Relational Database Systems 1
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1 Relational Database Systems 1 Wolf-Tilo Balke Benjamin Köhncke Institut für Informationssysteme Technische Universität Braunschweig
2 Overview Motivation Relational Algebra Query Optimization Advanced Realtional Algebra σ π Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 2
3 6.1 Motivation A data model needs three parts: Structural part Data structures which are used to create databases representing the objects modeled Integrity part Rules expressing the constraints placed on these data structures to ensure structural integrity Manipulation part Operators that can be applied to the data structures, to update and query the data contained in the database Relational Database Systems 1 Wolf-Tilo Balke Technische Universität Braunschweig 3
4 6.1 Motivation Last week we introduced the relational model Based on set theory A relation is a subset of the Cartesian product over a list of domains Relations can be written as tables: relation name attributes tuples PERSON firstname lastname sex Clark Joseph Kent m Louise Lane f Lex Luthor m Charles Xavier m Erik Magnus m Jeanne Gray f Ororo Munroe f domain values 4
5 6.1 Motivation How do you work with relations? Relational algebra! Proposed by Edgar F. Codd: A Relational Model for Large Shared Data Banks, Communications of the ACM, 1970 The theoretical foundation of all relational databases Describes how to manipulate relations and retrieve interesting parts of available relations Relational algebra is mandatory for advanced tasks like query optimization Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 5
6 6.1 Motivation The first thing that happens to a SQL query is that it gets translated into relational algebra. Query Parser & Translator Relational Algebra Expression Query Optimizer Query Result Evaluation Engine Execution Plan Access Paths Statistics Data Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 6
7 6.1 Motivation Queries need to be translated to an internal form Queries posed in a declarative DB language what should be returned, not how should it be returned Queries can be evaluated in different way Several relational algebra expressions might lead to the same results Each statement can be used for query evaluation But different statements might also result in vastly different performance! This is the area of query optimization, the heart of every database kernel Datenbanksysteme 2 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 7
8 6.1 Motivation Why do we need special query languages at all? Won t conventional languages like C or Java suffice to ask and answer any computational question about relations? The relational algebra is usefull, because it is less powerful than C or Java. 2 huge rewards: Ease of programming Ability of the compiler to produce highly optimized code [GUW09, 2] Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 8
9 6.2 Relational Algebra What is an algebra after all? It consists of operators and atomic operands. It allows to build expressions by applying operators to operands and / or other expressions of the algebra. Example: algebra of arithmetics Atomic operands: variables like x and constants like 15 Operators: addition, subtraction, multiplication and division Relational algebra: another example of an algebra Atomic operands: Variables, that stand for relations Constants, which are finite relations Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 9
10 6.2 Relational Algebra Elementary operations: Set algebra operations Set Union Set Intersection Set Difference Cartesian Product New relational algebra operations Selection ς Projection π Renaming ρ Additional derived operations (for convenience) All sorts of joins,,, Division EN 6 Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 10
11 6.2 Example Relations students matnr firstname lastname sex 1005 Clark Kent m 2832 Louise Lane f 4512 Lex Luther m 5119 Charles Xavier m 6676 Erik Magnus m 8024 Jeanne Gray f 9876 Logan m courses crsnr title 100 Intro. to being a Superhero 101 Secret Identities How to take over the world exams student course result EN 6 Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 11
12 6.2 Relational Algebra Selection ς Selects all tuples (rows) from a relation that satisfy some given Boolean predicate (condition) Selection = Create a new relation that contains exactly the satisfying tuples ς <condition> (R) Condition clauses: <attribute> θ <value> <attribute> θ <attribute> θ {=, <,,, >, } Clauses may be connected by, and (logical AND, OR, and NOT) EN 6 Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 12
