uniT 4 (1).pptx

1
Teaching Innovation - Entrepreneurial - Global
The Centre for Technology enabled Teaching & Learning , N Y S S, India
(Department for Technology Enhanced Learning)

DEPARTMENT OF COMPUTER TECHNOLOGY
VI-SEMESTER
2
CHAPTER NO.3
Relational Model

CHAPTER 2:- SYLLABUS
.
Structure of Relational Databases
1
The Relational Algebra
2
Tuple Relational Calculus
3
Relational Commercial Languages
4
3
SQL (DDL, DML, DCL)
5

CHAPTER-2 SPECIFIC OBJECTIVE / COURSE OUTCOME
Understand the Structure of Relational Databases.
1
Understand and Learned Relational Commercial Languages.
3
4
The student will be able to:
Fundamental Relational-Algebra-Operations.
2

Relational Model
 A particular way of structuring data (relations)
 Simple
 Mathematically based
Expressions (queries) can be analyzed by DBMS
Transformed to equivalent expressions automatically (query
optimization)
Optimizers have limits (=> programmer needs to know how
queries are evaluated and optimized)
LECTURE 1 Structure of Relational Database
5

Example of a Relation
6

Attribute Types
 Each attribute of a relation has a name
 The set of allowed values for each attribute is called the
domain of the attribute
 Attribute values are (normally) required to be atomic; that is,
indivisible
E.g. the value of an attribute can be an account
number,
but cannot be a set of account numbers
 Domain is said to be atomic if all its members are atomic
 The special value null is a member of every domain
 The null value causes complications in the definition of many
operations
We shall ignore the effect of null values in our main
7

Relation Schema
 Formally, given domains D1, D2, …. Dn a relation r is a subset
of
D1 x D2 x … x Dn
Thus, a relation is a set of n-tuples (a1, a2, …, an) where each
ai  Di
 Schema of a relation consists of
attribute definitions
name
type/domain
integrity constraints
8

Relation Instance
 The current values (relation instance) of a relation are
specified by a table
 An element t of r is a tuple, represented by a row in a table
 Order of tuples is irrelevant (tuples may be stored in an
arbitrary order)
Jones
Smith
Curry
Lindsay
customer_name
Main
North
North
Park
customer_street
Harrison
Rye
Rye
Pittsfield
customer_city
customer
attributes
(or columns)
tuples
(or rows)
9

Database
 A database consists of multiple relations
 Information about an enterprise is broken up into parts,
with each relation storing one part of the information
E.g.
account : information about accounts
depositor : which customer owns which account
customer : information about customers
10

The customer Relation
11

The depositor Relation
12

Why Split Information Across Relations?
 Storing all information as a single relation such as
bank(account_number, balance, customer_name, ..)
results in
repetition of information
e.g.,if two customers own an account (What gets
repeated?)
the need for null values
e.g., to represent a customer without an account
 Normalization theory (Chapter 7) deals with how to design
relational schemas
13

Keys
 Let K  R
 K is a superkey of R if values for K are sufficient to identify a unique
tuple of each possible relation r(R)
by “possible r ” we mean a relation r that could exist in
the enterprise we are modeling.
Example: {customer_name, customer_street} and
{customer_name}
are both superkeys of Customer, if no two customers
can possibly have the same name
In real life, an attribute such as customer_id would be used
instead of customer_name to uniquely identify customers,
but we omit it to keep our examples small, and instead
14

Keys (Cont.)
 K is a candidate key if K is minimal
Example: {customer_name} is a candidate key for
Customer, since it is a superkey and no subset of it is a
superkey.
 Primary key: a candidate key chosen as the principal
means of identifying tuples within a relation
Should choose an attribute whose value never,
or very rarely, changes.
E.g. email address is unique, but may change
15

Foreign Keys
 A relation schema may have an attribute that corresponds to
the primary key of another relation. The attribute is called a
foreign key.
E.g. customer_name and account_number
attributes of depositor are foreign keys to
customer and account respectively.
Only values occurring in the primary key attribute
of the referenced relation may occur in the
foreign key attribute of the referencing relation.
16

Schema Diagram
17

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”.)





