivalent:
1
2
L is star-free
the syntactic monoid of L is aperiodic
3 the minimal automaton of L is counter-free(McNaughton-Papert’71).
semantic
syntactic
Why is it interesting ? - Second Argument
• The proofs of the hard direction (2, 3 ) 1) is constructive.
• Assuming aperiodicity or counter-free, we have a generic wayto construct a star-free description of L by induction
) We get normal forms for star-free descriptions
A
B
E↵ective characterization of the star-free languages.
Generic construction of star-free descriptions.
) Altogether, we learn a lot about the star-free languages.
Schutzenberger’s Theorem (Schutzenberger’65)
Given a regular language L, the following are equivalent:
1
2
L is star-free
the syntactic monoid of L is aperiodic
3 the minimal automaton of L is counter-free(McNaughton-Papert’71).
semantic
syntactic
Why is it interesting ?
- Second Argument
• The proofs of the hard direction (2, 3 ) 1) is constructive.
• Assuming aperiodicity or counter-free, we have a generic wayto construct a star-free description of L by induction
) We get normal forms for star-free descriptions
A
B
E↵ective characterization of the star-free languages.
Generic construction of star-free descriptions.
) Altogether, we learn a lot about the star-free languages.
Why should we care about star-free languages ?
Complements to Schutzenberger’s Theorem
The class of star-free languages is important: it admits otherdefinitions.
Given a regular language L, the following properties are equivalent:
1
4
L is star-free.
the syntactic monoid of L is aperiodic.
5 the minimal automaton of L is counter-free(McNaughton-Papert’71).
2 L can be defined in first-order logic (FO)(McNaughton-Papert’71).
3 L can be defined in linear temporal logic (LTL)(Kamp’68).
semantic
syntactic
The benefits of the theorem apply to both FO and LTL .
Complements to Schutzenberger’s Theorem
The class of star-free languages is important: it admits otherdefinitions.
Given a regular language L, the following properties are equivalent:
1
4
L is star-free.
the syntactic monoid of L is aperiodic.
5 the minimal automaton of L is counter-free(McNaughton-Papert’71).
2 L can be defined in first-order logic (FO)(McNaughton-Papert’71).
3 L can be defined in linear temporal logic (LTL)(Kamp’68).
semantic
syntactic
The benefits of the theorem apply to both FO and LTL .
Complements to Schutzenberger’s Theorem
The class of star-free languages is important: it admits otherdefinitions.
Given a regular language L, the following properties are equivalent:
1
4
L is star-free.
the syntactic monoid of L is aperiodic.
5 the minimal automaton of L is counter-free(McNaughton-Papert’71).
2 L can be defined in first-order logic (FO)(McNaughton-Papert’71).
3 L can be defined in linear temporal logic (LTL)(Kamp’68).
semantic
syntactic
The benefits of the theorem apply to both FO and LTL .
Complements to Schutzenberger’s Theorem
The class of star-free languages is important: it admits otherdefinitions.
Given a regular language L, the following properties are equivalent:
1
4
L is star-free.
the syntactic monoid of L is aperiodic.
5 the minimal automaton of L is counter-free(McNaughton-Papert’71).
2 L can be defined in first-order logic (FO)(McNaughton-Papert’71).
3 L can be defined in linear temporal logic (LTL)(Kamp’68).
semantic
syntactic
The benefits of the theorem apply to both FO and LTL .
More about First-Order Logic (FO)
a b b b c a a a c a 2 A
⇤
0 1 2 3 4 5 6 7 8 9
A word is a sequence of labeled positions that can bequantified.
Given a 2 A, one can test if x is labeled by a with a(x).
One can test the linear order: x < y .
8x (a(x) ) 9y (b(y) ^ (y < x)))“for any a in the word, there is a b to its left”
Each sentence defines a language) FO defines a class of languages.
This class is the class of star-free languages.
More about First-Order Logic (FO)
a b b b c a a a c a 2 A
⇤
0 1 2 3 4 5 6 7 8 9
A word is a sequence of labeled positions that can bequantified.
Given a 2 A, one can test if x is labeled by a with a(x).
One can test the linear order: x < y .
8x (a(x) ) 9y (b(y) ^ (y < x)))“for any a in the word, there is a b to its left”
Each sentence defines a language) FO defines a class of languages.
This class is the class of star-free languages.
More about First-Order Logic (FO)
a b b b c a a a c a 2 A
⇤
0 1 2 3 4 5 6 7 8 9
A word is a sequence of labeled positions that can bequantified.
Given a 2 A, one can test if x is labeled by a with a(x).
One can test the linear order: x < y .
8x (a(x) ) 9y (b(y) ^ (y < x)))“for any a in the word, there is a b to its left”
Each sentence defines a language) FO defines a class of languages.
This class is the class of star-free languages.
Defining Languages with First-Order Logic (FO)
Why are all star-free languages FO definable ?
Finite and co-finite languages are FO-definable.
Union, Intersection and Complement correspond to booleanconnectives: _,^,¬.
Concatenation corresponds to existential quantification:
w 2 L1 · L2
There exists a position x in w such that:• the prefix made up of all positions y x belongs to L1.• the su�x made up of all positions y > x belongs to L2.
Proving the converse inclusion is more technical.
Defining Languages with First-Order Logic (FO)
Why are all star-free languages FO definable ?
Finite and co-finite languages are FO-definable.
Union, Intersection and Complement correspond to booleanconnectives: _,^,¬.Concatenation corresponds to existential quantification:
w 2 L1 · L2
There exists a position x in w such that:• the prefix made up of all positions y x belongs to L1.• the su�x made up of all positions y > x belongs to L2.
