Database Systems

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Database Systems

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Title: Database Systems

1
Database Systems
2
Major Content Grade

Introduction
The Relational Model
SQL
Transaction Management
Database Design (E-R)
Database Design (Normalization)

3
Part2 normalization

Introduction
Functional Dependencies
Normal Forms
Lossless Decompositions
Additional Design Considerations

4
Introduction

Normalization (???) is another approach to
logical design of a relational database.
E-R approach and normalization approach reinforce
each other.
Normalization starts with a real-world situation
to be modeled and lists the data items that are
candidates to become column names in relational
tables, together with a list of rules about the
relatedness of these data items.
The aim is to represent all these data items as
attributes of tables that obey restrictive
conditions associated with what we call normal
forms (??).
1NF --gt 2NF --gt 3NF --gt BCNF --gt 4NF --gt 5NF

5
Design of the Bank Database

branch (branch_name, branch_city, assets)
customer (customer_id, customer_name,
customer_street, customer_city)
loan (loan_number, amount)
account (account_number, balance)
employee (employee_id. employee_name,
telephone_number, start_date)
dependent_name (employee_id, dname)
account_branch (account_number, branch_name)
loan_branch (loan_number, branch_name)
borrower (customer_id, loan_number)
depositor (customer_id, account_number)
cust_banker (customer_id, employee_id, type)
works_for (worker_employee_id,
manager_employee_id)
payment (loan_number, payment_number,
payment_date, payment_amount)
savings_account (account_number, interest_rate)
checking_account (account_number,
overdraft_amount)

6
Design of the Bank Database

Suppose we combine borrow and loan to get
Schema bor_loan (customer_id, loan_number,
amount )
Instance

Result is possible repetition of information
For borrower is MN relationship

7
Design of the Bank Database

Consider combining loan_branch and loan
Schema loan_amt_br (loan_number, amount,
branch_name)
Instance

No repetition
For loan_branch is 1N relationship

8
Design of the Bank Database

Example (decompose, ??)

we cannot reconstruct the original employee
relation

9
Design of the Bank Database

combining loan_branch and loan into
Schema loan_amt_br (loan_number, amount,
branch_name)
Is a good relation schema
combine borrow and loan to get
Schema bor_loan (customer_id, loan_number,
amount )
Is NOT a good relation schema
Decide whether a particular relation R is in
good or NOT?
Suppose we had started with bor_loan. How would
we know to split up (decompose, ??) it into
borrower and loan?
Normalization theory is the tools used to solve
those questions.

10
A Running Example

Employee Information

From one up to a large number of skills useful to
the company
11
A Running Example

Employee Information

12
Anomalies of a Bad Database Design

Update Anomaly (????)
A table T is subject to an update anomaly when
changing a single attribute value for an entity
instance or relationship instance represented in
the table may require that several rows of T be
updated.

13
Anomalies of a Bad Database Design

Delete Anomaly
A table T is subject to a delete anomaly when
deleting some row of the table to reflect the
disappearance of some instance of an entity or
relationship can cause us to lose information
about some instance of a different entity or
relationship that we do not wish to forget.

14
Anomalies of a Bad Database Design

Insert Anomaly
We cannot represent information about some entity
or instance without including information about
some other instance of an entity or relationship
that does not exist.

15
Anomalies of a Bad Database Design

Redundant Data (????)
An entity instance or relationship instance
represented in a table T may account for several
rows of T.

16
Anomalies of a Bad Database Design

Normalize the relation

decompose
17
Functional Dependencies

Functional Dependencies (????)
The functional dependency
? ? ?
holds on R if and only if for any legal
relations r(R), whenever any two tuples t1 and t2
of r agree on the attributes ? (???????), they
also agree on the attributes ?.
That is,
t1? t2 ? ? t1? t2 ?
????????????,???????????(??????????,????????????)

