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Fourth normal form: 4NF 1

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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

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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

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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)

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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 (!!!)

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

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Fourth normal form: 4NF 7 1N relation

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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

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

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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?

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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

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

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Fourth normal form: 4NF 13 Bring to 1NF

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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)

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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

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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.

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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

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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

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

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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

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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

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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

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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

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Fourth normal form: 4NF 24 Example

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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

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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?

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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).

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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

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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

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