Data base normalization upto BCNF

1. DATABASE
NORMALIZATION
-VARSHA KUMARI

2. Content : Database Normalization:
 Functional dependencies
 Anomalies in database (Insert Update, Delete)
 Introduction to Normal forms based on primary keys
 First Normal Form
 Second Normal Form
 Third Normal Form
 Boyce Codd Normal Form
 De-normalization
 Lossless and Lossy Joins
 dependency preserving decomposition

3. Functional Dependancy (FDs)
 A functional dependency (FD) is a relationship
between two attributes, typically between the PK
and other non-key attributes within a table.

4. Functional Dependancy
A B
S 1
T 2
U 3
V 4
t1 ->
t2 ->
Let A and B are subset
of a Relation R
t1 ->
If t1(A) = t2(A)
then t1(B) = t2(B)
Then Functional Dependancy A-> B holds true:

5. Functional Dependancy
A B
S 1
S 2
U 3
V 4
t1 ->
t2 ->
Let A and B are subset
of a Relation R
t1 ->
If t1(A) = t2(A)
then t1(B) ! = t2(B)
Then Functional Dependancy A-> B does not holds true:

6. Functional Dependancy
A B
S 1
S 1
U 1
V 1
t1 ->
t2 ->
Let A and B are subset
of a Relation R
t1 ->
If t1(A) = t2(A)
then t1(B) = t2(B)
Then Functional Dependancy A-> B holds true:

7.  If A is unique then A-> B always holds true
 If values in B are all same then also A-> B always
holds true
 Now A and B can be a set of attributes

8. Functional Dependancy
University
Roll
S_name
U1 A
U2 A
U3 B
U4 C
University
Roll ->
S_name
True
S_name -> University Roll
False

9. Functional Dependancy
University
Roll
S_name
U1 A
U2 A
U3 B
U4 C
University
Roll ->
S_name
True
S_name -> University Roll
False

10. to check the determined attribute
 Given R( A, B, C, D, E)
 F = {A -> BC, DE ->C, B ->D}
 {A -> BC, C->DE, B ->D}
 Does A determine all other attributes?
 A->BC
 A-> ABC
 As B-> D so A->ABCD
 As c-> DE so A-> ABCDE
 Here we cannot determine E from A so A is not a
candidate key

11.  Given R( A, B, C, D, E)
 F = {A -> BC, DE ->C, B ->D}
 Is BE a key for R?
 BE -> BE
 As B-> D so BE-> BED
 As DE -> C so BE -> BEDC
 Here we cannot determine A from BE so A is not a
candidate key

12.  Given R( A, B, C, D, E)
 F = {A -> BC, DE ->C, B ->D}
 Is AE a candidate or super key for R?
 AE->AE
 As A-> BC so AE->ABCE
 As B->D so AE->ABCDE
 Here we can determine all the attributes of relation
R so AE is a candidate key
 Is ADE a candidate or super key for R?
 ADE is a superkey as ADE ⊃ AE

13. Various Axioms Rules of functional
dependency
Rule 1 Reflexivity
If A is a set of attributes and B is a subset of A, then A holds B. {
A → B }
Rule 2 Augmentation
If A hold B and C is a set of attributes, then AC holds BC. {AC
→ BC}
It means that attribute in dependencies does not change the
basic dependencies.
Rule 3 Transitivity
If A holds B and B holds C, then A holds C.
If {A → B} and {B → C}, then {A → C}
A holds B {A → B} means that A functionally determines B.
A. Primary Rules

14. B. Secondary Rules
Rule 1 Union
If A holds B and A holds C, then A holds BC.
If{A → B} and {A → C}, then {A → BC}
Rule 2 Decomposition
If A holds BC and A holds B, then A holds C.
If{A → BC} and {A → B}, then {A → C}
Rule 3 Pseudo Transitivity
If A holds B and BC holds D, then AC holds D.
If{A → B} and {BC → D}, then {AC → D}

15. Closure of Functional Dependencies
 Closure set F -> F+
 The set of all FDs that can be inferred from F
 We denote the closure of F by F+
 F+ is a superset of F

16.  Assume relation R (A, B, C)
 Given FDs : A → B, B → C, C → A
 What are the possible keys for R ?
 Step 1: find the closure of A , B, C
 A+ = AB =ABC
 B+ = BC =ABC
 C+ = CA =CAB
 Step 2: If X+ determines all the attributes then X
is a candidate key
 So all A, B and C are candidate keys for relation R.

