dependency preservation, 3nf revisited and bcnf 1

Dependency preservation, 3NF revisited and BCNF

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Dependency preservation,3NF revisited and BCNF


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Decomposition - more than one possibility

normalisation decomposition (non-loss) Modules(M_id, M_name, Type, Value)

solution #1 (3NF)• Modules_Descr(M_id, M_name, Type)

• Type_Val(Type, Val)

solution #2 (3NF)• Modules_Descr(M_id, M_name, Type)

• Module_Val(M_id, Val)

are they both non-loss? (apply Heath’s theorem) is there one better than the other?


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Decomposition - update anomalies

updates• u1: insert the fact that a 3 semester module is worth 1.5cu

• u2: modify 1 semester modules; they are not worth 0.5cu any longer, they are 0.75cu

• u3: change the type of a module but forget to change its value

solution #2 • u1 and u2 are impossible or very difficult to perform

• u3 is allowed

solution #1• u1 and u2 are straightforward

• u3 is not allowed


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Solution #1 vs solution #2

solution #1 more expressive

• certain facts cannot be expressed in solution #2; e.g. the value of a new type

• updates can be independently performed on the two component relations (i.e. all constraints are properly expressed)

• in solution #2: Type Value is lost, so this constraint must be enforced by the user by procedural code

independent projections • updates can be performed independently on each projection,

without the danger of ending with inconsistent data


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Independent projections

M-id Type Value

Solution #1 Solution #2

M_name

M-id TypeM_name

Type Value

M-id TypeM_name

M_id Value

all direct : intraall transitive : inter

one transitive : intraone direct : lost


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Independent projections - Risanen

R1 and R2 are two projections of R; R1 and R2 are independent if and only if

• every FD in R is a logical consequence of the FDs in R1 and R2

• the common attributes of R1 and R2 for a candidate key for at least one of R1 or R2

atomic relation• cannot be decomposed into independent projections


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Dependency preservation

R was decomposed (normalisation) into R1, …, Rn

S - the set of FDs for R S1, …, Sn - the set of FDs for R1, …, Rn (each Si refers to

only the attributes of Ri)

S’ = S1 … Sn (usually, S’ S)

the decomposition is dependency preserving if (not iff) S’+ = S+


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2NF and 3NF - more than one CK

2NF a relation is in 2NF if and only if it is in 1NF and all non-key

attributes are irreducibly dependent on the candidate keys

3NF (Zaniolo) R is a relation; X is any set of attributes of R; A is any single

attribute of R; consider the following conditions:• X contains A

• X contains a candidate key of R

• A is contained in a candidate key of R

if either of the three is true for every FD X A then R is in 3NF


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Example

Assume the supply department in a company is in charge of bringing parts from different manufacturers. A part is uniquely identified by its name and manufacturer; for convenience, a part is also given an id. A separate delivery is necessary for each type of part, from each manufacturer. At most one delivery is made in one day for one type of part from one manufacturer. A “transport” (e.g. van23) is associated with each delivery. Each transport has a unique driver. A driver can drive more than one “transports”.


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Relevant FDs

CK: (Type, Manufacturer, Date)CK: (Id, Date)(Type, Manufacturer) IdId (Type, Manufacturer)Transport DriverManufacturer AddressType Handling_req


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2NF

Type Man Id Date Add HR Qty Transp Driver

Type Man Id Date Qty Transp Driver

2NF?

2NF

Type HR

Man Add


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3NF

Type Man Id Date Qty Transp Driver

3NF?

Type Man Id Date Qty Transp

3NF

Transp Driver


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3NF

3NF is not free from update e.g. (Type, Manufacturer) Id

• exercise– insert– delete– update


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BCNF

a relation is in Boyce/Codd normal form (BCNF) if and only if every non-trivial irreducible FD has a candidate key as its determinant

any relation can be non-loss decomposed into an equivalent set of BCNF relations

BCNF 3NF 2NF 1NF BCNF is still not guaranteed to be free of any update

anomalies caused by FDs • example - later


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CKs: Id, Name, Photo (what do you think about this?), User_name

• draw the corresponding FD

overlapping CKs: (Name, Contest), (Contest, Position)

BCNF - examples

Type Man Id Id Date Qty Transp

previous example: one candidate key only

Id Name Photo Address Tel_no User_name

Name Contest Position Prize


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Zaniolo’s definitions

R is a relation; X is any set of attributes of R; A is any single attribute of R; consider the following conditions: X contains A X contains a candidate key of R A is contained in a candidate key of R

if either of the three is true for every FD X A then R is in 3NF

if either of the first two is true for every FD X A then R is in BCNF


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a patient is treated by a single doctor for a certain disease each doctor only treats one kind of disease a doctor can treat more than one patient

is this relation 3NF? is this relation BCNF? can you identify update anomalies? consider also (Patient, Disease, Doctor, Treatment)

with Patient, Disease Treatment

DiseaseDoctor

Patient

BCNF again

Patient Disease Doctor


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Possible decompositions

Patient Doctor

Patient Disease

Patient Doctor

Doctor Disease

Disease Doctor

Patient Disease

non-loss? (choose PKs)

non-loss? (choose PKs)

Heath’s theorem (choose PKs)


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BCNF vs dependency preservation

Patient Doctor Doctor Diseaseand

do not enforce a FD existing in the original specification, namely:

e.g. a patient can be given two doctors that treat the same disease (the system will not disallow this); the constraint would have to be maintained by procedural code

DiseaseDoctor

Patient


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BCNF vs dependency preservation

not every FD is expressible through normalisation when the relation was in 3NF

(Patient, Disease) Doctor was expressed• a doctor could not be assigned to more than one patient-disease

Doctor Disease was not expressed • generated update anomalies

in BCNF Doctor Disease was expressed (Patient, Disease) Doctor was not expressed

• generated update anomalies (refer to previous slide)

• this latter FD would not have been expressed even if the decomposition in all three 2-attribute relations had been considered


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Conclusions

normal forms : formalisation of common sense art engineering possibility for automation

BCNF always achievable not always free of update anomalies (recall previous

example), because it cannot always express all the FDs existing in the problem

dependency preservation, 3nf revisited and bcnf 1

Documents

nf bcnf

nf zaniolo r

projections of r r

dependency preservation

bcnf slide

nf modules

type type

candidate key of r