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Lecture 13: Relational Decomposition and Relational Algebra

Lecture 13: Relational Decomposition and Relational Algebra. February 5 th , 2003. Summary of Previous Discussion. FD’s are given as part of the schema. You derived keys from them. You can also specify keys directly (they’re just another FD)

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Lecture 13: Relational Decomposition and Relational Algebra

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  1. Lecture 13:Relational Decomposition and Relational Algebra February 5th, 2003

  2. Summary of Previous Discussion • FD’s are given as part of the schema. You derived keys from them. You can also specify keys directly (they’re just another FD) • First, compute the keys and the FD’s that hold on the relation you’re considering. • Then, look for violations of BCNF. There may be a few. Choose one.

  3. Summary (continued) • You may need to decompose the decomposed relations: • To decide that, you need to project the FD’s on the decomposed relations. • The end result may differ, depending on which violation you chose to decompose by. They’re all correct. • Relations with 2 attributes are in BCNF.

  4. Summary (continued –2) • If you have only a single FD representing a key, then the relation is in BCNF. • But, you can still have another FD that forces you to decompose. • This is all really pretty simple if you think about it long enough.

  5. Boyce-Codd Normal Form A simple condition for removing anomalies from relations: A relation R is in BCNF if: Whenever there is a nontrivial dependency A1, ..., An B in R , {A1, ..., An} is a key for R In English (though a bit vague): Whenever a set of attributes of R is determining another attribute, should determine all the attributes of R.

  6. Example Decomposition Person(name, SSN, age, hairColor, phoneNumber) SSN  name, age, hairColor age  hairColor Decompose in BCNF (in class): Step 1: find all keys ssn, phoneNumber SSN, name, age, hairColor SSN, phoneNumber Step 2: now decompose

  7. Other Example • R(A,B,C,D) A B, B C • Keys: • Violations of BCNF:

  8. Correct Decompositions A decomposition is lossless if we can recover: R(A,B,C) R1(A,B) R2(A,C) R’(A,B,C) should be the same as R(A,B,C) Decompose Recover R’ is in general larger than R. Must ensure R’ = R

  9. Correct Decompositions • Given R(A,B,C) s.t. AB, the decomposition into R1(A,B), R2(A,C) is lossless

  10. 3NF: A Problem with BCNF Unit Company Product FD’s: Unit  Company; Company, Product  Unit So, there is a BCNF violation, and we decompose. Unit Company Unit  Company Unit Product No FDs

  11. So What’s the Problem? Unit Company Unit Product Galaga99 UW Galaga99 databases Bingo UW Bingo databases No problem so far. All local FD’s are satisfied. Let’s put all the data back into a single table again: Unit Company Product Galaga99 UW databases Bingo UW databases Violates the dependency: company, product -> unit!

  12. Solution: 3rd Normal Form (3NF) A simple condition for removing anomalies from relations: A relation R is in 3rd normal form if : Whenever there is a nontrivial dependency A1, A2, ..., An Bfor R , then {A1, A2, ..., An } a super-key for R, or B is part of a key.

  13. Relational Algebra • Formalism for creating new relations from existing ones • Its place in the big picture: Declartivequerylanguage Algebra Implementation Relational algebraRelational bag algebra SQL,relational calculus

  14. Relational Algebra • Five operators: • Union:  • Difference: - • Selection: s • Projection: P • Cartesian Product:  • Derived or auxiliary operators: • Intersection, complement • Joins (natural,equi-join, theta join, semi-join) • Renaming: r

  15. 1. Union and 2. Difference • R1  R2 • Example: • ActiveEmployees  RetiredEmployees • R1 – R2 • Example: • AllEmployees -- RetiredEmployees

  16. What about Intersection ? • It is a derived operator • R1  R2 = R1 – (R1 – R2) • Also expressed as a join (will see later) • Example • UnionizedEmployees  RetiredEmployees

