The Relational Model
Lecture 3
INFS614, Fall 2008
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1
Relational Model
Relational Model = Structure + Operations

–
–
Structure: Relations (or Tables)
Operations: Relational Algebra, SQL.
Most widely implemented model.

–
Vendors: IBM DB2, Microsoft SQL Server, Oracle, etc.
Our design+implementation approach:

Step 1: ER design (ERD)
Step 2: Translate to Relational (Relational Schema)
Step 3: Querying over the relational model
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Relational Database: Definitions

Relational database: a set of relations

Relation: made up of 2 parts:
–
–
Instance : a table, with rows and columns.
#Rows = cardinality, #fields = degree / arity.
Schema : specifies name of relation, plus name
and type of each column.


E.G. Students(sid: string, name: string, login: string,
age: integer, gpa: real).
We can think of a relation as a set of rows
or tuples (i.e., all rows are distinct).
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Example: Instance of Students
Relation
sid
nam e
lo g in
age
gpa
53666
Jo n es
jo n es@ cs
18
3 .4
53688
S m ith
sm ith @ eecs
18
3 .2
53650
S m ith
sm ith @ m a th
19
3 .8
 Cardinality = 3, degree = 5, all rows distinct;
 The order in which the rows are listed is not
important;
 Do all columns in a relation instance have to
be distinct?
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Another Example:
Employees Relation


Employees Schema:
Employees(ssn:integer,name:string,rank:char,salary:float)
An instance of Employees:
ssn
633909767
674627883
193838904
534559257
123456789
354681756
123223665
231896598
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nam e
R ichard B oon
Adolfo Laurenti
W ill S mith
P hil C ollins
B rad Johnson
M artha S tewart
R alph R ama gashi
M ichael Jordan
rank
A
B
C
B
D
D
B
A
salary
75689.09
67890.00
50000.00
68901.00
45000.00
53500.00
69886.00
80999.00
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Example: Employees Relation (Contd.)
An instance of Employees = {
<633909767, Richard Boon, A, 75689.09>,
<674627883,Adolfo Laurenti, B, 67890.00>,
<193838904,Will Smith,C,50000.00>,…}
Set of tuples
(or rows)
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ssn
633909767
674627883
193838904
534559257
123456789
354681756
123223665
231896598
nam e
R ichard B oon
Adolfo Laurenti
W ill S mith
P hil C ollins
B rad Johnson
M artha S tewart
R alph R ama gashi
M ichael Jordan
rank
A
B
C
B
D
D
B
A
salary
75689.09
67890.00
50000.00
68901.00
45000.00
53500.00
69886.00
80999.00
6
Relational Database : Definitions

Instance : a set of tuples of the relation
A tuple : < a1:d1, …,an:dn >,
aj is an attribute name,
dj is the value of the attribute aj ,
dj either belongs to Domain(aj ) or is NULL
An instance of Employees = {
<ssn:633909767,name:Richard Boon, rank:A, salary:75689.09>,
<ssn:674627883,name:Adolfo Laurenti, rank:B, salary:67890.00>,
<ssn:193838904,name:Will Smith, rank:C, salary:50000.00>,
…
}
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Relational Database: Definitions


Relational database: a set of relations;
Relational database schema: the collection
of schemas for the relations in the
database;
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Example: A Company Database Schema
A First Schema:
Employees(ssn:integer,name:string,rank:integer,salary:float)
Projects(pid:integer,pname:string,budget:float)
Location(address:string,capacity:integer)
Departments(did:integer,dname:string,budget:float)
Manages(ssn:integer,did: integer,since:date)
Reports_To(ssnSubordinate:integer,ssnSupervisor:integer)
Works_for(ssn:integer,pid: integer,hours:float)
Works_in(ssn:integer,did: integer,address:string)
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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.