Proving the converse inclusion is more technical.
Defining Languages with First-Order Logic (FO)
Why are all star-free languages FO definable ?
Finite and co-finite languages are FO-definable.
Union, Intersection and Complement correspond to booleanconnectives: _,^,¬.Concatenation corresponds to existential quantification:
w 2 L1 · L2
There exists a position x in w such that:• the prefix made up of all positions y x belongs to L1.• the su�x made up of all positions y > x belongs to L2.
Proving the converse inclusion is more technical.
What is Next ?
After Schutzenberger’s Theorem
Schutzenberger’s Theorem
Solid understanding ofan important class ofregular word languages
Reproduce the theorem forother important classes of
regular word languages
First Direction
Second Direction
Reproduce the theorem formore general structures(infinite words, trees,...)
Second Direction
Reproduce the theorem formore general structures(infinite words, trees,...)
After Schutzenberger’s Theorem
Schutzenberger’s Theorem
Solid understanding ofan important class ofregular word languages
Reproduce the theorem forother important classes of
regular word languages
First Direction
Second Direction
Reproduce the theorem formore general structures(infinite words, trees,...)
Second Direction
Reproduce the theorem formore general structures(infinite words, trees,...)
After Schutzenberger’s Theorem
Schutzenberger’s Theorem
Solid understanding ofan important class ofregular word languages
Reproduce the theorem forother important classes of
regular word languages
First Direction
Second Direction
Reproduce the theorem formore general structures(infinite words, trees,...)
Second Direction
Reproduce the theorem formore general structures(infinite words, trees,...)
After Schutzenberger’s Theorem
Schutzenberger’s Theorem
Solid understanding ofan important class ofregular word languages
Reproduce the theorem forother important classes of
regular word languages
First Direction
Second Direction
Reproduce the theorem formore general structures(infinite words, trees,...)
Second Direction
Reproduce the theorem formore general structures(infinite words, trees,...)
General Objective: Membership Problem
Reproducing Schutzenberger’s Theorem for other classes oflanguages amounts to considering the following problem.
Given such a class C, the goal is to solve the associatedmembership problem:
L a regular language Does L belong to C ?
Given a particular class, there are two incremental goals to theproblem:
Objective 1: get an algorithm that decides it.
Objective 2: find a generic way to construct a sentence thatdefines L when it exists.
General Objective: Membership Problem
Reproducing Schutzenberger’s Theorem for other classes oflanguages amounts to considering the following problem.
Given such a class C, the goal is to solve the associatedmembership problem:
L a regular language Does L belong to C ?
Given a particular class, there are two incremental goals to theproblem:
Objective 1: get an algorithm that decides it.
Objective 2: find a generic way to construct a sentence thatdefines L when it exists.
Concatenation Hierarchiesand
Quantifier Alternation
What makes a star-free description complicated ?
Simple Language
A
⇤ · aba · A⇤
More Complicated
A
⇤ · a · (A⇤ · aba · A⇤ \ A
⇤ · bab · A⇤) · a · A⇤
Even More Complicated
A
⇤ · a · A⇤ · a · (A⇤ · aba · A⇤ \ A
⇤ · bab · A⇤) · a · A⇤ · b · A⇤
Complicated = alternation concatenation – complementationFormalized by concatenation hierarchies.
What makes a star-free description complicated ?
Simple Language
A
⇤ · aba · A⇤
More Complicated
A
⇤ · a · (A⇤ · aba · A⇤ \ A
⇤ · bab · A⇤) · a · A⇤
Even More Complicated
A
⇤ · a · A⇤ · a · (A⇤ · aba · A⇤ \ A
⇤ · bab · A⇤) · a · A⇤ · b · A⇤
Complicated = alternation concatenation – complementationFormalized by concatenation hierarchies.
What makes a star-free description complicated ?
Simple Language
A
⇤ · aba · A⇤
More Complicated
A
⇤ · a · (A⇤ · aba · A⇤ \ A
⇤ · bab · A⇤) · a · A⇤
Even More Complicated
A
⇤ · a · A⇤ · a · (A⇤ · aba · A⇤ \ A
⇤ · bab · A⇤) · a · A⇤ · b · A⇤
Complicated = alternation concatenation – complementationFormalized by concatenation hierarchies.
What makes a star-free description complicated ?
Simple Language
A
⇤ · aba · A⇤
More Complicated
A
⇤ · a · (A⇤ · aba · A⇤ \ A
⇤ · bab · A⇤) · a · A⇤
Even More Complicated
A
⇤ · a · A⇤ · a · (A⇤ · aba · A⇤ \ A
⇤ · bab · A⇤) · a · A⇤ · b · A⇤
Complicated = alternation concatenation – complementationFormalized by concatenation hierarchies.
Classifying Star-free Languages (1)
The idea was formalized by two (closely related) infinite hierarchieswhich classify star-free languages into levels.
0 12 1 3
2 2 52 3 7
2 4
1 one goes from level i to level i + 12 using concatenation.
2 one goes from level i + 12 to level i +1 using complementation.
Classifying Star-free Languages (2)
Level n + 1
from level n + 12 , close under:
• complementation,• union and intersection
Example for level 1:b · A⇤ \ A
⇤ · aa · A⇤ \ A
⇤ · bb · A⇤ \ A
⇤ · a
Level n + 12
from level n, close under:• concatenation:
L,K 7! L · K• union and intersection
Examples for level 12 :
A
⇤ · aa · A⇤,A⇤ · ab · A⇤ [ b · A⇤ · a,...
