CS 222 Database Management System

Spring 2010-11

Lecture 4 Database Design Theory

Korra Sathya BabuDepartment of Computer Science

NIT Rourkela

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Database Design Problem Redundancy and Anomaly

Functional Dependency Axioms, Logical Implications of FDs, Redundant FDs, Closure, Equivalence

of FDS, extraneous attributes, covers

Decomposition Rules of Decomposition, Test for Lossless Join and Dependency

Preservation

Normalization Normal Forms, Multivalued Dependency, Join Dependency,

Denormalization

Unit Overview

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Modeling

Movies Stars-In Stars

length filmType

title year

OwnsStudios

name address

name

address

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Automatic mappings from E/R to relations may not produce the best relational design possible

Suggested Design Strategy Real-world to E/R model to Relational schema to Better relational schema

to Relational DBMS

Database designers sometimes go directly from Real-world to Relational schema, in which case the relational design could be really bad.

Many problems may arise if the design is not careful

Relational Database Design

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Redundancy

Insertion Anomaly

Updation Anomaly

Deletion Anomaly

Problems using bad design

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Definition A functional dependency (FD) has the form XY, where X

and Y are sets of attributes in a relation R. Formally, XYmeans that whenever two tuples in R agree on all the attributes of X, they must also agree on all the attributes of Y.

Movies (title, year, length, filmtype,studioName, starName)

FDs we can reasonable assert are: Title, year length ; Title, year filmType; Title, year studioName

Trivial Dependency

Trivial: A fd A1A

2A

n B is said to be trivial if B is one of

the As. ex. title year title

Nontrivial: atleast one of the Bs not among As. ex. title year year length

Completely nontrivial: none of the Bs are part of As. ex. title year length

Functional Dependency

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An FD A1 A2 . . .An -> B1 B2 . . .Bm is trivial if the Bs are a subset of the As {B1,B2, . . . Bn} subset {A1,A2, . . . An}

Its non-trivial if at least one B is not among the As, i.e., {B1,B2, ... Bn} {A1,A2, ...An}

Its completely non-trivial if none of the Bs are among the As, {B1,B2, ... Bn} Intersect {A1,A2, ...An} =

Trivial dependency rule: The FD A1 A2 . . .An B1 B2 . . .Bm is equivalent to the FD A1 A2. . .An C1 C2 . . . Ck , where the Cs are those Bs that are not As, i.e., {C1, C2, . ., Ck} = {B1,B2, . ,Bm} {A1,A2,.. ,An}

Trivial Dependency

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Reflexivity

(If X Y, then X Y)

Augumentation (If X Y, then XZ YZ for any Z)

Transitivity (If X Y and Y Z, then X Z)

Armstrong Axioms

Armstrong axioms are sound and complete

Sound They generate only FDs in F+ when applied to a set of FDs

Complete They when repeatedly applied, these rules will generate all

FDs in F+

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Finding the closure of an FD Set may be tedious. So more rules may be derived from Armstrong Axioms

The Union Rule, Pseudotransitive Rule and Decompostion Rule are Sound but not complete

Union Rule (If X Y and X Z, then X YZ)

More Inference Axioms

Given xy, xzAugument x to xy and y to xz

xx xy ; xy yz x xy ; xy yz x yz [using transitive Axiom; x xy, xy yz]

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Pseudotransitive Rule (If X Y and YW Z, then XW Z)

Decomposition Rule (If X YZ then X Y and then X Z)

More Inference Axioms

Given xy, ywzAugument w to xy

xw yw xw z [using transitive Axiom; xw yw, yw z]

Lets prove from the back onwardsAssume xy and xz is givenTake x y and augument with x

xx yx We already have xz, So replace x with z in determinee xx yx xyz

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Let F be the following set of functional dependencies: {ABCD, BDE, CF, EG, AB}. Use Armstrongs axioms to show that {AFG} is logically implied by F

Logical Implications of FDs

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Used to determine if a relation R satisfies or doesnt satisfy a given FD: AB

Input: Relation R and an FD: A B

Output: TRUE if R satisfies A B, otherwise FALSE

The Satisfies Algorithm

The Satisfies Algorithm:Step 1: Sort the tuples of the relation R on the attribute(s) A (determinant) so that tuples with equal values under A are next to each otherStep 2: Check that tuples with equal values under A also have equal values under attribute(s) BStep 3: If any two tuples of R meet condition 1 but fail to meet condition 2 the output of the algorithm is FALSE. Otherwise, the relation satisfies the Functional Dependency and the output of the algorithm is TRUE

In short the satisfies algorithm can be stated as: The relation R satisfies the FD: AB if the following holds for every pair of tuples t1 and t2in R, if t1.A = t2.A then t1.B = t2.B

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Given a set F of FDs, a FD AB of F is said to be redundantw.r.t the FDs of F if and only if AB can be derived from the set of FDs F-{AB}

Eliminating Redundant FDs allows us to minimize the set of FDs

Membership Algorithm helps to determine the Redundant FDs.

