Marc Zeitoun Joint work with Thomas Place LaBRI, U ... · Marc Zeitoun Joint work with Thomas Place...
Transcript of Marc Zeitoun Joint work with Thomas Place LaBRI, U ... · Marc Zeitoun Joint work with Thomas Place...
Separating languages with no ambiguity
Marc Zeitoun
Joint work with Thomas Place
LaBRI, U. Bordeaux
ANR Delta@IRIF, March 2018
Concatenation and AlternationHierarchies
The membership problem for classes of languages
Membership problem for a class CI INPUT A (regular) language L.I QUESTION Does L belong to C?
Examples of classes C:I Languages definable in FO.I Languages of GSH k ≥ 1.I Languages of GSH 0 (called star-free, denoted SF).
(ab)+ =
a∅ ∩ ∅b ∩ ∅(aa+ bb)∅ ∩\
c 6=a;b
∅c∅
a
a
b
bb
c
a
a
c
aa
b
b
bb
c
a
a
c
a Does it belong to C?
Schützenberger ’65For L a regular language, the following are equivalent:I L is star-free.
semantic
I The minimal automaton of L is counter-free.
syntactic
2 / 29
The membership problem for classes of languages
Membership problem for a class CI INPUT A (regular) language L.I QUESTION Does L belong to C?
Examples of classes C:I Languages definable in FO.I Languages of GSH k ≥ 1.I Languages of GSH 0 (called star-free, denoted SF).
(ab)+ =
a∅ ∩ ∅b ∩ ∅(aa+ bb)∅ ∩\
c 6=a;b
∅c∅
a
a
b
bb
c
a
a
c
aa
b
b
bb
c
a
a
c
a Does it belong to C?
Schützenberger ’65For L a regular language, the following are equivalent:I L is star-free.
semantic
I The minimal automaton of L is counter-free.
syntactic
2 / 29
The membership problem for classes of languages
Membership problem for a class CI INPUT A (regular) language L.I QUESTION Does L belong to C?
Examples of classes C:I Languages definable in FO.I Languages of GSH k ≥ 1.I Languages of GSH 0 (called star-free, denoted SF).
(ab)+ =
a∅ ∩ ∅b ∩ ∅(aa+ bb)∅ ∩\
c 6=a;b
∅c∅
a
a
b
bb
c
a
a
c
aa
b
b
bb
c
a
a
c
a Does it belong to C?
Schützenberger ’65For L a regular language, the following are equivalent:I L is star-free. semanticI The minimal automaton of L is counter-free. syntactic
2 / 29
The membership problem for classes of languages
Membership problem for a class CI INPUT A (regular) language L.I QUESTION Does L belong to C?
Examples of classes C:I Languages definable in FO.I Languages of GSH k ≥ 1.I Languages of GSH 0 (called star-free, denoted SF).
(ab)+ = a∅
∩ ∅b ∩ ∅(aa+ bb)∅ ∩\
c 6=a;b
∅c∅
a
a
b
bb
c
a
a
c
aa
b
b
bb
c
a
a
c
a Does it belong to C?
Schützenberger ’65For L a regular language, the following are equivalent:I L is star-free. semanticI The minimal automaton of L is counter-free. syntactic
2 / 29
The membership problem for classes of languages
Membership problem for a class CI INPUT A (regular) language L.I QUESTION Does L belong to C?
Examples of classes C:I Languages definable in FO.I Languages of GSH k ≥ 1.I Languages of GSH 0 (called star-free, denoted SF).
(ab)+ = a∅ ∩ ∅b
∩ ∅(aa+ bb)∅ ∩\
c 6=a;b
∅c∅
a
a
b
bb
c
a
a
c
aa
b
b
bb
c
a
a
c
a Does it belong to C?
Schützenberger ’65For L a regular language, the following are equivalent:I L is star-free. semanticI The minimal automaton of L is counter-free. syntactic
2 / 29
The membership problem for classes of languages
Membership problem for a class CI INPUT A (regular) language L.I QUESTION Does L belong to C?
Examples of classes C:I Languages definable in FO.I Languages of GSH k ≥ 1.I Languages of GSH 0 (called star-free, denoted SF).
(ab)+ = a∅ ∩ ∅b ∩ ∅(aa+ bb)∅
∩\
c 6=a;b
∅c∅
a
a
b
bb
c
a
a
c
aa
b
b
bb
c
a
a
c
a Does it belong to C?
Schützenberger ’65For L a regular language, the following are equivalent:I L is star-free. semanticI The minimal automaton of L is counter-free. syntactic
2 / 29
The membership problem for classes of languages
Membership problem for a class CI INPUT A (regular) language L.I QUESTION Does L belong to C?
Examples of classes C:I Languages definable in FO.I Languages of GSH k ≥ 1.I Languages of GSH 0 (called star-free, denoted SF).
(ab)+ = a∅ ∩ ∅b ∩ ∅(aa+ bb)∅ ∩\
c 6=a;b
∅c∅
a
a
b
bb
c
a
a
c
aa
b
b
bb
c
a
a
c
a Does it belong to C?
Schützenberger ’65For L a regular language, the following are equivalent:I L is star-free. semanticI The minimal automaton of L is counter-free. syntactic
2 / 29
The membership problem for classes of languages
Membership problem for a class CI INPUT A (regular) language L.I QUESTION Does L belong to C?
Examples of classes C:I Languages definable in FO.I Languages of GSH k ≥ 1.I Languages of GSH 0 (called star-free, denoted SF).
(ab)+ =
a∅ ∩ ∅b ∩ ∅(aa+ bb)∅ ∩\
c 6=a;b
∅c∅
a
a
b
bb
c
a
a
c
aa
b
b
bb
c
a
a
c
a Does it belong to C?
Schützenberger ’65For L a regular language, the following are equivalent:I L is star-free. semanticI The minimal automaton of L is counter-free. syntactic
2 / 29
Why is Schützenberger’s theorem interesting?
1. Provides an effective characterization of SF.
2. Link with first-order logic FO.
Schützenberger ’65, McNaughton, Papert ’71For L a regular language, the following are equivalent:I L is star-free. semanticI L is FO-definable. semanticI The minimal automaton of L is counter-free. syntactic
SF FOA∗; ∅ True;False
∪; A∗\ ∨; ¬KaL ∃x a(x) ∧ ’<x
K (x) ∧ ’>xL (x)
3. Constructive proof ⇒ normal forms for star-free expressions, FO sentences.
3 / 29
Why is Schützenberger’s theorem interesting?
1. Provides an effective characterization of SF.
2. Link with first-order logic FO.
Schützenberger ’65, McNaughton, Papert ’71For L a regular language, the following are equivalent:I L is star-free. semanticI L is FO-definable. semanticI The minimal automaton of L is counter-free. syntactic
SF FOA∗; ∅ True;False
∪; A∗\ ∨; ¬KaL ∃x a(x) ∧ ’<x
K (x) ∧ ’>xL (x)
3. Constructive proof ⇒ normal forms for star-free expressions, FO sentences.
3 / 29
Why is Schützenberger’s theorem interesting?