  17. 3. Selection • Returns all tuples which satisfy a condition • Notation: sc(R) • Examples • sSalary > 40000(Employee) • sname = “Smithh”(Employee) • The condition c can be =, <, , >,, <>

  18. Find all employees with salary more than $40,000. s Salary > 40000(Employee)

  19. 4. Projection • Eliminates columns, then removes duplicates • Notation: P A1,…,An(R) • Example: project social-security number and names: • PSSN, Name (Employee) • Output schema: Answer(SSN, Name)

  20. PSSN, Name (Employee)

  21. 5. Cartesian Product • Each tuple in R1 with each tuple in R2 • Notation: R1  R2 • Example: • Employee  Dependents • Very rare in practice; mainly used to express joins

  22. Relational Algebra • Five operators: • Union:  • Difference: - • Selection: s • Projection: P • Cartesian Product:  • Derived or auxiliary operators: • Intersection, complement • Joins (natural,equi-join, theta join, semi-join) • Renaming: r

  23. Renaming • Changes the schema, not the instance • Notation: rB1,…,Bn (R) • Example: • rLastName, SocSocNo (Employee) • Output schema: Answer(LastName, SocSocNo)

  24. Renaming Example Employee Name SSN John 999999999 Tony 777777777 • LastName, SocSocNo (Employee) LastName SocSocNo John 999999999 Tony 777777777

  25. Natural Join • Notation: R1 ⋈ R2 • Meaning: R1 ⋈ R2 = PA(sC(R1  R2)) • Where: • The selection sC checks equality of all common attributes • The projection eliminates the duplicate common attributes

  26. Employee Dependents = PName, SSN, Dname(sSSN=SSN2(Employee x rSSN2, Dname(Dependents)) Natural Join Example Employee Name SSN John 999999999 Tony 777777777 Dependents SSN Dname 999999999 Emily 777777777 Joe Name SSN Dname John 999999999 Emily Tony 777777777 Joe

  27. Natural Join • R= S= • R ⋈ S=

  28. Natural Join • Given the schemas R(A, B, C, D), S(A, C, E), what is the schema of R ⋈ S ? • Given R(A, B, C), S(D, E), what is R ⋈ S ? • Given R(A, B), S(A, B), what is R ⋈ S ?

  29. Theta Join • A join that involves a predicate • R1 ⋈q R2 = sq (R1  R2) • Here q can be any condition

  30. Eq-join • A theta join where q is an equality • R1 ⋈A=B R2 = s A=B (R1  R2) • Example: • Employee ⋈SSN=SSN Dependents • Most useful join in practice

  31. Semijoin • R ⋉ S = PA1,…,An (R ⋈ S) • Where A1, …, An are the attributes in R • Example: • Employee ⋉ Dependents

  32. Semijoins in Distributed Databases • Semijoins are used in distributed databases Dependents Employee network Employee ⋈ssn=ssn (s age>71 (Dependents)) T = PSSNs age>71 (Dependents) R = Employee ⋉ T Answer = R ⋈ Dependents

  33. seller-ssn=ssn pid=pid buyer-ssn=ssn Complex RA Expressions Pname Person Purchase Person Product Pssn Ppid sname=fred sname=gizmo

  34. Operations on Bags A bag = a set with repeated elements All operations need to be defined carefully on bags • {a,b,b,c}{a,b,b,b,e,f,f}={a,a,b,b,b,b,b,c,e,f,f} • {a,b,b,b,c,c} – {b,c,c,c,d} = {a,b,b,d} • sC(R): preserve the number of occurrences • PA(R): no duplicate elimination • Cartesian product, join: no duplicate elimination Important ! Relational Engines work on bags, not sets ! Reading assignment: 5.3 – 5.4

  35. Finally: RA has Limitations ! • Cannot compute “transitive closure” • Find all direct and indirect relatives of Fred • Cannot express in RA !!! Need to write C program

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