Fd covers

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Covers for Functional

Dependencies

CIS-611. Victor Matos

Source:

Chapter 5, “The Theory of Relational Databases”, D. Maier, Ed. Comp. Sc. Press,

1983 ISBN 0-914894-42-0.

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Cleveland State University

Agenda

1. Covers and Equivalence

2. Non-redundant covers

3. Extraneous attributes

4. Canonical covers

5. Minimum covers

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Introduction

In this section we will explore ways of

representing FDs in a concise mode

Reducing the size of a set of FDs benefits

the computation of closures and other

related algorithms

Reduced sets of FDs help ensure database

consistency and correctness

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Overview

The overall strategy is to minimize the set

of FDs for a given relational schema in

order to – (1) save storage and,

– (2) reduce the number of compliance-tests performed

when the database records are modified

Optimization is done by throwing away

unwanted rules and unneeded attributes

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Example

Consider the database schema R(ABC) under the set

F1 = {A → B, B → C, A → C, A B → C, A → B C }

Again, the same database schema R(ABC) but the set

F2 = {A → B, B → C }

It is easy to see that F1 and F2 are equivalent.

Using the more compact set F2 will provide a better

representation of the semantic of data in R(ABC) as

well as better performance/storage.

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1. Covers and Equivalences

Two sets of FDs F1 and F2 over scheme R

are equivalent , written F1 F2 if F1+ = F2

+

If F1 F2 then F1 is a cover for F2

Example: The sets

F1= { AB → C, A → D, CD → E } and

F2= { A → BCE, A → ABC, CD → E }

are NOT equivalent

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Example

Consider the schema R(ABCDE) and the sets of FDs F1,

F2 and F3 listed below

F1= { A → BC, A → D, CD → E }

F2= { A → BCE, A → ABD, CD → E }

F3= { A → BCDE }

By observation of the sets we conclude that:

– F1 and F2 are equivalent (A+ =ABCDE, CD+ = CDE in both sets)

– F1 and F3 are not equivalent ( F1F3, but F3 cannot imply CD→E)

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Observation

A mechanical way of testing if F1 F2 is to

simply compare F1+ and F2

+

However:

Computing the closure F+ of a set of FDs is time

consuming.

We need a ‘better’ way of working (other than the

brute-force generation of all the possible

derivation sequences)

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Member Function

Determine if a single FD X →Y is derived from a set F of FDs

Input: A set of FDs and an FD X →YOutput: true if FX →Y, false otherwise

Member(F, X →Y)

beginif ( XF

+ Y )

then return (true)

else return (false);

end

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Lemma

Given sets F1 and F2 over schema R, F1 F2 if and only if F1F2 and F2 F1

Function DERIVES ( F1, F2)

Input: Two sets of FDs F1 and F2

Output: true if F1 F2, false otherwise

begin

v = true;

for each FD X Y in F2 do

v = v and Member ( X Y, F1 );

return ( v );

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Input: Two sets of FDs F1 and F2

Output: true if Fl F2, false otherwise

Function EQUIV ( Fl , F2)

v = DERIVES (F1 ,F2) and DERIVES ( F2, Fl);

return ( v );

end

How to check that two sets of dependencies are

equivalent ?

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Example: Consider the schema R(ABCDE) and the sets of

FDs F1 and F2 listed below

F1 = {A → B, B → C, A → C, A B → C, A → B C }

F2 = {A → B, B → C }

Observations:

1. Each dependency in F2 is in F1, therefore F1 DERIVES F2 .

2.1 Rule A → C in F1 is a member of F2

2.2 Rule A B → C in F1 is a member of F2

2.3 Rule A → B C in F1 is a member of F2

2.4 Therefore all dependencies in F1 are DERIVED from F2.

3. According to the EQUIV function F1 and F2 are equivalent.

How to check that two sets of dependencies are

equivalent ?

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A set F1 of FDs is non-redundant if there is no proper sub-set F' of F1 such that Fl F'

If such an F' exists then F1 is redundant.