Level 0
finite and co-finite
Dot-Depth Hierarchy(Brzozowski,Cohen)’71
Examples:A
⇤, {aa}, {b}, {aa, bb},...
Level n + 1
from level n + 12 , close under:
• complementation,• union and intersection
Level n + 12
from level n, close under:• marked concatenation:
L,K , a 7! L · a · K• union and intersection
Level 0
A
⇤ and ;
Straubing-Therien Hierarchy(Straubing)’81 (Therien)’81
Both hierarchies are know to be strict,(Brzozowski,Knast)’78
Classifying Star-free Languages (2)
Level n + 1
from level n + 12 , close under:
• complementation,• union and intersection
Example for level 1:b · A⇤ \ A
⇤ · aa · A⇤ \ A
⇤ · bb · A⇤ \ A
⇤ · a
Level n + 12
from level n, close under:• concatenation:
L,K 7! L · K• union and intersection
Examples for level 12 :
A
⇤ · aa · A⇤,A⇤ · ab · A⇤ [ b · A⇤ · a,...
Level 0
finite and co-finite
Dot-Depth Hierarchy(Brzozowski,Cohen)’71
Examples:A
⇤, {aa}, {b}, {aa, bb},...
Level n + 1
from level n + 12 , close under:
• complementation,• union and intersection
Level n + 12
from level n, close under:• marked concatenation:
L,K , a 7! L · a · K• union and intersection
Level 0
A
⇤ and ;
Straubing-Therien Hierarchy(Straubing)’81 (Therien)’81
Both hierarchies are know to be strict,(Brzozowski,Knast)’78
Classifying Star-free Languages (2)
Level n + 1
from level n + 12 , close under:
• complementation,• union and intersection
Example for level 1:b · A⇤ \ A
⇤ · aa · A⇤ \ A
⇤ · bb · A⇤ \ A
⇤ · a
Level n + 12
from level n, close under:• concatenation:
L,K 7! L · K• union and intersection
Examples for level 12 :
A
⇤ · aa · A⇤,A⇤ · ab · A⇤ [ b · A⇤ · a,...
Level 0
finite and co-finite
Dot-Depth Hierarchy(Brzozowski,Cohen)’71
Examples:A
⇤, {aa}, {b}, {aa, bb},...
Level n + 1
from level n + 12 , close under:
• complementation,• union and intersection
Level n + 12
from level n, close under:• marked concatenation:
L,K , a 7! L · a · K• union and intersection
Level 0
A
⇤ and ;
Straubing-Therien Hierarchy(Straubing)’81 (Therien)’81
Both hierarchies are know to be strict,(Brzozowski,Knast)’78
Classifying Star-free Languages (2)
Level n + 1
from level n + 12 , close under:
• complementation,• union and intersection
Example for level 1:b · A⇤ \ A
⇤ · aa · A⇤ \ A
⇤ · bb · A⇤ \ A
⇤ · a
Level n + 12
from level n, close under:• concatenation:
L,K 7! L · K• union and intersection
Examples for level 12 :
A
⇤ · aa · A⇤,A⇤ · ab · A⇤ [ b · A⇤ · a,...
Level 0
finite and co-finite
Dot-Depth Hierarchy(Brzozowski,Cohen)’71
Examples:A
⇤, {aa}, {b}, {aa, bb},...
Level n + 1
from level n + 12 , close under:
• complementation,• union and intersection
Level n + 12
from level n, close under:• marked concatenation:
L,K , a 7! L · a · K• union and intersection
Level 0
A
⇤ and ;
Straubing-Therien Hierarchy(Straubing)’81 (Therien)’81
Both hierarchies are know to be strict,(Brzozowski,Knast)’78
Classifying Star-free Languages (2)
Level n + 1
from level n + 12 , close under:
• complementation,• union and intersection
Example for level 1:b · A⇤ \ A
⇤ · aa · A⇤ \ A
⇤ · bb · A⇤ \ A
⇤ · a
Level n + 12
from level n, close under:• concatenation:
L,K 7! L · K• union and intersection
Examples for level 12 :
A
⇤ · aa · A⇤,A⇤ · ab · A⇤ [ b · A⇤ · a,...
Level 0
finite and co-finite
Dot-Depth Hierarchy(Brzozowski,Cohen)’71
Examples:A
⇤, {aa}, {b}, {aa, bb},...
Level n + 1
from level n + 12 , close under:
• complementation,• union and intersection
Level n + 12
from level n, close under:• marked concatenation:
L,K , a 7! L · a · K• union and intersection
Level 0
A
⇤ and ;
Straubing-Therien Hierarchy(Straubing)’81 (Therien)’81
Both hierarchies are know to be strict,(Brzozowski,Knast)’78
Classifying Star-free Languages (2)
Level n + 1
from level n + 12 , close under:
• complementation,• union and intersection
Example for level 1:b · A⇤ \ A
⇤ · aa · A⇤ \ A
⇤ · bb · A⇤ \ A
⇤ · a
Level n + 12
from level n, close under:• concatenation:
L,K 7! L · K• union and intersection
Examples for level 12 :
A
⇤ · aa · A⇤,A⇤ · ab · A⇤ [ b · A⇤ · a,...
Level 0
finite and co-finite
Dot-Depth Hierarchy(Brzozowski,Cohen)’71
Examples:A
⇤, {aa}, {b}, {aa, bb},...