13 6.2 Relational Algebra Example: students matnr firstname lastname sex 1005 Clark Kent m Select all female students ς sex= f (students) 2832 Louise Lane f 4512 Lex Luther m 5119 Charles Xavier m 6676 Erik Magnus m 8024 Jeanne Gray f 9876 Logan m ς sex= f (students) matnr firstname lastname sex 2832 Louise Lane f 8024 Jeanne Gray f EN 6 Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 13
14 Example: 6.2 Relational Algebra exams student course result Select exams within course 100 with a result of worse than 3.0 ς course=100 result>3.0 (exams) ς course=100 result>3.0 (exams) student course result EN 6 Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 14
15 6.2 Relational Algebra Projection π Retains only attributes (columns) with given names Again: Creates a new relation with only those attribute sets which are specified within some attribute list If tuples are identical after projection, duplicate tuples are removed π <attributelist> (R) courses crsnr title All courses titles : π title (courses) π title (courses) title 100 Intro. to being a Superhero 101 Secret Identities How to take over the world Intro. to being a Superhero Secret Identities 2 How to take over the world EN 6 Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 15
16 6.2 Relational Algebra Of course, these operations can be combined Retrieve first name and last name of all female students π firstname, lastname ς sex=ʼfʼ students students matnr firstname lastname sex 1005 Clark Kent m 2832 Louise Lane f 4512 Lex Luther m 5119 Charles Xavier m 6676 Erik Magnus m 8024 Jeanne Gray f 9876 Logan m ς sex= f students matnr firstname lastname sex 2832 Louise Lane f 8024 Jeanne Gray f π firstname, lastname ς sex=ʼfʼ students firstname lastname Loise Lane Jeanne Gray EN 6 Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 16
17 6.2 Relational Algebra Renaming operator ρ Renames a relation and/or its attributes Also denoted by ρ S(B1, B2,, Bn) (R) or ρ S (R) or ρ (B1, B2,, Bn) (R) courses crsnr Retrieve all course numbers contained in courses and name the resulting relation lectures lectures π crsnr courses title 100 Intro. to being a Superhero 101 Secret Identities How to take over the world lectures crsnr EN 6 Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 17
18 6.2 Relational Algebra Renaming operator ρ Select all result of course 100 from exams; the resulting relation should be called result and contain the attributes matno, crsno, and grade ρ results(matno, crsno, grade) ς course=100 exams ς course=100 exams student course result results matno crsno grade EN 6 Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 18
19 6.2 Relational Algebra Union, intersection, and set difference Operators work as in set theory Operands have to be union-compatible (i.e., they must consist of the same attributes) Written as R S, R S, and R S, respectively ς course=100 exams course title 100 Intro. to being a Superhero ς course=102 exams course title 102 How to take over the world ς course=100 exams ς course=102 exams course title 100 Intro. to being a Superhero 102 How to take over the world EN 6 Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 19
20 6.2 Relational Algebra Cartesian product Written as R S Also called cross product Creates a new relation by combining each tuple of the first relation with every tuple of the second relation Attribute set = all attributes of R plus all attributes of S The resulting relation contains exactly R S tuples EN 6 Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 20
21 6.2 Relational Algebra goodgrades females student course result matno lastname Lane Gray cross student course result matno lastname Lane Lane Lane Gray Gray Gray goodgrades ς result 3.0 exams females π matno, lastname ς sex=female students cross goodgrades females EN 6 Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 21
22 6.2 Relational Algebra π lastname, title, result ς matno=student course=crsno (females goodgrades courses) lastname course result Lane How to take over the world? 2.0 This type of statement is very important! Cartesian product, followed by selections/projections Selections/projections involve different source relations This kind of query is called a join EN 6 Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 22
23 6.2 Relational Algebra Theta join (θ) (θ is a Boolean condition) Sometimes also called inner join or just join Creates a new relation by combining related tuples from different source relations Written as R ς(condition) S or ς (condition) (R S) Theta joins are very similar to selections π lastname, title, result females ς(matno=student) goodgrades ς(course=crsno) courses π lastname, title, result ς matno=student course=crsno (females goodgrades courses ) lastname course result Lane How to take over the world? 2.0 EN 6 Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 23
