5nf and other normal forms 1. 2 outline n-decomposability 3d constraint join dependency 5nf non-5nf...

5NF and other normal forms

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Outline

n-decomposability 3D constraint join dependency 5NF non-5NF - update anomalies problems in bringing a relation to 5NF other normal forms


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Always two projections?

so far every relation was non-loss decomposable into two projections is this always possible?

n-decomposable relations


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Courses - tutors - levels (CTL)

Course Tutor Level

Databases M. Ursu Level3

Databases M. Marman Level2

Programming M. Ursu Level2



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CTL - 2 attribute projections

Course Tutor

Databases M. Ursu

Databases M. Harman

Programming M. Ursu

Tutor Level

M. Ursu Level3M. Harman Level2M. Ursu Level2

Course Level

Databases Level3

Databases Level2

Programming Level2

CT TL

CL


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CTL - 3-decomposable

the join of any two projections is not CTL; e.g:

Course Tutor Level






join(CT, TL)

Extra!


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Constraint 3D

Let R be a degree 3 relation. IF (a, b, x) R AND (a, y, c) R AND (z, b, c) R THEN (a, b, c) R


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Constraint 3D illustrated on the CTL relation

IF tutor t1 teaches subject s1 AND level l1 studies subject s1 AND tutor t1 teaches level l1 THEN tutor t1 teaches subject s1 for level l1

note: this constraint is not expressed in CTL


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Constraint 3D and Join Dependency

4NF does not express the constraint 3D the constraint 3D is a facet of a more general

constraint: join dependency


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Join dependency

Let R be a relation. Let A, B, ..., Z be arbitrary subsets of R’s attributes. R satisfies the JD

( A, B, ..., Z ) if and only if R is equal to the join of its

projections on A, B, ..., Z


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5 NF

R is in 5NF if and only if every join dependency in R is implied by the candidate keys of R

5NF is always achievable


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Explanation

a join dependency, (A, B, …, Z), is implied by the candidate keys, K1, …, Km of R

if the fact that K1, …, Km are candidate keys

for R determine the fact that R has the JD (A, B, …, Z)


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Illustration - positive example

consider R (S_id, S_name, Status, City) with S_id and S_name candidate keys

({S_id, S_name, Status}, {S_id, City}) is a JD because S_id is a candidate key in R

({S_id, S_name}, {S_id, Status}, {S_name, City}) is a JD because S_id and S_name are both candidate keys in R


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Illustration - negative example

consider CTL (Course, Tutor, Level) with (Course, Tutor, Level) - candidate key (and an extra constraint : constraint 3D)

({Course, Tutor}, {Course, Level}, {Tutor, Level})is a JD, but this is not due to the CK, but to the constraint 3D

if CTL had not had constraint 3D, would it have been in 5NF?


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Not 5NF - update anomalies

CTL satisfies ( {Course, Tutor}, {Tutor, Level}, {Course, Level} )

insert (Programming, M. Ursu, Level2) what else must be done?

Course Tutor Level




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CTL satisfies the same JD as before delete (Databases, M. Ursu, Level2) what else must be done?

Course Tutor Level





Not 5NF - update anomalies


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JDs and MVDs

Fagin’s theorem restated R ( A, B, C ) satisfies ( AB, AC ) if and only if it

satisfies the MVDs A B | C

JD is the most general form of dependency (read as determination) possible between the attributes of a relation (in the relational model)


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Activity

Is 4NF subsumed by 5NF? Can you prove this using Fagin’s theorem and the definitions for 4 and 5 NF?


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Problems in bringing a relation to 5NF

check whether all JDs are implied by the candidate keys

Fagin : provides an algorithm for doing this for any relation, given all its JDs and all candidate keys

discover all JDs for a given relation they do not have that intuitive meaning as FDs or MVDs


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Concluding remarks

5NF is the ultimate normal form with respect to projection / join

5NF is guaranteed to be free of all anomalies that can be eliminated via projections

determining whether a relation is in 4NF but not in 5NF is still fuzzy

very rare in practice


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Recap

JD - a more general constraint than MD a relation can be in 4NF and have un-expressed JDs

this results in update anomalies

such a relation can be decomposed (via projection) into an equivalent set of 5NF relations

a relation is 5NF if all its JDs are deducible from its candidate keys

for a relation in 4NF but not in 5NF, an unexpressed JD is a possible decomposition (towards 5NF)


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Other normal forms

FDs, MVDs or JDs are not used domain-key normal form

R is in DK/NF if and only if every constraint of R is a logical consequence of domain constraints and (candidate) key constraints

restriction-union normal form decomposing operator: restriction abusing the language it can be said that: this

normalisation theory is orthogonal on the “projection” normalisation theory

5nf and other normal forms 1. 2 outline n-decomposability 3d constraint join dependency 5nf non-5nf...

Documents

normal forms

c r slide

nf non

dependency slide

z slide

candidate keys of r

achievable slide

c r andz