–
–
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The key: precise semantics for relational
queries.
Allows the optimizer to extensively re-order
operations, and still ensure that the answer
does not change.
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The SQL Query Language
Developed by IBM (system R) in the 1970s
 Need for a standard since it is used by
many vendors
 Standards:

–
–
–
–
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SQL-86
SQL-89 (minor revision)
SQL-92 (major revision, current standard)
SQL-99 (major extensions)
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Creating Relations in SQL
Creates the Students
CREATE TABLE Students
(sid
CHAR(20),
relation. Observe that the
name CHAR(20),
type (domain) of each field
login CHAR(10),
is specified, and enforced by
age INTEGER,
the DBMS whenever tuples
gpa REAL)
are added or modified.
 As another example, the
CREATE TABLE Enrolled
Enrolled table holds
(sid
CHAR(20),
information about courses
cid
CHAR(20),
that students take.
grade CHAR(2))

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Adding and Deleting Tuples

We can insert a single tuple using:
INSERT
INTO Students (sid, name, login, age, gpa)
VALUES (53688, ‘Smith’, [email protected], 18, 3.2)

Can delete all tuples satisfying some
condition (e.g., name = Smith):
DELETE
FROM Students S
WHERE S.name = ‘Smith’
 Powerful variants of these commands are available; more later!
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Querying Relational Data

To find all 18 year old students, we can write:
SELECT *
FROM Students S
WHERE S.age=18
sid
Instance of Students:
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n am e
login
age
gp a
53666 Jon es
jon es@ cs
18
3.4
53688 Sm ith sm ith @ eecs
18
3.2
53650 Sm ith sm ith @ m ath
19
3.8
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Querying Relational Data
(Contd.)

The result is:
SELECT *
FROM Students S
WHERE S.age=18
s id
nam e
lo g in
age gpa
53666
Jo n e s
jo n e s @ c s
18
3 .4
53688
S m ith
s m ith @ e e
18
3 .2
•To find just names and logins, replace the first line:
SELECT S.name, S.login
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Updating Tuples

Can update tuples using:
UPDATE Students S
SET S.age = S.age + 1, S.gpa = S.gpa -1
WHERE S.sid = 53688
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Querying Multiple Relations

What does the following query compute?
SELECT S.name, E.cid
FROM Students S, Enrolled E
WHERE S.sid=E.sid AND E.grade=“A”
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Querying Multiple Relations
Instance of
Students:
sid
nam e
lo g in
age
gpa
53666
Jo n es
jo n es@ cs
18
3 .4
53688
S m ith
sm ith @ eecs
18
3 .2
53650
S m ith
sm ith @ m a th
19
3 .8
sid
Instance of
Enrolled:
cid
53831
C arn atic1 0 1
C
53831
R eg g ae2 0 3
B
53650
53666
T o p o lo g y 1 1 2
H isto ry 1 0 5
A
B
we get:
S .n am e
S m ith
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g rad e
E .cid
T o p o lo g y 1 1 2
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Destroying and Altering
Relations
DROP TABLE Students

Destroys the relation Students. The schema
information and the tuples are deleted.
ALTER TABLE Students
ADD COLUMN firstYear: INTEGER

The schema of Students is altered by
adding a new field; every tuple in the
current instance is extended with a null
value in the new field.
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Integrity Constraints (ICs)

IC: condition that must be true for any instance of
the database; e.g., domain constraints.
–
–

A legal instance of a relation is one that satisfies all
specified ICs.
–

ICs are specified when schema is defined.
ICs are checked when relations are modified.
DBMS should not allow illegal instances.
If the DBMS checks ICs, stored data is more
faithful to real-world meaning.
–
Avoids data entry errors, too!
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Primary Key Constraints

A set of fields is a (candidate) key for a relation
if :
1. No two distinct tuples can have same values in all key
fields, and
2. This is not true for any subset of the key.
– Part 2 false? A superkey.
– If there’s >1 candidate keys for a relation, one of the
keys is chosen (by DBA) to be the primary key.

E.g., sid is a key for Students. (What about
name?) The set {sid, gpa} is a superkey.
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Primary and Candidate Keys in
SQL



Possibly many candidate keys (specified using UNIQUE),
one of which is chosen as the primary key.
“For a given student and course, CREATE TABLE Enrolled
(sid CHAR(20)
there is a single grade.” vs.
cid CHAR(20),
“Students can take only one
grade CHAR(2),
course, and receive a single
PRIMARY KEY (sid,cid) )
grade for that course; further,
CREATE TABLE Enrolled
no two students in a course
(sid CHAR(20)
receive the same grade.”
cid CHAR(20),
Used carelessly, an IC can
grade CHAR(2),
prevent the storage of database
PRIMARY KEY (sid),
instances that arise in practice!
UNIQUE (cid, grade) )
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Foreign Keys, Referential
Integrity


