1. CS34311 Translating ER Schema to Relational Model

2. cs34312 Basic Mapping So Far Simple algorithm covers base case Each entity type = separate relation Each relationship type = separate relation

3. cs34313 Simple Algorithm: Example 3 PRIMARY KEY (Part) = PRIMARY KEY (Contains) = FOREIGN KEY Contains (superPart) REFERENCES Part (pNumber) FOREIGN KEY Contains (subPart) REFERENCES Part (pNumber) Part (pName, pNumber) Contains (superPart, subPart, quantity)

4. cs34314 Next Let’s consider constraints Let’s reduce number of relations

5. cs34315 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

6. cs34316 Simple Algorithm: Example 3 PRIMARY KEY (Part) = PRIMARY KEY (Contains) = FOREIGN KEY Contains (superPart) REFERENCES Part (pNumber) FOREIGN KEY Contains (subPart) REFERENCES Part (pNumber) Part (pName, pNumber) Contains (superPart, subPart, quantity)

7. cs34317 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.

8. cs34318 Example 1 If R contains entity type E whose max. cardinality is 1, then R may be represented as attributes of E.

9. cs34319 Example 1 Student (sNumber, sName, advisor, years) Professor (pNumber, pName) PRIMARY KEY (Student) = PRIMARY KEY (Professor) = FOREIGN KEY Student (advisor) REFERENCES Professor (pNumber) Question: Will Student.advisor attribute ever be NULL ? Answer: No ! So add the NOT NULL constraint.

10. cs343110 Example 2 Person (pNumber, pName, works-in, years) Dept (dNumber, dName) PRIMARY KEY (Person) = PRIMARY KEY (Dept) = FOREIGN KEY Person (worksin) REFERENCES Dept (dNumber) What about NULL attributes ? Worksin and years may be null for a person

11. cs343111 Remember the Simple Algorithm: Example 3 PRIMARY KEY (Part) = PRIMARY KEY (Contains) =  PRIMARY KEY (Contains) = FOREIGN KEY Contains (superPart) REFERENCES Part (pNumber) FOREIGN KEY Contains (subPart) REFERENCES Part (pNumber) Part (pName, pNumber) Contains (superPart, subPart, quantity)

12. cs343112 Example 3: The Reduced Case. Part (pNumber, pname, hasSuperPart, quantity) PRIMARY KEY (Part) = FOREIGN KEY Part (hasSuperPart) REFERENCES Part (pNumber) Note: hasSuperPart indicates the superpart of a part, and it may be null

13. cs343113 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 add “not null”

14. cs343114 Decreasing Number of Relations Technique 2: Even more radical : 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 one 1 relation.

15. cs343115 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 make all attributes of E1 and of R to become attributes of E2.

16. cs343116 Example 1 Student-BIG (sNumber, sName, pNumber, pName, years) PRIMARY KEY (Student) = ANY OTHER KEY ? CANDIDATE KEY (Student) = Note: pNumber, pName, and years can be null for students with no advisor

17. cs343117 Example 2 Student (sNumber, sName, pNumber, pName, years) PRIMARY KEY (Student) = CANDIDATE KEY (Student) = Note: pNumber cannot be null for any student.

18. cs343118 Decreasing Number of Relations Technique 2 For one-to-one relationship R with E cardinality (1, 1), then combine everything into 1 relation. While very compact, semantically may not be clearest choice. Therefore, not always recommended !

19. cs343119 ER Model: Complex Attributes Composite Attribute: address Student sNamesumer sAge statestreet address city Composite attribute in ER Include an attribute for every component of the composite attribute.

20. cs343120 ER Model: Complex Attributes Multivalued Attribute: major Mapping Multi-valued attribute in ER: Need separate relation for multi-valued attribute. Identify appropriate attributes, keys & foreign key constraints.

21. cs343121 Multi-valued attribute Student (sNumber, sName, sAge, street, city, state) PRIMARY KEY (Student) = StudentMajor (sNumber, major) PRIMARY KEY (StudentMajor) = FOREIGN KEY StudentMajor (sNumber) REFERENCES Student (sNumber)

22. cs343122 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