CS319-course-ArmstrongCompleteSound

8/7/2019 CS319-course-ArmstrongCompleteSound

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Dr. Alexandra I. Cristeahttp://www.dcs.warwick.ac.uk/~acristea/

CS 319: Theory of Databases: C3


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(provisionary) Content

1. Generalities DB2. Integrity constraints (FD revisited)

3. Relational Algebra (revisited)

4. Query optimisation

5. Tuple calculus6. Domain calculus

7. Query equivalence

8. LLJ, DP and applications

9. Temporal Data10. The Askew Wall


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previous

Proving with Armstrong axioms

(non)Redundancy of FDs


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FD Part 3

Soundness and Completeness ofArmstrongs axioms


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Armstrongs Axioms:

Sound & CompleteIngredients: Functional Dependency (reminder)

Definition

Inference rules

Closure of F : F*F+: Set of all FDs obtained by applying inferences rules on a basic set

of FDs

Issues and resolutions

Armstrongs Axioms (reminder) 3 inferences rules for obtaining the closure of F

Properties of the Armstrongs AxiomsThey are a sound an complete set of inference rules

Proof of the completeness


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Functional Dependency

- A functional dependency (FD) has the formXYwhereXand Yare sets of attributes in a relation R

XY iff any two tuples that agree onXvalue also agree on Yvalue

XYif and only if:

for any instance rof R

for any tuples t1 and t2 ofr

t1(X) = t2(X) t1(Y) = t2(Y)


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Inference of Functional

Dependencies Suppose R is a relation scheme, and F is a set of functionaldependencies for R

If X, Y are subsets of attributes of Attr(R) and if all instances r ofR which satisfy the FDs in F also satisfy X Y, then we say thatF entails (logically implies) X Y, written F | X Y

In other words:

there is no instance r of R that does not satisfy X Y

Example

if F = { A B, B C } then F | A Cif F = { S A, SI P } then

F | S I AP, F | SP SAP etc.

A B C

S A

PI


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Functional Dependency

Issue: How to represent set ofALL FDs for a relation R?

Solution Find a basic set of FDs ((canonical) cover)

Use axioms for inferring

Represent the set of all FDs as the set of FDs that can be inferredfrom a basic set of FDs

Axioms they must be a sound and complete set of inference rules


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Set of Functional Dependencies F*

Formal Definition of F*, the covercoverof F:

Informal Definitions F* is the set of all FDs logically implied by F

(entailed)

... usually F* is much too large even to enumerate!

if F is a set of FDs, then F* { X Y FX Y }


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F* Usually F* is much too large even to

enumerate! Example (3 attributes, 2 FDs and 43 entailed

dependencies)

Attr(R)=ABC and F ={ A B, B C } then F* isA S for all [subset of ABC] 8 FDsB BC, B B, B C, B 4 FDsC C, C , 3 FDs

AB S for all subsets S of ABC 8 FDsAC S for all subsets S of ABC 8 FDsBC BC, BC B, BC C, BC 4 FDsABC S for all subsets S of ABC 8 FDs

A B C


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Armstrongs Axioms

Axioms for reasoning about FDs

F1: reflexivity if Y X then X Y

F2: augmentation if X Y then XZ YZ

F3: transitivity if X Y and Y Z then X Z


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F+

Informal Definition

Formal Definition

F+(the closureclosure of F)is the set of dependencies which canbe deduced from Fby applying Armstrongs axioms

if F is a set of FDs, then F+ { X Y F |= X Y }


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Armstrongs Axioms

Theorem:Armstrongs axioms are a sound and complete

set of inference rules

Sound: all Armstrong axioms are valid (correct / hold) Complete: all fds that are entailed can be deduced with

the help of the Armstrong axioms

How to: Prove the soundness?


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Armstrongs Axioms

Theorem: Armstrongs axioms are a sound andcomplete set of inference rules Sound: the Armstrongs rules generate only FDs in F*

F+ F* Complete: the Armstrongs rules generate all FDs in F*

F* F +

If complete and sound then F+ = F*

Here

Proof of the completeness


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Armstrongs Theorems

Additional rules derived from axioms Union

ifA B and AC, then ABC

Decomposition

ifABC, then A B and A C

A B

C

AB

C

For the proof, we can use:


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Completeness of the

Armstrongs Axioms

Proving that Armstrongs axioms are acomplete set of inference rules

Armstrongs axioms generate all FDs inF*

F*

F +


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Completeness of the

Armstrongs Axioms First define X+, the closure of X with respect to F:

X+ is the set of attributes A such that X A can be deduced

from F with Armstrongs axioms

Note that we can deduce thatX Y for some set Y byapplying Armstrongs axioms if and only if Y X+

Attr(R)=LMNO

X=L

F={L M , M N, O N}then X+ = L+ = LMN

L M N

O

Example:


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X

Y F+

Y

X

+

Proof: We can deduce thatX Y for some set Y by applyingArmstrongs axioms if and only if Y X+

Y X+XY F+ Y X+and suppose that A Y then X A F+ (definition of X+) A Y: X A F+ X Y F+(union rule)

XY F+ Y X+ X Y F

+

and suppose that A Y then X A F+

(decomposition rule) A X+ (definition of X+) A Y: A X+ Y X+

C l t ( * ) f th


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Completeness (F* F +) of theArmstrongs Axioms

Completeness:

(R, X,Y Attr(R), F true in R : : X Y F* => X Y F +)

Idea: (A => B) (A v B) (B v A) (B =>A)

To establish completeness, it is sufficient to show:

if X Y cannot be deduced from F using Armstrongs axioms then also X Y is notlogically implied by F:

(R, X,Y Attr(R), F true in R : : X Y F+=> X Y F*)

(In other words) there is a relational instance rin R (rR) in which all thedependencies in F are true, but X Y does not hold

XXYY F*F* enough:enough:

Counter example!!Counter example!!