F1 is a non-redundant cover for F2 if1. F1 is a cover for F2 (F1 F2), and

2. F1 is a non-redundant set of FDs.

Redundant Covers

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Example. Universe is {A, B, C }. There are

three set of FDs

Fl= { AB →C, A → B, B → C, A → C }

F2 = {AB → C, A → B, B → C }

F3 = {A → B, B → C }

Observations

1. F1 is a redundant cover for F2

2. F3 is a non redundant cover for F2

Redundant Covers

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EXAMPLE:

Let R(ABC) and F= { A → B, B → A, B → C, A → C }

Observe that rule A → C is redundant. It could be obtained

by applying transitivity on A → B and B → C.

Therefore

F2 = { A → B, B → A, B → C } is a cover for F (it also

happens to be non-redundant).

1. Testing RedundancyHow to eliminate a rule X→Y from a set of FDs ?

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ALGORITM NonRedundant (F1)

Input: An arbitrary set F1 of FDs

Output: A non-redundant cover F2 for F1

begin

F2 = F1;

for each FD X→ Y in F1 do

if Member(X → Y, F2 - (X → Y) )

then F2 = F2 - ( X → Y );

return ( F2 );

end;


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EXAMPLE:

Let R(ABC) and F= { A → B, B → A, B → C, A → C }

After applying the previous function we obtain:

F2 = F – {A → C } = { A → B, B → A, B → C }

is a non-redundant cover for F.


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Let F be a set of FDs over schema R, and let X→Y be an FD

in F. Attribute A in R is extraneous in the rule X → Y if

either

LEFT. Assume X = A Z , X Z and

F ( F - { X → Y } ) { Z → Y }

RIGHT. Assume Y = AW, YW and

F ( F - {X →Y} ) {X →W}

Extraneous AttributesOur intention is to discard from F unnecessary attributes appearing

in individual FDs without changing the closure of F.

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Example

Consider schema R = { A, B, C, D } and FDs

F = { A → BC, B → C, AB → D }

Left Reduction. We look for FDs having more than one attribute on the

LHS. The only candidate is: AB → D

(1). Extract A from AB → D

(2). If (1) didn‟t work try taking B from AB → D

Right Reduction. Look for FDs with more than one attribute on the RHS.

The only candidate is: A → BC

(3). Try removing B from A → BC

(4). If (3) didn‟t work try deleting C from A → BC

Extraneous Attributes

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Input: A set G of FDs

Output: A left-reduced cover F for G

begin

F=G;

for each X → Y in G do

for each attribute A in X do

if Member (X-A → Y, F) then {

Remove A from X in X → Y;

Update F adding new rule X-A → Y;

}

return( F );

end;

Algorithm: LeftReduction (G)

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Input: A set G of FDs

Output: A right-reduced cover F for G

begin

F=G;

for each X → Y in G do

for each attribute A in Y do

if Member(X → Y, F-{X → Y}U {X → Y-A} ) {

Remove A from Y in X → Y;

Update F adding new rule X → Y-A

}

return( F );

end;

Algorithm: RightReduction (G)

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Consider the relation schema R subject to F

1. Create set F' by removing extraneous attributes

in the following order: first do left then right

reductions

2. Remove redundant functional dependencies.

3. The new set (R,F' ) is called a Canonical Cover

for (R, F)

Canonical CoverConstructing a „better‟ representation of F

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Example Let r(ABCDIJM) be subject to the FDs

F= { A → C, AC → J, AB → DE, AB → CDI, C → M, A → M }

(a) Left reduction

1. Test rule AC → J. Drop A, to make C → J. New set F2 is not

equivalent to F ( Member (AC → J, F2) = false ). Therefore A can

not be eliminated.

2. Try dropping C from AC → J. Make F2= F-{AC → J) + {A → C).

F2 is equivalent to F ( Member (AC+ J, F2) = true ). Consequently C

can be eliminated from AC → J.

