Relational model for Databases
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Relational model for Databases of different companies
Transcript of Relational model for Databases
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 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)
sNumber sName
1 Dave
2 Greg
Student
sNumber sName
2 Greg
1 Dave
Student
Cardinality = 2Arity/Degree = 2
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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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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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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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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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Keys: Example
sNumber sName address
1 Dave 144FL
2 Greg 320FL
Student
Primary Key: <sNumber>Candidate key: <sName>Some superkeys: {<sNumber, address>,
<sName>, <sNumber>, <sNumber, sName> <sNumber, sName, address>}
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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)
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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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Foreign Keys: ExamplesDept (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 and his/her father
Person (pNumber, pName, father)
PRIMARY KEY (Person) = <pNumber>FOREIGN KEY Person (father) REFERENCES Person (pNumber)
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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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Relational Model: Summary
Structures Relations (Tables) Attributes (Columns, Fields)
Constraints Key
Primary key, candidate key (unique) Super Key Foreign Key
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ER schema Relational schemaSimple 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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Simple algorithm: Example 1
Person (pNumber, pName)Dept (dNumber, dName)WorksFor (pNumber, dNumber, years)
Person
pNumber
pName
Dept
dNumber
dName
WorksFor
(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)
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Simple Algorithm: Example 2
Supplier
sName
sLoc
Consumer
cName
cLoc
Supply
price
Product
pName pNumber
qty
(1, *) (0, *)
(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)
Supplier (sName, sLoc)Consumer (cName, cLoc)Product (pName, pNumber)Supply (supplier, consumer,
product, price, qty)
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Simple Algorithm: Example 3
C onta ins
P art
pN am e pN um ber
subP artsuperP art
quantity
(0 , 1 )(0 , *)
PRIMARY KEY (Part) = <pNumber>PRIMARY KEY (Contains) = <subPart>
FOREIGN KEY Contains (superPart) REFERENCES Part (pNumber)FOREIGN KEY Contains (subPart) REFERENCES Part (pNumber)
Part (pName, pNumber)Contains (superPart, subPart, quantity)
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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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Example 1
Student
sNumber
sName
Professor
pNumber
pName
HasAdvisor
(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
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Example 2
Person
pNumber
pName
Dept
dNumber
dName
WorksFor
(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
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Example 3
C onta ins
P art
pN am e pN um ber
subP artsuperP art
quantity
(0 , 1 )(0 , *)
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
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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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Example 1
Student
sNumber
sName
Professor
pNumber
pName
HasAdvisor
(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
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Example 2
Student
sNumber
sName
Professor
pNumber
pName
HasAdvisor
(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.
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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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Composite and Multi-valued attributes in ER
Student
sNamesNumber
sAge
major
statestreet
address
city
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)
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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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Weak entity types: Example
Dept Offers Course
dNamedNumber cNamecNumber
(1, 1)(0, *)
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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ISA Relationship types: Method 1
Student
GradStudent
sNamesNumber
ISA ISA
UGStudent
programyear
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)
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ISA Relationship types: Method 2
Student
GradStudent
sNamesNumber
ISA ISA
UGStudent
programyear
Student (sNumber, sName, year, program)PRIMARY KEY (Student) = <sNumber>
Note: There will be null values in the relation.
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ISA Relationship types: Method 3
Student
GradStudent
sNamesNumber
ISA ISA
UGStudent
programyear
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.