18
Functional Dependencies

In the emp_info table, we get

19
Functional Dependencies

Analyze the following tables (suppose they are
valid)

T2 A ? B B ? A
20
Logical implications among functional dependencies

Inclusion Rule (????)
Given a table T with a specified heading Head(T).
If X and Y are sets of attributes contained in
Head(T), and Y ? X, then X?Y.
Proof. By def, need only demonstrate that if two
rows u and v agree on X they must agree on Y.
But Y is a subset of X, so seems obvious.
Trivial Dependency (???????)
A Trivial Dependency is a FD of the form X ?Y, in
a table T where X ? Y ? Head(T). That will hold
for any possible content of the table T.
(?????????trivial dependency)
Given a trivial dependency X ?Y in T, it must be
the case that Y ? X.
e.g. A ?A, AB ?A

21
Armstrongs Axioms

Armstrongs Axiom (??????? 1974)
A1 Inclusion rule(????) if Y ? X, then X?Y
????????RltU,Fgt?????r????????t?s,?tXsX,??Y
?X,?tYsY,?X?Y?
Example
customer_name, loan_number ? customer_name
customer_name ? customer_name

22
Armstrongs Axioms

Armstrongs Axiom (??????? 1974)
A2 Transitivity rule(????) if X ? Y and Y ? Z
, then X ? Z
????????RltU,Fgt ????? r????????
t?s,?tXsX,??X?Y,? tYsY??Y?Z,?tZsZ,
??F X?Z?
Example
For relation S( sno, sname, sdept, dept_manager
)
sno ? sdept, sdept ? dept_manager
THEN sno ? dept_manager

23
Armstrongs Axioms

Armstrongs Axiom (??????? 1974)
A3 Augmentation rule(????) if X ? Y, then
XZ ? YZ
????????RltU,Fgt?????r????????t?s,?tXZsXZ??t
XsX?tZsZ?X?Y,??tYsY??tYZsYZ,?
FXZ?YZ?
Example
For relation S( sno, sname, sdept, dept_manager
)
sno ? sdept
THEN (sno , sname) ? dept_manager , sname

24
Armstrongs Axioms

Some implications of Armstrongs Axiom
1 Union rule(????) if X ? Y and X ?Z, then
X ? YZ
?? (1) X?Y ??(P??)
(2) X?XY A2,(1)
(3) X?Z ??
(4) XY?YZ A2,(3)
(5) X?YZ A3,(2),(4)
? X?Y,X?Z X?YZ
Example S( sno, sname, sdept, dept_manager )
sno ? sname , sno ? sdept
THEN sno ? sname , sdept

25
Armstrongs Axioms

Some implications of Armstrongs Axiom
2 Decomposition rule(????) if X ? YZ, then
X ? Y and X ? Z
Example S( sno, sname, sdept, dept_manager )
sno ? sname , sdept
THEN sno ? sname , sno ? sdept
3 Pseudotransitivity rule(?????) if X ? Y and
WY ?Z, then XW ? Z
4 Set accumulation rule(??????) if X ?YZ and
Z ?W, then X? YZW
(????????,?)

26
Closure (??)

The set of all functional dependencies logically
implied by F is the closure of F, denoted by F
.
We can find all of F by applying Armstrongs
Axioms
if ? ? ?, then ? ? ?
(reflexivity)
if ? ? ?, then ? ? ? ? ?
(augmentation)
if ? ? ?, and ? ? ?, then ? ? ? (transitivity)
Armstrongs Axiom are often referred to as being
valid(sound, ???) and complete(???).

27
Closure

Given RltU, Fgt, UA, B, C, FA?B, B?C,
The closure of F
F ???, A??, A?A, , AB?A, //A1
A?B,A?AB,AB?B,,ABC?BC, //A2
B?C, AB?AC, //A2
A ?C //A3
notethere are 43 non-duplicate FDs.
The closure of functional dependency sets
includes all dependencies among attributes of a
relation.
drawbackits too hard to be managed.