1. Provides an effective characterization of SF.
2. Link with first-order logic FO.
Schützenberger ’65, McNaughton, Papert ’71For L a regular language, the following are equivalent:I L is star-free. semanticI L is FO-definable. semanticI The minimal automaton of L is counter-free. syntactic
SF FOA∗; ∅ True;False
∪; A∗\ ∨; ¬KaL ∃x a(x) ∧ ’<x
K (x) ∧ ’>xL (x)
3. Constructive proof ⇒ normal forms for star-free expressions, FO sentences.
3 / 29
Why is Schützenberger’s theorem interesting?
1. Provides an effective characterization of SF.
2. Link with first-order logic FO.
Schützenberger ’65, McNaughton, Papert ’71For L a regular language, the following are equivalent:I L is star-free. semanticI L is FO-definable. semanticI The minimal automaton of L is counter-free. syntactic
SF FOA∗; ∅ True;False
∪; A∗\ ∨; ¬KaL ∃x a(x) ∧ ’<x
K (x) ∧ ’>xL (x)
3. Constructive proof ⇒ normal forms for star-free expressions, FO sentences.
3 / 29
Concatenation hierarchies: Motivation
Definition of SF alternate Boolean operations and concatenationI SF = smallest class such that:
I ∅ ∈ SF and A∗ ∈ SF.I SF is closed under Boolean operations over A∗.I SF is closed under marked concatenation.
K;L ∈ SFa ∈ A
=⇒ KaL ∈ SF
GoalI Finer understanding of FO / SF.I Interplay between complement and concatenation.
4 / 29
Concatenation hierarchies: Motivation
Definition of SF alternate Boolean operations and concatenationI SF = smallest class such that:
I ∅ ∈ SF and A∗ ∈ SF.I SF is closed under Boolean operations over A∗.I SF is closed under marked concatenation.
K;L ∈ SFa ∈ A
=⇒ KaL ∈ SF
GoalI Finer understanding of FO / SF.I Interplay between complement and concatenation.
4 / 29
Classical hierarchies inside SFTwo classes built on top of CI Boolean closure Bool(C).I Polynomial closure Pol(C) = closure under marked concatenation + union
How many needed alternations between Boolean operations and concatenations?
Straubing-Thérien hierarchy ’81I ST [0] = {∅; A∗}.I ST
ˆn + 1
2
˜= Pol(ST [n]).
I ST [n + 1] = Bool(STˆn + 1
2
˜).
Brzozowski-Cohen hierarchy ’71I DD [0] = {∅; {"}; A+; A∗}.I DD
ˆn + 1
2
˜= Pol(DD [n]).
I DD [n + 1] = Bool(DDˆn + 1
2
˜).
0 12 1 3
2 2 52
PolBool
PolBool
Pol
5 / 29
Classical hierarchies inside SFTwo classes built on top of CI Boolean closure Bool(C).I Polynomial closure Pol(C) = closure under marked concatenation + union
How many needed alternations between Boolean operations and concatenations?
Straubing-Thérien hierarchy ’81I ST [0] = {∅; A∗}.I ST
ˆn + 1
2
˜= Pol(ST [n]).
I ST [n + 1] = Bool(STˆn + 1
2
˜).
Brzozowski-Cohen hierarchy ’71I DD [0] = {∅; {"}; A+; A∗}.I DD
ˆn + 1
2
˜= Pol(DD [n]).
I DD [n + 1] = Bool(DDˆn + 1
2
˜).
0 12 1 3
2 2 52
PolBool
PolBool
Pol
5 / 29
Classical hierarchies inside SFTwo classes built on top of CI Boolean closure Bool(C).I Polynomial closure Pol(C) = closure under marked concatenation + union
How many needed alternations between Boolean operations and concatenations?
Straubing-Thérien hierarchy ’81I ST [0] = {∅; A∗}.
I STˆn + 1
2
˜= Pol(ST [n]).
I ST [n + 1] = Bool(STˆn + 1
2
˜).
Brzozowski-Cohen hierarchy ’71I DD [0] = {∅; {"}; A+; A∗}.I DD
ˆn + 1
2
˜= Pol(DD [n]).
I DD [n + 1] = Bool(DDˆn + 1
2
˜).
0 12 1 3
2 2 52
PolBool
PolBool
Pol
5 / 29
Classical hierarchies inside SFTwo classes built on top of CI Boolean closure Bool(C).I Polynomial closure Pol(C) = closure under marked concatenation + union
How many needed alternations between Boolean operations and concatenations?
Straubing-Thérien hierarchy ’81I ST [0] = {∅; A∗}.I ST
ˆn + 1
2
˜= Pol(ST [n]).
I ST [n + 1] = Bool(STˆn + 1
2
˜).
Brzozowski-Cohen hierarchy ’71I DD [0] = {∅; {"}; A+; A∗}.I DD
ˆn + 1
2
˜= Pol(DD [n]).
I DD [n + 1] = Bool(DDˆn + 1
2
˜).
0 12 1 3
2 2 52
PolBool
PolBool
Pol
5 / 29
Classical hierarchies inside SFTwo classes built on top of CI Boolean closure Bool(C).I Polynomial closure Pol(C) = closure under marked concatenation + union
How many needed alternations between Boolean operations and concatenations?
Straubing-Thérien hierarchy ’81I ST [0] = {∅; A∗}.I ST
ˆn + 1
2
˜= Pol(ST [n]).
I ST [n + 1] = Bool(STˆn + 1
2
˜).
Brzozowski-Cohen hierarchy ’71I DD [0] = {∅; {"}; A+; A∗}.I DD
ˆn + 1
2
˜= Pol(DD [n]).
I DD [n + 1] = Bool(DDˆn + 1
2
˜).
0 12 1 3
2 2 52
PolBool
PolBool
Pol
5 / 29
Classical hierarchies inside SFTwo classes built on top of CI Boolean closure Bool(C).I Polynomial closure Pol(C) = closure under marked concatenation + union
How many needed alternations between Boolean operations and concatenations?
Straubing-Thérien hierarchy ’81I ST [0] = {∅; A∗}.I ST
ˆn + 1
2
˜= Pol(ST [n]).
I ST [n + 1] = Bool(STˆn + 1
2
˜).
Brzozowski-Cohen hierarchy ’71I DD [0] = {∅; {"}; A+; A∗}.I DD
ˆn + 1
2
˜= Pol(DD [n]).
I DD [n + 1] = Bool(DDˆn + 1
2
˜).
0 12 1 3
2 2 52
PolBool
PolBool
Pol
5 / 29
Classical hierarchies inside SFTwo classes built on top of CI Boolean closure Bool(C).I Polynomial closure Pol(C) = closure under marked concatenation + union
How many needed alternations between Boolean operations and concatenations?
Straubing-Thérien hierarchy ’81I ST [0] = {∅; A∗}.I ST
ˆn + 1
2
˜= Pol(ST [n]).
I ST [n + 1] = Bool(STˆn + 1
2
˜).
Brzozowski-Cohen hierarchy ’71I DD [0] = {∅; {"}; A+; A∗}.I DD
ˆn + 1
2
˜= Pol(DD [n]).
I DD [n + 1] = Bool(DDˆn + 1
2
˜).
0 12 1 3
2 2 52
PolBool
PolBool
Pol
5 / 29
Brzozowski-Cohen and Straubing-Thérien hierarchies
Natural questionsI Are the hierarchies strict?I Logical description of each level?I What is known about membership?