Input : A set F of FDs and a particular FD of F that is being tested

Output: FD is Redundant or not

Redundant FDs

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Assume F is a set of FDs with AB F

Redundant FDs

The Membership Algorithm:Step 1: Remove temporarily AB from F and initialize the set of FDs G to F. ie. Set G=F-{AB}. If G proceed to step 2; otherwise stop executing the algorithm since AB is non redundant

Step 2: Initialize the set of attributes Ti (with i=1) with the set of attribute(s) A(the determinant of the FD under consideration). ie. Set Ti = T1 = {A}. The set T1 is the current Ti

Step 3: In the set G search for FDs XY such that all the attributes of the determinant X are elements of the current set Ti. There are two possible outcomes

Step 3a: If such FD is found, add the attribute of Y (right hand side of FD) to set Tiand form a new Set Ti + 1= Ti U Y. The Set Ti + 1 is the current Ti . Check if all the attributes of B (the right hand side of FD under consideration) are members of Ti + 1. If this is the case, stop executing algorithm becos the FD:AB is redundant. If not all attributes of B are members of Ti + 1 , remove XY from G and repeat step 3Step 3a: If G= or there are no FDs in G that have all the attributes of its determinant in the current Ti then AB is not redundant

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Given the set F={x YW, XW Z, Z Y, XY Z}. Determine if the FD XY Z is redundant in F

Eliminate redundant FDs from F={XY, Y X, Y Z, Z Y, X Z, Z X} using the Membership algorithm

Find the redundant FDs in the set F={XYZ, ZW P, P Z, W XPQ, XYQ YW, WQ YZ}

Redundant FDs

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Definition

The set of all FDs implied by a given set F of FDs is called the closure of F, and denoted as F+

Armstrong Axioms can be applied repeatedly to infer all FDs implied by a set F of FDs

Closure of FD Set

Given R = ABCD and F = {A B, A C, CD A}. Compute F

+.

A+

F={ABC}

B+

F={B}

AB

+

F={ABC}

AC+

F={ABC}

ABC

+

F=

Given R = XYZ and F = {XY Z}. Compute F

+.

F+= {X X, Y Y, Z Z,

XY X, XY Y, XY XY, XY ZXZ X, XZ Z, XZ XZ, YZ Y, YZ Z, YZ YY, XYZ X, XYZ Y, XYZ Z,XYZ XY, XYZ XZ, XYZ YZ,XYZ XYZ,}

Consider a relation with schema R(A,B,C,D) and FD's F={AB C, C D, D A} Compute F

+

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Finding all the attributes in the relation that the current attribute can determine by using inference axioms and given FD set. Its denoted by {A}+

Given FDs set F={XYZ, ZW P, P Z, W XPQ, XYQ YW, WQ YZ}. Find the Closure of all the single attributes

Attribute Closure

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A unique minimal set of attribute(s) that determine the set of other attributes in a relation

Two properties of key are unique and minimalism

A superkey is a set of attributes that has the uniqueness property but is not necessarily minimal

If a relation has multiple keys, specify one to be the primary key

Convention: in a relational schema, underline the attributes of the primary key.

If a key has only one attribute A, we say

that A rather than {A} is a key.

Candidate Key

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Given a relation R(ABC) and FDs set F={ABC, B D, D B}. Find the candidate keys of the relation

Given a relation R(XYZWP) and FDs set F={Y Z, Z Y, Z W, Y P}. Find the number of candidate keys

Consider a schema R={S,T,V,C,P,D} and F= {S T, V SC, SD P}. Find keys for R

Given a relation R(XYZWP) and FDs set F={x Z, YZ W, Z Y}. Find the number of candidate keys

Candidate Key

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Given two sets F and G of FDs defined over same relational schema

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

A C follows from T = {A B, B C}.

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

S = {A B,B C,AC} and T={A B, B C} are equivalent

These notions are useful in deriving new FDs from a given set of FDs

Equivalence of set of FDs

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Two sets of FDs F and G defined over same relation schema are equivalent if

every FD in F can be inferred from G and

every FD in G can be inferred from F

F Covers G if every FD in G can be inferred from F (ie if G+ is subset of F+)

F and G are equivalent if F covers G and G covers F

If G covers F and no proper subset H of G exist such that H+ = G+ we say G is a non-redundant cover of F

Equivalence of set of FDs

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The non-redundant cover algorithm

