Functional Dependencies. FarkasCSCE 5202 Reading and Exercises Database Systems- The Complete Book: Chapter 3.1, 3.2, 3.3., 3.4 Following lecture slides.

Functional Dependencies

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Reading and Exercises

Database Systems- The Complete Book: Chapter 3.1, 3.2, 3.3., 3.4

Following lecture slides are modified from Jeff Ullman’s slides for Fall 2002 -- Stanford

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Database Design Goal:

Represent domain information Avoid anomalies Avoid redundancy

Anomalies: Update: not all occurrences of a fact are

changed Deletion: valid fact is lost when tuple is deleted

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Functional Dependencies FD: X A for relation R

X functional determines A, i.e., if any two tuples in R agree on attributes X, they must also agree on attribute A. X: set of attributes A: single attribute

If t1 and t2 are two tuples of r over R and t1[X]= t2[X] then t1[A]= t2[A]

What is the relation between functional dependencies and primary keys?

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Functional Dependency Example

Owner(Name, Phone) FD: Name Phone

Dog(Name, Breed, Age, Weight) FD: Name, Breed Age FD: Name, Breed Weight

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Example - FD

Name Breed Age Weight Date Kennel

Pepper G.S. 1 70 01/01/02 White Oak

Buddy Mix 4 50 03/04/01 Little Creek

Pepper G.S. 1 70 04/17/02 Little Creek

Panka Vizsla 12 40 02/14/02 White Oak

Functional Dependencies: Name,Breed AgeName,Breed Weight

Dog-Kennels(Name,Breed,Age,Weight,Date,Kennel)

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FD with Multiple Attributes

Right side: can be more than 1 attribute – splitting/combining rule E.g., FD: Name, Breed Age

FD: Name, Breed Weightcombine into:

FD: Name, Breed Age,Weight

Left side cannot be decomposed!

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FD EquivalenceLet S and T denote two sets of FDs. S and T are equivalent if the set of relation instances

satisfying S is exactly the same as the set of instances satisfying T.

A set of FDs S follows from a set of FDs T if every relation instance that satisfies all FDs in T also satisfies all FDs in S.

Two sets of FDs S and T are equivalent if S follows from T and T follows from S.

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Trivial FD

Given FD of the form A1,A2,…,AnB1,B2,…,Bk

FD is Trivial: if the Bs are subset of As Nontrivial: if at least one of the Bs is

not among As Completely nontrivial: if none of the

Bs is in As.

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Keys and FD

K is a (primary) key for a relation R if1. K functionally determines all attributes in R2. 1 does not hold for any proper subset of K

Superkey: 1 holds, 2 does not hold

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Example Dog-Kennels(Name,Breed,Age,Weight,Date,Kennel) Name,Breed,Date is a key:

K={Name,Breed,Date} functionally determines all other attributes

The above does not hold for any proper subset of K What are?

{Name,Breed,Kennel} {Name,Breed,Date,Kennel} {Name,Breed,Age} {Name,Breed,Age,Date}

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Where do Keys Come From?

Assert a key K, then only FDs areK A for all attributes A (K is the only key from FDs)

Assert FDs and deduce the keysE/R gives FDs from entity set keys and many-one relationships

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E/R and Relational Keys

E/R keys: properties of entities Relation keys: properties of tuples Usually: one tuple corresponds to one

entity Poor relational design: one entity

becomes several tuples

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Closure of Attributes

Let A1,A2,…,An be a set of attributes and S a set of FDs. The closure of A1,A2,…,An under S is the set of attributes B such that every relation that satisfies S also satisfies A1,A2,…,An B.

Closure of attributes A1,A2,…,An is denoted as {A1,A2,…,An}+

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Algorithm – Attribute Closure

Let X = A1,A2,…,An

Find B1,B2,…,Bk C such that B1,B2,…,Bk all in X but C is not in X

Add C to X Repeat until no more attribute can be

added to X X= {A1,A2,…,An}+

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Closures and Keys

{A1,A2,…,An}+ is a set of all attributes of a relation if and only if A1,A2,…,An is a superkey for the relation.

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Projecting FDs

Some FD are physical laws E.g., no two courses can meet in the

same room at the same time A professor cannot be at two places at

the same time.

How to determine what FDs hold on a projection of a relation?

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FD on Relation Relation schema design: which FDs

hold on relation Given: X1 A1, X2 A2, …, Xn An

whether Y B must hold on relations satisfying X1 A1, X2 A2, …, Xn An

Example: A B and B C, then A C must also hold

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Inference Test

Test whether Y B Assume two tuples agree on attributes Y Use FDs to infer these tuples also agree

on other attributes If B is one of the “other” attributes, then

Y B holds.

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Armstrong Axioms Reflexivity

If {A1,A2,…,Am} superset of {B1,B2,…,Bn} then A1,A2,…,Am B1,B2,…,Bn

Augmentation If A1,A2,…,Am B1,B2,…,Bn then A1,A2,…,Am,C1,

…,Ck B1,B2,…,Bn,C1,…,Ck

Transitivity If A1,A2,…,Am B1,B2,…,Bn and B1,B2,…,Bn

C1,C2,…,Ck then A1,A2,…,Am C1,C2,…,Ck

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FD Closure

Compute the closure of Y, denoted as Y+

Basis: Y+ = Y Induction: look FD, where left side X is

subset of Y+ . If FD is X A then add A to Y+ .

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Finding All FDs

Normalization: break a relation schema into two or more schemas

Example: R(A,B,C,D) FD: AB C, C D, D A Decompose into (A,B,C), (A,D)

FDs of (ABC): A,B C and C A

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Basic Idea

What FDs hold on a projections: Start with FDs Find all FDs that follow from given ones

Restrict to FDs that involve only attributes from a schema

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Algorithm

For each X compute X+ Add X A for all A in X+ - X Drop XY A if X A Use FD of projected attributes

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Tricks

Do not compute closure of empty set of the set of all attributes

If X+ = all attributes, do not compute closure of the superset of X

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Example ABC with A B, B C and projection

on AC: A+ = ABC, yields A B and A C B+ = BC, yields B C C+ = C, yields nothing BC+ = BC, yields nothing Resulting FDs: AB, AC, BC Projection AC: A C