Reasoning about Disclosure in Data Integration in the Presence of - - PowerPoint PPT Presentation

Reasoning about Disclosure in Data Integration in the Presence of Source Constraints

DIG Seminar 21/11/19

Michael Benedikt Pierre Bourhis Louis Jachiet Micha¨ el Thomazo

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Motivations

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Motivations

Schema+Constraints

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Motivations

֒ →

publication

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Motivations

Secret

֒ →

publication

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Motivations

Secret

֒ →

publication Secret leaked?

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Motivations

Secret

֒ →

Safe publication? Secret leaked?

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Example: Hospital setting

Patients Doctors Buildings Specialties Open hours

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Example: Hospital setting

DocSpec PatSpec PatDoc DocBldg PatBldg IsOpen

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Example

Database schema Predicate Meaning IsOpen(b, t) Building b is open on Date t PatBdlg(p, b) Patient p is present in Building b PatSpec(p, s) Patient p was treated for Specialty s PatDoc(p, d) Patient p was treated by Doctor d DocBldg(d, b) Doctor d is associated with Building b DocSpec(d, s) Doctor d is associated with Specialty s

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Example

Views OpenHours(b, t) = IsOpen(b, t) VisitingHours(p, t) = PatBdlg(p, b) ∧ IsOpen(b, t) DocList(d, s, b) = DocSpec(d, s) ∧ DocBldg(d, b)

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Example

Views OpenHours(b, t) = IsOpen(b, t) VisitingHours(p, t) = PatBdlg(p, b) ∧ IsOpen(b, t) DocList(d, s, b) = DocSpec(d, s) ∧ DocBldg(d, b) Constraints PatDoc(p, d) → ∃s PatSpec(p, s) ∧ DocSpec(d, s) PatBdlg(p, b) → ∃d PatDoc(p, d) ∧ DocBldg(d, b)

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Example

Views OpenHours(b, t) = IsOpen(b, t) VisitingHours(p, t) = PatBdlg(p, b) ∧ IsOpen(b, t) DocList(d, s, b) = DocSpec(d, s) ∧ DocBldg(d, b) Constraints PatDoc(p, d) → ∃s PatSpec(p, s) ∧ DocSpec(d, s) PatBdlg(p, b) → ∃d PatDoc(p, d) ∧ DocBldg(d, b) Secret ∃p, s PatSpec(p, s)?

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Example

OpenHour B1 Tuesday B2 Every day 10-17h VisitingHours Charline Tuesday DocList Alice Cancer B1 Alice Cancer B2 Bob Radiology B2 Daniel Cancer B1 Views OpenHours(b, t) = IsOpen(b, t) VisitingHours(p, t) = PatBdlg(p, b) ∧ IsOpen(b, t) DocList(d, s, b) = DocSpec(d, s) ∧ DocBldg(d, b)

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Example

OpenHour B1 Tuesday B2 Every day 10-17h VisitingHours Charline Tuesday DocList Alice Cancer B1 Alice Cancer B2 Bob Radiology B2 Daniel Cancer B1 Views OpenHours(b, t) = IsOpen(b, t) VisitingHours(p, t) = PatBdlg(p, b) ∧ IsOpen(b, t) DocList(d, s, b) = DocSpec(d, s) ∧ DocBldg(d, b)

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Example

OpenHour B1 Tuesday B2 Every day 10-17h VisitingHours Charline Tuesday DocList Alice Cancer B1 Alice Cancer B2 Bob Radiology B2 Daniel Cancer B1 Constraints PatBdlg(p, b) → ∃d PatDoc(p, d) ∧ DocBldg(d, b) Views OpenHours(b, t) = IsOpen(b, t) VisitingHours(p, t) = PatBdlg(p, b) ∧ IsOpen(b, t) DocList(d, s, b) = DocSpec(d, s) ∧ DocBldg(d, b)

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Example

OpenHour B1 Tuesday B2 Every day 10-17h VisitingHours Charline Tuesday DocList Alice Cancer B1 Alice Cancer B2 Bob Radiology B2 Daniel Cancer B1 Constraints PatBdlg(p, b) → ∃d PatDoc(p, d) ∧ DocBldg(d, b) Views OpenHours(b, t) = IsOpen(b, t) VisitingHours(p, t) = PatBdlg(p, b) ∧ IsOpen(b, t) DocList(d, s, b) = DocSpec(d, s) ∧ DocBldg(d, b)