Level n + 1
from level n + 12 , close under:
• complementation,• union and intersection
Level n + 12
from level n, close under:• marked concatenation:
L,K , a 7! L · a · K• union and intersection
Level 0
A
⇤ and ;
Straubing-Therien Hierarchy(Straubing)’81 (Therien)’81
Both hierarchies are know to be strict,(Brzozowski,Knast)’78
Classifying Star-free Languages (2)
Level n + 1
from level n + 12 , close under:
• complementation,• union and intersection
Example for level 1:b · A⇤ \ A
⇤ · aa · A⇤ \ A
⇤ · bb · A⇤ \ A
⇤ · a
Level n + 12
from level n, close under:• concatenation:
L,K 7! L · K• union and intersection
Examples for level 12 :
A
⇤ · aa · A⇤,A⇤ · ab · A⇤ [ b · A⇤ · a,...
Level 0
finite and co-finite
Dot-Depth Hierarchy(Brzozowski,Cohen)’71
Examples:A
⇤, {aa}, {b}, {aa, bb},...
Level n + 1
from level n + 12 , close under:
• complementation,• union and intersection
Level n + 12
from level n, close under:• marked concatenation:
L,K , a 7! L · a · K• union and intersection
Level 0
A
⇤ and ;
Straubing-Therien Hierarchy(Straubing)’81 (Therien)’81
Both hierarchies are know to be strict,(Brzozowski,Knast)’78
A natural objective
Objective
Precisely understand what can be defined with “simple” star-freedescriptions:
Solve membership for each level in both hierarchies.
Before talking about this objective, let us motivate it further:
The hierarchies also admit a natural logical definition.
A natural objective
Objective
Precisely understand what can be defined with “simple” star-freedescriptions:
Solve membership for each level in both hierarchies.
Before talking about this objective, let us motivate it further:
The hierarchies also admit a natural logical definition.
Connection with Logic: Quantifier Alternation
What is a simple FO sentence ?
8x (a(x) ) 9y (b(y) ^ (y < x)))
) ⇧2 sentence.
What is a complicated sentence ?
9x18x29x38x49x58x68x79x8 '(x1, x2, x3, x4, x5, x6, x7, x8)
) ⌃7 sentence (' quantifier-free)
Complicated = High Quantifier Alternation
Connection with Logic: Quantifier Alternation
What is a simple FO sentence ?
8x (a(x) ) 9y (b(y) ^ (y < x)))
) ⇧2 sentence.
What is a complicated sentence ?
9x18x29x38x49x58x68x79x8 '(x1, x2, x3, x4, x5, x6, x7, x8)
) ⌃7 sentence (' quantifier-free)
Complicated = High Quantifier Alternation
Connection with Logic: Quantifier Alternation
What is a simple FO sentence ?
8x (a(x) ) 9y (b(y) ^ (y < x)))
) ⇧2 sentence.
What is a complicated sentence ?
9x18x29x38x49x58x68x79x8 '(x1, x2, x3, x4, x5, x6, x7, x8)
) ⌃7 sentence (' quantifier-free)
Complicated = High Quantifier Alternation
FO Quantifier Alternation Hierarchy
Level n: ⌃n
For all n, a ⌃n sentence is (in prenex normal form)
9x1, . . . , xn18y1, . . . , yn2 · · · · · · '(x , y , . . . )n blocks (starting with 9) quantifier-free
⌃n is not closed under complement ) we get two other classes:
Level n: ⇧n
Negation of a ⌃n sentence:
8x1, . . . , xn19y1, . . . , yn2 · · · '
n blocks (starting with 8)
Level n: B⌃n
Boolean combinations of ⌃n
(and ⇧n) sentences.
FO Quantifier Alternation Hierarchy
Level n: ⌃n
For all n, a ⌃n sentence is (in prenex normal form)
9x1, . . . , xn18y1, . . . , yn2 · · · · · · '(x , y , . . . )n blocks (starting with 9) quantifier-free
⌃n is not closed under complement ) we get two other classes:
Level n: ⇧n
Negation of a ⌃n sentence:
8x1, . . . , xn19y1, . . . , yn2 · · · '
n blocks (starting with 8)
Level n: B⌃n
Boolean combinations of ⌃n
(and ⇧n) sentences.
FO Quantifier Alternation Hierarchy
⌃1
⇧1
B⌃1
⌃2
⇧2
B⌃2
⌃3
⇧3
B⌃3
⌃4
⇧4
B⌃4 FO
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
In the Straubing-Therien Hierarchy (Perrin-Pin’86),• level n is exactly B⌃n,• level n � 1
2 is exactly ⌃n
Same correspondence between dot-depth and a variantof FO with an enriched signature (Thomas’82)
FO Quantifier Alternation Hierarchy
⌃1
⇧1
B⌃1
⌃2
⇧2
B⌃2
⌃3
⇧3
B⌃3
⌃4
⇧4
B⌃4 FO
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
In the Straubing-Therien Hierarchy (Perrin-Pin’86),• level n is exactly B⌃n,• level n � 1
2 is exactly ⌃n
Same correspondence between dot-depth and a variantof FO with an enriched signature (Thomas’82)
FO Quantifier Alternation Hierarchy
⌃1
⇧1
B⌃1
⌃2
⇧2
B⌃2
⌃3
⇧3
B⌃3
⌃4
⇧4
B⌃4 FO
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
In the Straubing-Therien Hierarchy (Perrin-Pin’86),• level n is exactly B⌃n,• level n � 1
2 is exactly ⌃n
Same correspondence between dot-depth and a variantof FO with an enriched signature (Thomas’82)
Membership and the Hierarchies
Remark
1 From now on, I will use the logical point of view (⌃n, B⌃n).
2 Nowadays, people mostly focus on the Straubing-Therienhierarchy. This is what I will do.
(Straubing’85) For each level:
membership for the dot-depth hierarchycan be e↵ectively reduced
to membership for the Straubing-Therien Hierarchy
FO Quantifier Alternation: Membership State of the Art
⌃1
⇧1
B⌃1
⌃2
⇧2
B⌃2
⌃3
⇧3
B⌃3
⌃4
⇧4
B⌃4 FO
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(Schutzenberger)’65(McNaughton-Papert)’71
Solved
(Simon)’75
Solved
(Arfi)’87(Pin, Weil)’95
Solved
(P., Zeitoun)’14
Solved
(P.)’15
SolvedOpen Open
Why is progress on this question so slow ?Why are we still stuck ?
