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1 Lecture 07: Relational Algebra. 2 Outline Relational Algebra (Section 6.1)

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Presentation on theme: "1 Lecture 07: Relational Algebra. 2 Outline Relational Algebra (Section 6.1)"— Presentation transcript:

1 1 Lecture 07: Relational Algebra

2 2 Outline Relational Algebra (Section 6.1)

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

4 4 Relational Algebra Five operators: –Union:  –Difference: - –Selection:  –Projection:  –Cartesian Product:  Derived or auxiliary operators: –Intersection, complement –Joins (natural,equi-join, theta join, semi-join) –Renaming: 

5 5 1. Union and 2. Difference R1  R2 Example: –ActiveEmployees  RetiredEmployees R1 – R2 Example: –AllEmployees − RetiredEmployees

6 6 What about Intersection ? It is a derived operator R1  R2 = R1 – (R1 – R2) Also expressed as a join (will see later) Example –UnionizedEmployees  RetiredEmployees

7 7 3. Selection Returns all tuples which satisfy a condition Notation:  c (R) Examples –  Salary > (Employee) –  name = “Smith” (Employee) The condition c can be =,, , <> [in SQL: SELECT * FROM Employee WHERE Salary > 40000]

8 8 Find all employees with salary more than $40,000.  Salary > (Employee)

9 9 4. Projection Eliminates columns, then removes duplicates Notation:  A1,…,An (R) Example: project to social-security number and names: –  SSN, Name (Employee) – Output schema: Answer(SSN, Name) [In SQL: SELECT DISTINCT SSN, Name FROM Employee]

10 10  SSN, Name (Employee)

11 11 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]

12 12

13 13 Relational Algebra Five operators: –Union:  –Difference: - –Selection:  –Projection:  –Cartesian Product:  Derived or auxiliary operators: –Intersection, complement –Joins (natural,equi-join, theta join, semi-join) –Renaming: 

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

15 15 Renaming Example Employee NameSSN John Tony LastNameSocSocNo John Tony  LastName, SocSocNo (Employee)

16 16 Natural Join Notation: R1 ⋈ R2 Meaning: R1 ⋈ R2 =  A (  C (R1  R2)) Where: –The selection  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)]

17 17 Natural Join Example Employee NameSSN John Tony Dependents SSNDname Emily Joe NameSSNDname John Emily Tony Joe Employee Dependents =  Name, SSN, Dname (  SSN=SSN2 (Employee x  SSN2, Dname (Dependents))

18 18 Natural Join R= S= R ⋈ S= AB XY XZ YZ ZV BC ZU VW ZV ABC XZU XZV YZU YZV ZVW

19 19 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 ?

20 20 Theta Join A join that involves a predicate R1 ⋈  R2 =   (R1  R2) Here  can be any condition

21 21 Eq-join A theta join where  is an equality R1 ⋈ A=B R2 =  A=B (R1  R2) Example: –Employee ⋈ SSN=SSN Dependents Most useful join in practice (difference to natural join?)

22 22 Semijoin R ⋉ S =  A1,…,An (R ⋈ S) Where A 1, …, A n are the attributes in R Example: –Employee ⋉ Dependents

23 23 Semijoins in Distributed Databases Semijoins are used in distributed databases SSNName... SSNDnameAge... Employee Dependents network Employee ⋈ ssn=ssn (  age>71 (Dependents)) T =  SSN  age>71 (Dependents) R = Employee ⋉ T Answer = R ⋈ Dependents

24 24 Complex RA Expressions Person Purchase Person Product  name=fred  name=gizmo  pid  ssn seller-ssn=ssnpid=pidbuyer-ssn=ssn  name

25 25 Application: Query Rewriting for Optimization Reserves Sailors sid=sid bid=100 rating > 5 sname Reserves Sailors sid=sid bid=100 sname rating > 5 (Scan; write to temp T1) (Scan; write to temp T2) The earlier we process selections, less tuples we need to manipulate higher up in the tree (predicate pushdown) Disadvantages?

26 26 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 –  C AND C’ (R) =  C (  C’ (R)) =  C (R) ∩  C’ (R) –  C (R S) =  C (R) S When C involves only attributes of R Laws involving projections –  M (  N (R)) =  M,N (R)

27 27 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}  C (R): preserve the number of occurrences  A (R): no duplicate elimination Cartesian product, join: no duplicate elimination Important ! Relational Engines work on bags, not sets !

28 28 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 Name1Name2Relationship FredMaryFather MaryJoeCousin MaryBillSpouse NancyLouSister

29 29 Formulating queries in RA Consider a database for student enrollment for courses, and books used in the courses –STUDENT (SSN, Name, Major, Bdate) –COURSE (Course#, Cname, Dept) –ENROLL (SSN, Course#, Quarter, Grade) –BOOK_ADOPTION (Course#, Quarter, Book_ISBN) –TEXT (Book_ISBN, Book_Title, Publisher, Author)

30 30 Formulating queries in RA Specify the following queries in relational algebra –List the number of courses (Course#) taken by all students named ‘John Smith’ in Winter 1999 (i.e., Quarter = W99) –List any department which has all its adopted books published by ‘BC Publishing’

31 31 Formulating Queries in RA   Course# (  Quarter=W99 ((  Name= ‘John Smith’ (STUDENT) ⋈ ENROLL))  OtherDept =  Dept ((  Publisher <> ‘PS Publishers’ (BOOK_ADOPTION ⋈ TEXT)) ⋈ COURSE) AllDept =  Dept (BOOK_ADOPTION ⋈ COURSE) Answer = AllDept - OtherDept  And how will you express it in SQL? WHY?


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