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Example

OpenHour B1 Tuesday B2 Every day 10-17h VisitingHours Charline Tuesday DocList Alice Cancer B1 Alice Cancer B2 Bob Radiology B2 Daniel Cancer B1 Constraints PatBdlg(p, b) → ∃d PatDoc(p, d) ∧ DocBldg(d, b) PatDoc(p, d) → ∃s PatSpec(p, s) ∧ DocSpec(d, s) Views OpenHours(b, t) = IsOpen(b, t) VisitingHours(p, t) = PatBdlg(p, b) ∧ IsOpen(b, t) DocList(d, s, b) = DocSpec(d, s) ∧ DocBldg(d, b)

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Formalism

Data represented by databases R(1, 17), R(2, 42), S(23, 45), . . .

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Formalism

Data represented by databases R(1, 17), R(2, 42), S(23, 45), . . .

֒ →+ secret

Mappings and secrets are CQ V (x, z) := R(x, y) ∧ S(y, z)

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Formalism

Data represented by databases R(1, 17), R(2, 42), S(23, 45), . . .

֒ →+ secret

Mappings and secrets are CQ V (x, z) := R(x, y) ∧ S(y, z) Constraints are TGD R(x, y) → ∃z, S(y, z)

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Safe publication

Secret

֒ →

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Safe publication

Secret

֒ → ← ֓

No secret

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Safe publication

Secret

֒ → ← ֓

No secret View Problem Given (schema, constraints C, views V, secret S, visible ) do we have such that C( ), V( ) = and ¬S( )?

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Safe publication

Secret

֒ → ← ֓

No secret Schema Problem Given (schema, constraints C, views V, secret S) do we have for all an instance such that C( ), V( ) = V( ) and ¬S( )?

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Which configurations are decidable/tractable for the schema problem?

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Which configurations are decidable/tractable for the schema problem? Secrets Views Constraints

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Which configurations are decidable/tractable for the schema problem? Secrets Views Constraints CQ

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Which configurations are decidable/tractable for the schema problem? Secrets Views Constraints CQ CQ

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Which configurations are decidable/tractable for the schema problem? Secrets Views Constraints CQ CQ ProjMap AtomMap GuardedMap CQMap

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Which configurations are decidable/tractable for the schema problem? Secrets Views Constraints CQ CQ ProjMap AtomMap GuardedMap CQMap Let’s talk about constraints...

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Ontologies

Ontologies in a few words

An ontology represents entities and their relationship to each other.

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Ontologies in a few words

An ontology represents entities and their relationship to each other. Ontologies can be seen as a sort of expressive schema.

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Ontologies in a few words

An ontology represents entities and their relationship to each other. Ontologies can be seen as a sort of expressive schema. Ontologies allows to enrich data by inferring new facts from existing ones.

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An example of ontology

With plain words: All cats are mammals. All mammals are animals.

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An example of ontology

With plain words: All cats are mammals. All mammals are animals. With Tuple Generating Dependencies: CAT(x) → MAMMAL(x) MAMMAL(x) → ANIMAL(x)

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An example of ontology

With plain words: All cats are mammals. All mammals are animals. Database: {CAT( )}

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An example of ontology

With plain words: All cats are mammals. All mammals are animals. Database: {CAT( )} Query: Are there animals? (∃X, ANIMAL(X)?)

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An example of ontology

With plain words: All cats are mammals. All mammals are animals. Database: {CAT( )} Query: Are there animals? (∃X, ANIMAL(X)?) Answer: Yes: CAT( ) ⇒ MAMMAL( ) ⇒ ANIMAL( )

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More complex ontological rules

Foreign key constraint: SEMINAR(team, speaker, room, date) → ∃pers, RESERVED(room, date, pers)

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More complex ontological rules

Foreign key constraint: SEMINAR(team, speaker, room, date) → ∃pers, RESERVED(room, date, pers) Even more complex constraints: SEMINAR(team, speaker, room, date) → ∃pers, RESERVED(room, date, pers)∧MEMBER(pers, team)

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More complex ontological rules

Foreign key constraint: SEMINAR(team, speaker, room, date) → ∃pers, RESERVED(room, date, pers) Even more complex constraints: SEMINAR(team, speaker, room, date) → ∃pers, RESERVED(room, date, pers)∧MEMBER(pers, team) MEMBER(person, team) → ∃date, room, SEMINAR(team, person, room, date)

