Databases 2012 Normalization

Transcription

1 Databases 2012 Christian S. Jensen Computer Science, Aarhus University

2 Overview Review of redundancy anomalies and decomposition Boyce-Codd Normal Form Motivation for Third Normal Form Third Normal Form Fourth Normal Form 2

3 Redundancy Anomaly MeetingsOwners meetid what date slot userid name group office ddb csj Christian S. Jensen vip Turing ddb csj Christian S. Jensen vip Turing TA-meeting amoeller Anders Møller vip Turing-224 Information about owners is duplicated 3

4 Update Anomaly MeetingsOwners meetid what date slot userid name group office ddb csj Christian S. Jensen vip Turing ddb csj Christian S. Jensen vip Turing TA-meeting amoeller Anders Møller vip Turing-224 Christian now has conflicting office information 4

5 Deletion Anomaly MeetingsOwners meetid what date slot userid name group office ddb csj Christian S. Jensen vip Turing ddb csj Christian S. Jensen vip Turing TA-meeting amoeller Anders Møller vip Turing-224 When Anders meeting is deleted, we no longer know anything about him 5

6 Decomposition I Anomalies may be avoided by decomposition: Meetings meetid what date slot ddb ddb TA-meeting Loss of information. Owners userid name group office csj Christian S. Jensen vip Turing-216 amoeller Anders Møller vip Turing-224 6

7 Decomposition II Anomalies may be avoided by decomposition: Meetings meetid what date slot userid ddb csj ddb csj TA-meeting amoeller Lossless: R = R 1 R 2 Owners userid name group office csj Christian S. Jensen vip Turing-216 amoeller Anders Møller vip Turing-224 7

8 Lossless Decomposition Let R have schema a 1,...,a n,b 1,...,b k,c 1,...,c m Define R 1 = Define R 2 = a 1,...,a n,b 1,...,b k a 1,...,a n,c 1,...,c m (R) (R) If a 1,...,a n is a superkey for R 1 or for R 2 then this is a lossless decomposition Meetings and Owners form a lossless decomposition since userid is a (superkey) for Owners 8

9 Functional Dependencies A functional dependency in a table is of the form: a 1,...,a n b for some attributes a i and b It implies that the values of a 1,...,a n determine the value of b in any row Generalize notion of key Identify possible information loss from a given decomposition 9

10 Dependencies Cause Anomalies MeetingsOwners meetid what date slot userid name group office ddb csj Christian S. Jensen vip Turing ddb csj Christian S. Jensen vip Turing TA-meeting amoeller Anders Møller vip Turing-224 userid name userid group userid office 10

11 Obvious Dependencies Some functional dependencies always hold Trivial dependencies: a 1,...,a n a i for any set of attributes a 1,...,a n a 1,...,a n b for any b when a 1,...,a n is a superkey 11

12 Minimal Basis of a Set of Dependencies Assume a set F of dependencies on a table. We have already defined the closure of F, F +. Any set of dependencies equivalent to F is a basis for F. A minimal basis G satisfies these properties it is basis all dependencies are of the form a 1,...,a n b if any dependency is removed, G is no longer a basis if we remove any a i from a dependency, G is no longer a basis 12

13 BCNF We would like tables to have only obvious functional dependencies Boyce-Codd Normal Form (BCNF): For all nontrivial dependencies a 1,...,a n b, the attributes a 1,...,a n is a superkey. We can always obtain BCNF by systematic lossless decomposition 13

14 BCNF (1/3) Find a table R with a nonobvious, nonreducible functional dependency: a 1,...,a n b 1 (otherwise it is already in BCNF) Collect all others with the same left-hand side: a 1,...,a n b 2... a 1,...,a n b k 14

15 BCNF (2/3) Assume that the schema of R is a 1,...,a n,b 1,...,b k,c 1,...,c m Decompose R into: R 1 = R 2 = a 1,...,a n,b 1,...,b k a 1,...,a n,c 1,...,c m (R) (R) This is lossless since a 1,...,a n is a key for R 1 15

16 BCNF (3/3) We have now removed all nonobvious dependencies with left-hand side a 1,...,a n, since R 1 and R 2 have none of those Now, repeat this process all over again Termination is guaranteed, since there are only finitely many different left-hand sides 16

17 BCNF Example (1/3) Exams studentid date time vip room :00 csj Turing :30 csj Turing :30 amoeller Turing :30 eernst Ada :00 eernst Ada-017 Keys: (studentid, date) (vip, date, time) (room, date, time) 17

18 BCNF Example (2/3) Nonreducible functional dependencies: studentid, date time studentid, date vip studentid, date room vip, date, time studentid room, date, time studentid room, date, time vip vip, date room 18

19 BCNF Example (2/3) Nonreducible functional dependencies: studentid, date time studentid, date vip studentid, date room vip, date, time studentid room, date, time studentid room, date, time vip vip, date room 19

20 Rooms BCNF Example (3/3) Exams date vip room csj Turing amoeller Turing eernst Ada eernst Ada-017 studentid date time vip :00 csj :30 csj :30 amoeller :30 eernst :00 eernst 20