What is known1. Both hierarchies are strict (Brzozowski-Knast 1978 + interleaving),
(a · · · (a(ab)∗b)∗ · · · b)∗
2. Natural logical description wihin FO.3. Membership for BC reduces to membership for ST.4. Membership solved for only few levels.
6 / 29
Brzozowski-Cohen and Straubing-Thérien hierarchies
Natural questionsI Are the hierarchies strict?I Logical description of each level?I What is known about membership?
What is known1. Both hierarchies are strict (Brzozowski-Knast 1978 + interleaving),
(a · · · (a(ab)∗b)∗ · · · b)∗
2. Natural logical description wihin FO.3. Membership for BC reduces to membership for ST.4. Membership solved for only few levels.
6 / 29
Logical counterpart: quantifier alternation hierarchies
Intuition: marked concatenation corresponds to existential quantification.I Σi = ∃∗∀∗∃∗∀∗∃∗ · · ·| {z }
at most i blocks ∃∗ or ∀∗’, (’ quantifier free).
I BΣi = Finite Boolean combinations of Σi .
Quantifier Alternation Hierarchies
Σ1 BΣ1
12 1
Σ2 BΣ2
32 2
Σ3 BΣ3
52 3
Σ4
72
FO( ( ( ( ( ( (
Two versionsI Order signature: < and a().I Enriched signature: <, a(), +1, min(), max() and ".
Logical Correspondence Theorem (Thomas ’82, Perrin-Pin ’86)I Brzozowski-Cohen hierarchy = enriched quantifier alternation hierarchy.I Straubing-Thérien hierarchy = order quantifier alternation hierarchy.
7 / 29
Logical counterpart: quantifier alternation hierarchies
Intuition: marked concatenation corresponds to existential quantification.I Σi = ∃∗∀∗∃∗∀∗∃∗ · · ·| {z }
at most i blocks ∃∗ or ∀∗’, (’ quantifier free).
I BΣi = Finite Boolean combinations of Σi .
Quantifier Alternation Hierarchies
Σ1 BΣ1
12 1
Σ2 BΣ2
32 2
Σ3 BΣ3
52 3
Σ4
72
FO( ( ( ( ( ( (
Two versionsI Order signature: < and a().I Enriched signature: <, a(), +1, min(), max() and ".
Logical Correspondence Theorem (Thomas ’82, Perrin-Pin ’86)I Brzozowski-Cohen hierarchy = enriched quantifier alternation hierarchy.I Straubing-Thérien hierarchy = order quantifier alternation hierarchy.
7 / 29
Logical counterpart: quantifier alternation hierarchies
Intuition: marked concatenation corresponds to existential quantification.I Σi = ∃∗∀∗∃∗∀∗∃∗ · · ·| {z }
at most i blocks ∃∗ or ∀∗’, (’ quantifier free).
I BΣi = Finite Boolean combinations of Σi .
Quantifier Alternation Hierarchies
Σ1 BΣ1
12 1
Σ2 BΣ2
32 2
Σ3 BΣ3
52 3
Σ4
72
FO( ( ( ( ( ( (
Two versionsI Order signature: < and a().I Enriched signature: <, a(), +1, min(), max() and ".
Logical Correspondence Theorem (Thomas ’82, Perrin-Pin ’86)I Brzozowski-Cohen hierarchy = enriched quantifier alternation hierarchy.I Straubing-Thérien hierarchy = order quantifier alternation hierarchy.
7 / 29
Logical counterpart: quantifier alternation hierarchies
Intuition: marked concatenation corresponds to existential quantification.I Σi = ∃∗∀∗∃∗∀∗∃∗ · · ·| {z }
at most i blocks ∃∗ or ∀∗’, (’ quantifier free).
I BΣi = Finite Boolean combinations of Σi .
Quantifier Alternation Hierarchies
Σ1 BΣ1
12 1
Σ2 BΣ2
32 2
Σ3 BΣ3
52 3
Σ4
72
FO( ( ( ( ( ( (
Two versionsI Order signature: < and a().I Enriched signature: <, a(), +1, min(), max() and ".
Logical Correspondence Theorem (Thomas ’82, Perrin-Pin ’86)I Brzozowski-Cohen hierarchy = enriched quantifier alternation hierarchy.I Straubing-Thérien hierarchy = order quantifier alternation hierarchy.
7 / 29
The membership problem for BC and ST hierarchies
Σ1 BΣ1 Σ2 BΣ2 Σ3 BΣ3 Σ4 FO( ( ( ( ( ( (
Schützenberger’65McNaughton-Papert’71
Simon’75 Place,Z.’14
Arfi’87Pin, Weil’95
Place’15
Enrichment Theorem for membership (Straubing ’85 – Pin, Weil ’97)
Membership for a level in the enriched hierarchy (ie, BC)reduces to
Membership for the same level in the order hierarchy (ie, ST).
8 / 29
The membership problem for BC and ST hierarchies
Σ1 BΣ1 Σ2 BΣ2 Σ3 BΣ3 Σ4 FO( ( ( ( ( ( (
Schützenberger’65McNaughton-Papert’71
Simon’75
Place,Z.’14
Arfi’87Pin, Weil’95
Place’15
Enrichment Theorem for membership (Straubing ’85 – Pin, Weil ’97)
Membership for a level in the enriched hierarchy (ie, BC)reduces to
Membership for the same level in the order hierarchy (ie, ST).
8 / 29
The membership problem for BC and ST hierarchies
Σ1 BΣ1 Σ2 BΣ2 Σ3 BΣ3 Σ4 FO( ( ( ( ( ( (
Schützenberger’65McNaughton-Papert’71
Simon’75
Place,Z.’14
Arfi’87Pin, Weil’95
Place’15
Enrichment Theorem for membership (Straubing ’85 – Pin, Weil ’97)
Membership for a level in the enriched hierarchy (ie, BC)reduces to
Membership for the same level in the order hierarchy (ie, ST).
8 / 29
The membership problem for BC and ST hierarchies
Σ1 BΣ1 Σ2 BΣ2 Σ3 BΣ3 Σ4 FO( ( ( ( ( ( (
Schützenberger’65McNaughton-Papert’71
Simon’75 Place,Z.’14
Arfi’87Pin, Weil’95
Place’15
Enrichment Theorem for membership (Straubing ’85 – Pin, Weil ’97)
Membership for a level in the enriched hierarchy (ie, BC)reduces to
Membership for the same level in the order hierarchy (ie, ST).
8 / 29
The membership problem for BC and ST hierarchies
Σ1 BΣ1 Σ2 BΣ2 Σ3 BΣ3 Σ4 FO( ( ( ( ( ( (
Schützenberger’65McNaughton-Papert’71
Simon’75 Place,Z.’14
Arfi’87Pin, Weil’95
Place’15
Enrichment Theorem for membership (Straubing ’85 – Pin, Weil ’97)
Membership for a level in the enriched hierarchy (ie, BC)reduces to
Membership for the same level in the order hierarchy (ie, ST).
8 / 29
The membership problem for BC and ST hierarchies
Σ1 BΣ1 Σ2 BΣ2 Σ3 BΣ3 Σ4 FO( ( ( ( ( ( (
Schützenberger’65McNaughton-Papert’71
Simon’75 Place,Z.’14
Arfi’87Pin, Weil’95
Place’15
Enrichment Theorem for membership (Straubing ’85 – Pin, Weil ’97)
Membership for a level in the enriched hierarchy (ie, BC)reduces to
Membership for the same level in the order hierarchy (ie, ST).