LECTURE 2 Relational Algebra
18

Projection
 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??, what are the
consequences?)
 Note: real systems typically
don’t do duplicate elimination
unless the user explicitly asks
for it. (Why not?)
sname rating
yuppy 9
lubber 8
guppy 5
rusty 10
sname rating
S
,
( )
2
age
35.0
55.5
age S
( )
2
19

Selection
 Selects rows that satisfy
selection condition.
 Schema of result identical
to schema of (only) input
relation.
 Result relation can be the
input for another
relational algebra
operation! (Operator
composition.)
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
20

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?
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
31 lubber 8 55.5
58 rusty 10 35.0
S S
1 2

S S
1 2

22 dustin 7 45.0
S S
1 2

21

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 R
1 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:
22

Joins
 Condition Join:
 Result schema same as that of cross-product.
 Fewer tuples than cross-product. Filters tuples not satisfying the
join condition.
 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


. .

23

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
)
1
1
(
,..
,
,..,
R
S
sid
bid
age
sid



24

Division
 Not supported as a primitive operator, but useful for expressing
queries like:
Find sailors who have reserved all boats.
 Precondition: in A/B, the attributes in B must be included in the
schema for A. Also, the result has attributes A-B.
SALES(supId, prodId);
PRODUCTS(prodId);
Relations SALES and PRODUCTS must be built using
projections.
SALES/PRODUCTS: the ids of the suppliers supplying
ALL products.
25

Examples of Division A/B
sno pno
s1 p1
s1 p2
s1 p3
s1 p4
s2 p1
s2 p2
s3 p2
s4 p2
s4 p4
pno
p2
pno
p2
p4
pno
p1
p2
p4
sno
s1
s2
s3
s4
sno
s1
s4
sno
s1
A
B1
B2
B3
A/B1 A/B2 A/B3
26

Expressing A/B Using Basic Operators
 Division is not essential op; just a useful shorthand.
 (Also true of joins, but joins are so common that
systems implement joins specially. Division is NOT
implemented in SQL).
 Idea: For SALES/PRODUCTS, compute all products such that there
exists at least one supplier not supplying it.
x value is disqualified if by attaching y value from B,
we obtain an xy tuple that is not in A.
)
)
Pr
)
(
(( Sales
oducts
Sales
sid
sid
A 

 

The answer is sid(Sales) - A
27

Find names of sailors who’ve reserved boat
#103
• Solution 1:  
sname bid
serves Sailors
(( Re ) )
103


 Solution 2:  
( , Re )
Temp serves
bid
1
103

 ( , )
Temp Temp Sailors
2 1 

 sname Temp
( )
2
 Solution 3:  
sname bid
serves Sailors
( (Re ))
103


28

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!
29

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?
 
30

Find sailors who’ve reserved a red and a green
boat
Previous approach won’t work! Must identify
sailors who’ve reserved red boats, sailors
who’ve reserved green boats, then find the
intersection (note that sid is a key for Sailors):
  
( , ((
' '
) Re ))
Tempred
sid color red
Boats serves



 sname Tempred Tempgreen Sailors
(( ) )
 

  
( , ((
' '
) Re ))
Tempgreen
sid color green
Boats serves



31

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

.....
32

Tuple Relational Calculus
 A nonprocedural query language, where each query is of
the form
{t | P (t ) }
 It is the set of all tuples t such that predicate P is true for t
 t is a tuple variable, t [A ] denotes the value of tuple t on
attribute A
 t  r denotes that tuple t is in relation r
 P is a formula similar to that of the predicate calculus
LECTURE 3 Tuple Relational Calculus
33

Predicate Calculus Formula
1. Set of attributes and constants
2. Set of comparison operators: (e.g., , , , , , )
3. Set of connectives: and (), or (v)‚ not ()
4. Implication (): x  y, if x if true, then y is true
x  y x v y
5. Set of quantifiers:
 t  r (Q (t ))  ”there exists” a tuple in t in relation r
such that predicate Q (t ) is true
t r (Q (t )) Q is true “for all” tuples t in relation r
34