In addition to Students we have a second relation:
Enrolled(sid: string, cid: string, grade: string)
Only students listed in the Students relation should be
allowed to enroll for courses.
Enrolled
sid
cid
g rad e
53666
C arn atic1 0 1
C
53666
R eg g ae2 0 3
B
53650
53666
T o p o lo g y 1 1 2
H isto ry 1 0 5
A
B
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Students
sid
nam e
lo g in
age
gpa
53666
Jo n e s
jo n e s@ cs
18
3 .4
53688
S m ith
sm ith @ e e cs
18
3 .2
53650
S m ith
sm ith @ m a th
19
3 .8
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Foreign Keys, Referential
Integrity

Foreign key : Set of fields in one relation that
is used to `refer’ to a tuple in another relation.
(Must correspond to primary key of the second
relation.) Like a `logical pointer’.
 E.g. sid is a foreign key referring to Students:
–
–
–
Enrolled(sid: string, cid: string, grade: string)
If all foreign key constraints are enforced,
referential integrity is achieved, i.e., no dangling
references.
Can you name a data model w/o referential integrity?

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Links in HTML!
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Foreign Keys, Referential Integrity
Another Example :
– Only employees in the Employees Relation should
be allowed to be managers:
 ssn is a Foreign Key respect to Employees
– Only projects in the Project Relation should be
allowed to be managed :
 pid is a Foreign Key respect to Projects
Employees
ssn
633909767
674627883
193838904
534559257
123456789
354681756
123223665
231896598
nam e
R ichard B oon
Adolfo Laurenti
W ill S mith
P hil C ollins
B rad Johnson
M artha S tewart
R alph R amagashi
M ichael Jordan
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Projects
Managers
rank
A
B
C
B
D
D
B
A
salary
75689.09
67890.00
50000.00
68901.00
45000.00
53500.00
69886.00
80999.00
ssn
534559257
123456789
231896598
193838902
354681756
pid
1
1
53
18
18
h ou rs
2
56
8
36
46
pid
1
53
pname
XA011
Y
pbudget
5000000.00
7560000.00
18
X
250000.00
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Foreign Keys in SQL

Only students listed in the Students relation should
be allowed to enroll for courses.
CREATE TABLE Enrolled
(sid CHAR(20), cid CHAR(20), grade CHAR(2),
PRIMARY KEY (sid,cid),
FOREIGN KEY (sid) REFERENCES Students )
Enrolled
sid
cid
g rad e
53666
C arn atic1 0 1
C
53666
R eg g ae2 0 3
B
53650
53666
T o p o lo g y 1 1 2
H isto ry 1 0 5
A
B
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Students
sid
nam e
lo g in
age
gpa
53666
Jo n e s
jo n e s@ cs
18
3 .4
53688
S m ith
sm ith @ e e cs
18
3 .2
53650
S m ith
sm ith @ m a th
19
3 .8
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Foreign Keys, Referential Integrity



A Foreign Key must correspond to the primary key
of the referenced relation
A Foreign Key states a Referential IC between two
relations : a tuple in one relation that refers to
another must refer to an existing tuple in that
relation.
Referential Integrity Constraints is used to
maintain the consistency among tuples of two
related relations
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Enforcing Referential Integrity



Consider Students and Enrolled; sid in Enrolled is a foreign
key that references Students.
What should be done if an Enrolled tuple with a nonexistent student id is inserted? (Reject it!)
What should be done if a Students tuple is deleted?
–
–
–
–

Also delete all Enrolled tuples that refer to it.
Disallow deletion of a Students tuple that is referred to.
Set sid in Enrolled tuples that refer to it to a default sid.
(In SQL, also: Set sid in Enrolled tuples that refer to it to a special
value null, denoting `unknown’ or `inapplicable’.)
Similarly if primary key value of a Students tuple is
updated.
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Referential Integrity in
SQL/92

SQL/92 supports all 4 options CREATE TABLE Enrolled
on deletes and updates.
(sid CHAR(20),
– Default is NO ACTION
cid CHAR(20),
(delete/update is
grade CHAR(2),
rejected)
PRIMARY KEY (sid,cid),
– CASCADE (also delete all
FOREIGN KEY (sid)
tuples that refer to
deleted tuple)
REFERENCES Students
– SET NULL / SET DEFAULT
ON DELETE CASCADE
(sets foreign key value of
ON UPDATE NO ACTION )
referencing tuple)
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Where do ICs Come From?


