Boyce-Codd NF &
Lossless Decomposition
Professor Sin-Min Lee
Armstrong’s Axioms
For computing the set of FDs that follow a given FD, the
following rules called Armstrong’s axioms are useful:
1.
Reflexivity: If B  A, then A  B
2.
Augmentation: If A  B, then A  C  B  C
Note also that if A  B, then A  C  B for any set
of attributes C.
3.
Transitivity: If A  B and B  C then A  C
Projecting FDs
Given a relation R (A,B,C,D) and F(R) = {AB,
BC, CD}.
Suppose S is projected from R as S(A,C,D). What is
F(S).
To compute F(S), start by computing the closures of all
attributes
in S.
In R, A+ = {AB, AC, AD}
In S, A+ = {AC, AD}
C+ = {CD} and
D+ = {D}
Since A+ contains all attributes of S, it is not required to
compute
(AC)+, (AD)+ or (ACD)+.
Inference Rules for FD’s
A1, A2, …, An  B1, B2, …, Bm
Splitting rule
and
Combining rule
Is equivalent to
A1, A2, …, An  B1
A1, A2, …, An  B2
.....
A1, A2, …, An  Bm
A1
...
Am
B1
...
Bm
Inference Rules for FD’s
(continued)
Trivial Rule
A1, A2, …, An  Ai
where i = 1, 2, ..., n
A1
Why ?
…
Am
Inference Rules for FD’s
(continued)
Transitive Closure Rule
If
A1, A2, …, An  B1, B2, …, Bm
and
B1, B2, …, Bm  C1, C2, …, Cp
then
A1, A2, …, An  C1, C2, …, Cp
Why ?
A1
…
Am
B1
…
Bm
C1
...
Cp
Example (continued)
Start from the following FDs:
1. name  color
2. category  department
3. color, category  price
Infer the following FDs:
Inferred FD
4. name, category  name
5. name, category  color
6. name, category  category
7. name, category  color,
category
Which Rule
did we apply
?
Another Rule
Augmentation
If
A1, A2, …, An  B
then
A1, A2, …, An , C1, C2, …, Cp  B
Augmentation follows from trivial rules and transitivity
How ?
Problem: infer ALL FDs
Given a set of FDs, infer all possible FDs
How to proceed ?
 Try all possible FDs, apply all 3 rules


Drop trivial FDs, drop augmented FDs


E.g. R(A, B, C, D): how many FDs are possible
?
Still way too many
Better: use the Closure Algorithm (next)
Closure of a set of Attributes
Given a set of attributes A1, …, An
The closure, {A1, …, An}+ , is the set of attributes B
s.t. A1, …, An  B
Example:
name  color
category  department
color, category  price
Closures:
name+ = {name, color}
{name, category}+ = {name, category, color, department, price}
color+ = {color}
Closure Algorithm
Start with X={A1, …, An}.
Example:
Repeat until X doesn’t change do:
name  color
category  department
color, category  price
B1, …, Bn  C is a FD and
B1, …, Bn are all in X
then add C to X.
if
{name, category}+ =
{name, category, color,
department, price}
Example
R(A,B,C,D,E,F)
A, B 
A, D 
B

A, F 
C
E
D
B
Compute {A,B}+
X = {A, B,
}
Compute {A, F}+
X = {A, F,
}
Using Closure to Infer ALL FDs
Example:
A, B  C
A, D  B
B
 D
Step 1: Compute X+, for every X:
A+ = A, B+ = BD, C+ = C, D+ = D
AB+ = ABCD, AC+ = AC, AD+ = ABCD
ABC+ = ABD+ = ACD+ = ABCD (no need to compute– why ?)
BCD+ = BCD, ABCD+ = ABCD
Step 2: Enumerate all FD’s X  Y, s.t. Y  X+ and XY = :
AB  CD, ADBC, ABC  D, ABD  C, ACD  B
Problem: Finding FDs

Approach 1: During Database Design



Designer derives them from real-world
knowledge of users
Problem: knowledge might not be available
Approach 2: From a Database Instance



Analyze given database instance and find all
FD’s satisfied by that instance
Useful if designers don’t get enough
information from users
Problem: FDs might be artifical for the given
instance
Find All FDs
Student
Dept
Course
Room
Alice
CSE
C++
020
Bob
CSE
C++
020
Alice
EE
HW
040
Carol
CSE
DB
045
Dan
CSE
Java
050
Elsa
CSE
DB
045
Frank
EE
Circuits
020
Do all FDs
make sense
in practice ?
Answer
Course  Dept, Room
Dept, Room  Course
Student, Dept  Course, Room
Student, Course  Dept, Room
Student, Room  Dept, Course
Do all FDs
make sense
in practice ?
Keys



