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Translating ER Schema to Relational Model

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Basic Mapping So Far

Simple algorithms covers base case Idea:

Each entity type = separate relation Each relationship type = separate relation

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Simple Algorithm: Example 3

C onta ins

Part

pN am e pNum ber

subP artsuperPart

quantity

(0, 1 )(0 , *)

PRIMARY KEY (Part) = <pNumber>PRIMARY KEY (Contains) = <subPart,superPart>

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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Next

Let’s consider constraints Let’s reduce number of relations

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Refinement for Simple Mapping

Primary Key for R’ is defined as:

If the maximum cardinality of any Ei is 1, primary key for R’ = primary key for Ei

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Simple Algorithm: Example 3

C onta ins

Part

pN am e pNum ber

subP artsuperPart

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.

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Example 1Student

sNumber

sName

Professor

pNumber

pName

HasAdvisor

(1,1) (0, *)

years

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.

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Example 1Student

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)

Question: Will Student.advisor attribute ever be NULL ?Answer: No !

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Example 2Person

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)

What about NULL attributes ?Dept and years may be null for a person

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Remember the Simple Algorithm: Example 3

C onta ins

Part

pN am e pNum ber

subP artsuperPart

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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Example 3

C onta ins

Part

pN am e pN um ber

subP artsuperPart

quantity

(0 , 1 )(0 , *)

Part (pNumber, pname, superPart, quantity)

PRIMARY KEY (Part) = <pNumber>FOREIGN KEY Part (superPart) REFERENCES Part (pNumber)

Note: superPart indicates the superpart of a part, and it may be null

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Decreasing the Number of Relations

Summary of 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 cardinality of E is (1, 1), then no “new nulls” added If cardinality of E is (0, 1) then may add “new nulls”

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Decreasing the number of Relations

Technique 2

If relationship type R between E1 and E2 is one-to-one [1:1], and the cardinality of E1 or of E2 is (1, 1), then we can combine everything into 1 relation

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Decreasing the number of Relations

Technique 2 - Method Details

Let us assume the cardinality of E1 in R is (1, 1). Then we create one relation for entity E2 And, we move all attributes of E1 and for R to be

attributes of E2.

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Example 1

Student

sNumber

sName

Professor

pNumber

pName

HasAdvisor

(0,1) (1,1)

years

Student-BIG (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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Decreasing the Number of Relations

Technique 2

If relationship type R between E1 and E2 is one-to-one [1:1], and the cardinality of E1 or of E2 is (1, 1), then we can combine everything into 1 relation

Not always recommended! Thus use with care ! While very compact, semantically may not be clearest choice !

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ER Model: Complex AttributesComposite Attribute: address

sta testreet

address

city

Student

sNamesumer

sAge

statestreet

address

city

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Mapping details Composite attribute in ER

Include an attribute for every component of the composite attribute.

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ER Model: Complex Attributes

Multivalued Attribute: major

Student

sN am esN um ber

sA ge

m ajor

statestreet

address

city

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Mapping details 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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Example: Composite and Multi-valued attributes in ER

Student

sNamesNumber

sAge

major

statestreet

address

city

Student (sNumber, sName, sAge, street, city, state)PRIMARY KEY (Student) = <sNumber>

StudentMajor (sNumber, major)PRIMARY KEY (StudentMajor) = <sNumber, major>FOREIGN KEY StudentMajor (sNumber) REFERENCES Student (sNumber)

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Summary

Simple algorithms covers base case

Refinements : Reduce number of relations

Refinements: Consider constraints (not NULL)

Consider other ER constructs like complex and multi-valued attributes