1. The Relational Model
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2. Why Relational Model?
 Currently the most widely used
 Vendors: Oracle, Microsoft, IBM
 Older models still used
 IBM’s IMS (hierarchical model)
 Recent competitions
 Object Oriented Model: ObjectStore
 Implementation standard for relational Model
 SQL (Structured Query Language)
 SQL 3: includes object-relational extensions
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3. Relational Model
 Structures
 Relations (also called Tables)
 Attributes (also called Columns or Fields)
 Note: Every attribute is simple (not composite or
multi-valued)
Student
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 Constraints
 Key and Foreign Key constraints (More constraints later)
 Eg: Student Relation (The following 2 relations are
equivalent)
Student
sNumber sName
1 Dave
2 Greg
sNumber sName
2 Greg
1 Dave
Cardinality = 2
Arity/Degree = 2

4. Relational Model
 Schema for a relation
 Eg: Student (sNumber, sName)
 PRIMARY KEY (Student) = <sNumber>
 Schema for a database
 Schemas for all relations in the database
 Tuples (Rows)
 The set of rows in a relation are the tuples of that
relation
 Note: Attribute values may be null
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5. Primary Key Constraints
 A set of attributes is a key for a relation if:
 No two distinct tuples can have the same values
in all key fields
 A proper subset of the key attributes is not a key.
 Superkey: A proper subset of a superkey
may be a superkey
 If multiple keys, one of them is chosen to be
the primary key.
 Eg: PRIMARY KEY (Student) = <sNumber>
 Primary key attributes cannot take null values
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6. Candidate Keys (SQL: Unique)
 Keys that are not primary keys are candidate
keys.
 Specified in SQL using UNIQUE
 Attribute of unique key may have null values !
 Eg: Student (sNumber, sName)
PRIMARY KEY (Student) = <sNumber>
CANDIDATE KEY (Student) = <sName>
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7. Violation of key constraints
 A relation violates a primary key constraint if:
 There is a row with null values for any attribute of
primary key.
 (or) There are 2 rows with same values for all
attributes of primary key
 Consider R (a, b) where a is unique. R
violates the unique constraint if all of the
following are true
 2 rows in R have the same non-null values for a
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8. Keys: Example
Student
sNumber sName address
1 Dave 144FL
2 Greg 320FL
Primary Key: <sNumber>
Candidate key: <sName>
Some superkeys: {<sNumber, address>,
<sName>,
<sNumber>,
<sNumber, sName>
<sNumber, sName, address>}
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9. Foreign Key Constraints
 To specify an attribute (or multiple attributes)
S1 of a relation R1 refers to the attribute (or
attributes) S2 of another relation R2
 Eg: Professor (pName, pOffice)
Student (sNumber, sName, advisor)
PRIMARY KEY (Professor) = <pName>
FOREIGN KEY Student (advisor)
REFERENCES Professor
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(pName)

10. Foreign Key Constraints
 FOREIGN KEY R1 (S1) REFERENCES R2
(S2)
 Like a logical pointer
 The values of S1 for any row of R1 must be
values of S2 for some row in R2 (null values
are allowed)
 S2 must be a key for R2
 R2 can be the same as R1 (i.e., a relation
can have a foreign key referring to itself).
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11. Foreign Keys: Examples
Dept (dNumber, dName)
Person (pNumber, pName, dept)
PRIMARY KEY (Dept) = <dNumber>
PRIMARY KEY (Person) =
<pNumber>
FOREIGN KEY Person (dept)
REFERENCES Dept (dNumber)
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Persons working for Depts
Person and his/her father
Person (pNumber, pName, father)
PRIMARY KEY (Person) = <pNumber>
FOREIGN KEY Person (father)
REFERENCES Person (pNumber)

12. Violation of Foreign Key
constraints
 Suppose we have: FOREIGN KEY R1 (S1)
REFERENCES R2 (S2)
 This constraint is violated if
 Consider a row in R1 with non-null values for all
attributes of S1
 If there is no row in R2 which have these values
for S2, then the FK constraint is violated.
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13. Relational Model: Summary
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 Structures
 Relations (Tables)
 Attributes (Columns, Fields)
 Constraints
 Key
 Primary key, candidate key (unique)
 Super Key
 Foreign Key

