Dbms3

CS 222
Database Management System
Spring 2010-11
Lecture 4
Database Design Theory
Korra Sathya Babu
Department of Computer Science
NIT Rourkela

• 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
3/18/2022 Database Design 2
Unit Overview

Modeling
Movies Stars-In Stars
length filmType
title year
Owns
Studios
name address
name
address

• 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

• Redundancy
• Insertion Anomaly
• Updation Anomaly
• Deletion Anomaly
Problems using bad design

• Definition
• A functional dependency (FD) has the form XY, where X
and Y are sets of attributes in a relation R. Formally, XY
means 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 A1
A2
…An
 B is said to be trivial if B is one of
the A’s. ex. title year  title
• Nontrivial: atleast one of the B’s not among A’s. ex. title
year  year length
• Completely nontrivial: none of the B’s are part of A’s. ex.
title year  length
Functional Dependency

• An FD A1 A2 . . .An -> B1 B2 . . .Bm is trivial if the B’s are a
subset of the A’s {B1,B2, . . . Bn} subset {A1,A2, . . . An}
• It’s non-trivial if at least one B is not among the A’s, i.e.,
{B1,B2, ... Bn} − {A1,A2, ...An} ≠ Ø
• It’s completely non-trivial if none of the B’s are among the A’s,
{B1,B2, ... Bn} Intersect {A1,A2, ...An} = Ø
• Trivial dependency rule: The FD A1 A2 . . .An  B1 B2 . . .Bm
is equivalent to the FD A1 A2. . .An  C1 C2 . . . Ck , where
the C’s are those B’s that are not A’s, i.e., {C1, C2, . ., Ck} =
{B1,B2, . ,Bm} − {A1,A2,.. ,An}
Trivial Dependency

• 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+

• 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 xy, xz
Augument x to xy and y to xz
 xx  xy ; xy  yz
 x  xy ; xy  yz
 x  yz [using transitive Axiom; x  xy, xy  yz]

• 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 xy, ywz
Augument w to xy
 xw  yw
 xw  z [using transitive Axiom; xw  yw, yw  z]
Lets prove from the back onwards
Assume xy and xz is given
Take x  y and augument with x
 xx  yx
We already have xz, So replace x with z in determinee
 xx  yx
 xyz

• Let F be the following set of functional dependencies:
{AB→CD, B→DE, C→F, E→G, A→B}. Use Armstrong’s axioms
to show that {A→FG} is logically implied by F
Logical Implications of FDs

• Used to determine if a relation R satisfies or doesn’t satisfy a
given FD: AB
• 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 other
Step 2: Check that tuples with equal values under A also have equal values
under attribute(s) B
Step 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: AB if the following holds for every pair of tuples t1 and t2
in R, if t1.A = t2.A then t1.B = t2.B

• Given a set F of FDs, a FD AB of F is said to be redundant
w.r.t the FDs of F if and only if AB can be derived from the
set of FDs F-{AB}
• 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

• Assume F is a set of FDs with AB Є F
Redundant FDs
The Membership Algorithm:
Step 1: Remove temporarily AB from F and initialize the set of FDs G to F. ie. Set
G=F-{AB}. If G ≠ Ø proceed to step 2; otherwise stop executing the algorithm
since AB 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 XY 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 Ti
and 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:AB is
redundant. If not all attributes of B are members of Ti + 1 , remove XY from G
and repeat step 3
Step 3a: If G= Ø or there are no FDs in G that have all the attributes of its
determinant in the current Ti then AB is not redundant

• 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={X Y, Y X, Y Z, Z
Y, X Z, Z X} using the Membership algorithm
• Find the redundant FDs in the set F={X YZ, ZW P, P
Z, W XPQ, XYQ YW, WQ YZ}
Redundant FDs

• 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 →Z
XZ →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
+

• 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={X YZ, ZW P, P Z, W XPQ, XYQ YW,
WQ YZ}. Find the Closure of all the single attributes
Attribute Closure

• 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

• Given a relation R(ABC) and FDs set F={AB C, 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

• Given two sets F and G of FDs defined over same
relational schema
• A set of FD’s S ‘follows’ from a set of FD’s T if every
relation instance that satisfies all the FD’s in T also
satisfies all the FD’s in S
• A  C follows from T = {A B, B  C}.
• Two sets of FD’s S and T are ‘equivalent’ if and only
if S follows from T, and T follows from S
• S = {A B,B  C,AC} 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

• 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

The non-redundant cover algorithm
The Non-Redundant Cover Algorithm:
Step 1: Initialize G to F. i.e. set G=F
Step 2: Test every FD of G for redundancy using the Membership
Algorithm until there are no more FDs of G to be tested
Step 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
Problem
Find the non-redundant cover G for the set
F={X YZ, ZW P, P Z, W XPQ, XYQ YW, WQ YZ}

• 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

Left Reduced Algorithm
The Left-Reduced Algorithm:
Step 1: Initialize a set of FDs to F. i.e. set G=F
Step 2: For every A1
,A2
,…,Ai
,…,An
 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 3
Step 3: For each attribute A
i
in 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 2
Step 4: Test if all attributes of Y (the RHS of the FD under consideration)
are elements of the closure of A1
,A2
,…,An
(notice that we have removed
attribute A
i
from 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 A1
,A2
,…,An
then 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 T

• Remove any extraneous left attributes from
F={ABC, EC, DAEF, ABFBD}
• Reduce the set F={XZ, XYWP, XYZWQ, XZR}
by removing extraneous left attribute
• Reduce the set F={XWY, XWZ, ZY, XYZ} by
removing extraneous left attribute
Extraneous Attribute
Tip :
 There is no need to consider FDs with determinant that consist of single
attribute

• 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

• 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 XA in F with a dependency
YA, 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

• 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=(ABCD,AB  E, E  AB }.
Notice that G is reduced and minimum, but not
optimal.
Optimal Cover

• Find canonical cover of
1. F={XZ, XYWP, XYZWQ, XZR}
2. F={XYW, XWZ, ZY, XYZ}
3. F={ABC, EC, DAEF, ABFBD}
4. G = {A  C, A B  C, C DI, CD I, EC AB, EICC}
• Find minimal cover of the following FD sets
1. F={ABC, ABC, BC, AB}
2. F={AB, BA, BC, AC, CA}
3. F={ABC, AB, BA}
4. G = {ABCD  E, E  D, A  B, AC  D}
5. F={ABC, BAC, C  AB}
6. G={ABDC, CBE, AD  BF, BF}
Problems

• Compound Functional Dependency (CFD) and
Annular Covers
Seminar change?

• 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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