8 / 29
Generalizations in two directions
1. Proofs are ad hoc for DD and ST: obtain generic theorems.For given C, what about Pol(C), Bool(Pol(C)),. . .
2. Recent results via generalizations of membership: separation and covering.
9 / 29
Generic concatenation hierarchies
Generic pattern parametrized by the basisI C[0] (basis) Boolean algebra in REG closed under left/right quotients
C[n + 12 ]: close C[n] under K;L 7→ KaL and ∪.
C[n + 1]: close C[n + 12 ] under Boolean operations.
0 12 1 3
2 2 52
PolBool
PolBool
Pol
Examples
I Straubing-Thérien: C[0] = {∅; A∗}.I Brzozowski-Cohen: C[0] = {∅; {"}; A∗; A+}.I Pin-Margolis: C[0] = group languages.
10 / 29
Generic concatenation hierarchies
Generic pattern parametrized by the basisI C[0] (basis) Boolean algebra in REG closed under left/right quotients
C[n + 12 ]: close C[n] under K; L 7→ KaL and ∪.
C[n + 1]: close C[n + 12 ] under Boolean operations.
0 12 1 3
2 2 52
PolBool
PolBool
Pol
Examples
I Straubing-Thérien: C[0] = {∅; A∗}.I Brzozowski-Cohen: C[0] = {∅; {"}; A∗; A+}.I Pin-Margolis: C[0] = group languages.
10 / 29
Generic concatenation hierarchies
Generic pattern parametrized by the basisI C[0] (basis) Boolean algebra in REG closed under left/right quotients
C[n + 12 ]: close C[n] under K; L 7→ KaL and ∪.
C[n + 1]: close C[n + 12 ] under Boolean operations.
0 12 1 3
2 2 52
PolBool
PolBool
Pol
Examples
I Straubing-Thérien: C[0] = {∅; A∗}.I Brzozowski-Cohen: C[0] = {∅; {"}; A∗; A+}.I Pin-Margolis: C[0] = group languages.
10 / 29
Generic Hierarchies
Natural questionsI Are the hierarchies strict?I Logical description of each level?I What is known about membership?
11 / 29
Basic properties of generic hierarchies
Basic Structure (Place, Z. ’17)
For any hierarchy:I All levels are closed under quotient and in REG.I Full levels are closed under Bool.I Half levels are closed under union, intersection (and concatenation).
Quotienting: closed under u−1L = {x | ux ∈ L} and Lu−1:
Strictness Theorem (Place, Z. ’17)
Any hierarchy whose basis is finite is strict.
12 / 29
Basic properties of generic hierarchies
Basic Structure (Place, Z. ’17)
For any hierarchy:I All levels are closed under quotient and in REG.I Full levels are closed under Bool.I Half levels are closed under union, intersection (and concatenation).
Quotienting: closed under u−1L = {x | ux ∈ L} and Lu−1:
Strictness Theorem (Place, Z. ’17)
Any hierarchy whose basis is finite is strict.
12 / 29
Generic logical correspondence
Logical Correspondence Theorem (Place, Z. ’17)
For any basis C, there is a natural set S of first order predicates, st.
Concatenation hierarchy of basis C=
Quantifier alternation hierarchy over signature S
Generalizes the correspondences discovered for BC and ST hierarchies.
IntuitionFor each L ∈ C, add 4 predicates in addition to < and a(); b(); : : :
I w |= IL(x; y) when x < y and w ]x; y [ ∈ L (Infix).I w |= PL(y) when w [1; y [ ∈ L (Prefix).I w |= SL(x) when w ]x; n] ∈ L (Suffix).I w |= WL when w ∈ L (Whole word).
13 / 29
Generic logical correspondence
Logical Correspondence Theorem (Place, Z. ’17)
For any basis C, there is a natural set S of first order predicates, st.
Concatenation hierarchy of basis C=
Quantifier alternation hierarchy over signature S
Generalizes the correspondences discovered for BC and ST hierarchies.
IntuitionFor each L ∈ C, add 4 predicates in addition to < and a(); b(); : : :
I w |= IL(x; y) when x < y and w ]x; y [ ∈ L (Infix).
I w |= PL(y) when w [1; y [ ∈ L (Prefix).I w |= SL(x) when w ]x; n] ∈ L (Suffix).I w |= WL when w ∈ L (Whole word).
13 / 29
Generic logical correspondence
Logical Correspondence Theorem (Place, Z. ’17)
For any basis C, there is a natural set S of first order predicates, st.
Concatenation hierarchy of basis C=
Quantifier alternation hierarchy over signature S
Generalizes the correspondences discovered for BC and ST hierarchies.
IntuitionFor each L ∈ C, add 4 predicates in addition to < and a(); b(); : : :
I w |= IL(x; y) when x < y and w ]x; y [ ∈ L (Infix).I w |= PL(y) when w [1; y [ ∈ L (Prefix).I w |= SL(x) when w ]x; n] ∈ L (Suffix).
I w |= WL when w ∈ L (Whole word).
13 / 29
Generic logical correspondence
Logical Correspondence Theorem (Place, Z. ’17)
For any basis C, there is a natural set S of first order predicates, st.
Concatenation hierarchy of basis C=
Quantifier alternation hierarchy over signature S
Generalizes the correspondences discovered for BC and ST hierarchies.
IntuitionFor each L ∈ C, add 4 predicates in addition to < and a(); b(); : : :
I w |= IL(x; y) when x < y and w ]x; y [ ∈ L (Infix).I w |= PL(y) when w [1; y [ ∈ L (Prefix).I w |= SL(x) when w ]x; n] ∈ L (Suffix).I w |= WL when w ∈ L (Whole word).
13 / 29
Generic membership theorem
Generic membership Theorem (Place, Z. ’17, Place ’15)
For any finite basis C, levels 12 ; 1;
32 ;
52 have decidable membership.
Remember: decidability known up to level 72 for ST and BC hierarchies.
The alphabet trick
Languages in STˆ32
˜(Pin and Straubing ’85)
Languages of level STˆ32
˜are unions of languages of the form B∗
0a1B∗1 · · · anB∗
n
STˆ32
˜= level 1
2 with basis AT = {B∗ | B ⊆ A}.
ST[q] is also level (q − 1) in another hierarchy with finite basis.ST[q] = AT[q − 1].
Corollary (by Alphabet trick)
In ST
and BC
hierarchy, levels 12 ; 1;
32 ; 2;
52 ;
72 have decidable membership.
14 / 29
Generic membership theorem
Generic membership Theorem (Place, Z. ’17, Place ’15)
For any finite basis C, levels 12 ; 1;
32 ;
52 have decidable membership.
Remember: decidability known up to level 72 for ST and BC hierarchies.
The alphabet trick
Languages in STˆ32
˜(Pin and Straubing ’85)
Languages of level STˆ32
˜are unions of languages of the form B∗
0a1B∗1 · · · anB∗
n
STˆ32
˜= level 1
2 with basis AT = {B∗ | B ⊆ A}.