C l t f th


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Completeness of the

Armstrongs Axioms

Example for the proof idea for a given R, F, X:If X Y cannot be deduced using Armstrongs axioms: then thereis a relational instance for R in which all the dependencies in F are

true, but X Y does not hold

Counter example:

R=LMNO

X=L

F={L M , M N, O N}then X+ = LMN

L M N

O

L O cannot bededuced (so not in F+)

but also does not hold

(not in F*)


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What we want to prove thus:

(R, X,Y Attr(R), F true in R : : X Y F+ => X Y F*)

(In other words) Counterexample by construction:

for any R, any X, Y, any F true in R, with X->Y not inferable from the axioms

there is a relational instance rin R (rR) in which all the dependencies in F are true (F true in R ), but X Y does not hold

Completeness of the


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Completeness of the

Armstrongs Axioms

Suppose one can not deduce X Y from Armstrongsaxioms for an arbitrary R, F, X,Y; construct counter-ex. Consider the instance r

0for R with 2 tuples

(assuming Boolean attributes, or more generally, that the twotuples agree on X+ but disagree elsewhere)

Attributes of X+ Other Attributes

1 1 1 1 1 1

1 1 . 1 0 0 0

Relational instance r0 for R with 2 tuplesL+ = LMNL M N O

1 1 1 1

1 1 1 0

R=LMNO

X=L

then X+ = L+

L O cannot

be deduced

Example:


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Completeness of the

Armstrongs Axioms Check that all the dependencies in F are true in R:

Suppose that V W is a dependency in F If V is not a subset of X+, the dependency holds in r

0

If V is a subset of X+

, then both X V, and then X W can bededuced by Armstrongs axioms. This means that W is a

subset of X+, and thus V W holds in r0

Attributes of X+

Other Attributes1 1 1 1 1 1

1 1 . 1 0 0 0

Relational instance r0 for R with 2 tuples


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Completeness of the

Armstrongs Axioms

Check that all the dependencies in F are true

Extended Example (more tuples)

O N is a dependency in F but O is not a subset of X+,

the dependency holds in r0

M N is a dependency in F and M is a subset of X+, thenboth L M, and L N can be deduced by Armstrongs

axioms. This means that N is a subset of X+, and thus M N holds in r0

R=LMNO

X=L

F={L M , M N, O N}then X+ = LMN

Completeness of the


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Completeness of the

Armstrongs Axioms

Proof that X Y does not hold in r0:

Recall that we can deduce thatX Y for some set Y by

applying Armstrongs axioms if and only if Y X+

By assumption, we cant deduce that X Y holds in r0

Hence Y contains (at least) an attribute not in the subset X+,

confirming that X Y does not hold in r0

Attributes of X+ Other Attributes

1 1 1 1 1 1

1 1 . 1 0 0 0

Relational instance r0 for R with 2 tuples

(R, X,Y Attr(R), F true in R : : X -> Y F+ => X -> Y F*)


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Completeness of the

Armstrongs Axioms

We have proved the correctness (last module) andhere, the completeness of ArmstrongsAxioms: How can we prove the completeness of another set

of rules?

Repeat the proof for this set Deduce the Armstrongs Axioms from this set

How can we disprove the completeness of anotherset of rules?

By showing (via a counterexample) that someconsequence of Armstrong's rules cannot be deducedfrom them (see the proof technique for non-redundancy)


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Completeness of the

Armstrongs Axioms

Exercise Are the following set of rules a sound and complete set of

inference rules? (X, Y, Z, W R)S1: X XS2: if X Y then XZ YS3: if X Y, Y Z then XW ZW

(This is a typical exam question )


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Solution (soundness)

XX so by F1

YXif then by F2

XX

YZXZ

so by F1 YYZYZY

so by F3 YXZYYZYZXZ ,YXZ

YX

ZYYX ,if then by F3 ZX

so by F2 ZWXW ZWXW

ZYYX

,

S1

S2

S3


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Solution (completeness)

XY so {S1} YY,{S2}YZY

if XYZZ = ::then henceXY

YX {S1}XX, XY, so

{S3}XZYZ

if then hence

YZXZ

YX

A1

A2

A3ZYYX ,if

then {S3}hence

ZX

ZYYX

,00 // ZX


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Completeness of the

Armstrongs Axioms

Exercises Are the following set of rules a complete set of inference rules?

1111

111

11

11

A B AB

A

B

AB

}0{},,{)( AFBARAtt /==

Suppose

Apply above rules exhaustively

ABB

Hence: answer is NO

obtained from F1-3

but not from R1-3

R1: X X

R2: X Y then XZ YR3: X Y, Y Z then X Z


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Summary

We have learned how to prove that theArmstrong axiom set is complete (and

we already knew it is sound)

We can now prove the soundness andcompleteness of any other set of

axioms


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Extra Question: Can we have a sound and complete set of

inference rules consisting of only 2 rules? What about 1 rule?


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o follow

pendency preserving, Lossless join,rmal forms algorithms

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