28
Closure

Algorithm To compute the closure of a set of
functional dependencies F

begin F F repeat for each functional
dependency f in F apply inclusion and
augmentation rules on f add the resulting
functional dependencies to F for each pair of
functional dependencies f1 and f2 in F
if f1 and f2 can be combined using
transitivity then add the resulting functional
dependency to Funtil F does not change any
further End
29
Closure

Closure of attributes (??????)

Given a set of attributes a, define the closure
of a under F (denoted by aF) as the set of
attributes that are functionally determined by a
under F.
Algorithm to compute aF, the closure of a under
F .

result a while (changes to result) do for
each ? ? ? in F do begin if ? ? result then
result result ? ? end
30
Closure

Closure of attributes (??????)

Example1 Given RltU, Fgt, R (A, B, C, G, H, I)
F A ? B, A ? C, CG ? H, CG ? I, B ? H
(AG) ?
1) result AG
2) result AGBC (A ? B and A ? C)
3) result AGBCH (CG ? H and CG ? AGBC)
4) result AGBCHI (CG ? I and CG ? AGBCH)
Example2 Given RltU, Fgt, R (A, B, C, D, E)
FB?CD, AD?E, B?A
(BC) ?

31
Closure

Closure of attributes (??????)

There are several uses of the attribute closure
algorithm
1) Testing for superkey
To test if ? is a superkey, we compute ?, and
check if ? contains all attributes of R.
Example for relation RltU, Fgt, U A, B, C, D,
E, F AB?C, B?D, C?E, EC?B, AC?B IS AB a
superkey or not?
(AB)FABCDE U
So AB is a superkey

32
Closure

Closure of attributes (??????)

There are several uses of the attribute closure
algorithm

2) Testing functional dependencies
To check if a functional dependency ? ? ? holds
(or, in other words, is in F), just check if ? ?
?.
Example for relation RltU, Fgt, U A, B, C, D,
E, F AB?C, B?D, C?E, EC?B, AC?B
IS BE?CD implied by F?
For (BE)FBED , not include CD, so not
implied.
IS AB?E implied by F?
(Theorem)

33
Closure

Closure of attributes (??????)

There are several uses of the attribute closure
algorithm

3) Computing closure of F
For each ? ? R, we find the closure ?, and for
each S ? ?, we output a functional dependency ?
? S.
(Theorem)

34
Cover

FD Set Cover(????????)
A set F of FDs on a table T is said to cover
another set G of FDs on T, if the set G of FDs
can be derived by implication rules from the set
F, or in other words. If G?F.
If F covers G and G covers F, then the two sets
of FDs are said to be equivalent, and we write F
G.
If two FDs are equivalent, the have the same
implication of FDs.
Example
Consider the two sets of FDs on relaton R(ABCDE)
FB?CD, AD?E, B?A and
GB?CDE, B?ABC, AD?E
Is F G or NOT?

35
Database Systems
36
Cover

Sets of functional dependencies may have
redundant dependencies that can be inferred from
the others.
For example A ? C is redundant in A ? B,
B ? C
Parts of a functional dependency may be redundant
E.g. on RHS A ? B, B ? C, A ? CD
can be simplified to
A ? B, B ? C, A ? D
E.g. on LHS A ? B, B ? C, AC ? D
can be simplified to
A ? B, B ? C, A ? D
we need a cover of F is a minimal set of
functional dependencies equivalent to F, having
no redundant dependencies or redundant parts of
dependencies.

37
Minamal Cover

Minimal Cover(????,???????)
Step 1. Decomposition Right Hand Side of FDs
Create an equivalent set H of FDs, with only
single attributes on the right side.(????)
Step 2. Erase extraneous attributes on LHS
For ? ? ? in F Attribute A is extraneous in ?
if A ? ? and F logically implies (F ? ?
?) ? (? A) ? ?. Then replace ? ? ? with (?
A) ? ?
Step 3. Delete redundant FD
For ? ? ? in F, if (F ? ? ?) logically
implies ? ? ?, then delete ? ? ? from F.