21 Inferring Dependencies The BCNF algorithm assumes that we know all functional dependencies. We solved the problem of inferring all dependencies from a given set of dependencies last time. 21

22 Properties of a Decomposition A relational schema R with a set of functional dependencies F is decomposed into R 1 and R Lossless-join decomposition Test to see if at least one of the following dependencies are in F + R 1 R 2 R 1 R 1 R 2 R 2 If not, the decomposition may be lossy 2. Dependency preservation Let F i be the set of dependencies in F + that include only attributes in R i (Notation: F i = Ri (F + )) Test to see if (F 1 F 2 ) + = F + When a relation is modified, no other relations need to be checked to preserve dependencies. 3. No redundancy 22

23 R = (A, B, C) F = {A B, B C} R 1 = (A, B), R 2 = (A, C) An Example Lossless-join decomposition: yes R 1 R 2 = {A} and A AB Dependency preserving: no F + = {A B, A C, A AB, A BC, A AC, A ABC, AB C, AB AC, AB BC, AB ABC, AC BC, AC B, B C, B BC} {many trivial dependencies} F 1 = {A B, A AB} {many trivial dependencies} F 2 = {A C, A AC} {many trivial dependencies} (F 1 F 2 ) + F + Cannot check B C without computing R 1 R 2 23

24 Impact of Non-Dependency Preservation r 1 : r 2 : A B A C r 1 satisfies {A B}. r 2 satisfies {A C}. The programmer must maintain {B C} manually, which is difficult. (Say the programmer changed B in second row to 17 ) 24

25 Possible Dependency Loss in BCNF BCNF decomposition is not always dependency preserving. Example R = (C, S, Z) (City, State, Zip_code) F = {CS Z, Z S} There are two candidate keys: CS and CZ. R is not in BCNF. Any decomposition of R will fail to preserve CS Z. 25

26 Third Normal Form (3NF) Definition: A relation schema R is in third normal form (3NF) if for all X B F + at least one of the following holds: X B is trivial (i.e., B X), X is a superkey for R, or B is contained in a candidate key. If a relation is in BCNF it is in 3NF. 26

27 Example R = (C, S, Z) (City, State, and Zip code) F = {CS Z, Z S} There are two candidates keys: CS and CZ. R is in 3NF. CS Z Z S (CS is a superkey.) (S is contained in a key.) 3NF admits some redundancy C S Z Tucson Arizona Marana? 85718? = Arizona by Z S. Thus, all redundancy is not eliminated. 27

28 3NF Decomposition Algorithm Algorithm to decompose a relation schema R into a set of relation schemas {R 1, R 2,..., R n } such that each relation schema R is in 3NF, lossless-join decomposition, and decomposition is dependency preserving. Let G be a minimal basis for F m := 0 for each functional dependency X A G do m := m + 1 R m := XA if none of the schemas R j 1 j m contains a candidate key for R then m := m + 1 R m := any candidate key for R Combine R i and R j if: R i was obtained by using X i A i R j was obtained by using X i A j 28

29 What Can Be Achieved? 3NF decomposition Lossless Dependency preserving BCNF decomposition Lossless Not guaranteed to be dependency preserving 29

30 Fourth Normal Form (4NF) R is in Fourth Normal Form if whenever A 1 A 2 A n B 1 B 2 B m is a nontrivial MVD, {A 1 A 2 A n } is a superkey. Meaning that if a relation is in 4NF, then every nontrivial MVD is an FD with a superkey on the left. There are database schemas in BCNF and 3NF that do not seem to be sufficiently normalized. 30

31 Fourth Normal Form (4NF) Example R = Film(Title, Performer, TapeID), with no non-trivial FDs Title Performer TapeID True Lies Arnold 1 True Lies Arnold 2 True Lies Arnold 3 True Lies Jamie 1 True Lies Jamie 2 True Lies Jamie 3 R is in BCNF. Insertion anomalies exist, assume a new copy, 4, is made, two tuples need to be inserted (True Lies, Arnold, 4) (True Lies, Jamie, 4) 31

32 4NF, cont. It is better to decompose Film into two relations. Performer: Title Performer True Lies Arnold True Lies Jamie Tape: Title TapeID True Lies 1 True Lies 2 True Lies 3 These two relations are in 4NF. Normalize with respect to multi-valued dependencies, not functional dependencies. 32

33 Normal Form Relationships 1NF 2NF 3NF BCNF 4NF 5NF Higher normal form less redundancy less memory used 33

34 Denormalization Redundancy may be good sometimes speeds up data access requires fewer join operations reduces wait time during locking Denormalization is the careful introduction of redundancy in a database design Speeds up SELECT Slows down INSERT, UPDATE, DELETE 34

35 Review Properties of a good relational design No redundancy No anomalies Ability to represent all the information Converting a bad design to a good design Decompose large relation schemas into smaller ones (NB: Now we need to do more joins to answer queries.) Ensuring lossless join decomposition Specify semantic integrity constraints to be satisfied by all instances of the schemas. Use constraints to argue that decompositions are lossless. Normal forms Boyce-Codd Normal Form is the standard. Third Normal Form allows additional redundancy, but also guarantees dependency preservation. Fourth Normal Form is based on multivalued dependencies. 35