Informationslogistik Unit 7: Conceptual Design of Databases Normalization

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1 Informationslogistik Unit 7: Conceptual Design of Databases Normalization 17. V. 2011

2 Outline 1 Organization 2 SQL: Regular Expressions and Date Functions 3 Reminder and Overview 4 Normalization First Normal Form Second Normal Form Third Normal Form Boyce-Codd Normal Form Fourth Normal Form

3 Organization nächste Woche 2. Zwischentest (Stoff bis inkl. korrel. Subqueries) zusätzliche Unterlagen zu Normalisierung: Kapitel 2 von Andreas Meier: Relationale und postrelationale Datenbanken, Springer (innerhalb der Uni online verfügbar)

4 SQL-Lesson 7 Today: Self-Joins

5 Reminder and Overview In Unit 3 (The Relational Model) we already had some guidelines on how to design a relational database: start with entities, relations, and attributes put them in relational database (cf. slides of Unit 3) Now we deal with some refinements. The base for these refinements is the notion of functional dependency.

6 Normalization: Outlook We know about: functional dependencies (Unit 5) properties of good decompositions (Unit 6)

7 Normalization: Outlook We know about: functional dependencies (Unit 5) properties of good decompositions (Unit 6) We want: a good way to decompose a given relation R

8 Normalization: Outlook We know about: functional dependencies (Unit 5) properties of good decompositions (Unit 6) We want: a good way to decompose a given relation R We do this by: step-by-step normalization with each step we reach a higher level of normalization (1 to 4)

9 First Normal Form Outline 1 Organization 2 SQL: Regular Expressions and Date Functions 3 Reminder and Overview 4 Normalization First Normal Form Second Normal Form Third Normal Form Boyce-Codd Normal Form Fourth Normal Form

10 First Normal Form First Normal Form Definition (1NF) A relation is in 1NF, if all attributes are atomic. We have made this assumption already before. Example of table/relation not in 1NF: Father Mother Cildren Rene Mara {Kalle, Jakob} Ronald Elisabeth {Hannes, Kathrin} Andreas Marianne {Sophia}

11 First Normal Form First Normal Form An Example Example of table/relation not in 1NF: Father Mother Cildren Rene Mara {Kalle, Jakob} Ronald Elisabeth {Hannes, Kathrin} Andreas Marianne {Sophia} can easily be converted to 1NF: Father Mother Cild Rene Mara Kalle Rene Mara Jakob Ronald Elisabeth Hannes Ronald Elisabeth Kathrin Andreas Marianne Sophia

12 Second Normal Form Outline 1 Organization 2 SQL: Regular Expressions and Date Functions 3 Reminder and Overview 4 Normalization First Normal Form Second Normal Form Third Normal Form Boyce-Codd Normal Form Fourth Normal Form

13 Second Normal Form Second Normal Form Definition (2NF) A relation is in 2NF, if it is in 1NF and each non-key attribute is fully functionally dependent on each candidate key. (Recall: β is fully FD from α if α β and α minimal.) Intuition: Each attribute in the key matters.

14 Second Normal Form Second Normal Form An Example Consider the following table containing student information: (Assumption: only one study per student) fname sname mnr stkz study c_no course date mark List dependencies of non-key attributes: fname: sname: stkz: study: course: mark: table not in 2NF

15 Second Normal Form Second Normal Form An Example Consider the following table containing student information: (Assumption: only one study per student) fname sname mnr stkz study c_no course date mark List dependencies of non-key attributes: fname: mnr sname: stkz: study: course: mark: table not in 2NF

16 Second Normal Form Second Normal Form An Example Consider the following table containing student information: (Assumption: only one study per student) fname sname mnr stkz study c_no course date mark List dependencies of non-key attributes: fname: mnr sname: mnr stkz: study: course: mark: table not in 2NF

17 Second Normal Form Second Normal Form An Example Consider the following table containing student information: (Assumption: only one study per student) fname sname mnr stkz study c_no course date mark List dependencies of non-key attributes: fname: mnr sname: mnr stkz: mnr study: course: mark: table not in 2NF

18 Second Normal Form Second Normal Form An Example Consider the following table containing student information: (Assumption: only one study per student) fname sname mnr stkz study c_no course date mark List dependencies of non-key attributes: fname: mnr sname: mnr stkz: mnr study: mnr course: mark: table not in 2NF