38
Minamal Cover

Example for relation RltU, Fgt, U A, B, C, D,
E, FA?BC, BCD?E, B?D, A?D, E?Acompute the
minimal cover of F.
1) ????????
F1A?B, A?C, BCD?E, B?D, A?D, E?A
2) ????????
for (BC)FBCDEA, include E, so D in LHS of
BCD?E is extraneous.
F2A?B, A?C, BC?E, B?D, A?D, E?A
3) ????????
for A?D because of (A) F2-(A?D )ABCED, is
redundancy
Fmin A?B, A?C, BC?E, B?D, E?A

39
Canonical Cover(????)

A canonical cover for F is a set of dependencies
Fc such that
F logically implies all dependencies in Fc, and
Fc logically implies all dependencies in F, and
No functional dependency in Fc contains an
extraneous attribute, and
Each left side of functional dependency in Fc is
unique.
gather all FDs in Minimal Cover with equal
left-hand sides and use the union rule to create
an equivalent set of FDs where all left-hand
sides are unique, well get canonical cover .

40
Canonical Cover(????)

Example for relation RltU, Fgt, U A, B, C,
FA ? BC, B ? C, A ? B, AB ? Ccompute the
canonical cover of F.
1) ???????
F1 A ? B, A ? C, B ? C, A ? B, AB ? C
2) ????????
?AB ? C,(A)F1 ABC, ??C,??B ? C????
F1 A ? B, A ? C, B ? C
3) ????????
?A ? C , (A)F-( A ? C ) ABC, ??,???
4) ??????
Fc A ? B, B ? C

41
??

R (A, B, C, D, E, F)
F A ? BC, E ? CF, B ? E, CD ? EF
(AB) ?
(AD) ? Is AD ?F implied by F?
Page 307
7.6
7.7

42
KEY

K is a superkey for relation schema R if and only
if K ? R
K is a candidate key for R if and only if
K ? R, and
for no ? ? K, ? ? R
Prime attribute an attribute that appeared in
some candidate key
non-prime attribute an attribute that DO NOT
appeared in any candidate key

43
5.3 ?????????

???????
??
?????????RltU, Fgt,???????F?U??????????
L??? ?F???????????????
R??? ?F???????????????
LR??? ?????F??????????????
N??? ??F?????????????
??
? L????N?????????????
? R??????????????
? LR???????????????

44
5.3 ?????????

??
?????????RltU, Fgt,??U?????,F???????
(1) ???????F?R????????L??R??LR??N???,?X?L?N?????
?,Y?LR?????

??RltU, Fgt,?? UA,B,C,D,E,FA?BC, CD?E, B?D,
E?A,?R???????
? (1) R??L?N???,A, B,C,D,E??LR???,?X
?,YA,B,C,D,E
(2) XF ? ? U
(2) ?XFU,?X?R?????? (?),????,?(3)
45
5.3 ?????????

??
?????????RltU, Fgt,??U?????,F???????
(1)
(2)

??RltU, Fgt,?? UA,B,C,D,E,FA?BC, CD?E, B?D,
E?A,?R??????? ? (1) X ?,YA,B,C,D,E
(2)
(3)?A,?AFABCDEU,A????
?B?C?D,???????????U ?E,?EFABCDEU,E????
Y B,C,D
(3)???Y??????A,?(XA)FU,?XA???? ,?YY-A
,?(4)
46
5.3 ?????????