Banking Example
branch (branch_name, branch_city, assets )
customer (customer_name,
customer_street, customer_city )
account (account_number, branch_name,
balance )
loan (loan_number, branch_name, amount )
depositor (customer_name,
account_number )
borrower (customer_name, loan_number )
35

Example Queries
 Find the loan_number, branch_name, and amount
for loans of over $1200
 Find the loan number for each loan of an amount greater than $120
{t |  s loan (t [loan_number ] = s [loan_number ]  s [amount ]
1200)}
Notice that a relation on schema [loan_number ] is implicitly define
by
the query
{t | t  loan  t [amount ]  1200}
36

Example Queries
 Find the names of all customers having a loan, an
account, or both at the bank
{t | s  borrower ( t [customer_name ] = s [customer_name ])
 u  depositor ( t [customer_name ] = u [customer_name] )
 Find the names of all customers who have a loan and an
account
at the bank
{t | s  borrower ( t [customer_name ] = s [customer_name ])
 u  depositor ( t [customer_name ] = u [customer_name ])
37

Example Queries
Find the names of all customers having a loan
at the Perryridge branch
{t | s  borrower (t [customer_name ] = s [customer_name ]
 u  loan (u [branch_name ] = “Perryridge”
 u [loan_number ] = s [loan_number ]))
 not v  depositor (v [customer_name ] =
t [customer_name ])}
 Find the names of all customers who have a loan at the
Perryridge branch, but no account at any branch of the bank
{t | s  borrower (t [customer_name ] = s [customer_name ]
 u  loan (u [branch_name ] = “Perryridge”
 u [loan_number ] = s [loan_number ]))}
38

Example Queries
 Find the names of all customers having a loan from the
Perryridge branch, and the cities in which they live
{t | s  loan (s [branch_name ] = “Perryridge”
 u  borrower (u [loan_number ] = s [loan_number ]
 t [customer_name ] = u [customer_name ])
  v  customer (u [customer_name ] = v [customer_name ]
 t [customer_city ] = v [customer_city ])))}
39

Example Queries
Find the names of all customers who have an
account at all branches located in Brooklyn:
{t |  r  customer (t [customer_name ] = r [customer_name ]) 
(  u  branch (u [branch_city ] = “Brooklyn” 
 s  depositor (t [customer_name ] = s [customer_name ]
  w  account ( w[account_number ] = s [account_number ]
 ( w [branch_name ] = u [branch_name ]))))}
40

Safety of Expressions
 It is possible to write tuple calculus expressions that
generate infinite relations.
For example, { t |  t r } results in an infinite relation if
the domain of any attribute of relation r is infinite
 To guard against the problem, we restrict the set of
allowable expressions to safe expressions.
 An expression {t | P (t )} in the tuple relational calculus is
safe if every component of t appears in one of the
relations, tuples, or constants that appear in P
NOTE: this is more than just a syntax condition.
E.g. { t | t [A] = 5  true } is not safe --- it defines an
infinite set with attribute values that do not appear
in any relation or tuples or constants in P.
41

Domain Relational Calculus
 A nonprocedural query language equivalent in power to
the tuple relational calculus
 Each query is an expression of the form:
{  x1, x2, …, xn  | P (x1, x2, …, xn)}
x1, x2, …, xn represent domain variables
P represents a formula similar to that of the
predicate calculus
LECTURE 3 Domain Relational Calculus
42

Example Queries
 Find the loan_number, branch_name, and amount for
loans of over $1200
 Find the names of all customers who have a loan from the Perryridge branch and the
loan amount:
 { c, a  |  l ( c, l   borrower  b ( l, b, a   loan 
b = “Perryridge”))}
 { c, a  |  l ( c, l   borrower   l, “ Perryridge”, a   loan)}
{ c  |  l, b, a ( c, l   borrower   l, b, a   loan  a > 1200)}
 Find the names of all customers who have a loan of over $1200
{ l, b, a  |  l, b, a   loan  a > 1200}
43