24 6.2 Relational Algebra Equi-join (condition) Joins two relations only using equality conditions R (condition) S condition may only contain equality statements between attributes (A 1 =A 2 ) A special case of the theta join π lastname, title, result females matno=student goodgrades course=crsno courses π lastname, title, result ς matno=student course=crsno females goodgrades courses lastname course result Lane How to take over the world? 2.0 EN 6 Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 24
25 6.2 Relational Algebra Natural join A special case of the equi-join R (attributelist) S Implicit join condition Each attribute in attributelist must be contained in both source relations For each of these attributes, an equality condition is created All these conditions are connected by logical AND If attributelist is empty: Join attributes = All attributes that are shared between the two relations (that is, that have the same name) EN 6 Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 25
26 6.2 Relational Algebra goodgrades matno crsnr result females matno lastname 2832 Lane 8024 Gray courses crsnr title 100 Intro. to being a Superhero 101 Secret Identities How to take over the world π lastname, title, result females goodgrades courses lastname course result Lane How to take over the world? 2.0 EN 6 Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 26
27 Optimization Example Relational algebra usually allows for alternative, yet equivalent evaluation plans Respective execution costs may strongly differ Memory space, response time, etc. Idea: Find the best plan, before actually executing the query Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 27
28 Optimization Example students matnr firstname lastname sex 1005 Clark Kent m 2832 Lois Lane f 4512 Lex Luther m 5119 Charles Xavier m 6676 Erik Magnus m 8024 Jean Gray f 9876 Bruce Banner m Peter Parker m Raven Darkholme f 4 Byte 30 Byte 30 Byte 1 Byte courses crsnr title 100 Intro. to being a Superhero 101 Secret Identities How to take over the world exams 4 Byte matnr crsnr result Byte 4 Byte 4 Byte 8 Byte Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 28
29 Optimization Example SQL Statement SELECT lastname, result, title FROM students s, exams e, courses c WHERE e.result<=1.3 AND s.matnr=e.matnr AND e.crsnr=c.crsnr Canonical Relational Algebra Expression Expression directly mapped from the SQL query π lastname, result, title ς result 1.3 exams.crsnr=courses.crsnr students.matnr=exams.matnr students exams courses lastname result title Magnus 1.3 How to take over the world Kent 1.3 Intro. to being a Superhero Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 29
30 Optimization Example π lastname, result, title ς result 1.3 exams.crsnr = courses.crsnr students.matnr = exams.matnr students exams courses π lastname, result, title ς result 1.3 Create Canonical Operator Tree Operator tree visualized the order of primitive functions (Note: Illustration is not really canonical tree as selection is already separated) students ς exams.crsnr=courses.crsnr ς students.matnr=exams.matnr courses exams Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 30
31 Optimization Example How much space is needed for the intermediate results? 2 * 68B = 136B π lastname, result, title 2 * 115B = 230B ς result * 115B = 690B ς exams.crsnr=courses.crsnr 18 * 115B = 2,070B ς students.matnr=exams.matnr 162 * 115B = 18,630B 54 * 81B = 4,374B courses students exams 3 * 34B= 102B 9 * 65B = 585B 6 * 16B = 96B Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 31
32 Optimization Example Example: Final Operator Tree 2 * 68B = 136B π lastname, result, title 2 * 76B = 152B exams.crsnr=courses.crsnr 2 * 42B = 84B π lastname, result, crsnr 2 * 50B = 100B students.matnr=exams.matnr courses 3 * 34B= 102B 9 * 34B = 306B 2 * 16B = 32B π lastname, matnr ς result 1.3 students 9 * 65B = 585B exams 6 * 16B = 96B Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 32
33 Optimization Example Comparison of intermediate results 2 * 68B = 136B π lastname, result, title 2 * 115B = 230B ς result * 68B = 136B π lastname, result, title 6 * 115B = 690B ς exams.crsnr=courses.crsnr 2 * 76B = 152B exams.crsnr=courses.crsnr 18 * 115B = 2,070B ς students.matnr=exams.matnr 162 * 115B = 18,630B 2 * 42B = 84B 2 * 50B = 100B students.matnr=exams.matnr π lastname, matnr π lastname, result, crsnr 3 * 34B= 102B courses 9 * 34B = 306B 2 * 16B = 32B ς result * 81B = 4,374B courses students exams 3 * 34B= 102B students exams 9 * 65B = 585B 6 * 16B = 96B 9 * 65B = 585B 6 * 16B = 96B Intermediate result set sizes B versus 674 B Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 33