ICs are based upon the semantics of the realworld enterprise that is being described in the
database relations.
We can check a database instance to see if an IC
is violated, but we can NEVER infer that an IC is
true by looking at an instance.
–
–

An IC is a statement about all possible instances!
From example, we know name is not a key, but the
assertion that sid is a key is given to us.
Key and foreign key ICs are the most common;
more general ICs supported too.
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Logical DB Design: ER to
Relational

Entity sets to tables.
ssn
name
Employees
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lot
CREATE TABLE Employees
(ssn CHAR(11),
name CHAR(20),
lot INTEGER,
PRIMARY KEY (ssn))
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Relationship Sets to Tables

In translating a relationship
set to a relation, attributes CREATE TABLE Works_In(
of the relation must include: ssn CHAR(11),
did INTEGER,
since DATE,
– Keys for each
PRIMARY KEY (ssn, did),
participating entity set
FOREIGN KEY (ssn)
(as foreign keys).
REFERENCES Employees,
 This set of attributes
FOREIGN KEY (did)
forms a superkey for
REFERENCES Departments)
the relation.
– All descriptive attributes.
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Translating Ternary Relationship Set
name rank
ssn
salary
dname
dbudget
did
Employees
Works-In0
address
Locations
Departments
Capacity
Works_In0(ssn:integer,did:integer,address:string)
CREATE TABLE Works_In0
(ssn CHAR(11),
did INTEGER,
address CHAR(60),
PRIMARY KEY (ssn, did, address),
FOREIGN KEY (ssn) REFERENCES Employees,
FOREIGN KEY (did) REFERENCES Departments,
FOREIGN KEY (address) REFERENCES Locations)
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Review: Key Constraints

Each dept has at
most one manager,
according to the
key constraint on
Manages.
since
name
ssn
dname
lot
Employees
did
Manages
budget
Departments
Translation to
relational model?
1-to-1
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1-to Many
Many-to-1
Many-to-Many
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Translating ER Diagrams with Key
Constraints


Map relationship to a
table:
– Note that did is the
key now!
– Separate tables for
Employees and
Departments.
Since each department
has a unique manager,
we could instead
combine Manages and
Departments.
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CREATE TABLE Manages(
ssn CHAR(11),
did INTEGER,
since DATE,
PRIMARY KEY (did),
FOREIGN KEY (ssn) REFERENCES Employees,
FOREIGN KEY (did) REFERENCES Departments)
CREATE TABLE Dept_Mgr(
did INTEGER,
dname CHAR(20),
budget REAL,
ssn CHAR(11),
since DATE,
PRIMARY KEY (did),
FOREIGN KEY (ssn) REFERENCES Employees)
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Review: Participation
Constraints

Does every department have a manager?
– If so, this is a participation constraint: the participation of
Departments in Manages is said to be total (vs. partial).
 Every did value in Departments table must appear in a row of
the Manages table (with a non-null ssn value!)
since
name
ssn
dname
did
lot
Employees
Manages
budget
Departments
Works_In
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since
36

Participation Constraints in
SQL
We can capture participation constraints involving
one entity set in a binary relationship, but little
else (without resorting to CHECK constraints).
CREATE TABLE Dept_Mgr(
did INTEGER,
dname CHAR(20),
budget REAL,
ssn CHAR(11) NOT NULL,
since DATE,
PRIMARY KEY (did),
FOREIGN KEY (ssn) REFERENCES Employees,
ON DELETE NO ACTION)
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Participation Constraints

Is it possible to express this participation constraint using only
key and foreign key constraints?
NO
name rank
ssn
pname
pid
salary
Employees

hours
Works_For
Projects
Works_for( ssn: integer, pid: integer, hours: float)
CREATE TABLE Works_for
(ssn INTEGER,
pid INTEGER,
hours float,
PRIMARY KEY (ssn,pid),
FOREIGN KEY (ssn) REFERENCES Employees,
FOREIGN KEY (pid) REFERENCES Projects)
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Review: Weak Entities

A weak entity can be identified uniquely only by considering
the primary key of another (owner) entity.
– Owner entity set and weak entity set must participate in
a one-to-many relationship set (1 owner, many weak
entities).
– Weak entity set must have total participation in this
identifying relationship set.
name
ssn
lot
Employees
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cost
Policy
pname
age
Dependents
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Translating Weak Entity Sets

