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Review

Review. Exam Su 3:30PM - 6:30PM 2010/12/12 Room C9000. Algebra. Relational Algebra Operations. Basic operations Selection ( ) Selects a subset of rows from relation. Projection ( ) Deletes unwanted columns from relation.

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Review

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  1. Review

  2. Exam • Su 3:30PM - 6:30PM 2010/12/12 • Room C9000

  3. Algebra

  4. Relational Algebra Operations • Basic operations • Selection ( ) Selects a subset of rows from relation. • Projection( ) Deletes unwanted columns from relation. • Cartesian product( ) Combinetwo relations. • Set-difference ( ) Tuples in relation 1, but not in relation 2. • Union( ) Tuples in relation 1 or in relation 2.

  5. Functional Dependency

  6. Functional Dependencies (FDs) • A functional dependency XY holds over relation R if, for every allowable instance r of R: • given two tuples in r, if the X values agree, then the Y values must also agree. (X and Y are sets of attributes.) • An FD is a statement about all allowable relations. • Must be identified based on semantics of application. • Given some allowable instance r1 of R, we can check if it violates some FD f, but we cannot tell if f holds over R! • K is a candidate key for R means that KR • However, KR does not require K to be minimal!

  7. Wages Example (Contd.) • Problems due to R W : • Update anomaly: Can we change W in just the 1st tuple of SNLRWH? • Insertion anomaly: What if we want to insert an employee and don’t know the hourly wage for his rating? • Deletion anomaly: If we delete all employees with rating 5, we lose the information about the wage for rating 5! Hourly_Emps2

  8. Reasoning About FDs (Contd.) • Couple of additional rules (that follow from AA): • Union: If X  Y and X  Z, then X  YZ • Decomposition: If X  YZ, then X  Y and X  Z • Example: Contracts(cid,sid,jid,did,pid,qty,value), and: • C is the key: C  CSJDPQV • Project purchases a part using single contract: JPC • Dept purchases at most a part from a supplier:SD P • JP  C, C  CSJDPQV imply JP  CSJDPQV • SD  P implies SDJ  JP • SDJ  JP, JP  CSJDPQV imply SDJ  CSJDPQV

  9. Suppose that relation R contains attributes A1 ... An. A decompositionof R consists of replacing R by two or more relations such that: • Each new relation scheme contains a subset of the attributes of R (and no attributes that do not appear in R), and • Every attribute of R appears as an attribute of one of the new relations. • e.g., A  B, ABCD  E, EF  GH, ACDF  EG has the following minimal cover: • A  B, ACD  E, EF  G and EF  H

  10. Examples Does A B hold? Don’t know Does BC  A hold? Does not hold (first two tuples) Does B  C hold? Does not hold (last 2 tuples)

  11. Examples • Relation: ABCDE • Dependencies: • A  B • C  D • Decompose to?

  12. Examples • Relation: ABCD • Dependencies: • CD • CA • BC • What is the candidate key? • B

  13. Examples • Relation: ABCD • Dependencies: • BC • DA • What is the candidate key? • BD

  14. Examples • Relation: ABCD • Dependencies: • ABCD • DA • What is the candidate key? • ABC • BCD

  15. Examples • Relation: ABCD • Dependencies: • ABC • ABD • CA • DB • What is the candidate key? • AB • BC • CD • AD

  16. Examples • Suppose you are given a relation R(A,B,C,D) • Dependencies • ABC • CA • CD • What is the candidate key? • AB • BC • Decomposed into ACD and BC. Good? • Lossless. Can join back to ABCD. • Dependency ABC is not preserved.

  17. Examples • Suppose you are given a relation R(A,B,C,D) • Dependencies • ABC • CAD • What is the candidate key? • A • C • Decomposed into ABC and AD. Good? • No. A is key, so no need to decompose. • Dependency CAD is not preserved.

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