34 6.2 Relational Algebra Left semi-join and right semi-join Combination of a theta-join and projection R (condition) S π (list of all attributes in R) (R) (condition) S R (condition) S π (list of all attributes in S) (R) (condition) S Creates a copy of R (or S) while removing all those tuples that do not have a join partner A filtering operation Works like a natural join if condition is empty Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 34
35 6.2 Relational Algebra Left semi-join: students students exams matnr firstname lastname sex 1005 Clark Kent m 2832 Louise Lane f 4512 Lex Luther m 5119 Charles Xavier m exams matnr crsnr result students exams matnr firstname lastname sex 1005 Clark Kent m 2832 Louise Lane f All hard-trying students (those who did exams). Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 35
36 6.2 Relational Algebra Division Written R S S contains only attributes that are also present in R Division restricts R in terms of attributes and tuples Result s attributes = those in R but not in S Result s tuples = those in R such that any tuple in S is a join partner Useful for queries like Find the sailors who have reserved all boats. More formal: Let A R ={attributes of R}; A S ={attributes of S}; A S A R A = A R A S R S π A (R) π A ((π A (R) S) R) Why the name division? Because (R S) S = R. But: (R S) S = R is not true in general Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 36
37 6.2 Relational Algebra Division: SC (= students and courses) matnr name crsnr 1000 Clark Kent C Clark Kent C Louise Lane C Lex Luther C Lex Luther C Lex Luther C Charles Xavier C Charles Xavier C100 Result contains all those students who took the same courses as Clark Kent (and possibly some more). CCK (= courses of Clark Kent) CCK π crsnr ς name= Clark Kent SC crsnr C100 C102 SC CCK matnr name 1000 Clark Kent 1002 Lex Luther EN 6 Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 37
38 6.2 Summary Basic relational algebra Selection ς Projection π Renaming ρ Union, set difference Cartesian product Extended relational algebra Intersection Theta-join (θ-cond) Equi-join (=-cond) Natural join Left semi-join and right semi-join Division Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 38
39 Translation of Relational Algebra Expressions MOVIE(id, name not null, year, type, remark) COUNTRY(movie, country not null ) Π country, name (COUNTRY COUNTRY.movie=MOVIE.id ς year=1893 type= cinema MOVIE) From which countries are the cinema movies of the year 1893 and what are their names? Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 39
40 Translation of Relational Algebra Expressions MOVIE(id, name not null, year, type, remark) PERSON(id, name not null, sex) PLAYS(movie, person not null, role not null ) Π pe.id, pe.name ((ς m.name = Star Wars ) m.type= cinema MOVIE m) m.id=p.movie ς p.role= Killer (PLAYS p) p.person=pe.id ς pe.sex= female (PERSON pe))) Which female actors from Star Wars cinema movies played a killer? Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 40
41 Translation of Relational Algebra Expressions PERSON(id, name not null, sex) PLAYS(movie, person not null, role not null ) GENRE(movie, genre not null ) Π name (Π id,name (PERSON p p.id=ply.person ς ply.role= postman (PLAYS ply)) \ Π id,name (PERSON pe pe.id=pla.person PLAYS pla pla.movie=g.movie ς g.genre= Western (GENRE g)) Which actors have played a postman, but never participated in a Western? Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 41
42 6.3 Advanced Relational Algebra Advanced relational algebra Left outer join, Right outer join, Full outer join, Aggregation F These operations cannot be expressed with basic relational algebra Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 42
43 6.3 Advanced Relational Algebra Left outer join and right outer join Theta-joins and its specializations join exactly those tuples that satisfy the join condition Tuples without a matching join partner are eliminated and will not appear in the result All information about non-matching tuples gets lost Outer joins allow to keep all tuples of a relation Padding with NULL values if there is no join partner Left outer join : Keeps all tuples of the left relation Right outer join : Keeps all tuples of the right relation Full outer join : Keeps all tuples of both relations EN 6 Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 43
44 6.3 Advanced Relational Algebra Example: List students and their exam results π lastname, crsnr, result students exams students matnr firstname lastname sex 1005 Clark Kent m 2832 Louise Lane f 4512 Lex Luther m 5119 Charles Xavier m exams matnr crsnr result π lastname, crsnr, result students exams lastname crsnr result Kent Kent Lane Lex Luther and Charles Xavier are lost because they didn t take any exams! Also, information on student 9876 disappears Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 44
