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Relational Algebra Lecture 7 1 Outline Relational Algebra - PDF document

Relational Algebra Lecture 7 1 Outline Relational Algebra (Section 6.1) 2 1 Relational Algebra Formalism for creating new relations from existing ones Its place in the big picture: Declarative query Algebra Implementation


  1. Relational Algebra Lecture 7 1 Outline • Relational Algebra (Section 6.1) 2 1

  2. Relational Algebra • Formalism for creating new relations from existing ones • Its place in the big picture: Declarative query Algebra Implementation language Relational algebra SQL, 3 relational calculus 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 4 2

  3. 1. Union and 2. Difference • R1 ∪ R2 Example: – ActiveEmployees ∪ RetiredEmployees • R1 – R2 Example: – AllEmployees − RetiredEmployees 5 What about Intersection ? • It is a derived operator R1 ∩ R2 = R1 – (R1 – R2) • Also expressed as a join (will see later) Example – UnionizedEmployees ∩ RetiredEmployees 6 3

  4. 3. Selection • Returns all tuples which satisfy a condition • Notation: s c (R) • Examples – s Salary > 40000 (Employee) – s name = “Smith” (Employee) • The condition c can be =, <, ≤ , >, ≥ , <> [in SQL: SELECT * FROM Employee WHERE Salary > 40000] 7 Selection Example Employee SSN Name DepartmentID Salary 999999999 John 1 30,000 777777777 Tony 1 32,000 888888888 Alice 2 45,000 Find all employees with salary more than $40,000. s Salary > 40000 (Employee) SSN Name DepartmentID Salary 888888888 Alice 2 45,000 8 4

  5. 4. Projection • Eliminates columns, then removes duplicates • Notation: P A1,…,An (R) • Example: project to social-security number and names: – P SSN, Name (Employee) – Output schema: Answer(SSN, Name) [In SQL: SELECT DISTINCT SSN, Name FROM Employee] 9 Projection Example Employee SSN Name DepartmentID Salary 999999999 John 1 30,000 777777777 Tony 1 32,000 888888888 Alice 2 45,000 P SSN, Name (Employee) SSN Name 999999999 John 777777777 Tony 888888888 Alice 10 5

  6. 5. Cartesian Product • Combine each tuple in R1 with each tuple in R2 • Notation: R1 × R2 • Example: – Employee × Dependents • Very rare in practice; mainly used to express joins [In SQL: SELECT * FROM R1, R2] 11 Cartesian Product Example Employee Name SSN John 999999999 Tony 777777777 Dependents EmployeeSSN Dname 999999999 Emily 777777777 Joe Employee x Dependents Name SSN EmployeeSSN Dname John 999999999 999999999 Emily John 999999999 777777777 Joe Tony 777777777 999999999 Emily Tony 777777777 777777777 Joe 12 6

  7. 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 13 Renaming • Changes the schema, not the instance • Schema: R(A 1 , …, A n ) • Notation: r B1,…,Bn (R) • Example: – r LastName, SocSocNo (Employee) – Output schema: Answer(LastName, SocSocNo) [in SQL: SELECT Name AS LastName, SSN AS SocSocNo FROM Employee] 14 7

  8. Renaming Example Employee Name SSN John 999999999 Tony 777777777 r LastName, SocSocNo ( Employee ) LastName SocSocNo John 999999999 Tony 777777777 15 Natural Join • Notation: R1 �� R2 • Meaning: R1 �� R2 = P A ( s C (R1 × R2)) • Where: – The selection s C checks equality of all common attributes – The projection eliminates the duplicate common attributes [in SQL: SELECT DISTINCT R1.A, R1. B, R2.C FROM R1, R2 WHERE R1.B = R2.B Schema: R1(A,B), R2(B,C)] 16 8

  9. Natural Join Example Employee Name SSN John 999999999 Tony 777777777 Dependents SSN Dname 999999999 Emily 777777777 Joe Employee Dependents = P Name, SSN, Dname ( s SSN=SSN2 (Employee x r SSN2, Dname (Dependents)) Name SSN Dname John 999999999 Emily Tony 777777777 Joe 17 Natural Join • R= S= B C A B Z U X Y V W X Z Z V Y Z Z V A B C • R � S= X Z U X Z V Y Z U Y Z V Z V W 18 9

  10. 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 ? 19 Theta Join • A join that involves a predicate • R1 � q R2 = s q (R1 × R2) • Here q can be any condition 20 10

  11. 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 (difference to natural join?) 21 Semijoin • R � S = P A1,…,An (R � S) • Where A 1 , …, A n are the attributes in R • Example: – Employee � Dependents 22 11

  12. Semijoins in Distributed Databases • Semijoins are used in distributed databases Dependents Employee SSN Dname Age SSN Name . . . . . . network . . . . . . Employee � ssn=ssn ( s age>71 (Dependents)) T = P SSN s age>71 (Dependents) R = Employee �� T 23 Answer = R � Dependents Complex RA Expressions P name buyer-ssn=ssn pid=pid seller-ssn=ssn P ssn P pid s name=fred s name=gizmo Person Purchase Person 24 Product 12

  13. Application: Query Rewriting for Optimization sname sname rating > 5 bid=100 sid=sid (Scan; (Scan; sid=sid write to rating > 5 write to bid=100 temp T2) temp T1) Reserves Sailors Sailors Reserves The earlier we process selections, less tuples we need to manipulate higher up in the tree (predicate pushdown) Disadvantages? 25 Algebraic Laws (Examples) • Commutative and Associative Laws – R ∩ S = S ∩ R, R ∩ (S ∩ T) = (R ∩ S) ∩ T – R S = S R, R (S T) = (R S) T � � � � � � • Laws involving selection – s C AND C’ (R) = s C ( s C’ (R)) = s C (R) ∩ s C’ (R) – s C (R S) = s C (R) S • When C involves only attributes of R � � • Laws involving projections – P M ( P N (R)) = P M,N (R) 26 13

  14. 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} • s C (R): preserve the number of occurrences • P A (R): no duplicate elimination • Cartesian product, join: no duplicate elimination Important ! Relational Engines work on bags, not sets ! 27 Finally: RA has Limitations ! • Cannot compute “transitive closure” Name1 Name2 Relationship Fred Mary Father Mary Joe Cousin Mary Bill Spouse Nancy Lou Sister • Find all direct and indirect relatives of Fred • Cannot express in RA !!! Need to write C program 28 14

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