17.  Assume relation R (A, B, C,D)
 Given FDs : A → B, B → D, C → A
 What are the possible keys for R ?
 A+ = ABD
 B+ = BD
 C+ = CABD
 D+ = D

18. Anomalies
 There are three types of anomalies that occur when
the database is not normalized. These are –
Insertion, update and deletion anomaly. Let’s take
an example to understand this.

19. S_I
D
S_nam
e
C_I
d
C_nam
e
F_i
d
F_nam
e
Salar
y
S1 A C1 C F1 T 5K
S2 B C1 C F1 T 5K
S3 A C2 C++ F2 T 10K
S4 B C1 C F1 T 5K
C3 Java F3 S 8K

20. Anomalies
1. Updation Anomaly:
- if we want to update F1 salary to 7 K , we need to
perform updation of all redundant copies.
2. Deletion Anomaly:
- if we want to delete s3 tuple then we are loosing
the information of f2.
3. Insert Anomaly:
-Not possible to insert F3 information without Sid.

21. To avoid redundancy we use
the concept of decomposition
Fid Fna
me
Cid Cna
me
Sala
ry
F1 T C1 C 5K
F2 T C2 C++ 10K
Sid Sna
me
Cid
S1 A C1
S2 B C1
S3 A C2
S4 B C1

22. Normalization
 Normalization is a set of rules to systematically
achieve a good design.
 If these rules are followed, then the DB design is
guarantee to avoid several problems:
 Inconsistent data
 Anomalies: insert, delete and update
 Redundancy:

23. Normalization
 Normalization is a process of organizing the data
in database to avoid data redundancy, insertion
anomaly, update anomaly & deletion anomaly.
 Here are the steps for normalization:
 First normal form(1NF)
 Second normal form(2NF)
 Third normal form(3NF)
 Boyce & Codd normal form (BCNF)
 Fourthnormal form(4NF)
 Fifth normal form(5NF)

24. Types of Functional Dependencies upto
BCNF
 Trivial functional dependency:
 Non-trivial functional dependency:
 Transitive dependency:

25. Trivial Functional dependency:
 The Trivial dependency is a set of attributes which
are called a trivial if the set of attributes are
included in that attribute.
 So, X -> Y is a trivial functional dependency if Y is
a subset of X.

26. Example:
Emp_id Emp_name
AS555 Harry
AS811 George
AS999 Kevin
Consider this table with two columns Emp_id and Emp_name.
{Emp_id, Emp_name} -> Emp_id is a trivial functional dependency as
Emp_id is a subset of {Emp_id,Emp_name}.

27. Non trivial functional dependency
 Functional dependency which also known as a
nontrivial dependency occurs when A->B holds
true where B is not a subset of A.
 In a relationship, if attribute B is not a subset of
attribute A, then it is considered as a non-trivial
dependency.

28. Example:
Company CEO Age
Microsoft Satya Nadella 51
Google Sundar Pichai 46
Apple Tim Cook 57
(Company} -> {CEO} (if we know the Company, we
knows the CEO name)
But CEO is not a subset of Company, and hence
it's non-trivial functional dependency.

29. Transitive dependency:
 A transitive is a type of functional dependency
which happens when t is indirectly formed by two
functional dependencies.
Company CEO Age
Microsoft Satya Nadella 51
Google Sundar Pichai 46
Alibaba Jack Ma 54

30.  Company} -> {CEO} (if we know the compay,
we know its CEO's name)
 {CEO } -> {Age} If we know the CEO, we know
the Age
 Therefore according to the rule of rule of
transitive dependency:
 { Company} -> {Age} should hold, that makes
sense because if we know the company name, we
can know his age.
Note:
You need to remember that transitive
dependency can only occur in a relation of three
or more attributes.

