1. Fourth normal form: 4NF 1

2. 2 Normal forms desirable forms for relations in DB design eliminate redundancies avoid update anomalies enforce integrity constraints in one sentence: produce a good representation of the real world formally defined, based on semantic concepts not a panacea

3. Fourth normal form: 4NF 3 2NF, 3NF and BCNF defined based on FDs decomposition procedure : projection any relation can be brought to any (of the above) NF accompanied by normalisation procedures

4. Fourth normal form: 4NF 4 4NF based on another semantic concept (not FDs) (i.e. capturing different semantic aspects (constraints) of the real life system)

5. Fourth normal form: 4NF 5 Motivation - 4NF courses - tutors - reference books assumptions for a course - any number of tutors and any numbers of books a book can be used for any number of courses a tutor can teach any number of courses no link between tutors and books (!!!)

6. Fourth normal form: 4NF 6 Un-normalised relation (CTB) CTB

7. Fourth normal form: 4NF 7 1N relation

8. Fourth normal form: 4NF 8 Note IF(c1, t1, b1) and (c1, t2, b2) exist as tuples in the relation THEN (c1, t1, b2) and (c1, t2, b1) also exist in the relation

9. Fourth normal form: 4NF 9 Question what normal forms is this relation in? helping question what functional dependencies can you identify?

10. Fourth normal form: 4NF 10 Shortcomings redundancy therefore: update anomalies e.g. add a new tutor, C. Fox, for the Databases course: both tuples, (Databases, C. Fox, Introduction to DB) and (Databases, C. Fox, DB Design), have to be added could you think of a better solution? this is a problem BCNF relation what do you conclude about the studied normal forms?

11. Fourth normal form: 4NF 11 Common-sense solution separate independent repeating groups decompose the un-normalised relation bring it to 1NF the solution: shortcomings - removed non-loss decomposition

12. Fourth normal form: 4NF 12 Decompose the un-normalised relation

13. Fourth normal form: 4NF 13 Bring to 1NF

14. Fourth normal form: 4NF 14 Solution what is the theoretical basis? not FDs (there arent any) multi-valued dependencies (MVDs) generalisation of FDs e.g. Course Tutor each course has a well defined set of tutors in the relation CTB, Tutor depends on the value of Course alone (Book does not influence)

15. Fourth normal form: 4NF 15 Multi-valued dependency - simple Let R be a relation and A and B arbitrary sets of attributes of R. There is a multi-valued dependency from A to B A B if and only each A value exactly determines a set of B values, independently of the other attributes

16. Fourth normal form: 4NF 16 Multi-valued dependency - Date Let R be a relation; A, B and C are arbitrary sets of attributes of R A multi-determines B (B is multi-dependent on A) A B if and only if the set of B values matching an (A, C) pair depends only on the A value.

17. Fourth normal form: 4NF 17 Multi-valued dependency (Elmasri and Navathe, 2000, p. 514) Let R be a relation and X and Y arbitrary sets of attributes of R ; if for any 2 tuples (in any extension) t 1 and t 2 with the property t 1 [X] = t 2 [X] there exists 2 tuples t 3 and t 4 such that (Z = R - (X Y)) t 1 [X] = t 3 [X] = t 4 [X] t 1 [Y] = t 3 [Y] and t 2 [Y] = t 4 [Y] t 1 [Z] = t 2 [Z] and t 3 [Z] = t 4 [Z] then X Y

18. Fourth normal form: 4NF 18 Trivial MVDs R, is a relation; X and Y arbitrary sets of attributes of R X Y is trivial if and only if Y X or X Y = R a trivial MVD does not represent a constraint

19. Fourth normal form: 4NF 19 Questions is every FD a MVD? is every MVD a FD?

20. Fourth normal form: 4NF 20 Towards 4NF MVDs are a generalisation of FDs the existence of non-trivial MVDs that are not FDs causes problems AIM: to eliminate such situations

21. Fourth normal form: 4NF 21 Theorems R (A, B, C) and A B A C (A B | C) (Fagin) R (A, B, C) R = join ({A, B}, {A, C}) if and only if A B | C

22. Fourth normal form: 4NF 22 4NF R is in 4NF if and only if any MVD is either trivial, or it is also a FD (Connolly): A relation is in 4NF if and only if it is in BCNF and contains no non-trivial multi-valued dependencies can you spot the error in the definition? 4NF is always achievable

23. Fourth normal form: 4NF 23 Fagins theorem Let R = (A, B, C) and A ->> B | C then R = (A, B) join (A, C) similar in form with the variant for FDs

24. Fourth normal form: 4NF 24 Example

25. Fourth normal form: 4NF 25 Discussion R = (Patient, Disease, Doctor) a patient has a number of diseases and, independently of them is assigned a number of doctors; therefore Patient ->> Disease | Doctor each disease is treated by a certain number of doctors and, independently of this, there is a certain number of patients that suffer from each disease; therefore: Disease ->> Doctor | Patient a patient, for a certain disease, is assigned only one doctor; therefore: (Patient, Disease) -> Doctor and NO MVDs exist

26. Fourth normal form: 4NF 26 Discussion R = (Patient, Disease, Doctor) each patient suffers from a number of diseases (independently of doctors) and each disease is treated by a number of doctors (independently of the patients that suffer from it); can we say that Patient ->> Disease and Disease ->> Doctor ? have we not already discussed the above case?

27. Fourth normal form: 4NF 27 Discussion what about R = (Patient, Disease, Treatment, Nurse) where a patient can have a number of diseases (independently of treatments and nurses) a disease has a number of treatments (independently of patients and nurses); a treatment is administered by a number of nurses (independently of patients and diseases).

28. Fourth normal form: 4NF 28 Discussion what about R = (Patient, Disease, Doctor, Nurse) where a patient has a number of diseases independently of doctors and nurses a patient has a unique doctor for a given disease a doctor has (works with) a number of nurses, independently of patients and diseases

29. Fourth normal form: 4NF 29 Conclusion 2NF, 3NF and BCNF based on FDs there are other constraints that can be expressed through projection multi-valued dependency 4NF Fagin