(b) Right reduction.

1. Attribute D in AB → DE is redundant, and

2. Attribute C in AB → CDI is redundant, eliminate both.

(c ) Redundant Rules. Eliminate rule A → M

(d) Result: F= { A → CI, AB → EDI, C → M }

Canonical Covers

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Example. Let r(ABC) be subject to the FDs

F= { A → BC, B → C, A → B, AB → C }

(a) Left reduction.

(b) Right reduction.

(c) Redundant Rules.

(d) Result ?

Canonical Covers

Your turn …

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Observation: A schema R could have more than one non-

redundant cover. Our next goal is to explore the similarities

between them.

Definition

Two sets of attributes X and Y are equivalent under FDs F, written X ↔Y, if F XY and F YX.

Lemma. Given equivalent, non-redundant covers F and G,

for every left side X of an FD in F there is an equivalent left

side V of an FD in G.

The Structure of Non-Redundant Covers

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Example:

Let R(ABCDE) and FD sets

F = { A→BC, B→A, AD→E } and

G= { A → ABC, B → A, BD → E }Notice that

EF(A) = {A→BC, B→A } (observe that AF+= BF

+ = ABC)

EF(B) = {A→BC, B→A }

EF(AD) = { AD→E } (observe ADF+= ABCDE)

EG(A) = {A→BC, B→A } (observe that AG+= BG

+ = ABC)

EG(B) = {A→BC, B→A }

EG(BD) = { BD→E } (observe BDG+= ABCDE)

F and G are non-redundant and equivalent to each other (non redundant in the sense

that rules can not be removed, however there are superfluous attributes in G).

Observe that: AF ↔ AG , BF ↔ BG and ADF ↔ BDG


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Definition. Equivalent Classes

For a set of FDs F over scheme R and a set X R, let EF(X) be the set of

FDs in F with left sides equivalent to X.

Let E F be the set { EF(X) / X R and EF(X) Ø }

E F(X) is empty when no left side of any FD in F is equivalent to X.

E F is always a partition of F.

Example: Consider the previous example.


E F E G

EF(A) = {A→BC, B→A}

EF(B) = EF(A)

EF(AD) = {AD→E }

EG(A) = {A→ABC, B→A}

EG(B) = EG(A)

EG(BD) = {BD→E }

F = { A→BC, B→A, AD→E } and

G= { A → ABC, B → A, BD → E }

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Example: Consider the non-redundant set of FDs

F = { A→BC, B→A, AD→E, BD→I }

Let‟s find all the equivalent classes of F.

Find closures of left-hand-sides:

• A+ = ABC

• B+ = ABC same, therefore EF(A) = EF(B) = {A→BC, B→A }

• (AD)+ = ABCDEI

• (BD)+ = ABCDEI same EF(AD) = EF(BD) = { AD→E, BD→I }

Therefore

E F = { EF(A), EF(AD) }


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Observation. A non-redundant cover of a set G of

FDs does not necessarily have as few FDs as any

cover for G.

Definition. A set of FDs F is minimum if F has as

few FDs as any equivalent set of FDs.

A minimum set of FDs is also non-redundant.

Minimum Cover

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Example.

The set G is non-redundant but not minimum, since F is

equivalent to G but has fewer FDs.

F is a minimum cover for G.

(observe that G is already a Canonical cover)

G = { A→BC, B→A, AD→E, BD → I }

F = { A→BC, B→A, AD→E I }

Minimum Cover

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Unlike non-redundant covers, the definition of minimum covers provides

no guide for finding minimum covers or even for testing minimality.

The next concept on direct functional determination gives us the means to

compute minimum covers.

Definition. Given a set of FDs G, X directly determines Y under G, written X Y, if there is a non-redundant cover F for G with an F-based

DDAG H for X→Y such that U(H) ∩ EF(X) = Ø.

In other words, we can find a non-redundant cover F for G in which X→Y

can be derived using only FDs in { F - EF(X) }.

