2. Relational Data Model
• Two major strengths
• Three components
• Relation database based on Relational DM
3. The relational model
• Data is represented in the form of tables, and the model has 3
components
• Data structure – data are organised in the form of tables with
rows and columns
• Data manipulation – powerful operations (using the SQL
language) are used to manipulate data stored in the relations
• Data integrity – facilities are included to specify business rules
that maintain the integrity of data when they are manipulated
5. • EMP(ENO, ENAME, TITLE, SAL, PNO, RESP,
DUR)
• PROJ(PNO,PNAME, BUDGET)
• Each attribute in both relations has a domain
• Domains need not to be distinct
6. Database Table Keys
Definition:
A key of a relation is a subset of attributes with the
following attributes:
• Unique identification
• Non-redundancy
9. 9
In this chapter, you will learn:
• What normalization is and what role it plays in the
database design process
• About the normal forms 1NF, 2NF, 3NF, BCNF,
and 4NF
• How normal forms can be transformed from lower
normal forms to higher normal forms
• How normalization and ER modeling are used
concurrently to produce a good database design
• How some situations require denormalization to
generate information efficiently
10. 10
Database Tables and Normalization
• Normalization
– Process for evaluating and correcting table
structures to minimize data redundancies
• Reduces data anomalies
– Works through a series of stages called normal
forms:
• First normal form (1NF)
• Second normal form (2NF)
• Third normal form (3NF)
11. 11
Database Tables and Normalization
(continued)
• Normalization (continued)
– 2NF is better than 1NF; 3NF is better than 2NF
– For most business database design purposes, 3NF
is as high as we need to go in normalization
process
– Highest level of normalization is not always most
desirable
12. 12
The Need for Normalization
• Example: Company that manages building
projects
– Charges its clients by billing hours spent on each
contract
– Hourly billing rate is dependent on employee’s
position
– Periodically, report is generated that contains
information displayed in Table 5.1
15. 15
The Need for Normalization
(continued)
• Structure of data set in Figure 5.1 does not
handle data very well
• The table structure appears to work; report
generated with ease
• Unfortunately, report may yield different
results depending on what data anomaly has
occurred
16. 16
The Normalization Process
• Each table represents a single subject
• No data item will be unnecessarily stored in
more than one table
• All attributes in a table are dependent on the
primary key
18. 18
Conversion to First Normal Form
• Repeating group
– Derives its name from the fact that a group of
multiple entries of same type can exist for any
single key attribute occurrence
• Relational table must not contain repeating
groups
• Normalizing table structure will reduce data
redundancies
• Normalization is three-step procedure
19. 19
Conversion to First Normal Form
(continued)
• Step 1: Eliminate the Repeating Groups
– Present data in tabular format, where each cell
has single value and there are no repeating groups
– Eliminate repeating groups, eliminate nulls by
making sure that each repeating group attribute
contains an appropriate data value
21. 21
Conversion to First Normal Form
(continued)
• Step 2: Identify the Primary Key
– Primary key must uniquely identify attribute value
– New key must be composed
22. 22
Conversion to First Normal Form
(continued)
• Step 3: Identify All Dependencies
– Dependencies can be depicted with help of a
diagram
– Dependency diagram:
• Depicts all dependencies found within given table
structure
• Helpful in getting bird’s-eye view of all relationships
among table’s attributes
• Makes it less likely that will overlook an important
dependency
24. 24
Conversion to First Normal Form
(continued)
• First normal form describes tabular format in which:
– All key attributes are defined
– There are no repeating groups in the table
– All attributes are dependent on primary key
• All relational tables satisfy 1NF requirements
• Some tables contain partial dependencies
– Dependencies based on only part of the primary key
– Sometimes used for performance reasons, but should be
used with caution
– Still subject to data redundancies
25. 25
Conversion to Second Normal Form
• Relational database design can be improved
by converting the database into second
normal form (2NF)
• Two steps
26. 26
Conversion to Second Normal Form
(continued)
• Step 1: Write Each Key Component