The Non-Redundant Cover Algorithm:Step 1: Initialize G to F. i.e. set G=FStep 2: Test every FD of G for redundancy using the Membership Algorithm until there are no more FDs of G to be testedStep 3: The set G is a non-redundant cover of F

Note:Given a set F, there may be more than one non-redundant cover since the order in which the FDs are considered is irrelevant

ProblemFind the non-redundant cover G for the setF={X YZ, ZW P, P Z, W XPQ, XYQ YW, WQ YZ}

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Definition: A is extraneous in X Y if A can be removed from

the left side or right side of X Y without changing the closure of F. ex. Let G = {A B C, B C, A B D }

Attribute C is extraneous in the right side of A B C and attribute B is extraneous in the left side of A B D

The set G = {A B C, B C, A B D } is neither left-reduced nor right-reduced. G1 = {A B C, B C, A D } is left-reduced but not right-reduced, while G2 = (A B. B C, A B D} is right-reduced but not left-reduced. The set G3 = {A B, B C, A D} is left and right-reduced, hence reduced

One need to eliminate the extraneous attribute

Extraneous Attribute

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Left Reduced Algorithm

The Left-Reduced Algorithm:Step 1: Initialize a set of FDs to F. i.e. set G=FStep 2: For every A

1,A

2,,A

i,,A

n Y in G do step 3 until there are no

more FDS in G to which this step can be applied. The algorithm stops when all FDs of G have executed step 3Step 3: For each attribute A

iin the determinant of FD selected in the

previous step do step 4 until all attributes have been tested. After finishing testing of all attributes of a particular FD repeat step 2Step 4: Test if all attributes of Y (the RHS of the FD under consideration) are elements of the closure of A

1,A

2,,A

n(notice that we have removed

attribute Aifrom the determinant of the FD) with respect to the FDs of G. If

this is the case remove attribute Ai from the determinant of the FD undergoing testing becos Ai is an extraneous left attribute. If not all attributes of Y are elements of the closure of A

1,A

2,,A

nthen attribute Ai is

not an extraneous left attribute and should remain in the determinant of the FD under consideration

When the algorithm finishes the set G contains a left-reduced cover set of T25

Remove any extraneous left attributes from F={ABC, EC, DAEF, ABFBD}

Reduce the set F={XZ, XYWP, XYZWQ, XZR} by removing extraneous left attribute

Reduce the set F={XWY, XWZ, ZY, XYZ} by removing extraneous left attribute

Extraneous Attribute

Tip : There is no need to consider FDs with determinant that consist of single

attribute

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A set of FDs F is canonical if every FD in F is of the form X A and F is left-reduced and non-redundant

Since a canonical set of FDs is non-redundant and every FD has a single attribute on the right side, it is right-reduced. Since it is also left-reduced, it is reduced

Example: The set F = {A B, A C, A D, A E, B I J} is a canonical cover for G = {A B C E, A B D E, B I J}

Canonical Cover

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A set of FDs is minimal if it satisfies the following conditions every dependency in F has a single attribute for its RHS

we cannot remove any dependency from F and have a set of dependencies that is equivalent to F

we cannot replace any dependency XA in F with a dependency YA, where Y proper-subset-of X (Y subset-of X) and still have a set of dependencies that is equivalent to F

Every set of FDs has an equivalent minimal set

There can be several equivalent minimal sets

There is no simple algorithm for computing a minimal set of FDs that is equivalent to a set F of FDs

To synthesize a set of relations, we assume that we start with a set of dependencies that is minimal set

Minimal Cover

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We have been measuring our covers in terms of the number of FDs they contain. We can also measure them by the number of attribute symbols required to express them. example. (A B C, CD E, A C IJ> has size 10 under this measure

Defiition: A set of FDs F is optimal if there is no equivalent set of FDs with fewer attribute symbols than F

The set F = {EC D, AB E, E AB ) is an optimal cover for G=(ABCD,AB E, E AB }. Notice that G is reduced and minimum, but not optimal.

Optimal Cover

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Find canonical cover of 1. F={XZ, XYWP, XYZWQ, XZR}

2. F={XYW, XWZ, ZY, XYZ}

3. F={ABC, EC, DAEF, ABFBD}

4. G = {A C, A B C, C DI, CD I, EC AB, EICC}

Find minimal cover of the following FD sets1. F={ABC, ABC, BC, AB}

2. F={AB, BA, BC, AC, CA}

3. F={ABC, AB, BA}

4. G = {ABCD E, E D, A B, AC D}

5. F={ABC, BAC, C AB}

6. G={ABDC, CBE, AD BF, BF}

Problems

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Compound Functional Dependency (CFD) and Annular Covers

Seminar change?

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Good Design needs strategy

Armstrong Axioms are sound and complete

FDs are constraints

There may be a number of equivalent FD sets

FD sets may be minimized by checking the coverage

Summary

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