ST[q] is also level (q − 1) in another hierarchy with finite basis.ST[q] = AT[q − 1].
Corollary (by Alphabet trick)
In ST
and BC
hierarchy, levels 12 ; 1;
32 ; 2;
52 ;
72 have decidable membership.
14 / 29
Generic membership theorem
Generic membership Theorem (Place, Z. ’17, Place ’15)
For any finite basis C, levels 12 ; 1;
32 ;
52 have decidable membership.
Remember: decidability known up to level 72 for ST and BC hierarchies.
The alphabet trick
Languages in STˆ32
˜(Pin and Straubing ’85)
Languages of level STˆ32
˜are unions of languages of the form B∗
0a1B∗1 · · · anB∗
n
STˆ32
˜= level 1
2 with basis AT = {B∗ | B ⊆ A}.
ST[q] is also level (q − 1) in another hierarchy with finite basis.ST[q] = AT[q − 1].
Corollary (by Alphabet trick)
In ST
and BC
hierarchy, levels 12 ; 1;
32 ; 2;
52 ;
72 have decidable membership.
14 / 29
Generic membership theorem
Generic membership Theorem (Place, Z. ’17, Place ’15)
For any finite basis C, levels 12 ; 1;
32 ;
52 have decidable membership.
Remember: decidability known up to level 72 for ST and BC hierarchies.
The alphabet trick
Languages in STˆ32
˜(Pin and Straubing ’85)
Languages of level STˆ32
˜are unions of languages of the form B∗
0a1B∗1 · · · anB∗
n
STˆ32
˜= level 1
2 with basis AT = {B∗ | B ⊆ A}.
ST[q] is also level (q − 1) in another hierarchy with finite basis.ST[q] = AT[q − 1].
Corollary (by Alphabet trick)
In ST
and BC
hierarchy, levels 12 ; 1;
32 ; 2;
52 ;
72 have decidable membership.
14 / 29
Generic membership theorem
Generic membership Theorem (Place, Z. ’17, Place ’15)
For any finite basis C, levels 12 ; 1;
32 ;
52 have decidable membership.
Remember: decidability known up to level 72 for ST and BC hierarchies.
The alphabet trick
Languages in STˆ32
˜(Pin and Straubing ’85)
Languages of level STˆ32
˜are unions of languages of the form B∗
0a1B∗1 · · · anB∗
n
STˆ32
˜= level 1
2 with basis AT = {B∗ | B ⊆ A}.
ST[q] is also level (q − 1) in another hierarchy with finite basis.ST[q] = AT[q − 1].
Corollary (by Alphabet trick)
In ST
and BC
hierarchy, levels 12 ; 1;
32 ; 2;
52 ;
72 have decidable membership.
14 / 29
Generic membership theorem
Generic membership Theorem (Place, Z. ’17, Place ’15)
For any finite basis C, levels 12 ; 1;
32 ;
52 have decidable membership.
Remember: decidability known up to level 72 for ST and BC hierarchies.
The alphabet trick
Languages in STˆ32
˜(Pin and Straubing ’85)
Languages of level STˆ32
˜are unions of languages of the form B∗
0a1B∗1 · · · anB∗
n
STˆ32
˜= level 1
2 with basis AT = {B∗ | B ⊆ A}.
ST[q] is also level (q − 1) in another hierarchy with finite basis.ST[q] = AT[q − 1].
Corollary (by Alphabet trick)
In ST and BC hierarchy, levels 12 ; 1;
32 ; 2;
52 ;
72 have decidable membership.
14 / 29
Recap
I Generic construction process for concatenation hierarchies.I Generic logical correspondence.I Generic strictness theorem.I Generic membership theorem.
Recent results required solving harder problems than membership.
15 / 29
Recap
I Generic construction process for concatenation hierarchies.I Generic logical correspondence.I Generic strictness theorem.I Generic membership theorem.
Recent results required solving harder problems than membership.
15 / 29
Beyond Membership:Separation Problems
Beyond membership: Separation
Recent results required solving harder problems than membership.
Motivation:
I Classe C with decidable membership.I Class Op(C) built on top of C with undecidable membership.
Nice idea, Henckell and Rhodes ’88
Prove more on C to recover membership decidability for Op(C).
Nice statement, Almeida ’96Almeida’96: a problem introduced by Henckell can be formulated as separation.
16 / 29
Beyond membership: Separation
Recent results required solving harder problems than membership.
Motivation:
I Classe C with decidable membership.I Class Op(C) built on top of C with undecidable membership.
Nice idea, Henckell and Rhodes ’88
Prove more on C to recover membership decidability for Op(C).
Nice statement, Almeida ’96Almeida’96: a problem introduced by Henckell can be formulated as separation.
16 / 29
Beyond membership: Separation
Decide the following problem:
Take 2 regular languages L1; L2
a
a
a
a b b b
a
Take 2 regular languages L1; L2
a
a
L1
L2
a
b a b b
b
a
a
Can L1 be separated from L2
with a language from C?
L1L2
A∗
in C
Can L1 be separated from L2
with a language from C?
L1
L2
A∗
in CC-separable from complement⇔in C
Membership can be formally reduced to separation
17 / 29
Beyond membership: Separation
Decide the following problem:
Take 2 regular languages L1; L2
a
a
a
a b b b
a
Take 2 regular languages L1; L2
a
a
L1
L2
a
b a b b
b
a
a
Can L1 be separated from L2
with a language from C?
L1L2
A∗
in C
Can L1 be separated from L2
with a language from C?
L1L2
A∗
in CC-separable from complement⇔in C
Membership can be formally reduced to separation
17 / 29
Beyond membership: Separation
Decide the following problem:
Take 2 regular languages L1; L2
a
a
a
a b b b
a
Take 2 regular languages L1; L2
a
a
L1
L2
a
b a b b
b
a
a
Can L1 be separated from L2
with a language from C?
L1L2
A∗
in C
Can L1 be separated from L2
with a language from C?
L1L2
A∗
in C
C-separable from complement⇔in C
Membership can be formally reduced to separation
17 / 29
Beyond membership: Separation
Decide the following problem:
Take 2 regular languages L1; L2
a
a
a
a b b b
a
Take 2 regular languages L1; L2
a
a
L1
L2
a
b a b b
b
a
a
Can L1 be separated from L2
with a language from C?
L1L2
A∗
in C
Can L1 be separated from L2
with a language from C?
L2 = A∗ \ L1 L1
L2
A∗
in C
C-separable from complement⇔in C
Membership can be formally reduced to separation
17 / 29
Separation for classical hierarchiesSe
para
tion
Mem
bers
hip
Σ1 BΣ1 Σ2 BΣ2 Σ3 BΣ3 Σ4 FO( ( ( ( ( ( (
Schützenberger’65McNaughton-Papert’71
Henckell’88Henckell, Rhodes, Steinberg’10
Place,Z.’14
Simon’75
Almeida,Z.’97Czerwinski,Martens,Masopust’13
Place,Van Rooijen,Z.’13
Place,Z.’14Arfi’87
Pin, Weil’95
Place,Z.’14
Place,Z.’17
Place’15
Place’15
Some membership algorithms come from separation algorithms for simpler levels
18 / 29
Separation for classical hierarchiesSe
para
tion
Mem
bers
hip
Σ1 BΣ1 Σ2 BΣ2 Σ3 BΣ3 Σ4 FO( ( ( ( ( ( (
Schützenberger’65McNaughton-Papert’71
Henckell’88Henckell, Rhodes, Steinberg’10
Place,Z.’14
Simon’75
Almeida,Z.’97Czerwinski,Martens,Masopust’13
Place,Van Rooijen,Z.’13
Place,Z.’14Arfi’87
Pin, Weil’95
Place,Z.’14
Place,Z.’17
Place’15
Place’15
Some membership algorithms come from separation algorithms for simpler levels
18 / 29
Separation results for generic hierarchies
All what we know for ST and BC follow from:
Generic Separation Theorem (Place, Z. ’17, Place ’15)
Levels 12 ; 1 and 3
2 have decidable separation for any hierarchy whose basis is finite.
