Spring 2012 Instructor: Hassan Khosravi

1. Relational Algebra Spring 2012 Instructor: Hassan Khosravi

2. Querying relational databases • Lecture given by Dr. Widom on querying Relational Models

3. 2.1 An Overview of Data Models • 2.1.1 What is a Data Model? • 2.1.2 Important Data Models • 2.1.3 The Relational Model in Brief • 2.1.4 The Semi-structured Model in Brief • 2.1.5 Other Data Models • 2.1.6 Comparison of Modeling Approaches

4. 2.1.1 What is a Data Model? Data model is a notion for describing data or information. Real World  Math Model: • Structure of the data (tuples) • Operations on the data –queries to retrieve and modify information • Constraints on the data – year has to be integer, name is string . • Important data models • The relational Model • The semi-structured data model XML

5. Relational Model in Brief • Relational model is based on tables • Operations: query, modify • Constraints: year is Integer between 1930-2012 • The structure may appear to resemble an array of structs in C where the column headers are the field names and each row represent the values of one struct in the array. • Distinction in scales of relations • Not normally implemented as main-memory structure • Take into consideration to access relations on hard drive

6. The Semi-structured Model in Brief • Semi structure data resembles trees or graphs rather than tables or arrays. • Operations usually involve following in the tree. • Find the movies with the comedy genre. • Constraints often involve data types of values associated with a tag. • Values associated with the length tag are integers <Movies> <Movie title=“Gone with the Wind”> <Year>1939</Year> <Length>231</Length> <Genre>drama</Genre> </Movie> <Movie title=“Star Wars”> <title= Wars > <Year>1977</Year> <Length>124</Length> <Genre>sciFi</Length> </Movie> <Movie title=“Wayne’s World”> <Year>1992</Year> <Length>95</Length> <Genre>comedy</Genre> </Movie> </Movies>

7. Comparison of Modeling Approaches • Semi-structured models have more flexibility than relations. However, the relational model is still preferred in DBMS’s. • Efficiency of access to data and efficiency of modifications to that data are more important than flexibility • ease of use is more important than flexibility. • SQL enables the programmer to express their wishes at very high level. The strongly limited set of operations can be optimized to run very fast

8. Basics of the Relational model • Attributes: columns of a relation are named attributes. • Schema: the name of the relation and the set of attributes • Movies(title, year, length, genre) • Tuples: The rows of a relation, other than the header • Domains: the value for each attribute must be atomic (can not be structure). Each attribute has a domain of values.

9. Equivalent Representations of a Relation Relations are sets of tuples not lists of tuples. The order of tuples does not matter. Attributes could be reordered too. How many different ways can we present the given relation?

10. Relation Instances and Keys • A set of tuples for a given relation is called an instance of that relation. It is expected for the instance of the relation to change over time. • New movies are added to the table • It is less common for the schema of a relation to change. It is hard to add a new value for all the current tuples if a new attribute is added to the schema. • Keyes of relations • Key constraints: A set of attributes form a key if we do not allow two tuples in a relation instance to have the same value. • We indicate the attributes that form a key by underlining them • Movies(title, year, length, genre) • Key most be true for all possible instances of a relation not a specific instance. • Genre is not a key • What if our data does not have a key? • Generate artificial ID. Student Number

11. Database Schema about Movies MovieExec ( name: string, address : string cert# : integer netWorth : integer ) Studio ( name: string, address : string pressC# : integer ) Movies( title: string; Year : integer, Length : integer, Genre : string, studioName : string, producerC# : integer ) Moviestar ( name : string, address : string, gender : char, birthdate : date ) StarsIn ( MovieTitle: string, Movieyear : integer Starname : string )

12. Defining a Relation Schema in SQL 2.3.1 Relations in SQL 2.3.2 Data Types 2.3.3 Simple Table Declarations 2.3.4 Modifying Relation Schemas 2.3.5 Default Values 2.3.6 Declaring Keys 2.3.7 Exercises for Section 2.3

13. 2.3.1 Relations in SQL SQL also pronounced (sequel) is the principal language used to describe and manipulate relational database SQL makes a distinction between three kinds of relations Stored relations (tables): this relations are tables that exist in the database we can query and modify Views: are relations defined by a computation. They are not stored but constructed. We just query them (chapter 8) Temporary tables: are constructed by SQL language processor during optimization. These are not stored nor seen by the user

14. Data Types Char(n): a fixed-length string of up to n characters. Char(5) of foo is stored “foo ” Varchar(n): a variable-length string of up to n characters Varchar(5) of foo is stored “foo” Bit(n), Varbit(n) fixed and variable string of upto n bits. Boolean: True False and although it would surprise George Boole Unknown Int or Integer: typical integer values Float or real: typical real values Decimal(6,2) could be 0123.45 Date and time: essentially char strings with constraints.