14. ER schema ® Relational
schema
Simple Algorithm
 Entity type E ® Relation E’
 Attribute of E ® Attribute as E’
 Key for E ® Primary Key for E’
 For relationship type R between E1, E2, …, En
 Create separate relation R’
 Attributes of R’ are primary keys of E1, E2, …, En and
attributes of R
 Primary Key for R’ is defined as:
 If the maximum cardinality of any Ei is 1, primary key for R’ =
primary key for Ei
 Else, primary key for R’ = primary keys for E1, E2, …, En
 Define “appropriate” foreign keys from R’ to E1, E2, …, En
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15. Simple algorithm: Example 1
Person (pNumber, pName)
Dept (dNumber, dName)
WorksFor (pNumber, dNumber, years)
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Person
pNumber
pName
Dept
dNumber
dName
Works
For
(1, *) (0, *)
years
PRIMARY KEY (Person) = <pNumber>
PRIMARY KEY (Dept) = <dNumber>
PRIMARY KEY (WorksFor) = <pNumber, dNumber>
FOREIGN KEY WorksFor (pNumber) REFERENCES Person (pNumber)
FOREIGN KEY WorksFor (dNumber) REFERENCES Dept (dNumber)

16. Simple Algorithm: Example 2
pName pNumber
(1, *) (0, *)
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Supplier
sName
sLoc
Supplier (sName, sLoc)
Consumer (cName, cLoc)
Product (pName, pNumber)
Supply (supplier, consumer,
Consumer
product, price, qty)
cName
cLoc
Supply
price
Product
qty
(0, *)
PRIMARY Key (Supplier) = <sName> PRIMARY Key (Consumer) = <cName>
PRIMARY Key (Product) = <pName>
PRIMARY Key (Supply) = <supplier, consumer, product>
FOREIGN KEY Supply (supplier) REFERENCES Supplier (sName)
FOREIGN KEY Supply (consumer) REFERENCES Consumer (cName)
FOREIGN KEY Supply (product) REFERENCES Product (pName)

17. Simple Algorithm: Example 3
p N a m e p N u m b e r
s u p e r P a r t s u b P a r t
( 0 , * ) ( 0 , 1 )
Part (pName, pNumber)
Contains (superPart, subPart, quantity)
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P a r t
C o n t a i n s
q u a n t i t y
PRIMARY KEY (Part) = <pNumber>
PRIMARY KEY (Contains) = <subPart>
FOREIGN KEY Contains (superPart) REFERENCES Part (pNumber)
FOREIGN KEY Contains (subPart) REFERENCES Part (pNumber)

18. Decreasing the number of
Relations
Technique 1
 If the relationship type R contains an entity type,
say E, whose maximum cardinality is 1, then R
may be represented as attributes of E.
 If the cardinality of E is (1, 1), then no “new nulls” are
introduced
 If the cardinality of E is (0, 1) then “new nulls” may be
introduced.
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19. Murali Mani
Example 1
Student
sNumber
sName
Professor
pNumber
pName
Has
Advisor
(1,1) (0, *)
years
Student (sNumber, sName, advisor, years)
Professor (pNumber, pName)
PRIMARY KEY (Student) = <sNumber>
PRIMARY KEY (Professor) = <pNumber>
FOREIGN KEY Student (advisor) REFERENCES Professor (pNumber)
Note: advisor will never be null for a student

20. Murali Mani
Example 2
Person
pNumber
pName
Dept
dNumber
dName
Works
For
(0,1) (0, *)
years
Person (pNumber, pName, dept, years)
Dept (dNumber, dName)
PRIMARY KEY (Person) = <pNumber>
PRIMARY KEY (Dept) = <dNumber>
FOREIGN KEY Person (dept) REFERENCES Dept (dNumber)
Dept and years may be null for a person

21. p N a m e p N u m b e r
P a r t
s u p e r P a r t s u b P a r t
( 0 , * ) ( 0 , 1 )
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Example 3
C o n t a i n s
q u a n t i t y
Part (pNumber, pname, superPart, quantity)
PRIMARY KEY (Part) = <pNumber>
FOREIGN KEY Part (superPart) REFERENCES Part (pNumber)
Note: superPart gives the superpart of a part, and it may be null