19 Second Normal Form Second Normal Form An Example Consider the following table containing student information: (Assumption: only one study per student) fname sname mnr stkz study c_no course date mark List dependencies of non-key attributes: fname: mnr sname: mnr stkz: mnr study: mnr course: c_no mark: table not in 2NF

20 Second Normal Form Second Normal Form An Example Consider the following table containing student information: (Assumption: only one study per student) fname sname mnr stkz study c_no course date mark List dependencies of non-key attributes: fname: mnr sname: mnr stkz: mnr study: mnr course: c_no mark: mnr, c_no, date table not in 2NF

21 Second Normal Form Second Normal Form An Example fname sname mnr stkz study c_no course date mark Table can be turned into 2NF by summarizing attributes that depend on the same key attributes into a new table: mnr fname surname stkz study c_no course mnr c_no date_exam mark

22 Third Normal Form Outline 1 Organization 2 SQL: Regular Expressions and Date Functions 3 Reminder and Overview 4 Normalization First Normal Form Second Normal Form Third Normal Form Boyce-Codd Normal Form Fourth Normal Form

23 Third Normal Form Third Normal Form Definition (3NF) A relation is in 3NF, if it is in 1NF and for each FD α B at least one of the following conditions holds: 1 B α, i.e. the FD α B is trivial, 2 B is contained in a candidate key, 3 α is a superkey. In particular, a relation is not in 3NF, if there is a FD between attributes that are not part of a (candidate/super-) key.

24 Third Normal Form Third Normal Form An Example Example. mnr fname surname stkz study c_no course mnr c_no date_exam mark

25 Third Normal Form Third Normal Form An Example Example. mnr fname surname stkz study c_no course mnr c_no date_exam mark The functional dependency stkz study does not satisfy any of the above conditions.

26 Third Normal Form Third Normal Form An Example Example. mnr fname surname stkz stkz study c_no course mnr c_no date_exam mark

27 Third Normal Form The Synthesis Algorithm for 3NF Given: relation R, set F of FDs We want: lossless decomposition R 1,..., R n that conserves FDs, such that each R i is in 3NF Algorithm: 1 Determine canonical covering F c of F (see algorithm in Unit 6) 2 Create for each FD α β in F c a relation R α := α β (and determine corresponding set of FDs F α ) 3 If one of the R α s contains candidate key of R, we are done. Otherwise create relation R κ for a candidate key κ of R (with F κ = ). 4 Delete relations R α that are contained in other relations R α, i.e. R α R α

28 Boyce-Codd Normal Form Outline 1 Organization 2 SQL: Regular Expressions and Date Functions 3 Reminder and Overview 4 Normalization First Normal Form Second Normal Form Third Normal Form Boyce-Codd Normal Form Fourth Normal Form

29 Boyce-Codd Normal Form Boyce-Codd Normal Form Definition (BCNF) A relation is in BCNF, if it is in 1NF and for each FD α β at least one of the following two conditions holds: 1 β α, i.e. the FD α β is trivial, 2 α is a superkey. Difference to 3NF: β is not allowed to be part of candidate key

30 Boyce-Codd Normal Form Boyce-Codd Normal Form An Example Example. mnr c_no c_term c_term_abbrv date_exam mark WS 2008/09 W08 As the relation between c_term and c_term_abbrv is 1:1, another candidate key is { mnr, c_no, c_term, date_exam }. We have FDs: {c_term} {c_term_abbrv} and {c_term_abbrv} {c_term} However, neither attribute c_term and c_term_abbrv is superkey, nor is the FD trivial. table is not in BCNF

31 Boyce-Codd Normal Form Boyce-Codd Normal Form An Example Obvious solution: new table for c_term mnr c_no c_term_abbrv date_exam mark c_term_abbrv c_term

32 Boyce-Codd Normal Form Decomposition Algorithm for BCNF Given: relation R, set F of FDs We want: decomposition R 1,..., R n in BCNF Decomposition Algorithm: Initialize Z := {R}; while (there is R i in Z not in BCNF) do{ 1 find FD α β for R i with α β = and α R i ; 2 decompose R i into R i1 := α β and R i2 := R i β; 3 replace R i in Z with R i1 and R i2, i.e. Z := (Z {R i }) {R i1, R i2 }.

