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The Relational Model. 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

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why relational model
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

Murali Mani

relational model
Relational Model
  • Structures
    • Relations (also called Tables)
    • Attributes (also called Columns or Fields)
      • Note: Every attribute is simple (not composite or multi-valued)
  • Constraints
    • Key and Foreign Key constraints (More constraints later)
  • Eg: Student Relation (The following 2 relations are equivalent)

Student

Student

Cardinality = 2

Arity/Degree = 2

Murali Mani

relational model4
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

Murali Mani

primary key constraints
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

Murali Mani

candidate keys sql unique
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>

Murali Mani

violation of key constraints
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

Murali Mani

keys example
Keys: Example

Student

Primary Key: <sNumber>

Candidate key: <sName>

Some superkeys: {<sNumber, address>,

<sName>,

<sNumber>,

<sNumber, sName>

<sNumber, sName, address>}

Murali Mani

foreign key constraints
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 (pName)

Murali Mani

foreign key constraints10
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).

Murali Mani

foreign keys examples
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)

Persons working for Depts

Person (pNumber, pName, father)

PRIMARY KEY (Person) = <pNumber>

FOREIGN KEY Person (father)

REFERENCES Person (pNumber)

Person and his/her father

Murali Mani

violation of foreign key constraints
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.

Murali Mani

relational model summary
Relational Model: Summary
  • Structures
    • Relations (Tables)
    • Attributes (Columns, Fields)
  • Constraints
    • Key
      • Primary key, candidate key (unique)
    • Super Key
    • Foreign Key

Murali Mani

er schema relational schema
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

Murali Mani

simple algorithm example 1
Simple algorithm: Example 1

Person (pNumber, pName)

Dept (dNumber, dName)

WorksFor (pNumber, dNumber, 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)

Murali Mani

simple algorithm example 2
Simple Algorithm: Example 2

Supplier (sName, sLoc)

Consumer (cName, cLoc)

Product (pName, pNumber)

Supply (supplier, consumer,

product, price, qty)

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)

Murali Mani

simple algorithm example 3
Simple Algorithm: Example 3

Part (pName, pNumber)

Contains (superPart, subPart, quantity)

PRIMARY KEY (Part) = <pNumber>

PRIMARY KEY (Contains) = <subPart>

FOREIGN KEY Contains (superPart) REFERENCES Part (pNumber)

FOREIGN KEY Contains (subPart) REFERENCES Part (pNumber)

Murali Mani

decreasing the number of relations
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.

Murali Mani

example 1
Example 1

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

Murali Mani

example 2
Example 2

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

Murali Mani

example 3
Example 3

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

Murali Mani

decreasing the number of relations22
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.

Murali Mani

example 123
Example 1

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

Murali Mani

example 224
Example 2

Student (sNumber, sName, pNumber, pName, years)

PRIMARY KEY (Student) = <sNumber>

CANDIDATE KEY (Student) = <pNumber>

Note: pNumber cannot be null for any student.

Murali Mani

other details
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.

Murali Mani

composite and multi valued attributes in er
Composite and Multi-valued attributes in ER

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)

Murali Mani

weak entity types
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.

Murali Mani

weak entity types example
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)

Murali Mani

isa relationship types method 1
ISA Relationship types: Method 1

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)

REFERENCES Student (sNumber)

An UGStudent will be represented in both Student relation as well as UGStudent relation (similarly GradStudent)

Murali Mani

isa relationship types method 2
ISA Relationship types: Method 2

Student (sNumber, sName, year, program)

PRIMARY KEY (Student) = <sNumber>

Note: There will be null values in the relation.

Murali Mani

isa relationship types method 3
ISA Relationship types: Method 3

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.

Murali Mani