on a Separate Line
– Write each key component on separate line, then
write original (composite) key on last line
– Each component will become key in new table
27. 27
Conversion to Second Normal Form
(continued)
• Step 2: Assign Corresponding Dependent
Attributes
– Determine those attributes that are dependent on
other attributes
– At this point, most anomalies have been
eliminated
29. 29
Conversion to Second Normal Form
(continued)
• Table is in second normal form (2NF) when:
– It is in 1NF and
– It includes no partial dependencies:
• No attribute is dependent on only portion of primary
key
30. 30
Conversion to Third Normal Form
• Data anomalies created are easily eliminated
by completing three steps
• Step 1: Identify Each New Determinant
– For every transitive dependency, write its
determinant as PK for new table
• Determinant
– Any attribute whose value determines other values within a
row
31. 31
Conversion to Third Normal Form
(continued)
• Step 2: Identify the Dependent Attributes
– Identify attributes dependent on each
determinant identified in Step 1 and identify
dependency
– Name table to reflect its contents and function
32. 32
Conversion to Third Normal Form
(continued)
• Step 3: Remove the Dependent Attributes
from Transitive Dependencies
– Eliminate all dependent attributes in transitive
relationship(s) from each of the tables that have
such a transitive relationship
– Draw new dependency diagram to show all tables
defined in Steps 1–3
– Check new tables as well as tables modified in
Step 3 to make sure that each table has
determinant and that no table contains
inappropriate dependencies
34. 34
Conversion to Third Normal Form
(continued)
• A table is in third normal form (3NF) when
both of the following are true:
– It is in 2NF
– It contains no transitive dependencies
35. Formal Relational Query Languages
• Two mathematical Query Languages form the
basis for “real” languages (e.g. SQL), and for
implementation:
– Relational Algebra: More operational, very
useful for representing execution plans.
– Relational Calculus: Lets users describe what
they want, rather than how to compute it.
(Non-operational, declarative.)
36. Example Instances
sid sname rating age
22 dustin 7 45.0
31 lubber 8 55.5
58 rusty 10 35.0
sid sname rating age
28 yuppy 9 35.0
31 lubber 8 55.5
44 guppy 5 35.0
58 rusty 10 35.0
sid bid day
22 101 10/10/96
58 103 11/12/96
R1
S1
S2
• “Sailors” and “Reserves”
relations for our
examples.
• We’ll use positional or
named field notation,
assume that names of
fields in query results are
`inherited’ from names of
fields in query input
relations.
37. Relational Algebra
• Basic operations:
– Selection ( ) Selects a subset of rows from relation.
– Projection ( ) Deletes unwanted columns from relation.
– Cross-product ( ) Allows us to combine two relations.
– Set-difference ( ) Tuples in reln. 1, but not in reln. 2.
– Union ( ) Tuples in reln. 1 and in reln. 2.
• Additional operations:
– Intersection, join, division, renaming: Not essential, but (very!)
useful.
• Since each operation returns a relation, operations can be composed!
(Algebra is “closed”.)
38. Projection
sname rating
yuppy 9
lubber 8
guppy 5
rusty 10
sname rating
S
,
( )2
age
35.0
55.5
age S( )2
• Deletes attributes that are not in
projection list.
• Schema of result contains exactly the
fields in the projection list, with the
same names that they had in the
(only) input relation.
• Projection operator has to eliminate
duplicates! (Why??)
– Note: real systems typically don’t
do duplicate elimination unless
the user explicitly asks for it.
(Why not?)
39. Selection
rating
S
8
2( )
sid sname rating age
28 yuppy 9 35.0
58 rusty 10 35.0
sname rating
yuppy 9
rusty 10
sname rating rating
S
,
( ( ))
8
2
• Selects rows that satisfy
selection condition.
• No duplicates in result!
(Why?)
• Schema of result identical
to schema of (only) input
relation.
• Result relation can be the
input for another
relational algebra
operation! (Operator
composition.)
40. Union, Intersection, Set-Difference
• All of these operations take
two input relations, which
must be union-compatible:
– Same number of fields.
– `Corresponding’ fields have
the same type.
• What is the schema of result?
sid sname rating age
22 dustin 7 45.0
31 lubber 8 55.5
58 rusty 10 35.0
44 guppy 5 35.0
28 yuppy 9 35.0
sid sname rating age
31 lubber 8 55.5
58 rusty 10 35.0
S S1 2
S S1 2
sid sname rating age
22 dustin 7 45.0
S S1 2
41. Cross-Product
• Each row of S1 is paired with each row of R1.
• Result schema has one field per field of S1 and R1, with
field names `inherited’ if possible.