Corollary (by Alphabet trick + Enrichment)
Levels 12 ; 1,
32 , 2 and 5
2 have decidable separation in ST and BC hierarchies.
Jump Theorem for quotienting lattices (Place, Z. ’15)
For any basis, membership for level n + 12 reduces to separation for level n − 1
2 .
Enrichment Theorem (Place, Z. ’15)
Separation for a level in the enriched hierarchy (ie, BC)reduces to
Separation for the same level in the order hierarchy (ie, ST).
19 / 29
Separation results for generic hierarchies
All what we know for ST and BC follow from:
Generic Separation Theorem (Place, Z. ’17, Place ’15)
Levels 12 ; 1 and 3
2 have decidable separation for any hierarchy whose basis is finite.
Corollary (by Alphabet trick + Enrichment)
Levels 12 ; 1,
32 , 2 and 5
2 have decidable separation in ST and BC hierarchies.
Jump Theorem for quotienting lattices (Place, Z. ’15)
For any basis, membership for level n + 12 reduces to separation for level n − 1
2 .
Enrichment Theorem (Place, Z. ’15)
Separation for a level in the enriched hierarchy (ie, BC)reduces to
Separation for the same level in the order hierarchy (ie, ST).
19 / 29
Separation results for generic hierarchies
All what we know for ST and BC follow from:
Generic Separation Theorem (Place, Z. ’17, Place ’15)
Levels 12 ; 1 and 3
2 have decidable separation for any hierarchy whose basis is finite.
Corollary (by Alphabet trick + Enrichment)
Levels 12 ; 1,
32 , 2 and 5
2 have decidable separation in ST and BC hierarchies.
Jump Theorem for quotienting lattices (Place, Z. ’15)
For any basis, membership for level n + 12 reduces to separation for level n − 1
2 .
Enrichment Theorem (Place, Z. ’15)
Separation for a level in the enriched hierarchy (ie, BC)reduces to
Separation for the same level in the order hierarchy (ie, ST).
19 / 29
Separation results for generic hierarchies
All what we know for ST and BC follow from:
Generic Separation Theorem (Place, Z. ’17, Place ’15)
Levels 12 ; 1 and 3
2 have decidable separation for any hierarchy whose basis is finite.
Corollary (by Alphabet trick + Enrichment)
Levels 12 ; 1,
32 , 2 and 5
2 have decidable separation in ST and BC hierarchies.
Jump Theorem for quotienting lattices (Place, Z. ’15)
For any basis, membership for level n + 12 reduces to separation for level n − 1
2 .
Enrichment Theorem (Place, Z. ’15)
Separation for a level in the enriched hierarchy (ie, BC)reduces to
Separation for the same level in the order hierarchy (ie, ST).
19 / 29
The Jump Theorem
Jump Theorem for quotienting lattices (Place, Z. ’15)
For any basis, membership for level n + 12 reduces to separation for level n − 1
2 .
The Jump Theorem on Automata
A regular language is in Cˆn+ 1
2
˜iff its minimal automaton has no pattern:
p q
not final final
w w
where Lp;q is not Cˆn− 1
2
˜-separable from Lp;p ∩ Lq;q
Lp;q = {w | p w−−−−→ q}
20 / 29
The Jump Theorem
Jump Theorem for quotienting lattices (Place, Z. ’15)
For any basis, membership for level n + 12 reduces to separation for level n − 1
2 .
The Jump Theorem on Automata
A regular language is in Cˆn+ 1
2
˜iff its minimal automaton has no pattern:
p q
not final final
w w
where Lp;q is not Cˆn− 1
2
˜-separable from Lp;p ∩ Lq;q
Lp;q = {w | p w−−−−→ q}
20 / 29
The Jump Theorem
Jump Theorem for quotienting lattices (Place, Z. ’15)
For any basis, membership for level n + 12 reduces to separation for level n − 1
2 .
The Jump Theorem on Automata
A regular language is in Cˆn+ 1
2
˜iff its minimal automaton has no pattern:
p q
not final final
w w
where Lp;q is not Cˆn− 1
2
˜-separable from Lp;p ∩ Lq;q
Lp;q = {w | p w−−−−→ q}
20 / 29
The Jump Theorem
Jump Theorem for quotienting lattices (Place, Z. ’15)
For any basis, membership for level n + 12 reduces to separation for level n − 1
2 .
The Jump Theorem on Automata
A regular language is in Cˆn+ 1
2
˜iff its minimal automaton has no pattern:
p q
not final final
w w
where Lp;q is not Cˆn− 1
2
˜-separable from Lp;p ∩ Lq;q
Lp;q = {w | p w−−−−→ q}
20 / 29
Recap
Current knowledge is captured by few generic results:
1. Separation theoremC finite ⇒ separation decidable for Pol(C), BPol(C) and Pol(BPol(C)).In particular, cannot yet deal with 2 levels of complement.
2. Jump theoremC-separation decidable ⇒ Pol(C)-membership decidable.
3. Enrichment theorem (see also Varun’s talk).
21 / 29
Unambiguous Polynomial Closure
History and motivations
I Schützenberger, 1976: UL = Unambiguous Polynomial closure(AT).
AT = {B∗ | B ⊆ A}:
I Robust: FO2(<), ∆2(<), UTL, TL(rankers), 2way turtle DFAs, DA. . .
I Pin, 1980: Generic investigation of unambiguous polynomial closure
I Margolis Pin Thérien, 1988 UPol(C) membership reduces to C-membershipAlgebraic proofs (relational morphisms, categories, bilateral kernels, etc.)
22 / 29
History and motivations
I Schützenberger, 1976: UL = Unambiguous Polynomial closure(AT).
AT = {B∗ | B ⊆ A}:
I Robust: FO2(<), ∆2(<),
UTL, TL(rankers), 2way turtle DFAs, DA. . .
I Pin, 1980: Generic investigation of unambiguous polynomial closure
I Margolis Pin Thérien, 1988 UPol(C) membership reduces to C-membershipAlgebraic proofs (relational morphisms, categories, bilateral kernels, etc.)
22 / 29
History and motivations
I Schützenberger, 1976: UL = Unambiguous Polynomial closure(AT).
AT = {B∗ | B ⊆ A}:
I Robust: FO2(<), ∆2(<), UTL, TL(rankers),
2way turtle DFAs, DA. . .
I Pin, 1980: Generic investigation of unambiguous polynomial closure
I Margolis Pin Thérien, 1988 UPol(C) membership reduces to C-membershipAlgebraic proofs (relational morphisms, categories, bilateral kernels, etc.)