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Tuple Generating Dependencies

∀ X, Y ϕ( X, Y ) ⇒ ∃ Z ψ( Y , Z)

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Tuple Generating Dependencies

∀ X, Y ϕ( X, Y ) ⇒ ∃ Z ψ( Y , Z) Often Omitted

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Tuple Generating Dependencies

ϕ( X, Y ) ⇒ ∃ Z ψ( Y , Z) Body Head

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Tuple Generating Dependencies

ϕ( X, Y ) ⇒ ∃ Z ψ( Y , Z) Frontier

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Open World Query Answering

Open World Query Answering (OWQA)

Open World Query Answering

• A set of facts F
• A set of TGD constraints C
• A conjunctive query Q

Do we have, for all : (F ⊆ ∧ C( )) ⇒ Q( )?

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Open World Query Answering

OWQA(F, C, Q) asks if Q is true in all completions of F (respecting C).

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Open World Query Answering

OWQA(F, C, Q) asks if Q is true in all completions of F (respecting C). The Chase algorithms build a universal model.

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Open World Query Answering

OWQA(F, C, Q) asks if Q is true in all completions of F (respecting C). The Chase algorithms build a universal model. OWQA(F, C, Q) ⇔ Chase(F, C) Q

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The Chase

Intuitively the chase simply “applies” the constraints.

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The Chase

Intuitively the chase simply “applies” the constraints. With:

• F = CAT(

)

• C = {CAT(X) → MAMMAL(X), MAMMAL(X) →

ANIMAL(X)} We obtain:

• 1. F1 = {CAT(

)}

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The Chase

Intuitively the chase simply “applies” the constraints. With:

• F = CAT(

)

• C = {CAT(X) → MAMMAL(X), MAMMAL(X) →

ANIMAL(X)} We obtain:

• 1. F1 = {CAT(

)}

• 2. F2 = {CAT(

), MAMMAL( )}

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The Chase

Intuitively the chase simply “applies” the constraints. With:

• F = CAT(

)

• C = {CAT(X) → MAMMAL(X), MAMMAL(X) →

ANIMAL(X)} We obtain:

• 1. F1 = {CAT(

)}

• 2. F2 = {CAT(

), MAMMAL( )}

• 3. F3 = {CAT(

), MAMMAL( ), ANIMAL( )}

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The Chase

Intuitively the chase simply “applies” the constraints.

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The Chase

Intuitively the chase simply “applies” the constraints. With:

• F = {PERSON(alice)}
• C = {PERSON(X) → ∃Y , PARENT(X, Y )}

We obtain:

• 1. F1 = {PERSON(alice)}

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The Chase

Intuitively the chase simply “applies” the constraints. With:

• F = {PERSON(alice)}
• C = {PERSON(X) → ∃Y , PARENT(X, Y )}

We obtain:

• 1. F1 = {PERSON(alice)}
• 2. F2 = {PERSON(alice), PARENT(alice, Y )}

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The Chase

Intuitively the chase simply “applies” the constraints. With:

• F = {PERSON(alice)}
• C = {PERSON(X) → ∃Y , PARENT(X, Y )}

We obtain:

• 1. F1 = {PERSON(alice)}
• 2. F2 = {PERSON(alice), PARENT(alice, Y )}

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The Chase

With:

• F = {PERSON(alice)}
• C = {PERSON(X) → ∃Y , PARENT(X, Y )

PARENT(X, Y ) → ∃PERSON(Y )} We obtain:

• 1. F1 = {PERSON(alice)}

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The Chase

With:

• F = {PERSON(alice)}
• C = {PERSON(X) → ∃Y , PARENT(X, Y )

PARENT(X, Y ) → ∃PERSON(Y )} We obtain:

• 1. F1 = {PERSON(alice)}
• 2. F2 = {PERSON(alice), PARENT(alice, Y )}

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The Chase

With:

• F = {PERSON(alice)}
• C = {PERSON(X) → ∃Y , PARENT(X, Y )

PARENT(X, Y ) → ∃PERSON(Y )} We obtain:

• 1. F1 = {PERSON(alice)}
• 2. F2 = {PERSON(alice), PARENT(alice, Y )}
• 3. F3 = {PERSON(alice), PARENT(alice, Y ), PERSON(Y )}

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The Chase

With:

• F = {PERSON(alice)}
• C = {PERSON(X) → ∃Y , PARENT(X, Y )

PARENT(X, Y ) → ∃PERSON(Y )} We obtain:

• 1. F1 = {PERSON(alice)}
• 2. F2 = {PERSON(alice), PARENT(alice, Y )}
• 3. F3 = {PERSON(alice), PARENT(alice, Y ), PERSON(Y )}
• 4. F4 = {PERSON(alice), PARENT(alice, Y ), PERSON(Y ),

PARENT(Y , Y ′)}

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The Chase

With:

• F = {PERSON(alice)}
• C = {PERSON(X) → ∃Y , PARENT(X, Y )

PARENT(X, Y ) → ∃PERSON(Y )} We obtain:

• 1. F1 = {PERSON(alice)}
• 2. F2 = {PERSON(alice), PARENT(alice, Y )}
• 3. F3 = {PERSON(alice), PARENT(alice, Y ), PERSON(Y )}
• 4. F4 = {PERSON(alice), PARENT(alice, Y ), PERSON(Y ),

PARENT(Y , Y ′)} . . .

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The Chase

The Chase model is not always finite.

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The Chase

The Chase model is not always finite. When it is not finite it sometimes has a regularity that allows for decidable OWQA.

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The Chase

The Chase model is not always finite. When it is not finite it sometimes has a regularity that allows for decidable OWQA. And in general the OWQA is undecidable. . .

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Decidable Classes of TGD

• UID, the foreign key constraint with one variable

A(x, Y ) → ∃Z, B(x, Z)

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Decidable Classes of TGD

• UID, the foreign key constraint with one variable

A(x, Y ) → ∃Z, B(x, Z)

• IncDep, the foreign key constraint

A( X, Y ) → ∃Z, B( X, Z)

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Decidable Classes of TGD

• UID, the foreign key constraint with one variable

A(x, Y ) → ∃Z, B(x, Z)

• IncDep, the foreign key constraint

A( X, Y ) → ∃Z, B( X, Z)

• LTGD, the foreign key constraint with repetition of atoms

A(x, x, y) → ∃Z, B(x, y, y, z)

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Decidable Classes of TGD

• UID, the foreign key constraint with one variable

A(x, Y ) → ∃Z, B(x, Z)

• IncDep, the foreign key constraint

A( X, Y ) → ∃Z, B( X, Z)

• LTGD, the foreign key constraint with repetition of atoms

A(x, x, y) → ∃Z, B(x, y, y, z)

• GTGD, one atom in the body guards all variables

A(x, y, z) ∧ B(x) ∧ C(y, z) → ∃w, D(x, y, w)

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Decidable Classes of TGD

• UID, the foreign key constraint with one variable

A(x, Y ) → ∃Z, B(x, Z)

• IncDep, the foreign key constraint

A( X, Y ) → ∃Z, B( X, Z)

• LTGD, the foreign key constraint with repetition of atoms

A(x, x, y) → ∃Z, B(x, y, y, z)

• GTGD, one atom in the body guards all variables

A(x, y, z) ∧ B(x) ∧ C(y, z) → ∃w, D(x, y, w)

• FGTGD, one atom in the body guards all the frontier variables

A(w, y, y) ∧ B(x) ∧ C(y, z) → ∃u, D(x, y, u)

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Decidable Classes of TGD

• UID, the foreign key constraint with one variable

A(x, Y ) → ∃Z, B(x, Z)

• IncDep, the foreign key constraint

A( X, Y ) → ∃Z, B( X, Z)

• LTGD, the foreign key constraint with repetition of atoms

A(x, x, y) → ∃Z, B(x, y, y, z)

• GTGD, one atom in the body guards all variables

A(x, y, z) ∧ B(x) ∧ C(y, z) → ∃w, D(x, y, w)

• FGTGD, one atom in the body guards all the frontier variables

A(w, y, y) ∧ B(x) ∧ C(y, z) → ∃u, D(x, y, u)

• Fr1LTGD, one atom in the body guards the only frontier

variable A(w, y, y) ∧ B(x) ∧ C(y, z) → ∃u, D(x, u)

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Another approach: query rewriting

With

• MAMMAL(X) → ANIMAL(X)
• CAT(X) → MAMMAL(X)

And the query ∃X, ANIMAL(X), we obtain:

• ANIMAL(X)

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Another approach: query rewriting