FO Quantifier Alternation: Membership State of the Art
⌃1
⇧1
B⌃1
⌃2
⇧2
B⌃2
⌃3
⇧3
B⌃3
⌃4
⇧4
B⌃4 FO
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(Schutzenberger)’65(McNaughton-Papert)’71
Solved
(Simon)’75
Solved
(Arfi)’87(Pin, Weil)’95
Solved
(P., Zeitoun)’14
Solved
(P.)’15
SolvedOpen Open
Why is progress on this question so slow ?Why are we still stuck ?
FO Quantifier Alternation: Membership State of the Art
⌃1
⇧1
B⌃1
⌃2
⇧2
B⌃2
⌃3
⇧3
B⌃3
⌃4
⇧4
B⌃4 FO
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(Schutzenberger)’65(McNaughton-Papert)’71
Solved
(Simon)’75
Solved
(Arfi)’87(Pin, Weil)’95
Solved
(P., Zeitoun)’14
Solved
(P.)’15
SolvedOpen Open
Why is progress on this question so slow ?Why are we still stuck ?
FO Quantifier Alternation: Membership State of the Art
⌃1
⇧1
B⌃1
⌃2
⇧2
B⌃2
⌃3
⇧3
B⌃3
⌃4
⇧4
B⌃4 FO
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(Schutzenberger)’65(McNaughton-Papert)’71
Solved
(Simon)’75
Solved
(Arfi)’87(Pin, Weil)’95
Solved
(P., Zeitoun)’14
Solved
(P.)’15
SolvedOpen Open
Why is progress on this question so slow ?Why are we still stuck ?
FO Quantifier Alternation: Membership State of the Art
⌃1
⇧1
B⌃1
⌃2
⇧2
B⌃2
⌃3
⇧3
B⌃3
⌃4
⇧4
B⌃4 FO
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(Schutzenberger)’65(McNaughton-Papert)’71
Solved
(Simon)’75
Solved
(Arfi)’87(Pin, Weil)’95
Solved
(P., Zeitoun)’14
Solved
(P.)’15
SolvedOpen Open
Why is progress on this question so slow ?Why are we still stuck ?
FO Quantifier Alternation: Membership State of the Art
⌃1
⇧1
B⌃1
⌃2
⇧2
B⌃2
⌃3
⇧3
B⌃3
⌃4
⇧4
B⌃4 FO
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(Schutzenberger)’65(McNaughton-Papert)’71
Solved
(Simon)’75
Solved
(Arfi)’87(Pin, Weil)’95
Solved
(P., Zeitoun)’14
Solved
(P.)’15
SolvedOpen Open
Why is progress on this question so slow ?Why are we still stuck ?
A deeper look at membership questions
Quick Remark
The original approach uses an algebraic point of view: thecentral object is the syntactic monoid.
I will use an automata point of view: the central object is theminimal automaton.
Membership Problem: General approach
For a level C in the hierarchy, we search for a membershipalgorithm for C which decides the following problem.
L a regular language Does L belong to C?
This is not how the question is approached.
Membership Problem: General approach
For a level C in the hierarchy, we search for a membershipalgorithm for C which decides the following problem.
L a regular language Does L belong to C?
This is not how the question is approached.
The problem we are really considering
L a regular language
It has a minimal automaton
ac
b
bb
c
d
a
ac
a
The initial and final statescan be changed
) set of accepted languages
L in C
All accepted languages in C
The set is finite and has a structure.
This structure is connected to our building operations:boolean operations and concatenation.
L is built-up from the languages in the set.
Algorithms (and their proofs) are based on this structure.
The problem we are really considering
L a regular language
It has a minimal automaton
ac
b
bb
c
d
a
ac
a
The initial and final statescan be changed
) set of accepted languages
L in C
All accepted languages in C
The set is finite and has a structure.
This structure is connected to our building operations:boolean operations and concatenation.
L is built-up from the languages in the set.
Algorithms (and their proofs) are based on this structure.
The problem we are really considering
L a regular language
It has a minimal automaton
ac
b
bb
c
d
a
ac
a
The initial and final statescan be changed
) set of accepted languages
L in C
All accepted languages in C
The set is finite and has a structure.
This structure is connected to our building operations:boolean operations and concatenation.
L is built-up from the languages in the set.
Algorithms (and their proofs) are based on this structure.
The problem we are really considering
L a regular language
It has a minimal automaton
ac
b
bb
c
d
a
ac
a
The initial and final statescan be changed
) set of accepted languages
L in C
All accepted languages in C
The set is finite and has a structure.
This structure is connected to our building operations:boolean operations and concatenation.
L is built-up from the languages in the set.
Algorithms (and their proofs) are based on this structure.
The problem we are really considering
L a regular language
It has a minimal automaton
ac
b
bb
c
d
a
ac
a
The initial and final statescan be changed
) set of accepted languages
L in C
All accepted languages in C
The set is finite and has a structure.