??
?????????RltU, Fgt,??U?????,F???????
(1)
(2)
(3)

??RltU, Fgt,?? UA,B,C,D,E,FA?BC, CD?E, B?D,
E?A,?R??????? ?(3) ,A???? ,E????
Y B,C,D
(4)?Y????????? (BC)FBCDEAU,BC????
(BD)F???????U,BC????? (CD)FCDEABU,CD????
(4) ???Y????????????????XZ,?XZ??????????,????F
???(XZ)F,?(XZ)F U,?XZ?????????Y????????,?????
47
5.3 ?????????
??RltU, Fgt,?? UA,B,C,D,E,FA?BC, CD?E, B?D,
E?A,?R??????? ?(3) ,A???? ,E????
Y B,C,D (4)?Y????????? , BC????
, CD????
(5) BCD?????????BC,BCD???,??? ???R?????A,E,BC,
CD?
48
Normal Forms -- 1NF

A relational schema R is in first normal form if
the domains of all attributes of R are atomic.
NO composite attributes, such ascustomer(
customer-id, name(first-name, middle-initial,
last-name), date-of-birth )
Each attribute as an unit, even they have several
part that have individual information.
Example Strings would normally be considered
indivisible. For student number 130711, 13
is department number, but you cannot use. For
doing so is a bad idea leads to encoding of
information in application program rather than in
the database.

49
Normal Forms -- 1NF

A schema R not in 1NF, then its NOT a relational
schema.
A relation R is in 1NF is not good enough.
For relationEmployee( emp_id, emp_name,
emp_phone, dept_name, dept_phone, dept_mgrname,
skill_id, skill_name, skill_date, skill_lvl )
Is in 1NF
Has Insert Anomaly, Delete Anomaly, Update
Anomaly and Data Redundancy .

50
Normal Forms -- 2NF

Second normal form (2NF) A relation schema R
with FD set F is said to be in 2NF, if for any
functional dependency X?A implied by F that lies
in R, where A is a single attribute that is not
in X and is non-prime(????,?????????????), X is
not a proper subset(???) of any key K of R.
Or there are NO non-prime attributes dependent on
Candidate Key partially in 2NF.(??????????????)
ExampleR(A, B,C,D), F AB ? C, AC ? BD
Candidate Key AB, AC
AB ? D, AC? D is FULL dependency
R?2NF

51
Normal Forms -- 2NF

For example

Is relation schema emp_info ?2NF ?
Candidate Key?
Non-Prime attributes?
Test all FD according the definition of Normal
Form.

52
Database Systems
53
Normal Forms -- 2NF

emp_info ( emp_id, emp_name, epm_phone,
dept_name, dept_phone, dept_mgrname, skill_id,
skill_name, skill_date, skill_lvl )
F emp_id ? emp_name, epm_phone, dept_name,
dept_name ? dept_phone, dept_mgrname,
skill_id ? skill_name, emp_id,
skill_id ? skill_date, skill_lvl
Decomposition(????)
emp (emp_id, emp_name, epm_phone, dept_name,
dept_phone, dept_mgrname )
skill ( skill_id, skill_name )
emp_skill ( emp_id, skill_id, skill_date,
skill_lvl )

?2NF
?2NF
?2NF
54
Normal Forms -- 2NF

For relationbor_loan (customer_id, loan_number,
amount )
F loan_number ? amount
CK ( customer_id, loan_number )
bor_loan is NOT in 2NF
For borrower is MN relationship
Merging a MN relationship with an entity it
associated induces a NON-2NF relation schema.

55
Normal Forms -- 2NF

A relation R is in 2NF is not good enough.
For relation emp (emp_id, emp_name, epm_phone,
dept_name, dept_phone, dept_mgrname )
?2NF
Has Insert Anomaly, Delete Anomaly, Update
Anomaly and Data Redundancy .

56
Normal Forms -- 3NF

A relation schema R is in third normal form (3NF)
if for all ? ? ? in F at least one
of the following holds
? ? ? is trivial (i.e., ? ? ?) (not exist in
canonical cover )
? is a superkey for R
Each attribute A in ? ? is contained in a
candidate key for R.(or for canonical cover, A in
? is Prime attribute)
For example
????SJP(S, J, P) ?,S???,J????,P??????????
??????????????,???????
FD (S, J)?P,(J, P)?S
CK (S, J), (J, P)
LHS of each FD is superkey, SPJ is in 3NF.

