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Lecture 07: Relational Algebra

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1 Lecture 07: Relational Algebra

2 Outline Relational Algebra (Section 6.1)

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

4 Relational Algebra Five operators: Derived or auxiliary 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

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

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

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

9 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]

10 P SSN, Name (Employee)

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]


13 Relational Algebra Five operators: Derived or auxiliary 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

14 Renaming Changes the schema, not the instance Schema: R(A1, …, An )
Notation: r B1,…,Bn (R) Example: rLastName, SocSocNo (Employee) Output schema: Answer(LastName, SocSocNo) [in SQL: SELECT Name AS LastName, SSN AS SocSocNo FROM Employee]

15 LastName, SocSocNo (Employee)
Renaming Example Employee Name SSN John Tony LastName, SocSocNo (Employee) LastName SocSocNo John Tony

16 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 [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 Natural Join Example Employee Name SSN John Tony Dependents SSN Dname Emily Joe Employee Dependents = PName, SSN, Dname(s SSN=SSN2(Employee x rSSN2, Dname(Dependents)) Name SSN Dname John Emily Tony Joe

18 Natural Join R= S= R ⋈ S= A B X Y Z V B C Z U V W A B C X Z U V Y W

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

21 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?)

22 Semijoin R ⋉ S = P A1,…,An (R ⋈ S)
Where A1, …, An are the attributes in R Example: Employee ⋉ Dependents

23 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 Answer = R ⋈ Dependents

24 Complex RA Expressions
P name buyer-ssn=ssn pid=pid seller-ssn=ssn P ssn P pid sname=fred sname=gizmo Person Purchase Person Product

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) temp T2) - If predicates for selection have significant overlap, then it will require unnecessary computation and memory consumption. The earlier we process selections, less tuples we need to manipulate higher up in the tree (predicate pushdown) Disadvantages?

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 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 PM(PN(R)) = PM,N(R)

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} 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 !

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 Name1 Name2 Relationship Fred Mary Father Joe Cousin Bill Spouse Nancy Lou Sister

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 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 Formulating Queries in RA
PCourse# (s Quarter=W99 ((s Name= ‘John Smith’ (STUDENT) ⋈ ENROLL)) OtherDept = PDept ((s Publisher <> ‘PS Publishers’ (BOOK_ADOPTION ⋈ TEXT)) ⋈ COURSE) AllDept = PDept (BOOK_ADOPTION ⋈ COURSE) Answer = AllDept - OtherDept And how will you express it in SQL? WHY?

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