– Conflict: Both S1 and R1 have a field called sid.
( ( , ), )C sid sid S R1 1 5 2 1 1
(sid) sname rating age (sid) bid day
22 dustin 7 45.0 22 101 10/10/96
22 dustin 7 45.0 58 103 11/12/96
31 lubber 8 55.5 22 101 10/10/96
31 lubber 8 55.5 58 103 11/12/96
58 rusty 10 35.0 22 101 10/10/96
58 rusty 10 35.0 58 103 11/12/96
Renaming operator:
42. Joins
• Combine information from two or more tables
• Example: students enrolled in courses:
S1 S1.sid=E.studidE
Sid name gpa
50000 Dave 3.3
53666 Jones 3.4
53688 Smith 3.2
53650 Smith 3.8
53831 Madayan 1.8
53832 Guldu 2.0
cid grade studid
Carnatic101 C 53831
Reggae203 B 53832
Topology112 A 53650
History 105 B 53666
S1
E
43. Joins
sid name gpa cid grade studid
53666 Jones 3.4 History105 B 53666
53650 Smith 3.8 Topology112 A 53650
53831 Madayan 1.8 Carnatic101 C 53831
53832 Guldu 2.0 Reggae203 B 53832
Sid name gpa
50000 Dave 3.3
53666 Jones 3.4
53688 Smith 3.2
53650 Smith 3.8
53831 Madayan 1.8
53832 Guldu 2.0
cid grade studid
Carnatic101 C 53831
Reggae203 B 53832
Topology112 A 53650
History 105 B 53666
S1
E
44. Joins
• Condition Join:
• Result schema same as that of cross-product.
• Fewer tuples than cross-product, might be able to
compute more efficiently
• Sometimes called a theta-join.
R c S c R S ( )
(sid) sname rating age (sid) bid day
22 dustin 7 45.0 58 103 11/12/96
31 lubber 8 55.5 58 103 11/12/96
S R
S sid R sid
1 1
1 1
. .
45. Joins
• Equi-Join: A special case of condition join where the condition c
contains only equalities.
• Result schema similar to cross-product, but only one copy of
fields for which equality is specified.
• Natural Join: Equijoin on all common fields.
sid sname rating age bid day
22 dustin 7 45.0 101 10/10/96
58 rusty 10 35.0 103 11/12/96
S R
sid
1 1
46. Division
• Not supported as a primitive operator, but useful for expressing
queries like:
Find sailors who have reserved all boats.
• Let A have 2 fields, x and y; B have only field y:
– A/B =
– i.e., A/B contains all x tuples (sailors) such that for every y tuple
(boat) in B, there is an xy tuple in A.
– Or: If the set of y values (boats) associated with an x value
(sailor) in A contains all y values in B, the x value is in A/B.
• In general, x and y can be any lists of fields; y is the list of fields in B,
and x y is the list of fields of A.
x x y A y B| ,
48. Expressing A/B using Basic Operators
• For A/B, compute all x values that are not
`disqualified’ by some y value in B.
– x value is disqualified if by attaching y value
from B, we obtain an xy tuple that is not in A.
50. Find names of sailors who’ve reserved a red
boat
• Information about boat color only available in
Boats; so need an extra join:
sname color red
Boats serves Sailors((
' '
) Re )
A more efficient solution:
sname sid bid color red
Boats s Sailors( ((
' '
) Re ) )
A query optimizer can find this, given the first solution!
51. Find sailors who’ve reserved a red or a green boat
• Can identify all red or green boats, then find
sailors who’ve reserved one of these boats:
( ,(
' ' ' '
))Tempboats
color red color green
Boats
sname Tempboats serves Sailors( Re )
Can also define Tempboats using union! (How?)
What happens if is replaced by in this query?
52. Find the names of sailors who’ve reserved all boats
• Uses division; schemas of the input relations to /
must be carefully chosen:
( , (
,
Re ) / ( ))Tempsids
sid bid
serves
bid
Boats
sname Tempsids Sailors( )
To find sailors who’ve reserved all ‘Interlake’ boats:
/ (
' '
)
bid bname Interlake
Boats
.....
54. Summary
• The relational model has rigorously defined query
languages that are simple and powerful.
• Relational algebra is more operational; useful as
internal representation for query evaluation plans.
• Several ways of expressing a given query; a query
optimizer should choose the most efficient version.
