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Functional Dependencies

Why FD'sMeaning of FD’s

Keys and SuperkeysInferring FD’s

Source: slides by Jeffrey Ullman

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Improving Relation Schemas

Set of relation schemas obtained by translating from an E/R diagram might need improvement modeling with E/R diagrams is an art, not a

science; relies on experience and intuition multiple alternative designs are possible, how to

choose? Problems caused by redundant storage of info

wasted space anomalies when updating, inserting or deleting

tuples Basic idea: replace a relation schema with a

collection of "smaller" schemas. Called decomposition

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Improving Relation Schemas

There is a theory for systematically guiding the improvement of relational designs, called normalization

Normalization uses the notion of "constraints" on the information functional dependencies multi-valued dependencies referential integrity constraints

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Functional Dependencies

X -> A is an assertion about a relation R that whenever two tuples of R agree on all the attributes of X, then they must also agree on the attribute A. Say “X -> A holds in R.” Convention: …, X, Y, Z represent sets of

attributes; A, B, C,… represent single attributes.

Convention: no set formers in sets of attributes, just ABC, rather than {A,B,C }.

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Example

Consumers(name, addr, candiesLiked, manf, favCandy)

Reasonable FD’s to assert: name -> addr name -> favCandy candiesLiked -> manf

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Example Data

name addr candiesLiked manf favCandyJaneway Voyager Twizzlers Hershey SmartiesJaneway Voyager Smarties Nestle SmartiesSpock Enterprise Twizzlers Hershey Twizzlers

Because name -> addr Because name -> favCandy

Because candiesLiked -> manf

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FD’s With Multiple Attributes

No need for FD’s with > 1 attribute on right. But sometimes convenient to combine

FD’s as a shorthand. Example: name -> addr and

name -> favCandy become name -> addr favCandy

> 1 attribute on left may be essential. Example: store candy -> price

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Keys of Relations

K is a superkey for relation R if K functionally determines all of R.

K is a key for R if K is a superkey, but no proper subset of K is a superkey.

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Example

Consumers(name, addr, candiesLiked, manf, favCandy)

{name, candiesLiked} is a superkey because together these attributes determine all the other attributes. name -> addr favCandy candiesLiked -> manf

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Example, Cont.

{name, candiesLiked} is a key because neither {name} nor {candiesLiked} is a superkey. name doesn’t -> manf; candiesLiked

doesn’t -> addr. There are no other keys, but lots of

superkeys. Any superset of {name, candiesLiked}.

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E/R and Relational Keys

Keys in E/R concern entities. Keys in relations concern tuples. Usually, one tuple corresponds to

one entity, so the ideas are the same.

But --- in poor relational designs, one entity can become several tuples, so E/R keys and Relational keys are different.

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Example Data

name addr candiesLiked manf favCandyJaneway Voyager Twizzlers Hershey SmartiesJaneway Voyager Smarties Nestle SmartiesSpock Enterprise Twizzlers Hershey Twizzlers

Relational key = {name candiesLiked}But in E/R, name is a key for Consumers, and candiesLikedis a key for Candies.Note: 2 tuples for Janeway entity and 2 tuples for Twizzlersentity.

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Discovering Keys

Suppose schema was obtained from an E/R diagram.

If relation R came from an entity set E, then key for R is the keys of E

If R came from a binary relationship from E1 to E2:

many-many: key for R is the keys of E1 and E2 many-one: key for R is the keys for E1 (but not

the keys for E2) one-one: key for R is the keys for E1; another

key for R is the keys for E2

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Key Example

Con-sumers

CandiesLikes

Likes(consumer, candy)Favorite

Favorite(consumer, candy)

Married

husband

wife

Married(husband, wife)

name addr name manf

Buddies

1 2

Buddies(name1, name2)

key: consumer candy

key: name1 name2

key: consumer

keys: husband or wife

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More FD’s From Application

Example: “no two courses can meet in the same room at the same time” tells us: hour room -> course.

Ultimately, FD's and keys come from the semantics of the application.