31. Normalization
 Normalization is a process of organizing the data
in database to avoid data redundancy, insertion
anomaly, update anomaly & deletion anomaly.
 Here are the steps for normalization:
 First normal form(1NF)
 Second normal form(2NF)
 Third normal form(3NF)
 Boyce & Codd normal form (BCNF)
 Fourthnormal form(4NF)
 Fifth normal form(5NF)
However (1NF, 2NF,
3NF) are sufficient for
normalization.

32. First normal form (1NF)
 Relation R is in 1NF only if
 an attribute (column) of a R does not contain multiple
values.
OR
 An attribute of R should hold only atomic values

33. Consider the student table
S_id S_name Course
S1 A C
S2 B C++/java
S3 C C++/python
Multi valued
attribute
Here , Relation student is not in 1NF as each attribute of a
table must have atomic (single) values and course attribute
does not satisfies.

34. Convert student into 1NF
S_id S_name Course
S1 A C
S2 B C++
S2 B java
S3 C C++
S3 C python
single
valued
attribute
Now , Relation student is in 1NF

35. Disadvantages
 Relation student still suffering from redundancy
problem.
 Find the functional dependancy from student table
?
 Sid->Sname T
 Sid,Course ->Sname T
 Sid,Sname-> course F
 Sname->Sid T

36. Second normal form (2NF)
 Relation R is in 2NF only if
 R is in 1NF (First normal form)
 No non-prime(non key) attribute is dependent on the
proper subset of any candidate key of table.
OR
 R should not contain any partial dependancy.
OR
 All non key attribute are fully dependant on candidate
key of the table.

37. Prime(key) and Non prime(Non key)
attributes
 Suppose Candidate key for relation R(A,B,C,D,E)
is AE
 Then prime attribute are : A, E
 Then Non-prime attribute are : B,C,D

38. Partial Dependancy
 Suppose Candidate key for relation R(A,B,C,D,E)
is AE
 If A-> C , here A is the subset of candidate key AE
and C is non prime attribute this is called partial
dependancy .
 If AE-> C , here AE is the candidate key and C is
non prime attribute this is called fully dependancy.

39. student in 1NF
S_id S_nam
e
Course
S1 A C
S2 B C++
S2 B java
S3 C C++
S3 C python
S_id -> S_name
S_id,Course ->
S_name
Here S_id,Course is
candidate key
Non key attribute =
S_name
And also
S_id -> S_name
So , Relation student is not in 2NF decompose the
relation

40. Convert student table into R1 and R2
S_id S_name
S1 A
S2 B
S3 C
S_id Course
S1 C
S2 C++
S2 java
S3 C++
S3 python
R1(Sid->Sname)
CK= Sid
R2(S_id,course->S_id,course)
CK = S_id,course
No partial dependency so R1 and R2 are in 2NF

41. Third Normal Form
 Relation R is in 3NF only if
 R is in 2NF (First normal form)
 Transitive functional dependency of non-prime
attribute on any super key should be removed.
OR
 R should not contain any transitive dependency.
OR
 For each non trivial functional dependency X->Y then
either X must be candidate key or super key or Y
must be prime attribute.

42. Transitive Dependency
 Let R be the relational schema with non trivial
functional dependency X->Y is transitive
dependency if
 1. X is not a candidate key
OR
 2. Y is non-prime attibute.
 Eg : Mob_no,name->name

43. Example of to check transitive
dependency.
 Relation R(A,B,C,D)
 And FD’s {A->B , B-> C, C-> D , D-> A}
 Here candidate keys are A,B,C,D
 So no transitive dependency.