15. 2.3.3 Simple Table Declarations CREATE TABLE Movie ( title VARCHAR(255), year INTEGER, length INTEGER, inColor CHAR(1), studioName CHAR(50), producerC# INTEGER, ); CREATE TABLE MOVIESTAR ( NAME CHAR(30), ADDRESS VARCHAR2(50), GENDER CHAR(6) , BIRTHDATE DATE ); Movies( title: string; Year : integer, Length : integer, Genre : string, studioName : string, producerC# : integer ) Moviestar ( name : string, address : string, gender : char, birthdate : date )

16. Modifying Relation Schemas • We can delete a table R by the following SQL command • Drop table R; • We can modify a table by the command • Alter Table MovieStar ADD phone CHAR(16); • Alter Table MovieStar Drop birthdate; • Defaults values • To use the default character ? As the default for an unknown gender. • Earliest possible date for Unknown Birthdate. DATE ‘0000-00-00’ • Gender CHAR(1) DEFAULT ‘?’, • Birthdate DATE DEFAULT DATE ‘0000-00-00’, • ALTER TABLE MovieStar ADD phone CHAR (16) DEFAULT ‘ unlisted’;

17. 2.3.6 Declaring Keys Two ways to declare keys in CRATE table statement Primary key can not be null Unique can be null Replace primary with unique in examples to get the example with unique CREATE TABLE MovieStar ( name CHAR (30) Primary Key, address VARCHAR (255), gender CHAR(1), birthdate DATE ); CREATE TABLE MovieStar ( name CHAR (30), address VARCHAR (255), gender CHAR(1), birthdate DATE PRIMARY KEY (name) );

18. Example 2.7 • The Relation Movie, whose key is the pair of attributes ‘title and year’ must be declared like this CREATE TABLE Movies( title CHAR(100), year INTEGER, length INTEGER, genre CHAR(10), studiName CHAR(30), producerC# INTEGER, PRIMARY KEY (title,year) );

19. Quick summary • Lecture given by Dr. Widom on Relational Model definition

20. 2.4 An Algebraic Query Language 2.4.1 Why Do We Need a Special Query Language? 2.4.2 What is an Algebra? 2.4.3 Overview of Relational Algebra 2.4.4 Set Operations on Relations 2.4.5 Projection 2.4.6 Selection 2.4.7 Cartesian Product 2.4.8 Natural Joins 2.4.9 Theta-Joins 2.4.10 Combining Operations to Form Queries 2.4.11 Naming and Renaming 2.4.12 Relationships Among Operations 2.4.13 A Linear Notation for Algebraic Expressions 2.4.14 Exercises for Section 2.4

21. Why Do We Need a Special Query Language? Why not just use C or java instead of introducing relational algebra ? Relational algebra is useful because it is less powerful than C and Java. One of the only areas where non-Turing-complete languages make sense. Relational algebra CANNOT determine whether the number of tuples are odd or even Being less powerful is helpful because Ease of programming Ease of compilation Ease of optimization

22. Projection • The Projection operator applied to a relation R, produces a new relation with a subset of R’s columns. • Duplicate tuples are eliminated. ∏Title,year,length (Movies) ∏genre (Movies)

23. Selection and Projection • Lecture given by Dr. Widom on selection and projection

24. 2.4.6 Selection The selection operator applied to a relation R, produces a new relation with a subset of R’s tuples. σ length >= 100(Movie)

25. Example for Selection • σ Length >= 100 AND studioName = ‘Fox’ (Movies) • Set tuples in the relation movies that represent Fox Movies at least 100 minutes long.

26. 2.4.7 Cartesian Product The Cartesian Product of two sets R and S is the set of pairs that can be formed by choosing the first element from R and the second from S. If R and S have some attribute in common, we need to invent new name for the identical attributes. Relation R X S Relation S Relation R

27. Cartesian Product • Lecture given by Dr. Widom on duplicates , cross product

29. 2.4.8 Natural Joins The Natural join of two sets R and S is the set of pairs that agree in whatever attributes are common to the schemas of R and S. Let A1,A2, …, An be attributes in both R and S. a tuple r from R and s from S are successfully paired if and only if r and s agree on A1,A2, …, An that can be formed by choosing the first element from R and the second from S. Relation S Relation R ⋈ S Relation R