22 / 29
History and motivations
I Schützenberger, 1976: UL = Unambiguous Polynomial closure(AT).
AT = {B∗ | B ⊆ A}:
I Robust: FO2(<), ∆2(<), UTL, TL(rankers), 2way turtle DFAs,
DA. . .
I Pin, 1980: Generic investigation of unambiguous polynomial closure
I Margolis Pin Thérien, 1988 UPol(C) membership reduces to C-membershipAlgebraic proofs (relational morphisms, categories, bilateral kernels, etc.)
22 / 29
History and motivations
I Schützenberger, 1976: UL = Unambiguous Polynomial closure(AT).
AT = {B∗ | B ⊆ A}:
I Robust: FO2(<), ∆2(<), UTL, TL(rankers), 2way turtle DFAs, DA. . .
I Pin, 1980: Generic investigation of unambiguous polynomial closure
I Margolis Pin Thérien, 1988 UPol(C) membership reduces to C-membershipAlgebraic proofs (relational morphisms, categories, bilateral kernels, etc.)
22 / 29
History and motivations
I Schützenberger, 1976: UL = Unambiguous Polynomial closure(AT).
AT = {B∗ | B ⊆ A}:
I Robust: FO2(<), ∆2(<), UTL, TL(rankers), 2way turtle DFAs, DA. . .
I Pin, 1980: Generic investigation of unambiguous polynomial closure
I Margolis Pin Thérien, 1988 UPol(C) membership reduces to C-membershipAlgebraic proofs (relational morphisms, categories, bilateral kernels, etc.)
22 / 29
History and motivations
I Schützenberger, 1976: UL = Unambiguous Polynomial closure(AT).
AT = {B∗ | B ⊆ A}:
I Robust: FO2(<), ∆2(<), UTL, TL(rankers), 2way turtle DFAs, DA. . .
I Pin, 1980: Generic investigation of unambiguous polynomial closure
I Margolis Pin Thérien, 1988 UPol(C) membership reduces to C-membershipAlgebraic proofs (relational morphisms, categories, bilateral kernels, etc.)
22 / 29
History and motivations
I Schützenberger, 1976: UL = Unambiguous Polynomial closure(AT).
AT = {B∗ | B ⊆ A}:
I Robust: FO2(<), ∆2(<), UTL, TL(rankers), 2way turtle DFAs, DA. . .
I Pin, 1980: Generic investigation of unambiguous polynomial closure
I Margolis Pin Thérien, 1988 UPol(C) membership reduces to C-membershipAlgebraic proofs (relational morphisms, categories, bilateral kernels, etc.)
22 / 29
Unambiguous polynomial closure
I KL is unambiguous if each w ∈ KL has a single factorization in K × L.Generalizes to n languages (and so to).
I A∗aA∗? (A \ {a})∗aA∗?
I UPol(C) = closure of C underI unambiguous concatenation,I disjoint union.
I ExamplesI UPol({∅; A∗})?I UPol(AT) = UL from Schützenberger.
I Not even clear whether UPol(C) is closed under union.
23 / 29
Unambiguous polynomial closure
I KL is unambiguous if each w ∈ KL has a single factorization in K × L.Generalizes to n languages (and so to).
I A∗aA∗?
(A \ {a})∗aA∗?
I UPol(C) = closure of C underI unambiguous concatenation,I disjoint union.
I ExamplesI UPol({∅; A∗})?I UPol(AT) = UL from Schützenberger.
I Not even clear whether UPol(C) is closed under union.
23 / 29
Unambiguous polynomial closure
I KL is unambiguous if each w ∈ KL has a single factorization in K × L.Generalizes to n languages (and so to).
I A∗aA∗? (A \ {a})∗aA∗?
I UPol(C) = closure of C underI unambiguous concatenation,I disjoint union.
I ExamplesI UPol({∅; A∗})?I UPol(AT) = UL from Schützenberger.
I Not even clear whether UPol(C) is closed under union.
23 / 29
Unambiguous polynomial closure
I KL is unambiguous if each w ∈ KL has a single factorization in K × L.Generalizes to n languages (and so to).
I A∗aA∗? (A \ {a})∗aA∗?
I UPol(C) = closure of C underI unambiguous concatenation,I disjoint union.
I ExamplesI UPol({∅; A∗})?I UPol(AT) = UL from Schützenberger.
I Not even clear whether UPol(C) is closed under union.
23 / 29
Unambiguous polynomial closure
I KL is unambiguous if each w ∈ KL has a single factorization in K × L.Generalizes to n languages (and so to).
I A∗aA∗? (A \ {a})∗aA∗?
I UPol(C) = closure of C underI unambiguous concatenation,I disjoint union.
I ExamplesI UPol({∅; A∗})?
I UPol(AT) = UL from Schützenberger.
I Not even clear whether UPol(C) is closed under union.
23 / 29
Unambiguous polynomial closure
I KL is unambiguous if each w ∈ KL has a single factorization in K × L.Generalizes to n languages (and so to).
I A∗aA∗? (A \ {a})∗aA∗?
I UPol(C) = closure of C underI unambiguous concatenation,I disjoint union.
I ExamplesI UPol({∅; A∗})?I UPol(AT) = UL from Schützenberger.
I Not even clear whether UPol(C) is closed under union.
23 / 29
Unambiguous polynomial closure
I KL is unambiguous if each w ∈ KL has a single factorization in K × L.Generalizes to n languages (and so to).
I A∗aA∗? (A \ {a})∗aA∗?
I UPol(C) = closure of C underI unambiguous concatenation,I disjoint union.
I ExamplesI UPol({∅; A∗})?I UPol(AT) = UL from Schützenberger.
I Not even clear whether UPol(C) is closed under union.
23 / 29
Alternating polynomial closure
I KaL is left deterministic if KaA∗ ∩K = ∅.Note. Does not depend on L.
I Right deterministic: symmetric
I Unambiguous concatenation which is not left neither right deterministic?(A \ {a})∗aA∗b(A \ {b})∗.
I ADet(C) = closure of C under left+right deterministic marked concatenation
Easy remark
ADet(C) ⊆ UPol(C)
24 / 29
Alternating polynomial closure
I KaL is left deterministic if KaA∗ ∩K = ∅.Note. Does not depend on L.
I Right deterministic: symmetric
I Unambiguous concatenation which is not left neither right deterministic?(A \ {a})∗aA∗b(A \ {b})∗.
I ADet(C) = closure of C under left+right deterministic marked concatenation
Easy remark
ADet(C) ⊆ UPol(C)
24 / 29
Alternating polynomial closure
I KaL is left deterministic if KaA∗ ∩K = ∅.Note. Does not depend on L.
I Right deterministic: symmetric
I Unambiguous concatenation which is not left neither right deterministic?
(A \ {a})∗aA∗b(A \ {b})∗.
I ADet(C) = closure of C under left+right deterministic marked concatenation
Easy remark
ADet(C) ⊆ UPol(C)
24 / 29
Alternating polynomial closure
I KaL is left deterministic if KaA∗ ∩K = ∅.Note. Does not depend on L.
I Right deterministic: symmetric
I Unambiguous concatenation which is not left neither right deterministic?(A \ {a})∗aA∗b(A \ {b})∗.