44. Example to check 3NF
 Relation R(A,B,C,D) and FD’s ={AB->C, C->D}
 Here candidate key AB
 In AB->C, here AB is candidate key
 In C->D, here C is not a candidate key and D is
non prime attribute
 Here Transitive dependency exist so relation R is
not in 3NF

45. Solution: Decompose the relation
 R1(A,B,C) R2(C,D)
 FD’s={AB->C} FD’s={C->D}
 Ck=AB CK= C
 Now both relations are in 3NF.

46. Boyce & Codd normal form (BCNF)
 Relation R is in BCNF only if
 it is in 3NF
 and for every functional dependency X->Y,
X should be the candidate key or super key of the
table.
 It is an advance version of 3NF that’s why it is also
referred as 3.5NF. Also BCNF is stricter than 3NF.

47. Example to check BCNF
 Relation R(A,B,C,D) and FD’s ={AB->C, C->D}
 Here candidate key AB
 In AB->C, here AB is candidate key
 In C->D, here C is not a candidate key
 R is not in BCNF so decompose the relation

48. Solution: Decompose the relation
 R1(A,B,C) R2(C,D)
 FD’s={AB->C} FD’s={C->D}
 Ck=AB CK= C
 Now both relations R1 and R2 are in BCNF

49. Check the highest Normal Form
Example 1
 Consider a relation R(A,B,C,D,E)
 and FD’s ={AB->C, C->D, D->E, E->A, D->B}
 Step 1. Identify the Candidate key.
 Step 2. make a table to check NF from BCNF to 1NF

50. Check the highest Normal Form
 Consider a relation R(A,B,C,D,E)
 and FD’s ={AB->C, C->D, D->E, E->A, D->B}
 Step 1. Identify the Candidate key.
 Candidate keys: AB,C, D, EB
 AB+ = ABC=ABCD= ABCDE
 C+ = CD= CDE= CDEA=CDEAB
 D+ = DEB=DEBA = DEBAC
 E+ = EA , but EB+ = EAB= EABC= EABCD

51.  Step 2. make a table to check NF from BCNF to
1NF
 Candidate keys: AB,C, D, EB
AB->C C->D D->E E->A D->B
BCNF X
3NF
2NF
1NF
Relation R is in 3 NF as E is not a candidate key but A is a
prime attribute

52. Check the highest Normal Form
Example 2
 Consider a relation R(A,B,C,D,E,F)
 and FD’s ={AB->CD, D->E, E->F, E->A}
 Step 1. Identify the Candidate key.
 Step 2. make a table to check NF from BCNF to 1NF

53. Check the highest Normal Form
 Consider a relation R(A,B,C,D,E)
 and FD’s ={AB->CD, D->E, E->F, E->A}
 Step 1. Identify the Candidate key.
 Candidate keys: AB,EB,BD
 AB+ = ABCD= ABCDEF
 D+ = DE=DEFA , but BD+ = BDEFAC
 E+ = EFA , but EB+ = EFAB= EFABCD

54.  Step 2. make a table to check NF from BCNF to
1NF
 Candidate keys: AB,EB,BD
AB->CD D->E E->F E-> A check
BCNF X X X Candidate
key(LHS)
3NF X LHS =CK or
RHS prime
2NF X Partial
dependency
1NF
Relation R is in 1 NF as assumed it does not contain multi
valued attribute

55. Question
 Find the highest normal form of a
relation R(A,B,C,D,E) with FD set as {BC->D,
AC->BE, B->E}

56. Question
 Find the highest normal form of a
relation R(A,B,C,D,E) with FD set {A->D, B->A,
BC->D, AC->BE}

57. Question
 Find the highest normal form of a
relation R(A,B,C,D,E) with FD set {B->A, A->C,
BC->D, AC->BE}
 B+
 A+
 BC+
 AC+

58. Denormalization
 Normalization is the technique of dividing the data
into multiple tables to reduce the data redundancy
and inconsistency and to achieve data integrity.
 Denormalization increases redundancy as it is used
to combine multiple table data into one so that it
can be queried quickly.
 It is an optimization technique in which we add
redundant data to one or more tables.