57
Normal Forms -- 3NF

Another define A relation R is in 3NF if there
are no nonprime attributes which transitively
dependent on a key for R.
(3NF ???????????????)
For example loan_b (loan_number,
branch_name, branch_city, assets)
F loan_number ? branch_name ,
branch_name ? branch_city, assets
loan_number ? branch_name, branch_name ?
branch_city
so nonprime attribute branch_city is
transitively dependent on candidate key
loan_number
SPJ is NOT in 3NF

58
Normal Forms -- 3NF

The two definations are equivalent
A relation schema R is in third normal form (3NF)
if for all
? ? ? in F at least one of the
following holds
? ? ? is trivial (i.e., ? ? ?) (not exist in
canonical cover )
? is a superkey for R
Each attribute A in ? ? is contained in a
candidate key for R.(or for canonical cover, A in
? is Prime attribute)
Another define A relation R is in 3NF if there
are no nonprime attributes which transitively
dependent on a key for R.

59
Normal Forms -- 3NF

For example
emp (emp_id, emp_name, epm_phone, dept_name,
dept_phone, dept_mgrname ) ?2NF
F emp_id ? emp_name, epm_phone, dept_name,
dept_name ? dept_phone, dept_mgrname
dept_name is NOT a superkey
emp_name NOT in any candidate key
emp is NOT in 3NF
Nonprime attribute dept_phone is transitively
dependent on candidate key emp_id. So emp is NOT
in 3NF.

60
Normal Forms -- 3NF

For example
emp (emp_id, emp_name, epm_phone, dept_name,
dept_phone, dept_mgrname )
F emp_id ? emp_name, epm_phone, dept_name,
dept_name ? dept_phone, dept_mgrname
emp is NOT in 3NF

Decomposition
emp (emp_id, emp_name, epm_phone, dept_name ) F
emp_id ? emp_name, epm_phone, dept_name
emp?3NF.
dept (dept_name, dept_phone, dept_mgrname )F
dept_name ? dept_phone, dept_mgrname
dept ?3NF.

61
Normal Forms -- 3NF

A relation R is in 3NF is not good enough.
For relation STC( S, T, C) SStudent,
TTeacher, C--Course
F (S,C)?T, (S,T)?C, T?C
There is no nonprime attribute. STC is IN 3NF.
The first two FD, LHS is SuperKeyC in T?C is
prime attribute
STC is IN 3NF.
Has Insert Anomaly, Delete Anomaly, Update
Anomaly and Data Redundancy .

62
Normal Forms -- BCNF

A relation schema R is in BCNF(Boyce-Codd Normal
Form) with respect to a set F of functional
dependencies if for all functional dependencies
in F of the form ??? ?where ? ?
R and ? ? R, at least one of the following holds
?? ? ? is trivial (i.e., ? ? ?)
? is a superkey for R
For example
bor_loan ( customer_id, loan_number, amount )
F loan_number ? amount
bor_loan is not in BCNF, for loan_number is not
a superkey
bor_loan is not in 2NF, it just in 1NF.

63
Normal Forms -- BCNF

example1
????SJP(S, J, P) ?,S???,J????,P?????????????????
???????,???????
FD (S, J)?P,(J, P)?S
CK (S, J), (J, P)
LHS of each FD is superkey, SPJ is in BCNF.
example2
STC( S, T, C) F (S,C)?T, (S,T)?C, T?C
There is no nonprime attribute. STC is IN 3NF.
For T?C, T is not a superkeySTC is NOT in BCNF.

64
Normal Forms

Theorem
1NF ? 2NF ? 3NF ? BCNF
To determine a relation in nNF, one should give
the highest Normal Form.