45 6.3 Advanced Relational Algebra Left outer join: List students and their exam results π lastname, crsnr, result students exams students exams matnr firstname lastname sex matnr crsnr result 1005 Clark Kent m 2832 Louise Lane f 4512 Lex Luther m 5119 Charles Xavier m π lastname, crsnr, result students exams lastname crsnr result Kent Kent Lane Luther NULL NULL Now, you could easily count all non-null results to get student statistics without looking into the students table. Kent: 2; Lane: 1; Luther: 0; Xavier: 0 Xavier NULL NULL Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 45
46 6.3 Advanced Relational Algebra Right outer join: π lastname, crsnr, result students exams students exams matnr firstname lastname sex matnr crsnr result 1005 Clark Kent m Louise Lane f Lex Luther m Charles Xavier m π lastname, crsnr, result students exams lastname crsnr result Kent Kent Lane NULL All exam result with student names if known. Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 46
47 6.3 Advanced Relational Algebra Full outer join: π lastname, crsnr, result students exams students exams matnr firstname lastname sex 1005 Clark Kent m 2832 Louise Lane f 4512 Lex Luther m 5119 Charles Xavier m lastname crsnr result Kent Kent Lane Luther NULL NULL NULL Xavier NULL NULL matnr crsnr result π lastname, crsnr, result students exams Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 47
48 6.3 Aggregation Aggregation operator: Typically used in simple statistical computations Merges tuples into groups Computes (simple) statistics for each group Examples: Compute the average exam score For each student, count the total number of exams he/she has taken so far Find the highest exam score ever Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 48
49 6.3 Aggregation Written as <grouping attributes> F <function list> (R) F is called script F Creates a new relation: All tuples in R that have identical values with respect to all grouping attributes, are grouped together All functions in the function list are applied to each group For each group, a new tuple is created, which contains the result values of all the functions The schema of the resulting relation looks as follows: The grouping attributes and, for each function, one new attribute The name of each new attribute is function name + argument name Example: courseid F average(grade) (ExamResults) Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 49
50 6.3 Aggregation Attention: Duplicates are usually NOT eliminated when grouping Some available functions: Sum Sum of all non-null values Average Mean of all non-null values Maximum Maximum value of all non-null values Minimum Minimum value of all non-null values Count Number of tuples having a non-null value Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 50
51 6.3 Aggregation Example (without grouping): If there are no grouping attributes, the whole input relation is treated as one group F sum(student), average(result), min(result), max(result) (exams) exams student course result sum student average result min result max result Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 51
52 6.3 Aggregation Example: For each course, count results and compute the average score. coursef average(result), count(result) (exams) exams student course result course average result count result Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 52
53 6.3 Aggregation Example: For each student, count the exams and compute average result. ρ overview(name, #exams, avgresult) ( lastnamef count(crsnr), avg(result) (π lastname, crsnr, result (students exams))) students matnr firstname lastname sex 1005 Clark Kent m 2832 Louise Lane f 4512 Lex Luther m 5119 Charles Xavier m overview exams name #exams avgresult Kent Lane Luther 0 NULL Xavier 0 NULL matnr crsnr result Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 53
54 6.3 Conflicting Attribute Names In most previous examples, all relations had different attribute names The real world is different In case of conflicting names, the full name of relation R s attribute A is R.A Still, attribute names need to be unique within each relation Example: ς students.matno = grade.matno (students grade) Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 54
55 Next Lecture Relational tuple calculus SQUARE & SEQUEL Domain tuple calculus Query-by-example (QBE) Relational Database Systems 1 Wolf-Tilo Balke Institut für Informationssysteme TU Braunschweig 55
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