This structure is connected to our building operations:boolean operations and concatenation.
L is built-up from the languages in the set.
Algorithms (and their proofs) are based on this structure.
The problem we are really considering
L a regular language
It has a minimal automaton
ac
b
bb
c
d
a
ac
a
The initial and final statescan be changed
) set of accepted languages
L in C
All accepted languages in C
The set is finite and has a structure.
This structure is connected to our building operations:boolean operations and concatenation.
L is built-up from the languages in the set.
Algorithms (and their proofs) are based on this structure.
The problem we are really considering
L a regular language
It has a minimal automaton
ac
b
bb
c
d
a
ac
a
The initial and final statescan be changed
) set of accepted languages
L in C
All accepted languages in C
The set is finite and has a structure.
This structure is connected to our building operations:boolean operations and concatenation.
L is built-up from the languages in the set.
Algorithms (and their proofs) are based on this structure.
The problem we are really considering
L a regular language
It has a minimal automaton
ac
b
bb
c
d
a
ac
a
The initial and final statescan be changed
) set of accepted languages
L in C
All accepted languages in C
The set is finite and has a structure.
This structure is connected to our building operations:boolean operations and concatenation.
L is built-up from the languages in the set.
Algorithms (and their proofs) are based on this structure.
The problem we are really considering
L a regular language
It has a minimal automaton
ac
b
bb
c
d
a
ac
a
The initial and final statescan be changed
) set of accepted languages
L in C
All accepted languages in C
The set is finite and has a structure.
This structure is connected to our building operations:boolean operations and concatenation.
L is built-up from the languages in the set.
Algorithms (and their proofs) are based on this structure.
With this approach alone, we get the following results
⌃1
⇧1
B⌃1
⌃2
⇧2
B⌃2
⌃3
⇧3
B⌃3
⌃4
⇧4
B⌃4 FO
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(Schutzenberger)’65(McNaughton-Papert)’71
Solved
(Simon)’75
(Arfi)’87(Pin, Weil)’95
Solved
What is the missing ingredient to go higher ?
Issues with the general approach
Reminder: the problem we are really considering
L a regular language
It has a minimal automaton
ac
b
bb
c
d
a
ac
a
We ask whether all acceptedlanguages belong to C
L in C
All accepted languages in C
Benefits
Algorithms (and their proofs) are based on the structure of theset of accepted languages.
In particular, this is essential for the inductive construction ofsentences.
Reminder: the problem we are really considering
L a regular language
It has a minimal automaton
ac
b
bb
c
d
a
ac
a
We ask whether all acceptedlanguages belong to C
L in C
All accepted languages in C
Benefits
Algorithms (and their proofs) are based on the structure of theset of accepted languages.
In particular, this is essential for the inductive construction ofsentences.
Why is this a problem for the Hierarchy ?
A ⌃n sentence is layered: Consider a ⌃3 sentence
9x1
9x2 9x3
8y1 8y2
8y3
8y49z1 9z2 9z3 9z4
⌃1 layer
⇧2 layer
⌃3 layer
Reminder: For C-membership the hard part is gettinga generic way to build C-sentences (when possible)
• Building ⌃3 sentences by induction for allaccepted languages is also a layered construction.
• This requires to first build ⇧2 sentences.
) We first need to ask a “⇧2 question”on our input automaton.
• In general this “⇧2 question” cannot be:“are all accepted languages in ⇧2?”
) The “⇧2 question” must involvea more general problem than membership
Why is this a problem for the Hierarchy ?
A ⌃n sentence is layered: Consider a ⌃3 sentence
9x1
9x2 9x3
8y1 8y2
8y3
8y49z1 9z2 9z3 9z4
⌃1 layer
⇧2 layer
⌃3 layer
Reminder: For C-membership the hard part is gettinga generic way to build C-sentences (when possible)
• Building ⌃3 sentences by induction for allaccepted languages is also a layered construction.
• This requires to first build ⇧2 sentences.
) We first need to ask a “⇧2 question”on our input automaton.
• In general this “⇧2 question” cannot be:“are all accepted languages in ⇧2?”
) The “⇧2 question” must involvea more general problem than membership
Why is this a problem for the Hierarchy ?
A ⌃n sentence is layered: Consider a ⌃3 sentence
9x1
9x2 9x3
8y1 8y2
8y3
8y49z1 9z2 9z3 9z4
⌃1 layer
⇧2 layer
⌃3 layer
Reminder: For C-membership the hard part is gettinga generic way to build C-sentences (when possible)
• Building ⌃3 sentences by induction for allaccepted languages is also a layered construction.
• This requires to first build ⇧2 sentences.
) We first need to ask a “⇧2 question”on our input automaton.
• In general this “⇧2 question” cannot be:“are all accepted languages in ⇧2?”
) The “⇧2 question” must involvea more general problem than membership
Why is this a problem for the Hierarchy ?
A ⌃n sentence is layered: Consider a ⌃3 sentence
9x1
9x2 9x3
8y1 8y2
8y3
8y49z1 9z2 9z3 9z4
⌃1 layer
⇧2 layer
⌃3 layer
Reminder: For C-membership the hard part is gettinga generic way to build C-sentences (when possible)
• Building ⌃3 sentences by induction for allaccepted languages is also a layered construction.
• This requires to first build ⇧2 sentences.
) We first need to ask a “⇧2 question”on our input automaton.
• In general this “⇧2 question” cannot be:“are all accepted languages in ⇧2?”
) The “⇧2 question” must involvea more general problem than membership
Why is this a problem for the Hierarchy ?