22. Decreasing the number of
Relations
Technique 2 (not recommended)
 If the relationship type R between E1 and E2 is
1:1, and the cardinality of E1 or E2 is (1, 1), then
we can combine everything into 1 relation.
 Let us assume the cardinality of E1 is (1, 1). We
have one relation for E2, and move all attributes
of E1 and for R to be attributes of E2.
 If the cardinality of E2 is (1, 1), no “new nulls” are
introduced
 If the cardinality of E2 is (0, 1) then “new nulls” may be
introduced.
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23. Murali Mani
Example 1
Student
sNumber
sName
Professor
pNumber
pName
Has
Advisor
(0,1) (1,1)
years
Student (sNumber, sName, pNumber, pName, years)
PRIMARY KEY (Student) = <sNumber>
CANDIDATE KEY (Student) = <pNumber>
Note: pNumber, pName, and years can be null for students with
no advisor

24. Murali Mani
Example 2
Student
sNumber
sName
Professor
pNumber
pName
Has
Advisor
(1,1) (1,1)
years
Student (sNumber, sName, pNumber, pName, years)
PRIMARY KEY (Student) = <sNumber>
CANDIDATE KEY (Student) = <pNumber>
Note: pNumber cannot be null for any student.

25. Other details
 Composite attribute in ER
 Include an attribute for every component of the
composite attribute.
Multi-valued attribute in ER
 We need a separate relation for any multi-valued
attribute.
 Identify appropriate attributes, keys and foreign
key constraints.
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26. Composite and Multi-valued
attributes in ER
sNumber sName
Student
address
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sAge
major
street city
state
Student (sNumber, sName, sAge, street, city, state)
StudentMajor (sNumber, major)
PRIMARY KEY (Student) = <sNumber>
PRIMARY KEY (StudentMajor) = <sNumber, major>
FOREIGN KEY StudentMajor (sNumber) REFERENCES Student (sNumber)

27. Weak entity types
 Consider weak entity type E
 A relation for E, say E’
 Attributes of E’ = attributes of E in ER + keys for
all indentifying entity types.
 Key for E’ = the key for E in ER + keys for all the
identifying entity types.
 Identify appropriate FKs from E’ to the identifying
entity types.
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28. Weak entity types: Example
Dept (dNumber, dName)
Course (cNumber, dNumber, cName)
PRIMARY KEY (Dept) = <dNumber>
PRIMARY KEY (Course) = <cNumber, dNumber>
FOREIGN KEY Course (dNumber) REFERENCES Dept (dNumber)
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29. ISA Relationship types:
Method 1
year program
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sNumber sName
Student
ISA ISA
GradStudent
UGStudent
Student (sNumber, sName)
UGStudent (sNumber, year)
GradStudent (sNumber, program)
PRIMARY KEY (Student) = <sNumber>
PRIMARY KEY (UGStudent) = <sNumber>
PRIMARY KEY (GradStudent) = <sNumber>
FOREIGN KEY UGStudent (sNumber)
REFERENCES Student (sNumber)
FOREIGN KEY UGStudent (sNumber)
An UGStudent will be represented REFERENCES Student (sNumber)
in both Student relation as well as
UGStudent relation (similarly
GradStudent)

30. ISA Relationship types:
Method 2
sNumber sName
Student
ISA ISA
year program
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GradStudent
UGStudent
Student (sNumber, sName, year, program)
PRIMARY KEY (Student) = <sNumber>
Note: There will be null values in the relation.

31. ISA Relationship types:
Method 3
year program
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sNumber sName
Student
ISA ISA
GradStudent
UGStudent
Student (sNumber, sName)
UGStudent (sNumber, sName, year)
GradStudent (sNumber, sName, program)
UGGradStudent (sNumber, sName,
year, program)
PRIMARY KEY (Student) = <sNumber>
PRIMARY KEY (UGStudent) = <sNumber>
PRIMARY KEY (GradStudent) = <sNumber>
PRIMARY KEY (UGGradStudent) = <sNumber>
Any student will be represented in
only one of the relations as
appropriate.