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An exercise in proving undecidability

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An exercise in proving undecidability Balder ten Cate Bertinoro 15/12/2006 Query answering under GAV mappings Input: a GAV mapping m: S T a source instance I a ... – PowerPoint PPT presentation

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Title: An exercise in proving undecidability


1
An exercise in proving undecidability
  • Balder ten Cate
  • Bertinoro 15/12/2006

2
Query answering under GAV mappings
  • Input a GAV mapping m S?T
  • a source instance I
  • a target query ?
  • Output the certain answers
  • ?(I,J) m J(?)

3
Complexity
  • For conjunctive queries ?, the problem is in
    LOGSPACE (by unfolding)
  • For FO queries ?, its undecidable.
  • This talk There a fixed FO query ? for which
    computing the certain answers is undecidable.

(Corrolary CERT(m, ?) is not definable in
FO/datalog/...)
4
More precisely
  • Fact There is a GAV mapping m S?T and a Boolean
    FO query ? over T such that the following is
    undecidable
  • Given a source instance I, is Yes a certain
    answer to ? ?
  • Proof by reduction from an undecidable tiling
    problem.

5
Periodic tiling
  • An undecidable problem
  • Given a finite set of tile types
  • Can we tile any n ? n square with these tiles so
    that (a) neighboring tiles match, (b) the first
    and last column coincide, and (c) the first and
    last row coincide (n gt 1) ?

...
6
Reduction to GAV answering
  • Basic idea
  • The source instance I specifies the set of tile
    types
  • The GAV mapping m (which is fixed) simply copies
    all the information
  • The FO query ? (which is fixed) describes a
    periodic tiling with the given tile types.
  • Yes is a certain answer to ?? on source
    instance I iff the set of tile types specified by
    I admits no periodic tiling.

7
First attempt
  • Source schema
  • A unary relation TT listing tile types
  • Binary relations COMPH and COMPV specifying
    horizontal and vertical compatibility
  • The GAV mapping
  • ?x (TTx ? TTx)
  • ?x (RHx ? RH x)
  • ?x (RVx ? RV x)
  • Before we continue
  • What is wrong with this attempt?

8
Bug fix
  • We need to make sure that ...
  • no compatibilities are added in the target
  • Solution represent incompatibilities
  • no new tile types are added in the target
  • Solution use extra relations so that tampering
    can be detected

9
The correct reduction
  • Source schema
  • A unary relation TT listing tile types
  • Binary relations INCOMPH and INCOMPV specifying
    horizontal and vertical incompatibility
  • Two binary relations coding a linear ordering of
    the tile types and a corresponding successor
    relation.
  • The GAV mapping copies everything (as before)
  • The target query ? describes a periodic tiling
    using the given tile types (homework exercise,
    for the solution see Börger- Grädel-Gurevich) .

10
Added in print
  • Prof. Kolaitis found a simpler and more elegant
    proof by reduction of the undecidable embedding
    problem for finite semi-groups given a partial
    binary function, can it be extended to a
    semi-group (over a possible larger but finite
    carrier set)?
  • Source schema a single ternary relation R
  • Target schema a single ternary relation R
  • GAV mapping ?xyz (Rxyz ? Rxyz)
  • The target query ? expresses that R is an
    associative total function (this can be expressed
    in FO logic, even using only ??-formulas).
  • Yes is a certain answer to ?? on source
    instance I iff the I(R) cannot be extended to a
    finite semi-group.
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