30. Example for Natural Join • A more complicated example for natural join Result U ⋈ V Relation U Relation V

31. Lecture given by Dr. Widom on Natural Join

34. Theta-Joins U ⋈ A < D V Relation U Relation V • It is sometimes desirable to pair tuples on other conditions except all the common attributes being equal. The notation for a theta-join of relation R and S based on condition C is R ⋈C S • The result is constructed as follows: • Take product of R and S • Select tuples that satisfy C

35. Example on Theta-Joins Relation U Relation V U ⋈ A < D AND U.B <> V.B V • U and V that has more complex condition : • We require for successful pairing not only that the A component of U-tuple be less than D component of the V-tuple, but that the two tuples disagree on their respective B components

36. Combining Operations to Form Queries Example: “ What are the titles and years of movies made by Fox that are at least 100 minutes long” ∏ Title,year ∩ σStudioName =‘Fox’ σlength >=100 Movies Movies • ∏ Title,year (σlength >=100 (Movies) ∩ σStudioName =‘Fox’ (Movies) ∏Title,year (σ length >=100ANDStudioName =‘Fox’ (Movies)

37. Relational algebra Algebra in general consists of operators and atomic operands Algebra of arithmetic operands are variables and constants and operators are (+, -, *, /). Any algebra allows us to build expressions by applying an operator to operands and other expressions. (x+y)/z Relation R Relation S

38. Operations of relational algebra Union (R S): the set of elements that are in R, or S or both. Appears only once in the union. Relation R Relation S

39. Operations of relational algebra • Intersection (R S): the set of elements that are in both R and S.Appears only once in the intersection. Relation R Relation S

40. Operations of relational algebra • The Difference (R-S): the set of elements that are in R and not in S.Appears only once in the difference. Relation R Relation S

41. Lecture given by Dr. Widom on union, difference, intersection

43. 2.4.11 Naming and Renaming Operator to explicitly rename attributes in relations. PS(A1,A2, …, An ) (R) results in a relation S that has exactly the same tuples as R but the attributes names are A1,A2, …, An starting from the left most attribute. Relation S Relation R R X ρ s (X,C,D) (S)

44. Lecture given by Dr. Widom on Renaming

45. Relationships Among Operations Intersection can be expressed as difference. RS = R –(R –S) See video Theta join can be expressed by product and selection R ⋈ CS= C(RS) Natural join can be rewritten by product, selection, projection Example Result U ⋈ V = ∏A,U.B, U.C, D(U.B=V.B AND U.C=V.B (UV)) These are the only redundancies ( union, difference, selection, projection, product, renaming) form an independent set. Relation U Relation V Result U ⋈ V

46. 2.5 Constraints on Relations 2.5.1 Relational Algebra as a Constraint Language 2.5.2 Referential Integrity Constraints 2.5.3 Key Constraints 2.5.4 Additional Constraint Examples 2.5.5 Exercises for Section 2.5 2.6 Summary of Chapter 2 2.7 References for Chapter 2

47. Referential Integrity Constraints Referential Integrity Constraints A value appearing in one context also appears in another, related context StarsIn(movietitle, movieYear,starName) Movie(title,year,length,studioName, producerC#) ∏movieTitle, movieYear (StarsIn) ⊆ ∏title,year(Movies) Movie(title,year,length,genre,studioName, producerC#) MovieExec(name,address,cert#,netWorth) ∏producerC# (Movies) ⊆ ∏cert# (MocvieExec)

48. Key Constraints • Recall that name is the key for relation • MovieStar(name,address,gender,birthdate) • The requirement can be expressed by the algebraic expression • σMS1.name = MS2.name AND MS1.address ≠ MS2.address(MS1 x MS2) = ∅ • MS1 in the product MS1 x MS2 is shorthand for the remaining • ρ MS1(name,address,gender,birthdate) (MovieStar)

49. Example 2.24 • The only legal value for Gender attribute is ‘F’ and ‘M’. We can express the gender attribute of MovieStaralegrabically by: • σMgender ≠‘F’ AND gender ≠‘M’(MovieStar) = ∅

50. Example 2.25 • If one must have networth of at least $100,000,000 to be president of movie studio. FROM • MovieExec(name,address,cert#,networth) • Studio(name,address, presC#) • First we have to perform theta-join on this two relations. • σnetworth < 100000000(Studio ⋈ presC# = cert# MovieExec) = ∅ • Second way • ∏TpressC#(Studio) ⊆ ∏cert#(σnetworth < 100000000(MovieExec)) • Which one is more efficient?