A ⌃n sentence is layered: Consider a ⌃3 sentence
9x1
9x2 9x3
8y1 8y2
8y3
8y49z1 9z2 9z3 9z4
⌃1 layer
⇧2 layer
⌃3 layer
Reminder: For C-membership the hard part is gettinga generic way to build C-sentences (when possible)
• Building ⌃3 sentences by induction for allaccepted languages is also a layered construction.
• This requires to first build ⇧2 sentences.
) We first need to ask a “⇧2 question”on our input automaton.
• In general this “⇧2 question” cannot be:“are all accepted languages in ⇧2?”
) The “⇧2 question” must involvea more general problem than membership
The main idea behind membership for B⌃2, ⌃3
Two steps:
1 solve a deeper problem for ⌃2 independently.
2 reuse the answer to this problem in the membership algorithmfor B⌃2 or ⌃3.
What is a “more general problem than membership”?
What we want is an “approximation problem”:
to build ⌃3 sentences, we first want compute the “best possible⇧2-approximation of our languages”.
We have several “approximation problems” (each one tailored to aparticular logic). However they are all based on a common one:
the separation problem
What is a “more general problem than membership”?
What we want is an “approximation problem”:
to build ⌃3 sentences, we first want compute the “best possible⇧2-approximation of our languages”.
We have several “approximation problems” (each one tailored to aparticular logic). However they are all based on a common one:
the separation problem
More General: Separation is Here
Given a level C in the hierarchy, decide the following problem:
L1, L2 two regular languages
a
a
a
a bb b
a
L1
L2
Can L1 be separated from L2
with a sentence of C?
L1
L2
A
⇤
C-definabledefinable in C
Membership can be formally
reduced to separation
More General: Separation is Here
Given a level C in the hierarchy, decide the following problem:
L1, L2 two regular languages
a
a
a
a bb b
a
L1
L2
Can L1 be separated from L2
with a sentence of C?
L1L2
A
⇤
C-definabledefinable in C
Membership can be formally
reduced to separation
More General: Separation is Here
Given a level C in the hierarchy, decide the following problem:
L1, L2 two regular languages
a
a
a
a bb b
a
L1
L2
Can L1 be separated from L2
with a sentence of C?
L1L2
A
⇤
C-definable
definable in C
Membership can be formally
reduced to separation
More General: Separation is Here
Given a level C in the hierarchy, decide the following problem:
L1, L2 two regular languages
a
a
a
a bb b
a
L1
L2
Can L1 be separated from L2
with a sentence of C?
L2 = A
⇤ \ L1 L1
L2
A
⇤
C-definabledefinable in C
Membership can be formally
reduced to separation
More General: Separation is Here
Given a level C in the hierarchy, decide the following problem:
L1, L2 two regular languages
a
a
a
a bb b
a
L1
L2
Can L1 be separated from L2
with a sentence of C?
L2 = A
⇤ \ L1 L1
L2
A
⇤
C-definable
definable in C
Membership can be formally
reduced to separation
A more General Problem than Membership
a
a
a
a bb b
a
L1
L2
a bb
a
b
b
a
a
bb
a
aa
b
Problem: Among the languages accepted by Awhich ones can be separated by C?
Product Automaton Aaccepts them both
Membership
1) Can we define everythingabout the input with C?
2) If yes, compute sentences.
Separation
1) How well can the languagesbe approximated with C?
2) Compute sentences realizingthis best approximation.
A more General Problem than Membership
a
a
a
a bb b
a
L1
L2
a bb
a
b
b
a
a
bb
a
aa
b
Problem: Among the languages accepted by Awhich ones can be separated by C?
Product Automaton Aaccepts them both
Membership
1) Can we define everythingabout the input with C?
2) If yes, compute sentences.
Separation
1) How well can the languagesbe approximated with C?
2) Compute sentences realizingthis best approximation.
A more General Problem than Membership
a
a
a
a bb b
a
L1
L2
a bb
a
b
b
a
a
bb
a
aa
b
Problem: Among the languages accepted by Awhich ones can be separated by C?
Product Automaton Aaccepts them both
Membership
1) Can we define everythingabout the input with C?
2) If yes, compute sentences.
Separation
1) How well can the languagesbe approximated with C?
2) Compute sentences realizingthis best approximation.
The example of ⌃3
Theorem ((P.,Zeitoun)’14)
For all n � 1,
membership for ⌃n+1 reduces to separation for ⌃n
Using this theorem, an algorithm for ⌃n+1 membership works inthree steps,
1 compute the syntactic monoid of the input L.2 among the pairs of languages recognized by this syntactic
monoid, compute which ones are ⌃n-separable.(this requires an independent ⌃n-separation algorithm).
3 ⌃n+1 is now equivalent to a syntactic criterion on thesyntactic which depends on the above pairs.
The proof for ⌃n+1-membership is constructive provided that theindependent ⌃n-separation algorithm is constructive as well.
The example of ⌃3
Theorem ((P.,Zeitoun)’14)
For all n � 1,
membership for ⌃n+1 reduces to separation for ⌃n
Using this theorem, an algorithm for ⌃n+1 membership works inthree steps,
1 compute the syntactic monoid of the input L.
2 among the pairs of languages recognized by this syntacticmonoid, compute which ones are ⌃n-separable.(this requires an independent ⌃n-separation algorithm).
3 ⌃n+1 is now equivalent to a syntactic criterion on thesyntactic which depends on the above pairs.
The proof for ⌃n+1-membership is constructive provided that theindependent ⌃n-separation algorithm is constructive as well.