59. Desirable Properties of Decomposition
 if we combine the decomposed table (de-
normalization), it should give the original table int
terms of rows and columns.
 the following two properties are described as:
 Lossless Join Decomposition Property
 Dependency Preserving Property

62. Lossless vs. Lossy Decomposition
 Consider relation R is divided into R1 and R2
 Lossless Decomposition
 R1 natural join R2 should create exactly R
 Lossy Decomposition
 R1 natural join R2 adds more records (or delete
records ) from R

65. To ensure lossless decomposition
 The common columns must be candidate key in
one of the two relations

66. Dependency preserving
 every dependency in original table must be
preserved or say, every dependency must be
satisfied by at least one decomposed table.

67. Dependency preserving
 Consider R be the original relational schema
having FD set F. Let R1 and R2 having FD set F1
and F2 respectively, are the decomposed sub-
relations of R.
 The decomposition of R is said to be preserving if
 F1 ∪ F2 ≡ F {Dependency Preserving}
 If F1 ∪ F2 ⊂ F {NOT Preserving Dependency}
 and F1 ∪ F2 ⊃ F {this is not possible}

68. Question 1
 Consider R(ABC) has following FD's
 F = {A→B, B → C, C → A}
 D = {AB,BC}
 check whether decomposition is dependency
preserving or not

69. Decomposed relations
AB(R1) BC(R2)
A+: A→A , A → B
B+: B →B , B→C,
B→A
AB+: AB->AB
B+: B→B , B→C
C+: C→C, C→A, C→B
BC+: BC→BC
F = {A→B, B → C, C → A}
F1 ∪ F2 ∪ F3 = A → B, B→A, B→C, C→B
To check C →A,find closure of C in F1 ∪ F2 ∪ F3
C + : CBA, C->A exist so dependency. preserving

70. Decomposed relations
AB(R1) BC(R2)
A+: A → B
B+: B→A
B+: B→C
C+: C→B
F = {A→B, B → C, C → A}
F1 ∪ F2 = A → B, B→A, B→C, C→B
To check C →A,find closure of C in F1 ∪ F2
C + : CBA, C->A exist so dependency. preserving

71. Question 2
 Consider R(ABCD) has following FD's
 F = {A→B, B → C, C → D,D → B}
 D = {AB,BC,BD}
 check whether decomposition is dependency
preserving or not

72. Decomposed relations
AB(R1) BC(R2) BD(R3)
A+: A→A , A
→ B
B+: B →B ,
B→C
AB+: AB->AB
B+: B→B ,
B→C
C+: C→C,
C→A, C→B
BC+: BC→BC
B+: B→B ,
B→C, B→D
D+: D→D,D→B
D→C
BD+: BD→BD
F = {A→B, B → C, C → D, D → B}
F1 ∪ F2 ∪ F3 = A → B, B→C, C→B, B→D,D→B
To check C → D,find closure of C in F1 ∪ F2 ∪ F3
C + : CBD, C->D exist so dependency. preserving

73. Decomposed relations
AB(R1) BC(R2) BD(R3)
A+: A → B B+:, B→C
C+: C→B
B+: B→D
D+: D→B
F = {A→B, B → C, C → D, D → B}
F1 ∪ F2 ∪ F3 = A → B, B→C, C→B, B→D,D→B
To check C → D,find closure of C in F1 ∪ F2 ∪ F3
C + : CBD, C->D exist so dependency. preserving

74. Question 3
 Consider R(ABCD) has following FD's
 F = {AB→CD, D→ A}
 D = {AD,BCD}
 check whether decomposition is dependency
preserving or not

75. Decomposed relations
AD(R1) BCD(R2)
A+: A→A , A → D
D+: D →D , D→A
AD+: AD->AD
B+: B→B
C+: C→C
D+: D→D
BC+: BC→BC
CD+: CD→CD
BC+: BC→BC
BD+: BD→BD, BD→C
F = {AB→CD, D→ A}
F1 ∪ F2 = D→A , BD→C
To check AB→CD, find closure of AB in F1 ∪ F2
AB + : AB , AB→CD cannot be determined so not preserving