65
Normal Forms

Relation Database
emp (emp_id, emp_name, epm_phone, dept_name ) F
emp_id ? emp_name, epm_phone, dept_name
emp?BCNF.
dept (dept_name, dept_phone, dept_mgrname )F
dept_name ? dept_phone, dept_mgrname
dept ?BCNF.
skill ( skill_id, skill_name )F skill_id ?
skill_name
skill ?BCNF.
emp_skill ( emp_id, skill_id, skill_date,
skill_lvl )F emp_id, skill_id ? skill_date,
skill_lvl
emp_skill ?BCNF.

66
Normal Forms (4NF)

Multivalued dependency
Let R be a relation schema and let ? ? R and ? ?
R. The multivalued dependency(MVD, ????)
? ?? ?
holds on R if in any legal relation r(R), for
all pairs for tuples t1 and t2 in r such that
t1? t2 ?, there exist tuples t3 and t4 in r
such that
t1? t2 ? t3 ? t4 ? t3?
t1 ? t3R ? t2R ? t4 ?
t2? t4R ? t1R ?

X Y Z
t1 x y1 z1
t2 x y2 z2
t3 x y1 z2
t4 x y2 z1
67
Normal Forms

For example
WSC(W,S,C)
Wwarehouse
Ssafeguard
Ccargo
MVD
W??S
W??C

W S C
w1 s1 c1
w1 s1 c2
w1 s1 c3
w1 s2 c1
w1 s2 c2
w1 s2 c3
w2 s3 c4
w2 s3 c5
w2 s4 c4
w2 s4 c5
68
Normal Forms

Consider a database
classes (course, teacher, book )

MVD
course ?? teacher, course ?? book

69
Normal Forms

Consider a database
classes (course, teacher, book )

Therefore, it is better to decompose classes into

70
Normal Forms -- 4NF

Fourth normal form (4NF) A relation schema R is
in 4NF with respect to a set D of functional and
multivalued dependencies if for all multivalued
dependencies in D of the form ? ?? ?, where ? ?
R and ? ? R, at least one of the following hold
? ?? ? is trivial (i.e., ? ? ? or ? ? ? R)
? is a superkey for schema R
Where the closure D of D is the set of all
functional and multivalued dependencies logically
implied by D.

If a relation is in 4NF, it is in BCNF

71
Normal Forms -- 4NF

Normal forms 4NF
WSC(W,S,C)
W??S
W??C

CTB(course, teacher, book)
course ?? teacher
course ?? book
The above formal definition is supposed to
formalize the notion that given a particular
value of X (course) it has associated with it a
set of values of Y (teacher) and a set of values
of Z (book), and these two sets are in some sense
independent of each other

WSC??NF CTB ??NF
72
Normal Forms

Normal forms 4NF
WSC(W,S,C)
W??S
W??C
Anomalies
Decomposition
WS (W,S)
W??S
WSC(W, C)
W??C

W S C
w1 s1 c1
w1 s1 c2
w1 s1 c3
w1 s2 c1
w1 s2 c2
w1 s2 c3
w2 s3 c4
w2 s3 c5
w2 s4 c4
w2 s4 c5
W S
w1 s1
w1 s2
w2 s3
w2 s4
W C
w1 c1
w1 c2
w1 c3
w2 c4
w2 c5
WS ?4NF
WC ?4NF
73
Decompositions

For relation RltU, Fgt , a decomposition(????) of R
into k relatons ? R1ltU1, F1gt, R2ltU2, F2gt, ,
RkltUk, Fkgt with two properties
(1) For each relation Ri, Ui is a proper subset
of U
(2) U U1? U2 ? ? Uk , Ui nUj f
Given any specific instance r of R, the rows of r
are projected onto the columns of each Ui as a
result of the decomposition.

decomposition
74
Lossless Decompositions
AB BC
AB JOIN BC
ABC
A B
a1 100
a2 200
a3 300
a4 200
B C
100 c1
200 c2
300 c3
200 c4
A B C
a1 100 c1
a2 200 c2
a2 200 c4
a3 300 c3
a4 200 c2
a4 200 c4
A B C
a1 100 c1
a2 200 c2
a3 300 c3
a4 200 c4
?? ?
?? ?

