functional dependency in dbms

8/13/2019 Functional Dependency in DBMS

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FUNCTIONAL D EPENDENCIES AND S CHEMA R EFINEMENT

Science is the knowledge ofconsequences, and dependenceof one fact upon another.

Thomas Hobbes(1588-1679 )


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R EVIEW: D ATABASE DESIGN

Requirements Analysis user needs; what must database do?

Conceptual Design

high level descr (often done w/ER model)Logical Design

translate ER into DBMS data model

Schema Refinement consistency,normalizationPhysical Design - indexes, disk layoutSecurity Design - who accesses what


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FUNCTIONAL DEPENDENCIES (FDS)


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A functional dependency X Y holds overrelation schema R if, for every allowableinstance r of R:

t1 r, t2 r, p X (t1 ) = p X (t2 )implies p Y (t1 ) = p Y (t2 )(where t1 and t2 are tuples;X and Y are sets of attributes)

In other words: X Y meansGiven any two tuples in r , if the X valuesare the same, then the Y values must alsobe the same. (but not vice versa)


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EXAMPLE: CONSTRAINTS ON ENTITY SET

Consider relation obtained from Hourly_Emps:Hourly_Emps ( ssn, name, lot, rating, wage_per_hr ,hrs_per_wk )

We sometimes denote a relation schema by listing the

attributes: e.g., SNLRWHThis is really the set of attributes {S,N,L,R,W,H}.Sometimes, we refer to the set of all attributes of arelation by using the relation name. e.g.,

Hourly_Emps

for SNLRWH

What are some FDs on Hourly_Emps (Given)?ssn is the key: S SNLRWHrating determines wage_per_hr : R W

lot determines lot : L L (

trivial

dependnency)


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F UNCTIONAL D EPENDENCIES E XAMPLE

For Student(name, matric, dob)

we write:{matric} {name, dob}

If two tuples in the relation student have the same matricnumber then they must be the same tuple ( no duplicate )


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F UNCTIONAL D EPENDENCIES E XAMPLE

For Student(name, matric, dob)

we write:{matric, dob} {name}


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F UNCTIONAL D EPENDENCIES T YPE

Trivial FDsNon-Trivial FDsCompletely Non-Trivial FDs


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TRIVIAL FD S

X Y

Y X

Eg. {name, address} {name}


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N ON -T RIVIAL FD S

X Y Y X

{name, address} {dob} {name, address} {name, dob}


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COMPLETELY N ON -T RIVIAL FD S

X Y

Y X =

{name, address} {dob}


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S UPERKEYS

A set of attributes whose knowledge determinesthe value of the entire tuple is a superkey

Example:Student(NRIC, matric, name, dob, )

e.g.

superkey {matric, name, dob}superkey {NRIC, name}


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C ANDIDATE K EYS

A minimal set of attributes whose knowledgedetermines the value of the entire tuple is acandidate key

Example:Student(NRIC, matric, name, dob, )

e.g. candidate key {matric}e.g. candidate key {NRIC}


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P RIMARY K EYS

The designer chooses a minimal set of attributeswhose knowledge determines the value of theentire tuple to be the primary key


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REASONING ABOUT F UNCTIONAL DEPENDENCIES

It is sometimes possible to infer new functionaldependencies from a set of given functionaldependencies


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REASONING ABOUT F UNCTIONAL DEPENDENCIES

For example:

From

{ssn } {first_name}and

{ssn} {last_name}

We can infer{ssn } {first_name, last_name}


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A RMSTRONG S A XIOMS

Be X, Y, Z be subset of the relation scheme of arelation R

Reflexivity :If Y X, then X Y

Augmentation :If X Y , then X ZY Z

Transitivity :If XY and YZ, then XZ


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For a relation scheme:employee(name, office_number,telephone_number, age)

Reflexivity :{name} {name,office_number}, then{name,office_number }{name}


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Augmentation :{name} { office_number} , then{name,age }{office_number,age}


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Transitivity :{name}{ telephone_number} and{telephone_number }{office_number},then {name}{ office_number}


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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 deducedwith the help of the Armstrong axioms

How to:Prove the soundness?

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A RMSTRONG S A XIOMS Armstrongs axioms are sound

for example: TransitivityLet X, Y, Z be subsets of the relation R

If XY and YZ, then XZ

Proof:

if two tuples T1 and T2 of |R| are such that, for allattributes Ax in X, T1. Ax = T2.Ax,

since XY then for all Ay in Y, T1. Ay = T2.Ay since YZ and for all Ay in Y, T1. Ay = T2.Ay then

for all Az in Z, T1. Az = T2.Az


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Armstrongs axioms are sound

For example: Consider the scheme {name, room,tel} with the set of functional dependencies:

{{room} { tel}, {tel } {name}}

Then we can deduct that the following functionaldependency hold:

{room} {name}

Proof:By Transitivity


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Armstrongs axioms are sound

For example: Weak-AugmentationLet X, Y, Z be subsets of the relation R

If X Y , then X ZY

ProofIf X Y

(1) Then by Augmentation X ZY Z(2) And by Reflexivity Y Z Y because Y Y Z(3) Then by Transitivity of (1) and (2) we have X ZY

Q.E.D.