With

• MAMMAL(X) → ANIMAL(X)
• CAT(X) → MAMMAL(X)

And the query ∃X, ANIMAL(X), we obtain:

• ANIMAL(X)
• ANIMAL(X) ∨ MAMMAL(X)

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Another approach: query rewriting

With

• MAMMAL(X) → ANIMAL(X)
• CAT(X) → MAMMAL(X)

And the query ∃X, ANIMAL(X), we obtain:

• ANIMAL(X)
• ANIMAL(X) ∨ MAMMAL(X)
• ANIMAL(X) ∨ MAMMAL(X) ∨ CAT(X)

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Solving our problem

Back to our problems

View Problem Given (schema, constraints C, views V, secret S, visible ) do we have such that C( ), V( ) = and ¬S( )? Schema Problem Given (schema, constraints C, views V, secret S) do we have for all an instance such that C( ), V( ) = V( ) and ¬S( )?

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Solving the schema problem

The critical instance The instance

C contains one fact per relation, with one

constant: C.

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Solving the schema problem

The critical instance The instance

C contains one fact per relation, with one

constant: C. We note

C = V( C) its view image.

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Solving the schema problem

The critical instance The instance

C contains one fact per relation, with one

constant: C. We note

C = V( C) its view image.

Reduction for schema problem SchemaProblem(C, V, S) reduces to ViewProblem(

C, C, V, S)

From Querying Visible and Invisible Information. LICS 2016

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Reduction to Open World Query Answering (OWQA)

Open World Query Answering

• A set of facts F
• A set of TGD constraints C
• A query Q

Do we have, for all : (F ⊆ ∧ C( )) ⇒ Q( )?

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Reduction to Open World Query Answering

Encoding ViewProblem(

C, C, V, S) as OWQA

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Reduction to Open World Query Answering

Encoding ViewProblem(

C, C, V, S) as OWQA

• The query is S

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Reduction to Open World Query Answering

Encoding ViewProblem(

C, C, V, S) as OWQA

• The query is S
• The initial facts encode the forward constraints

C ⊆ V(

)

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Reduction to Open World Query Answering

Encoding ViewProblem(

C, C, V, S) as OWQA

• The query is S
• The initial facts encode the forward constraints

C ⊆ V(

)

• The constraints are the original constraints C.

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Reduction to Open World Query Answering

Encoding ViewProblem(

C, C, V, S) as OWQA

• The query is S
• The initial facts encode the forward constraints

C ⊆ V(

)

• The constraints are the original constraints C.

But we also need to encode the backward constraints V( ) ⊆

C!

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Reduction to Open World Query Answering

Encoding ViewProblem(

C, C, V, S) as OWQA

• The query is S
• The initial facts encode the forward constraints

C ⊆ V(

)

• The constraints are the original constraints C.

But we also need to encode the backward constraints V( ) ⊆

C!

For this we use that adom(V( )) = {C}

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Lower bounds

Various reductions from:

• OWQA

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Lower bounds

Various reductions from:

• OWQA
• Query evaluation

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Lower bounds

Various reductions from:

• OWQA
• Query evaluation
• Alternating Turing Machines

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Complexity results

Constraints Views ProjMap AtomMap GuardedMap CQMap IncDep PSpace ExpTime 2ExpTime 2ExpTime LTGD ExpTime ExpTime 2ExpTime 2ExpTime GTGD 2ExpTime 2ExpTime 2ExpTime 2ExpTime FGTGD 2ExpTime 2ExpTime 2ExpTime 2ExpTime

Table 1: Complexity of disclosure

⇒ all bounds are tight!

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Complexity results

Constraints Views ProjMap AtomMap GuardedMap CQMap IncDep NP NP ExpTime 2ExpTime LTGD NP NP ExpTime 2ExpTime GTGD ExpTime ExpTime ExpTime 2ExpTime FGTGD 2ExpTime 2ExpTime 2ExpTime 2ExpTime

Table 2: Complexity of disclosure in bounded arity

⇒ all bounds are tight!

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Complexity results

In PTime:

• CQ secret, foreign keys constraints and projection views

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Complexity results

In PTime:

• CQ secret, foreign keys constraints and projection views
• bounded CQ secret, ProjMap, LTGD

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Future Works

• Implement model checker for publication methods.

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Future Works

• Implement model checker for publication methods.
• How to synthesize publications automatically?

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Thank you!

Questions?

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