Observe that X X always holds, that X Y implies X→Y, and that

EF(X) can be empty.

Direct Determination

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Re-phrasing

Definition. Given a set of FDs G, X directly determines Y under G, written X Y, if there is a non-redundant cover F for G with an F-based

DDAG H for X→Y such that U(H) ∩ EF(X) = Ø.

Equivalent to :

X Y if { F – EF(X) } X→Y


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Example. Let G = F = { A→CD, AB→E, B → I, DI → J }.Then AB J under G, as the DDAG H in next figure shows.

Here the „use set‟ for the DDAG H drawn for AB →J is:

U(H) = { A→D, B → I, DI → J } and

Observe that

EF(AB) = { AB → E }

Their intersection is empty, therefore AB J


A D

B I

J

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Example.

Let F = {A → BC, BC → A, AD→E, AD → I, E→B }.

Observe that BCD ↔ A D ( observe that BCD+=AD+=ABCDEI )

Figure below shows an F-based DDAG for BCD→AD that uses no FDs

from EF(AD) and no FDs in EF(BCD).

U(H) = { BC → A }EF(BCD) = ØEF(AD) = { AD→E, AD→I }

observe that U(H) EF(BCD) = U(H) EF(AD) =Ø therefore BCD ADsimilarly AD BCD

Computing Minimum Covers

B

D

C

A

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Theorem. Let F be a non-redundant set of FDs that is not minimum.

There is some EF(X) containing distinct FDs Y→U and Z→V such thatY Z.

Observation

We know the set F = { A→BC, B→A, AD→E, BD → I } is non-redundant,

and we have computed its equivalence classes already.

EF(A) = EF(B) = {A→BC, B→A } and

EF(AD) = EF(BD) = { AD→E, BD→I }

Consider the set EF(BD). According to previous theorem AD BD as well as

BDAD. See the DDAG and its use set for BD → AD

Use set U(H) = { B→A }notice that U(H) ∩ EF(BD) = Øtherefore BDAD


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Observation (continuation…)

Therefore the non-redundant cover

F = { A→BC, B→A, AD→E, BD → I } is not a minimum set.

As we will see it could be reduced to the smaller minimum cover

F2 = { A→BC, B→A, BD→EI }

or equivalently

F3 = { A→BC, B→A, AD→EI }


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Algorithm. MINIMIZE

Input: A set of FDs G.

Output: A minimum cover for G.

MINIMIZE(G)

begin

F : = NONREDUN(G);

Find the sets of EF

for each EF(X) in EF do

for each Y → M in EF(X) dofor each Z → V Y → M in EF(X) do

If DDERIVES(F, Y → Z) then

replace Y → M and Z → V by Z → MV in F,

return(F)

end.

MINIMIZE can be implemented to have time complexity O(np) on inputs of length n with p FDs.


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Example

Set F = { A→BC, B→A, AD→E, BD → I } from previous examples is

known to be non-redundant, we also have computed its equivalence classes:

EF(A) = EF(B) = {A→BC, B→A } and

EF(AD) = EF(BD) = { AD→E, BD→I }

MINIMIZE algorithm will consider the set EF(A) and conclude that neither

A B nor BA holds.

However, when working with the set EF(AD) the MINIMIZE algorithm will discover that BDAD and AD BD. Therefore the rules

{AD→E, BD→I} will be re-written as {AD→EI}.

This contraction will reduce the set F to a minimum cover

F2 = { A→BC, B→A, AD→EI }


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David Maier, "The Theory of Relational Databases", Comp. Sc. Press,

1983.

H. Mannila, K. Raiha, “The Design of Relational Databases”. Addison-

Wesley, 1992.

Elmasri, Navathe, “Fundamentals of Database Systems”, 5th Ed.

Addisson-Wesley, 2007.

Silberschatz, Korth, Surdarshan, “Database System Concepts”, McGraw-

Hill, 2011

Where to get more information?

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Fd covers

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