I ADet(C) = closure of C under left+right deterministic marked concatenation
Easy remark
ADet(C) ⊆ UPol(C)
24 / 29
Alternating polynomial closure
I KaL is left deterministic if KaA∗ ∩K = ∅.Note. Does not depend on L.
I Right deterministic: symmetric
I Unambiguous concatenation which is not left neither right deterministic?(A \ {a})∗aA∗b(A \ {b})∗.
I ADet(C) = closure of C under left+right deterministic marked concatenation
Easy remark
ADet(C) ⊆ UPol(C)
24 / 29
Quasi separation wrt. a surjective morphism
A∗ M
¸s
t
¸−1(s)
¸−1(t)
¸−1(s)
∈ C
¸−1(s); ¸−1(t) are C-separable by some ¸−1(P )⇔
(s; t) is not a quasi C-pair
25 / 29
Quasi separation wrt. a surjective morphism
A∗ M
¸s
t
¸−1(s)
¸−1(t)
¸−1(s)
∈ C
¸−1(s); ¸−1(t) are C-separable⇔
(s; t) is not a C-pair
¸−1(s); ¸−1(t) are C-separable by some ¸−1(P )⇔
(s; t) is not a quasi C-pair
25 / 29
Quasi separation wrt. a surjective morphism
A∗ M
¸s
t
¸−1(s)
¸−1(t)
¸−1(s)
∈ C
¸−1(s); ¸−1(t) are C-separable by some ¸−1(P )⇔
(s; t) is not a quasi C-pair25 / 29
Quasi C-pairs
(s; t) quasi C-pair = ¸−1(s); ¸−1(t) not C-separable by language recognized by ¸
I C-membership decidable =⇒ quasi C-pairs computable.
I C-pair ⇒ quasi C-pair.
I The “C-pair” relation is not transitive.Ex. for C =AT: L1 = {ab}, L2 = {ba; cd}, L3 = {dc}.
I The “quasi C-pair” relation is a congruence.
I Quasi C-pair relation = transitive closure of C-pair relation.
26 / 29
Quasi C-pairs
(s; t) quasi C-pair = ¸−1(s); ¸−1(t) not C-separable by language recognized by ¸
I C-membership decidable =⇒ quasi C-pairs computable.
I C-pair ⇒ quasi C-pair.
I The “C-pair” relation is not transitive.Ex. for C =AT: L1 = {ab}, L2 = {ba; cd}, L3 = {dc}.
I The “quasi C-pair” relation is a congruence.
I Quasi C-pair relation = transitive closure of C-pair relation.
26 / 29
Quasi C-pairs
(s; t) quasi C-pair = ¸−1(s); ¸−1(t) not C-separable by language recognized by ¸
I C-membership decidable =⇒ quasi C-pairs computable.
I C-pair ⇒ quasi C-pair.
I The “C-pair” relation is not transitive.Ex. for C =AT: L1 = {ab}, L2 = {ba; cd}, L3 = {dc}.
I The “quasi C-pair” relation is a congruence.
I Quasi C-pair relation = transitive closure of C-pair relation.
26 / 29
Quasi C-pairs
(s; t) quasi C-pair = ¸−1(s); ¸−1(t) not C-separable by language recognized by ¸
I C-membership decidable =⇒ quasi C-pairs computable.
I C-pair ⇒ quasi C-pair.
I The “C-pair” relation is not transitive.Ex. for C =AT: L1 = {ab}, L2 = {ba; cd}, L3 = {dc}.
I The “quasi C-pair” relation is a congruence.
I Quasi C-pair relation = transitive closure of C-pair relation.
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Quasi C-pairs
(s; t) quasi C-pair = ¸−1(s); ¸−1(t) not C-separable by language recognized by ¸
I C-membership decidable =⇒ quasi C-pairs computable.
I C-pair ⇒ quasi C-pair.
I The “C-pair” relation is not transitive.Ex. for C =AT: L1 = {ab}, L2 = {ba; cd}, L3 = {dc}.
I The “quasi C-pair” relation is a congruence.
I Quasi C-pair relation = transitive closure of C-pair relation.
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Membership resultsGeneric Separation Theorem (Place, Z. ’18)
Let C be a quotienting Boolean algebra of regular languages.Let L ⊆ A∗ be regular and ¸ : A∗ → M be its syntactic morphism. TFAE:
1. L ∈ UPol(C).2. L ∈ ADet(C).3. L ∈ Pol(C) ∩ co-Pol(C).
4. For all C-pairs (s; t) ∈ M2, we have s!+1 = s!ts!.
5. For all quasi C-pairs (s; t) ∈ M2, we have s!+1 = s!ts!.
2 =⇒ 1 =⇒ 4 ⇐⇒ 3 =⇒ 5 =⇒ 2
Corollary1. If C is a quotienting Boolean algebra, so is UPol(C).2. UPol(C) membership reduces to C membership.
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Membership resultsGeneric Separation Theorem (Place, Z. ’18)
Let C be a quotienting Boolean algebra of regular languages.Let L ⊆ A∗ be regular and ¸ : A∗ → M be its syntactic morphism. TFAE:
1. L ∈ UPol(C).2. L ∈ ADet(C).3. L ∈ Pol(C) ∩ co-Pol(C).
4. For all C-pairs (s; t) ∈ M2, we have s!+1 = s!ts!.
5. For all quasi C-pairs (s; t) ∈ M2, we have s!+1 = s!ts!.
2 =⇒ 1 =⇒ 4 ⇐⇒ 3 =⇒ 5 =⇒ 2
Corollary1. If C is a quotienting Boolean algebra, so is UPol(C).2. UPol(C) membership reduces to C membership.
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Membership resultsGeneric Separation Theorem (Place, Z. ’18)
Let C be a quotienting Boolean algebra of regular languages.Let L ⊆ A∗ be regular and ¸ : A∗ → M be its syntactic morphism. TFAE:
1. L ∈ UPol(C).2. L ∈ ADet(C).3. L ∈ Pol(C) ∩ co-Pol(C).
4. For all C-pairs (s; t) ∈ M2, we have s!+1 = s!ts!.
5. For all quasi C-pairs (s; t) ∈ M2, we have s!+1 = s!ts!.
2 =⇒ 1 =⇒ 4 ⇐⇒ 3 =⇒ 5 =⇒ 2
Corollary1. If C is a quotienting Boolean algebra, so is UPol(C).2. UPol(C) membership reduces to C membership.
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Separation result
Generic Separation Theorem (Place, Z. ’18)
If C is a finite quotienting Boolean algebra, separation for UPol(C) is decidable.
Notes
1. This result is stronger than the one for membership!2. Solved by considering a more general problem: covering.
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Separation result
Generic Separation Theorem (Place, Z. ’18)
If C is a finite quotienting Boolean algebra, separation for UPol(C) is decidable.
Notes
1. This result is stronger than the one for membership!2. Solved by considering a more general problem: covering.
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Summary of results
I C finite+. . . : separation OK for Pol(C), BPol(C), Pol(BPol(C)), UPol(C).I Via transfer results, imply all known algorithms.
Thanks!
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Summary of results
I C finite+. . . : separation OK for Pol(C), BPol(C), Pol(BPol(C)), UPol(C).I Via transfer results, imply all known algorithms.
Thanks!
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