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CLOSURE OF A S ET OF F UNCTIONAL DEPENDENCIES : E XAMPLE

Consider the relation scheme R(A,B,C,D)

F = {{A} {B},{B,C} {D}}

F + = {{A} {A}, {B}{B}, {C}{C}, {D}{D},[],{A}{B}, {A,B}{B}, {A,D}{B,D},{ A,C}{B,C},{A,C,D}{B,C,D}, {{A} {A,B },{ A,B}{A,B},{A,D}{A,B,D}, {A,C}{A,B,C },{ A,C,D}{A,B,C,D}, {B,C} {D}, [], {A,C} {D},[]}


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E QUIVALENCE OF S ETS OF F UNCTIONAL DEPENDENCIES

Two sets of functional dependencies F and G areequivalent if and only if

F+ = G+


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Armstrongs axioms are complete


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BOGUS RULES

Disprove that if X and Y are sets of attributes ofa relational schema R, and the fd X Y holdsin R, then Y X also holds in R.

Disprove that if X, Y and Z are sets of attributesof a relational schema R, and the fds X Yand Y Z hold in R, then Z X also holds inR.

Disprove that if X, Y and Z are sets of attributesof a relational schema R, and the fd XY Zholds in R, then X YZ also holds in R.


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BOGUS RULES

Disprove that if X and Y are sets ofattributes of a relational schema R, and thefd X Y holds in R, then Y X also holdsin R.

Solution:Consider the following relation instance,where we use singletons for X and Y:

We see that X Y holds, but not Y XX Y

1 0

0 0


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BOGUS RULES

Disprove that if X, Y and Z are sets ofattributes of a relational schema R, and thefds X Y and Y Z hold in R, then Z Xalso holds in R.

Solution:Consider the following relation instance, wherewe use singletons for X, Y, and Z: We see thatboth X Y and Y Z hold But not Z X.

X Y Z

0 0 0

1 0 0


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BOGUS RULES

Disprove that if X, Y and Z are sets ofattributes of a relational schema R, andthe fd XY Z holds in R, then X YZalso holds in R.

Solution:Consider the following relationinstance where we use singletons for X,

Y, and Z: We see that XY Z holds,but not X YZ.

X Y Z

0 0 0

0 1 0


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F INDING K EYS : E XAMPLE

Example: Consider the relation schemeR(A,B,C,D)

with functional dependencies{ A}{C} and {B}{D}.

Is {A,B} a candidate key?


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F INDING K EYS : E XAMPLE

Example: {A,B} is a superkey.Indeed from Armstrongs Axioms we can infer:

{ A}{C} {A,B}{A,B,C} (augmentation by AB){B}{D} {A,B,C}{A,B,C,D} (augmentation by

A,B,C)We obtain {A,B}{A,B,C,D} (transitivity)


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CLOSURE OF A S ET OF A TTRIBUTES

For a set A of attributes,we call the closure of A (with respect to a set of

functional dependencies F) , noted A+, the maximum setof attributes such that AA+ (as a consequence of F)


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CLOSURE OF A S ET OF A TTRIBUTES :E XAMPLE

Consider the relation scheme R(A,B,C,D) withfunctional dependencies

{A }{C} and {B}{D }.

{A}+ = {A,C}

{B}+ = {B,D}

{A,B}+ = {A,B,C,D}


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CLOSURE OF A S ET OF A TTRIBUTES : A LGORITHM 1

Input:R a relation schemeF a set of functional dependenciesX R

Output:X+ the closure of X w.r.t. F


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CLOSURE OF A S ET OF A TTRIBUTES : A LGORITHM 1


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R = {A,B,C,D,E,G}F = { {A,B}{C}, {C}{A}, {B,C}{D },{ A,C,D}{B}, {D}{E,G}, {B,E}{C },{C,G}{B,D}, {C,E}{A,G}} X = {B,D}


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A TTRIBUTE CLOSURE ( EXAMPLE)

R = {A, B, C, D, E}F = { B CD, D E, B A, E C, AD B }Is B E in F + ?B+ = BB+ = BCDB+ = BCDAB+ = BCDAE Yes!

and B is a key for R too! Is D a key for R?D+ = DD+ = DED+ = DEC

Nope!

Is AD a key for R? AD+ = AD

AD+ = ABD and B is a key, so Yes!

Is AD a candidate keyfor R? A + = A

A not a key, nor is D so Yes! Is ADE a candidate key

for R?No! AD is a key, so ADE is a

superkey, but not a cand. key


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A TTRIBUTE CLOSURE ( EXAMPLE)


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E QUIVALENCE OF S ETS OF F UNCTIONAL DEPENDENCIES

Given two sets of functional dependencies F and G

Option 1Generate FD closures for both Set and Check, F+ = G+

Option 2 For each functional dependency YZ in F

Compute Y+ with respect to GCheck whether Z Y+If yes then YZ is in G +

And vice versa


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M INIMAL S ET OF D EPENDENCIES

A set of dependencies F is minimal if andonly if:

1. Every right-hand side is a single attribute

2. For no XA in F and proper subset Z of X is F {XA} {ZA} equivalent to F

3. For no X A in F is the set F {XA } equivalent toF


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M INIMAL C OVER : E XAMPLE

F = { {A,B}{C}, {C}{A}, {B,C}{D }, { A,C,D}{B},{D}{E,G}, {B,E}{C }, {C,G}{B,D}, {C,E}{A,G}}


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M INIMAL C OVER : E XAMPLE (1)


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functional dependency in dbms

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