A key is a set of attributes A1, ..., An s.t. for
any other attribute B, we have A1, ..., An  B
A minimal key is a set of attributes which is a
key and for which no subset is a key
Note: book calls them superkey and key
Computing Keys



Compute X+ for all sets X
If X+ = all attributes, then X is a
key
List only the minimal keys
Note: there can be many minimal
keys !
 Example: R(A,B,C), ABC, BCA
Minimal keys: AB and BC
Examples of Keys

Product(name, price, category, color)
name, category  price
category  color
Keys are:

{name, category} and all supersets
Enrollment(student, address, course, room, time)
student  address
room, time  course
student, course  room, time
Keys are:
Relational Schema Design
(or Logical Schema Design)
Main idea:
 Start with some relational schema
 Find out its FD’s
 Use them to design a better
relational schema
Data Anomalies
When a database is poorly designed we get
anomalies:
Redundancy: data is repeated
Update anomalies: need to change in several
places
Delete anomalies: may lose data when we
don’t want
Relational Schema Design
Example: Persons with several phones
Name
SSN
PhoneNumbe
r
City
Fred
123-45-6789
206-5551234
Seattle
Fred
123-45-6789
206-5556543
Seattle
Joe
Westfield
SSN 
Name, City987-65-4321
but not908-555SSN

PhoneNumber
2121
Anomalies:
• Redundancy
= repeat data
• Update anomalies = Fred moves to “Bellevue”
• Deletion anomalies = Joe deletes his phone number:
what is his city ?
Relation Decomposition
Break the relation into two:
Name
SSN
PhoneNumber
City
Fred
123-45-6789
206-555-1234
Seattle
Fred
123-45-6789
206-555-6543
Seattle
Joe
987-65-4321
908-555-2121
Westfield
Name
SSN
City
Fred
123-45-6789
Seattle
Joe
987-65-4321
Westfield
Anomalies have gone:
SSN
PhoneNumbe
r
123-45-6789
206-5551234
123-45-6789
206-5556543
• No more repeated data
987-65-4321
• Easy to move Fred to “Bellevue” (how ?)
• Easy to delete all Joe’s phone number (how ?)
908-5552121
Relational Schema Design
name
Conceptual Model:
Product
price
Relational Model:
plus FD’s
Normalization:
Eliminates anomalies
Person
buys
name
ssn
Decompositions in General
R(A1, ..., An, B1, ..., Bm, C1, ..., Cp)
R1(A1, ..., An, B1, ..., Bm)
R2(A1, ..., An, C1, ..., Cp)
R1 = projection of R on A1, ..., An, B1, ..., Bm
R2 = projection of R on A1, ..., An, C1, ..., Cp
Decomposition

Sometimes it is correct:
Name
Price
Category
Gizmo
19.99
Gadget
OneClick
24.99
Camera
Gizmo
19.99
Camera
Name
Price
Name
Category
Gizmo
19.99
Gizmo
Gadget
OneClick
24.99
OneClick
Camera
Gizmo
19.99
Gizmo
Camera
Lossless decomposition
Incorrect Decomposition

Sometimes it is not:
Name
Price
Category
Gizmo
19.99
Gadget
OneClick
24.99
Camera
Gizmo
19.99
Camera
What’s
incorrect ??
Name
Category
Price
Category
Gizmo
Gadget
19.99
Gadget
OneClick
Camera
24.99
Camera
Gizmo
Camera
19.99
Camera
Lossy decomposition
Decompositions in General
R(A1, ..., An, B1, ..., Bm, C1, ..., Cp)
R1(A1, ..., An, B1, ..., Bm)
R2(A1, ..., An, C1, ..., Cp)
If A1, ..., An  B1, ..., Bm
Then the decomposition is lossless
Note: don’t need necessarily A1, ..., An  C1, ..., Cp
Example: name  price, hence the first decomposition is lossless
Normal Forms
First Normal Form = all attributes are atomic
Second Normal Form (2NF) = old and obsolete
Third Normal Form (3NF) = this lecture
Boyce Codd Normal Form (BCNF) = this lecture
Others...
R
F
(J, K, L)
= (JK  L, L  K)
Two
R
candidate keys: JK and JL
is in 3NF

JK  L
JK is a superkey

LK
K is prime
BCNF
decomposition yields:

R1 (L,K), R2 (L,J)

testing for JK  L requires a join
There
is some redundancy in R
Boyce-Codd Normal Form
A simple condition for removing anomalies from relations:
A relation R is in BCNF if:
If A1, ..., An  B is a non-trivial dependency
in R , then {A1, ..., An} is a key for R
In English (though a bit vague):
Whenever a set of attributes of R is determining another attribute,
it should determine all the attributes of R.
BCNF Decomposition Algorithm
Repeat
choose A1, …, Am  B1, …, Bn that violates the BNCF condition
split R into R1(A1, …, Am, B1, …, Bn) and R2(A1, …, Am, [others])
continue with both R1 and R2
Until no more violations
B’s
R1
A’s
Others
R2
Is there a
2-attribute
relation that is
not in BCNF ?
Example
Name
SSN
PhoneNumbe
r
City
Fred
123-45-6789
206-5551234
Seattle
Fred
123-45-6789
206-5556543
Seattle
Joe
987-65-4321
908-5552121
Westfield
Joe
987-65-4321
908-5551234
Westfield
What are the dependencies?
SSN  Name, City
What are the keys?
{SSN, PhoneNumber}
Is it in BCNF?
Decompose it into BCNF
Name
SSN
City
Fred
123-45-6789
Seattle
Joe
987-65-4321
Westfield
SSN
PhoneNumbe
r
123-45-6789
206-5551234
123-45-6789
206-5556543
987-65-4321
908-5552121
987-65-4321
908-5551234
SSN  Name, City
Let’s check anomalies:
• Redundancy ?
• Update ?
• Delete ?
Summary of BCNF Decomposition
Find a dependency that violates the BCNF condition:
A1, A2, …, An  B1, B2, …, Bm
Heuristics: choose B1 , B2, … Bm“as large as possible”
Decompose:
Others
A’s
B’s
Continue until
there are no
BCNF violations
left.
2-attribute
relations are BCNF
R1
R2
Example Decomposition
Person(name, SSN, age, hairColor, phoneNumber)
SSN  name, age
age  hairColor
Decompose in BCNF (in class):
Step 1: find all keys (How ? Compute S+, for various sets S)
Step 2: now decompose
Other Example





R(A,B,C,D)
A  B,
BC
Key: AD
Violations of BCNF: A  B, A C,
ABC
Pick A BC: split into R1(A,BC)
R2(A,D)
What happens if we pick A  B first ?
Lossless Decompositions
A decomposition is lossless if we can recover:
R(A,B,C)
Decompose
R1(A,B)
R2(A,C)
R’(A,B,C)
Recover
should
be the same as
R(A,B,C)
R’ is in general larger than R. Must ensure R’ = R
Lossless Decompositions

Given R(A,B,C) s.t. AB, the
decomposition into R1(A,B),
R2(A,C) is lossless
3NF: A Problem with BCNF
Unit
Company
Product
FD’s: Unit  Company;
Company, Product  Unit
So, there is a BCNF violation, and we decompose.
Unit
Company
Unit
Product
Unit  Company
No FDs
Notice: we loose the FD: Company, Product  Unit
So What’s the Problem?
Unit
Company
Unit
Galaga99
Bingo
UW
UW
Galaga99
Bingo
Product
databases
databases
No problem so far. All local FD’s are satisfied.
Let’s put all the data back into a single table again (anomalies?):
Unit
Galaga99
Bingo
Company
UW
UW
Product
databases
databases
Violates the dependency: company, product -> unit!
Solution: 3rd Normal Form (3NF)
A simple condition for removing anomalies from relations:
A relation R is in 3rd normal form if :
Whenever there is a nontrivial dependency A1, A2, ..., An  B
for R , then {A1, A2, ..., An } is a key for R,
or B is part of a key.
Tradeoff:
BCNF = no anomalies, but may lose some FDs
3NF = keeps all FDs, but may have some anomalies
Purpose of Normalization
 To
reduce the chances for
anomalies to occur in a database.
 normalization prevents the
possible corruption of databases
stemming from what are called
“insertion anomalies," "deletion
anomalies," and "update
anomalies."
Insertion Anomaly
A
failure to place a new
database entry into all the
places in the database where
that new entry needs to be
stored.
 In a properly normalized
database, a new entry needs to
be inserted into only one place
in the database
Deletion Anomaly
A
failure to remove an existing
database entry when it is time
to remove that entry.
 In a properly normalized
database, an old, to-be-gottenrid-of entry needs to be deleted
from only one place in the
database
Update anomaly

An update of a database involves
modifications that may be additions,
deletions, or both. Thus "update
anomalies" can be either of the
kinds of anomalies discussed above.