FD's and keys apply to the schema, not to specific instantiations of the relations

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Manipulating FD's

Need to be able to reason about FD's in order to support the normalization process (improving relational schemas)

Some preliminaries: can split FD X -> A B into X -> A, X -> B can combine FD's X -> A, X -> B into FD X ->

A B Since A -> A is trivially true, no point in

having any attribute on the RHS that is also on the LHS

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Closure of a Set of FD's

If we have FD's A -> B and B -> C, then it is also true that A -> C. Ex: If name -> address and address -

> phone, then name -> phone. What about a chain of such

deductions? Called closure

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FD Closure Algorithm

Input: a set of attributes {A1,…,An} and a set of FD's S

1. Z := {A1,…,An}2. find an FD in S of the form X -> C such

that all the attributes in X are in Z but C is not in Z. Add C to Z

3. repeat step 2 until there is nothing more that can be put in Z

4. return Z as the closure of {A1,…,An}

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Example of Closure Algorithm

Given relation with attributes A, B, C, D, E, F and FD's AB -> C BC -> A, BC -> D D -> E CF -> B

Compute closure of {A,B}. Answer:

Z := {A,B} add C add A and D add E final answer is Z = {A,B,C,D,E}

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Use of Closure Algorithm

Now we can check if a particular FD A1 … An -> B follows from a set of FD's S: compute {A1,…,An}+ using S if B is in the closure, then the FD

follows otherwise it does not

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Example

Given same set of attributes and FD's as in previous example:

Does AB -> D follow? compute {A,B}+ = {A,B,C,D,E}.

Since D is in {A,B}+, YES. Does D -> A follow?

compute {D}+ = {D,E}. Since A is not in {D}+, NO.

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Finding All Implied FD’s Motivation: “normalization,” the

process where we break a relation schema into two or more schemas.

Example: ABCD with FD’s AB ->C, C ->D, and D ->A. Decompose into ABC, AD. What FD’s

hold in ABC ? Not only AB ->C, but also C ->A !

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Why?

a1b1cABC

ABCD

a2b2c

Thus, tuples in the projectionwith equal C’s have equal A’s;C -> A.

a1b1cd1 a2b2cd2

comesfrom

d1=d2 becauseC -> D

a1=a2 becauseD -> A

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Basic Idea

Start with given FD’s and find all nontrivial FD’s that follow from the given FD’s.

Nontrivial = left and right sides disjoint.

Restrict to those FD’s that involve only attributes of the projected schema.

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Simple, Exponential Algorithm

For each set of attributes X, compute X +.

Add X ->A for all A in X + - X. However, drop XY ->A whenever we

discover X ->A. Because XY ->A follows from X ->A in

any projection. Finally, use only FD’s involving

projected attributes.

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A Few Tricks

No need to compute the closure of the empty set or of the set of all attributes.

If we find X + = all attributes, so is the closure of any superset of X.

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Example

ABC with FD’s A ->B and B ->C. Project onto AC. A +=ABC ; yields A ->B, A ->C.

• We do not need to compute AB + or AC +.

B +=BC ; yields B ->C. C +=C ; yields nothing. BC +=BC ; yields nothing.

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Example --- Continued

Resulting FD’s: A ->B, A ->C, and B ->C.

Projection onto AC : A ->C. Only FD that involves a subset of

{A,C }.

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A Geometric View of FD’s

Imagine the set of all instances of a particular relation.

That is, all finite sets of tuples that have the proper number of components.

Each instance is a point in this space.

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Example: R(A,B)

{(1,2), (3,4)}

{}

{(1,2), (3,4), (1,3)}

{(5,1)}

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An FD is a Subset of Instances

For each FD X -> A there is a subset of all instances that satisfy the FD.

We can represent an FD by a region in the space.

Trivial FD = an FD that is represented by the entire space.

Example: A -> A.

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Example: A -> B for R(A,B)

{(1,2), (3,4)}

{}

{(1,2), (3,4), (1,3)}

{(5,1)}A -> B

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Representing Sets of FD’s

If each FD is a set of relation instances, then a collection of FD’s corresponds to the intersection of those sets. Intersection = all instances that

satisfy all of the FD’s.

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Example

A->BB->C

CD->A

Instances satisfyingA->B, B->C, andCD->A

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Implication of FD’s

If an FD Y -> B follows from FD’s X1 -> A1,…,Xn -> An , then the region in the space of instances for Y -> B must include the intersection of the regions for the FD’s Xi -> Ai . That is, every instance satisfying all

the FD’s Xi -> Ai surely satisfies Y -> B. But an instance could satisfy Y -> B,

yet not be in this intersection.

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Example

A->B B->CA->C