Weak entity set and identifying relationship set
are translated into a single table.
– When the owner entity is deleted, all owned
weak entities must also be deleted.
CREATE TABLE Dep_Policy (
pname CHAR(20),
age INTEGER,
cost REAL,
ssn CHAR(11),
PRIMARY KEY (pname, ssn),
FOREIGN KEY (ssn) REFERENCES Employees,
ON DELETE CASCADE)
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name
ssn
Review: ISA Hierarchies
hourly_wages
in C++, or other PLs,
attributes are inherited.
If we declare A ISA B, every
A entity is also considered to
be a B entity.
lot
Employees
hours_worked
ISA
As


contractid
Hourly_Emps
Contract_Emps
Overlap constraints: Can Joe be an Hourly_Emps as well
as a Contract_Emps entity? (Allowed/disallowed)
Covering constraints: Does every Employees entity also
have to be an Hourly_Emps or a Contract_Emps entity?
(Yes/no)
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Translating ISA Hierarchies to
Relations

General approach:
–

3 relations: Employees, Hourly_Emps and Contract_Emps.
 Hourly_Emps: Every employee is recorded in Employees. For
hourly emps, extra info recorded in Hourly_Emps
(hourly_wages, hours_worked, ssn); must delete Hourly_Emps
tuple if referenced Employees tuple is deleted).
 Queries involving all employees easy, those involving just
Hourly_Emps require a join to get some attributes.
Alternative: Just Hourly_Emps and Contract_Emps.
–
–
Hourly_Emps: ssn, name, lot, hourly_wages, hours_worked.
Each employee must be in one of these two subclasses.
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Translating ISA Hierarchies to
Relations

General approach:
CREATE TABLE Hourly_Emps (
hourly_wages REAL,
hours_worked REAL,
ssn CHAR(11),
PRIMARY KEY (ssn),
FOREIGN KEY (ssn)
REFERENCES Employees,
ON DELETE CASCADE)
Similarly for Contract_Emps TABLE
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Translating Aggregations
name rank
ssn
salary
Employees
Monitors
Departments
did
dbudget
until
Projects
Sponsors
since
pid
pname
dname
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Translating Aggregations
Sponsors(did, pid, since)
CREATE TABLE Sponsors
(did INTEGER,
pid INTEGER,
since DATE,
PRIMARY KEY (did,pid),
FOREIGN KEY (did) REFERENCES Departments,
FOREIGN KEY (pid) REFERENCES Projects)
Monitors(ssn, did, pid,until)
CREATE TABLE Monitors
(ssn INTEGER,
did INTEGER,
pid INTEGER,
until DATE,
PRIMARY KEY (ssn, did, pid),
FOREIGN KEY (did,pid) REFERENCES Sponsors,
FOREIGN KEY (ssn) REFERENCES Employees)
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Translating Aggregations
If: Every sponsored project has a
monitor, and
 the attribute “since” is not required for
Sponsors ….

Every possible instance of the Sponsors
relationship is obtained by looking at the
set of pairs <pid,did> in the relation
Monitors
Therefore, we can omit the Sponsors relation
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Review: Binary vs. Ternary
Relationships
name
ssn

If each policy is
owned by just 1
employee:
–

Key constraint
on Policies
would mean
policy can only
cover 1
dependent!
What are the
additional
constraints in the
2nd diagram?
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pname
lot
Employees
Dependents
Covers
Bad design
Policies
policyid
cost
name
ssn
age
pname
lot
age
Dependents
Employees
Purchaser
Better design
policyid
Beneficiary
Policies
cost
47
Binary vs. Ternary Relationships
(Contd.) CREATE TABLE Policies (


policyid INTEGER,
The key
constraints allow cost REAL,
ssn CHAR(11) NOT NULL,
us to combine
PRIMARY KEY (policyid),
Purchaser with
FOREIGN KEY (ssn) REFERENCES Employees,
Policies and
ON DELETE CASCADE)
Beneficiary with
Dependents.
CREATE TABLE Dependents (
pname CHAR(20),
Participation
constraints lead age INTEGER,
policyid INTEGER,
to NOT NULL
PRIMARY KEY (pname, policyid),
constraints.
FOREIGN KEY (policyid) REFERENCES Policies,
ON DELETE CASCADE)
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Relational Model: Summary



A tabular representation of data.
Simple and intuitive, currently the most widely
used.
Integrity constraints can be specified by the DBA,
based on application semantics. DBMS checks for
violations.
–
–


Two important ICs: primary and foreign keys
In addition, we always have domain constraints.
Powerful and natural query languages exist.
Rules to translate ER to relational model
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The ER and Relational Models