Normalization
Reading Assignments

Database Systems The Complete Book: Chapters 3.6, 3.7,
3.8

Following lecture slides are modified from Jeff Ullman’s
slides for Fall 2002 -- Stanford
Farkas
CSCE 520
2
Closures
 Closure of attributes (A+): find keys
 Closure of FDs (S+): projection of FDs
to decompositions of schema
 Canonical cover: minimize the
number of functional dependencies
 Important for updates
Farkas
CSCE 520
3
Canonical Cover
 Combine FDs if possible and eliminate
extraneous attributes: Given a set of
FDs S, and an FD XY
 Attribute A is extraneous in X if S
logically implies (S-{XY})  {(XA)Y}.
 Attribute A is extraneous in Y if the set of
FDs (S-{XY})  {X(Y-A)} logically
implies S.
Farkas
CSCE 520
4
Canonical Cover
 No functional dependency in Sc
contains an extraneous attribute
 Each left side of a functional
dependency in Sc is unique.
Farkas
CSCE 520
5
Example Canonical Cover
 Given:
1.
2.
3.
4.




Farkas
A  BC
BC
AB
AB  C
Combine 1 and 3 into: A BC
From 2 and 4: A is extraneous in 4: BC
C is extraneous in 1: A  B
Result: AB, BC
CSCE 520
6
Problems of Relational Database
Design
 Loss of information (lossless-join)
 Inability to represent certain
information (dependency
preservation)
 Repetition of information (normal
forms)
Farkas
CSCE 520
7
Desirable Properties of
Decomposition – Lossless-join
 Lossless-Join:
Let R be a relation schema, S a set of FDs
on R, R1 and R2 a decomposition of R.
R1 and R2 form a lossless-join
decomposition if at least one of the
following functional dependencies are in S+
 R1  R2  R1
 R1  R2  R2
Farkas
CSCE 520
8
Create Lossless-Join Decomposition
 Let R be the DB Schema and f: X  Y
an FD in S+
 Decompose R into two relations
 R1(X,Y)
 R2(R-Y)
 Continue for each decomposed
relation until cannot be decomposed
any more
Farkas
CSCE 520
9
Desirable Properties of Decomposition
– Dependency Preserving
 Dependency Preservation: all
dependencies that hold on the
original schema should be able to test
on individual schemas after
decomposition.
 Testing dependency preservations on
projections is straight forward.
Farkas
CSCE 520
10
Testing dependency preservations
on decomposition
 Let S be the original set of FDs on R,
and S’ the union of FDs on
projections R1, R2,…,Rn.
 A decomposition is dependency
preserving if S’+=S+
Farkas
CSCE 520
11
Desirable Properties of Decomposition
– Avoid Redundancy
Owner
Name
Owner
Address
Owner Phone
Dog Name
Dog
Breed
Dog
Color
Dog
Age
Viki
13 Lake Rd.
Irmo, SC
1-803-777-8989
Pepper
G.S.
Black
3
Viki
13 Lake Rd.
Irmo, SC
1-803-777-8989
Buddy
G.S.
Brown
1
Viki
13 Lake Rd.
Irmo, SC
1-803-777-8989
Alexandra
Mix
Gray
5
Viki
13 Lake Rd.
Irmo, SC
1-803-777-8989
Missy
Labrador
Brown
5
Pete
300 Main St.
Columbia, SC
1-803-657-5678
Hunter
Vizsla
Brown
8
Paul
99 Apple St
Irmo, SC
1-803-899-8524
Giant
Kuvasz
White
4
FD:
Farkas
O.Name  O.Address, O.Phone
D.Name, D.Breed  D.Color, D.Age
CSCE 520
12
Decomposition and FDs
 Functional dependencies: can be used
in designing a relational database to
remove the undesired properties
 Normalization using FDs:
 Boyce-Codd Normal Form (BCNF)
 3rd Normal Form (3NF)
Farkas
CSCE 520
13
Boyce-Codd Normal Form
 A relation is in Boyce-Codd Normal Form if
for all FDs X  A in S+ over R at least one
of the followings hold:
 X  A is a trivial FD
 if X  A is a nontrivial FD then X is a superkey
for schema R
 Example:
 R(Name,Breed,Date, Kennel)
 FD: Name,Breed,Date  Kennel
 R is in BCNF
Farkas
CSCE 520
14
Decomposition into BCNF
 Compute S+ (for FDs in S);
 if there is a R that is not BCNF then
- let XA a non-trivial FD on R s.t.
XR is not in S+ and X  A = ;
- Decompose R into:
 R1(R-A)
 R2(X,A)
Farkas
CSCE 520
15
BCNF
 Not every BCNF decomposition is
dependency preserving
 Example:
 FDs: AB  C and C B
 Keys: {A,B} and {A,C}
 C  B is BCNF violation, therefore need
to decompose to R1=(AC) and R2=(BC)
 Decomposition cannot enforce AB
C!
Farkas
CSCE 520
16
Third Normal Form
 Modifies BCNF conditions so no need
to decompose in this problem
situation
 An attribute is prime if it is member
of any key
 X  A violates 3NF if and only if X is
not a superkey and also A is not a
prime.
Farkas
CSCE 520
17
3NF decomposition
 Find minimal basis G for FDs
 For every FD X  Y in G, use XY is
the schema of one of the
decomposition
 If none of the relation schemas is a
super key for the original relation,
add another relation with the schema
as a key
Farkas
CSCE 520
18
3NF Conditions
 A relation is in 3NF if for all FDs X 
A in S+ over R at least one of the
followings hold:
 X  A is a trivial FD
 if X  A is a nontrivial FD then X is a
superkey for schema R
 Each attribute B in A-X is contained in a
candidate key for R
Farkas
CSCE 520
19
3NF and BCNF
 BCNF gives
 Lossless-join
 No-redundancy
 Dependency preservation not always possible
 3NF gives
 Lossless-join
 Dependency preservation
 May have null values (transitive dependencies)
Farkas
CSCE 520
20