Example Queries
 Find the names of all customers having a loan, an account, or
both at the Perryridge branch:
{ c  |  s,n ( c, s, n   customer) 
 x,y,z ( x, y, z   branch  y = “Brooklyn”) 
 a,b ( x, y, z   account   c,a   depositor)}
 Find the names of all customers who have an account at
all
branches located in Brooklyn:
{ c  |  l (  c, l   borrower
  b,a ( l, b, a   loan  b = “Perryridge”))
  a ( c, a   depositor
  b,n ( a, b, n   account  b = “Perryridge”))}
44

Relational Query Languages
 A major strength of the relational model: supports simple,
powerful querying of data.
 Queries can be written intuitively, and the DBMS is
responsible for efficient evaluation.
 The key: precise semantics for relational queries.
 Allows the optimizer to extensively re-order
operations, and still ensure that the answer does not
change.
45

Relational Query Languages
 Query languages: Allow manipulation and retrieval of
data from a database.
 Relational model supports simple, powerful QLs:
 Strong formal foundation based on logic.
 Allows for much optimization.
 Query Languages != programming languages!
 QLs not expected to be “Turing complete”.
 QLs not intended to be used for complex calculations.
 QLs support easy, efficient access to large data sets.
46
LECTURE 3 Relational Commercial
Languages

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.)
47
Languages

The SQL Query Language
 Developed by IBM’s System-R Project in the 1970s
 Need for a standard since it is used by many vendors
 Standards:
 SQL-86
 SQL-89 (minor revision)
 SQL-92 (major revision)
 SQL-99 (major extensions, current standard)
 Find Out what is SQL-MM!!
48
Languages

 SQL consists of:
DDL (Data Definition Language)
Create conceptual schema
DML (Data Manipulation Language)
Relational operators
Insert, Delete, Update
VDL (View Definition Language)
Specify user views & their mapping to the conceptual
schema (in most DBMSs, done by DDL)
SDL (Storage Definition Language)
File organization
Indexes 49
Languages

 SDL is being removed from SQL
 DMLs
High-level or nonprocedural (declarative)
Can be entered at the SQL > or can be embedded in a
general purpose programming language
Can specify & retrieve many records uin a single DML
statement & are hence called Set-at-a-time DMLs
Low-level or procedural
Must be embedded in a general purpose programming
language
Typically retrieves individual records or objects from
the DB & processes each separately
Therefore it need PL constructs like looping
Record-at-a-time DMLs 50
Languages

DMLs
 Whenever DML statements are embedded in a PL, that language is
called as the host language and the DML is called the Data
Sublanguage
 In object DBs, the host language & data sublanguage form one
integrated language – for eg. C++ with some extensions to support
database functionality
 Some RDBMSs also provide integrated languages – for eg. ORACLE’s
PL/SQL.
51
Languages

Data Definition Language (DDL) statements are used to define
the database structure or schema. Some examples:
* CREATE - to create objects in the database
* ALTER - alters the structure of the database
* DROP - delete objects from the database
* TRUNCATE - remove all records from a table, including all
spaces allocated for the records are removed
* COMMENT - add comments to the data dictionary
* RENAME - rename an object
52
LECTURE 4
Data Definition Language (DDL)
Relational Commercial
Languages

53
LECTURE 4
Data Manipulation Language (DML) statements are used for
managing data within schema objects. Some examples:
* SELECT - retrieve data from the a database
* INSERT - insert data into a table
* UPDATE - updates existing data within a table
* DELETE - deletes all records from a table, the space for the
records remain
* MERGE - UPSERT operation (insert or update)
* CALL - call a PL/SQL or Java subprogram
* EXPLAIN PLAN - explain access path to data
* LOCK TABLE - control concurrency
Data Manipulation Language (DML)
Languages

© Prof. Navneet Goyal, BITS, Pilani 54
LECTURE 4
Data Control Language (DCL) statements. Some examples:
* GRANT - gives user's access privileges to database
* REVOKE - withdraw access privileges given with the GRANT
command
Data Control Language (DCL)
Languages

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.
55
LECTURE 4

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