The example of ⌃3
Theorem ((P.,Zeitoun)’14)
For all n � 1,
membership for ⌃n+1 reduces to separation for ⌃n
Using this theorem, an algorithm for ⌃n+1 membership works inthree steps,
1 compute the syntactic monoid of the input L.2 among the pairs of languages recognized by this syntactic
monoid, compute which ones are ⌃n-separable.(this requires an independent ⌃n-separation algorithm).
3 ⌃n+1 is now equivalent to a syntactic criterion on thesyntactic which depends on the above pairs.
The proof for ⌃n+1-membership is constructive provided that theindependent ⌃n-separation algorithm is constructive as well.
The example of ⌃3
Theorem ((P.,Zeitoun)’14)
For all n � 1,
membership for ⌃n+1 reduces to separation for ⌃n
Using this theorem, an algorithm for ⌃n+1 membership works inthree steps,
1 compute the syntactic monoid of the input L.2 among the pairs of languages recognized by this syntactic
monoid, compute which ones are ⌃n-separable.(this requires an independent ⌃n-separation algorithm).
3 ⌃n+1 is now equivalent to a syntactic criterion on thesyntactic which depends on the above pairs.
The proof for ⌃n+1-membership is constructive provided that theindependent ⌃n-separation algorithm is constructive as well.
The example of ⌃3
Theorem ((P.,Zeitoun)’14)
For all n � 1,
membership for ⌃n+1 reduces to separation for ⌃n
Our ⌃3-membership algorithm is based on an (independent)⌃2-separation algorithm.
Important Remark
Separation is harder than membership. The theorem above doesnot solve the whole hierarchy.
The example of ⌃3
Theorem ((P.,Zeitoun)’14)
For all n � 1,
membership for ⌃n+1 reduces to separation for ⌃n
Our ⌃3-membership algorithm is based on an (independent)⌃2-separation algorithm.
Important Remark
Separation is harder than membership. The theorem above doesnot solve the whole hierarchy.
The example of B⌃2
A similar theorem holds between ⌃2 and B⌃2. However,
1 the theorem is specific to level n = 2.
2 the connection is based on a problem even more generalthan separation (which we call “pointed covering”).
Current state of the art: Approximation ProblemsSep
aration
(andmore
general)
Mem
bership
⌃1 B⌃1 ⌃2 B⌃2 ⌃3 B⌃3 ⌃4 FO( ( ( ( ( ( (
(Schutzenberger)’65(McNaughton-Papert)’71
(Henckell)’88(P.,Zeitoun)’14
(Simon)’75
(Almeida,Zeitoun)’97(Czerwinski,Martens,Masopust)’13
(P.,Van Rooijen,Zeitoun)’13
(P.,Zeitoun)’14(Arfi)’87
(Pin, Weil)’95
(P.,Zeitoun)’14
(P., Zeitoun)(unpublished yet)
(P.)’15
(P.)’15
Current state of the art: Approximation ProblemsSep
aration
(andmore
general)
Mem
bership
⌃1 B⌃1 ⌃2 B⌃2 ⌃3 B⌃3 ⌃4 FO( ( ( ( ( ( (
(Schutzenberger)’65(McNaughton-Papert)’71
(Henckell)’88(P.,Zeitoun)’14
(Simon)’75
(Almeida,Zeitoun)’97(Czerwinski,Martens,Masopust)’13
(P.,Van Rooijen,Zeitoun)’13
(P.,Zeitoun)’14(Arfi)’87
(Pin, Weil)’95
(P.,Zeitoun)’14
(P., Zeitoun)(unpublished yet)
(P.)’15
(P.)’15
Conclusion
Conclusion
What we learned:
We have to consider more general problems than membership.
Current and future work:
the family of “approximation problems” is a jungle.
our original separation algorithms are non-constructive.
We think, we have found the solution to these issues with a newproblem called “the covering problem”.
higher levels and other structures.
The big problem right now:
No transfer result seems possible for approximation problems
Conclusion
What we learned:
We have to consider more general problems than membership.
Current and future work:
the family of “approximation problems” is a jungle.
our original separation algorithms are non-constructive.
We think, we have found the solution to these issues with a newproblem called “the covering problem”.
higher levels and other structures.
The big problem right now:
No transfer result seems possible for approximation problems
Conclusion
What we learned:
We have to consider more general problems than membership.
Current and future work:
the family of “approximation problems” is a jungle.
our original separation algorithms are non-constructive.
We think, we have found the solution to these issues with a newproblem called “the covering problem”.
higher levels and other structures.
The big problem right now:
No transfer result seems possible for approximation problems
Conclusion
What we learned:
We have to consider more general problems than membership.
Current and future work:
the family of “approximation problems” is a jungle.
our original separation algorithms are non-constructive.
We think, we have found the solution to these issues with a newproblem called “the covering problem”.
higher levels and other structures.
The big problem right now:
No transfer result seems possible for approximation problems
ConclusionSep
aration
(andmore
general)
Mem
bership
⌃1 B⌃1 ⌃2 B⌃2 ⌃3 B⌃3 ⌃4 FO( ( ( ( ( ( (
(Schutzenberger)’65(McNaughton-Papert)’71
(Henckell)’88(P.,Zeitoun)’14
(Simon)’75
(Almeida,Zeitoun)’97(Czerwinski,Martens,Masopust)’13
(P.,Van Rooijen,Zeitoun)’13
(P.,Zeitoun)’14(Arfi)’87
(Pin, Weil)’95
(P.,Zeitoun)’14
(P., Zeitoun)’16(unpublished yet)
(P.)’15
(P.)’15
Thank You