33 Boyce-Codd Normal Form 3NF vs. BCNF The Synthesis Algorithm guarantees decomposition that is lossless, conserves FDs is in 3NF The Decomposition Algorithm guarantees decomposition that is lossless, is in BCNF

34 Boyce-Codd Normal Form 3NF vs. BCNF The Synthesis Algorithm guarantees decomposition that is lossless, conserves FDs is in 3NF The Decomposition Algorithm guarantees decomposition that is lossless, is in BCNF Unfortunately: In some cases a BCNF decomposition does not conserve FDs. be content with 3NF in these cases

35 Fourth Normal Form Outline 1 Organization 2 SQL: Regular Expressions and Date Functions 3 Reminder and Overview 4 Normalization First Normal Form Second Normal Form Third Normal Form Boyce-Codd Normal Form Fourth Normal Form

36 Fourth Normal Form Fourth Normal Form and Multivalued Dependencies An Example: We want to store language and sports courses for students: matrnr language sports Italian skiing Italian fencing Italian soccer Spanish skiing Spanish fencing Spanish soccer

37 Fourth Normal Form Multivalued Dependencies The dependencies {matrnr} {language} and {matrnr} {sports} are so-called multivalued dependencies (MVDs). Definition (MVD) β is multivalued dependent from α (Notation: α β), if for each tuples with identical α-values there are further tuples with swapped β-values. (MVDs are generalization of FDs.)

38 Fourth Normal Form Multivalued Dependencies Definition (MVD) β is multivalued dependent from α (Notation: α β), if for each tuples with identical α-values there are further tuples with swapped β-values. In example: If there are rows , Italian, fencing and , Spanish, soccer, there also must be rows , Spanish, fencing and , Italian, soccer.

39 Fourth Normal Form Fourth Normal Form Idea: no table contains more than one multivalued relation Definition (4NF) A relation R is in 4NF, if it is in 1NF and for each MVD α β at least one of the following two conditions holds: 1 the MVD α β is trivial, i.e. either β α or β = R α, 2 α is a superkey of R. Note: Each relation in 4NF is also in BCNF.

40 Fourth Normal Form Fourth Normal Form An Example matrnr language sports Italian skiing Italian fencing Italian soccer Spanish skiing Spanish fencing Spanish soccer MVDs: {matrnr} {language} and {matrnr} {sports} table not in 4NF

41 Fourth Normal Form Fourth Normal Form An Example Example. matrnr language Italian Spanish matrnr sports skiing fencing soccer

42 Fourth Normal Form Generalized Decomposition Algorithm for 4NF Given: relation R, set F of MVDs We want: decomposition R 1,..., R n in 4NF Generalized Decomposition Algorithm: Initialize Z := {R}; while (there is R i in Z not in 4NF) do{ 1 find MVD α β for R i with α β = and α R i ; 2 decompose R i into R i1 := α β and R i2 := R i β; 3 replace R i in Z with R i1 and R i2, i.e. Z := (Z {R i }) {R i1, R i2 }. (As each relation in 4NF is in BCNF, in some cases 4NF decomposition does not conserve FDs.)

43 Fourth Normal Form Another Example Human resources database: pers_no name salary dept boss project course FDs: {pers_no} {name, salary, dept, boss} {dept} {boss} {boss} {dept} MVDs: {pers_no} {project} {pers_no} {course}

44 Fourth Normal Form Another Example Apply decomposition algorithm: FD {pers_no} {name, salary, dept, boss} gives decomposition pers_no name salary dept boss pers_no project course

45 Fourth Normal Form Another Example Apply decomposition algorithm: FD {dept} {boss} gives decomposition pers_no name salary dept dept boss pers_no project course

46 Fourth Normal Form Another Example Apply decomposition algorithm: MVDs {pers_no} {project} and {pers_no} {course} give decomposition pers_no name salary dept dept boss pers_no project pers_no course

47 Fourth Normal Form Another Example Apply decomposition algorithm: MVDs {pers_no} {project} and {pers_no} {course} give decomposition pers_no name salary dept dept boss pers_no project pers_no course is in 4NF

48 Fourth Normal Form Another Example When there is more information about projects and courses to store: pers_no name salary dept dept boss pers_no project_no pers_no course_no project_no name... course_no name...