A decomposition of a relation R with an
associated set F of FDs is said to be a lossless
decomposition, or sometimes a lossless-join
decomposition (????) if, for any possible
instance r of R guarantee that

75
Lossless Decompositions

For the case of R (R1, R2), we require that for
all possible relations r on schema R

r ?R1 (r ) ?R2 (r )

Theorem A decomposition of R into R1 and R2 is
lossless join if and only if at least one of the
following dependencies is in F
R1 ? R2 ? R1
R1 ? R2 ? R2

76
Dependency Preservation

Let Fi be the set of dependencies F that
include only attributes in Ri.
A decomposition is dependency preserving
(??????), if
(F1 ? F2 ? ? Fn ) F
If it is not, then checking updates for violation
of functional dependencies may require computing
joins, which is expensive.

77
Decompositions

Examples
R ( A, B, C ) F A ? B, B ? C
Decomposition1 R1 (A, B), R2 (B, C)
IS Lossless-join ?
R1 ? R2 B and B ? BC
IS Dependency preserving?
( F1 ? F2 ) A ? B, B ? C F
Decomposition2 R1 (A, B), R2 (A, C)
IS Lossless-join ?
R1 ? R2 A and A ? AB
IS Dependency preserving?
( F1 ? F2 ) A ? B, A ? C can not imply B
? C , is non Dependency preserving

78
Goals of Normalization

Let R be a relation scheme with a set F of
functional dependencies.
Decide whether a relation scheme R is in good
form.
In the case that a relation scheme R is not in
good form, decompose it into a set of relation
scheme R1, R2, ..., Rn such that
each relation scheme is in good form
the decomposition is a lossless-join
decomposition
Preferably, the decomposition should be
dependency preserving.

79
BCNF Decomposition Algorithm
result R done falsecompute Fwhile
(not done) do if (there is a schema Ri
in result that is not in BCNF) then
begin let ?? ? ? be a nontrivial
functional dependency that holds on Ri such
that ?? ? Ri is not in F, and ? ? ? ?
result (result Ri ) ? (Ri ?) ? (?, ?
) end else done true Note
each Ri is in BCNF, and decomposition is
lossless-join.

R (A, B, C ) F A ? B, B ? C
Key A
R is not in BCNF (B ? C but B is not superkey)
Decomposition R1 (B, C), R2 (A,B)

80
BCNF Decomposition Algorithm

Original relation R and functional dependency F
R (branch_name, branch_city, assets,
customer_name, loan_number, amount )
F branch_name ? assets, branch_city
loan_number ? amount, branch_name
Key loan_number, customer_name
Decomposition
For FD branch_name ? assets, branch_city,
decomposition
R1 (branch_name, branch_city, assets )
R2 (branch_name, customer_name, loan_number,
amount )
For FD in R2 loan_number ? amount, branch_name
R21 (branch_name, loan_number, amount )
R22 (customer_name, loan_number )
Final decomposition R1, R21, R22

81
BCNF Decomposition Algorithm

emp_info ( emp_id, emp_name, epm_phone,
dept_name, dept_phone, dept_mgrname, skill_id,
skill_name, skill_date, skill_lvl )
F emp_id ? emp_name, epm_phone, dept_name,
dept_name ? dept_phone, dept_mgrname,
skill_id ? skill_name, (emp_id, skill_id) ?
skill_date, skill_lvl
CK ( emp_id, skill_id )
Decomposition to BCNF
For emp_id ? emp_name, epm_phone, dept_name,
dept_phone, dept_mgrname
R1 (emp_id, emp_name, epm_phone, dept_name,
dept_phone, dept_mgrname)F1 emp_id ?
emp_name, epm_phone, dept_name, dept_name ?
dept_phone, dept_mgrname
R2 (emp_id, skill_id, skill_name, skill_date,
skill_lvl )F2 skill_id ? skill_name,( emp_id,
skill_id) ? skill_date, skill_lvl