Multi Valued Dependencies & 4NF - Good

8/2/2019 Multi Valued Dependencies & 4NF - Good

1/30

Multi-valued Dependencies and

4

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n

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N rm l F rm

Louis D. Nel 2010 1Design with Normal Forms 8 -COMP 3005


2/30

Multi-valued Dependencies

Consider the situation of bank customers whopossibly have multiple addresses and multipleaccounts

LOANSLOAN NO CUSTOMER STREET CITY_

101 Sue Elgin Ottawa

101 Sue Eagleson Kanata

112 Frank Bank Ottawa



3/30

Suppose Sue takes out another loan (#113) wecould add the following tuple to the LOANS table

LOANS

LOAN_NO CUSTOMER STREET CITY




Problem: Which of Sues addresses do we enter inthe table? (Can we insist that both appear?)



4/30

We want to consider the table shown below to bein a bad state potentially dangerous

LOANS





How can this table be fixed?

What is the key for this table?



5/30

Solution 1) add another tuple

LOANS









6/30

Solution 2) Decompose the relation

LOANS

LOAN_NO CUSTOMER

101 Sue

ADDRESS

CUSTOMER STREET CITY

112 Frank

113 SueSue Eagleson Kanata

Frank Bank Ottawa



7/30

Decomposing The Table

What would guide the decomposition

Which of the functional dependency based normalforms does the table appear to satisfy?

LOANS








8/30

Decomposing The Table

1nf because there is a key

2nf because there is not attribute partiallydependent on the key

determining a non-key attribute

BCNF because no part of the key appears toepen on anot er attr ute

LOANS

LOAN NO CUSTOMER STREET CITY_

101 Sue Elgin Ottawa101 Sue Eagleson Kanata





9/30

Couldnt we just require?

CUSTOMER -> ADDRESSCUSTOMER -> LOAN

No because customers can have multiple addressesand multiple loans



10/30

What is the problem?

The problem seems to be occurring because theta e represents two n epen ent :

relationships being represented in the same table.

Customers can have multi le loans and alsoindependently, customers can have multipleaddresses.

relationship appears in each table can fix theproblem, but what formal constraint would guide



11/30

The multi-valued dependency

CUSTOMER ->-> ADDRESS

specifies that a customer can have multiple

It does not rule out the possibility of multiple

addresses, instead specifies that if a tuplecontains one address, other tuples may need tobe added with the other



12/30

Multi-valued dependencies are a consequence ofrst orma orm -attr utes cannot e mu t -

valued

If we do have two multi-valued attributes in arelation (e.g. loan, address) we have to repeatever value of one with ever value of the other ifwe want to keep things consistent

-says that, in effect, that each loan must appearwith each address



13/30

Defn Multi-valued Dependency

For a relation r(R) with attribute subsets X and Y,the multi-valued de endenc X->-> Y re uires that

if two tuples t1 and t2 exist with t1[X] = t2[X] thentwo tuples t3 and t4 must also exist with

t3[X] = t4[X] = t1[X] = t2[X]t3[Y] = t1[Y] and t4[Y] = t2[Y]t3 R- XY = t2 R- XY and t4 R- XY = t1 R- XY

t1 t2 t3 t4 need not be distinct



14/30

Defn Multi-valued Dependency

If t1 and t2 exist with t1[X] = t2[X] then t3 and t4must exist witht3[X] = t4[X] = t1[X] = t2[X]t3[Y] = t1[Y] and t4[Y] = t2[Y]t3[R-(XY)] = t2[R-(XY)] and t4[R-(XY)] = t1[R-(XY)]

ADDRESS 1LOAN 1

CUSTOMER

_

ADDRESS_2LOAN_2

_

X

ADDRESS_3LOAN_3Y R - (XY)



15/30

Example

If t1 and t2 exist with t1[X] = t2[X] then t3 and t4 must existwitht3[X] = t4[X] = t1[X] = t2[X]

= =

t3[R-(XY)] = t2[R-(XY)] and t4[R-(XY)] = t1[R-(XY)]

Sue

Ottawa

113

101Given tuples

Then so must

Sue

Kanata113

exist



16/30

Does CUSTOMER ->-> ADDRESS hold here?

LOANS

_

101 Sue Elgin Ottawa101 Sue Eagleson Kanata


t1

t4


There is no t3 for which agrees with t1 address

If t1 and t2 exist with t1[X] = t2[X] then t3 and t4 must exist

and t2 loan -so no

wt3[X] = t4[X] = t1[X] = t2[X]t3[Y] = t1[Y] and t4[Y] = t2[Y]t3 R- XY = t2 R- XY and t4 R- XY = t1 R- XY



17/30


LOANS

LOAN N CUSTOME STREET CITY

t1t4

_


101 Sue Eagleso Kanata


t2t3113 Sue Elgin Ottawa

113 Sue Eagleso Kanata

111 John Riverside Ottawas1 s3

If t1 and t2 exist with t1[X] = t2[X] then t3 and t4 must exist

wt3[X] = t4[X] = t1[X] = t2[X]t3[Y] = t1[Y] and t4[Y] = t2[Y]t3 R- XY = t2 R- XY and t4 R- XY = t1 R- XY



18/30

Whenever two independent 1:N relationships X:Yand X:Z are mixed in the same relation a multi-valued de endenc could arise

e.g. a customers address is independent of the,several of each



19/30

Trivial Multi-valued Dependencies

If Y is a subset of X, X->->Y

= , - -

These are called trivial because they hold in anylegal relation R (and so dont specify anyadditional constraint



20/30

LOANS



101 Sue Ea leson Kanata




Which should we specify?

- - ,CUSTOMER ->-> LOAN

oes not matter ecause one mp es t e ot er

Theorem: If X->->Y then X->-> R - X - Y



21/30

Inference Rules for Functional and Multi-valued Dependencies

For R=(A1,A2, ... ,An) and W, X, Y, Z all subsets of R, thefollowing inference rules hold

-

2) If X->Y then XZ -> YZ (augmentation)3) If X->Y and Y->Z, then X->Z (transitive)

4) If X->->Y, then X->->(R - X - Y) (complementation)

5) If X->->Y and Z is a subset of W->-> mv augmen a on

6) If X->->Y, Y->->Z then X->->(Z-Y) (mv transitive)

- - -,

8)If X->-> Y and there is a W such thatW INTERSECT Y is empty, W->Z, and


s a su se o , en -> coa escence


22/30

Exercise

Let R=(A, B, C, G, H, I) withMVD = { A->-> B,B ->-> HICG -> H }

A ->-> CGHI (hint: complementation rule)A ->-> HI (hint: transitivity)

-



23/30

Do Functional Dependency normal forms help


LOANS

_





This table is in BCNF because no functionaldependencies apply

Still it is undesirable because of re eated

information



24/30

Fourth Normal Form

A relation schema R is in 4NF with respect to a setof dependencies F if, for every nontrivial MVDX->->Y in F+ X is a su erke of R



25/30

Fourth Normal Form

LOANS



101 Sue Ea leson Kanata




F = { CUSTOMER->->STREET,CITYCUSTOMER->->LOAN_NO }

Violates 4NF because CUSTOMER is not asuperkey

(and CUSTOMER->->LOAN_NO is non-trivial)



26/30

Decomposition to 4NF

LOANS




112 Frank Bank Ottawa113 Sue Elgin Ottawa

->-> n

X->Y not in F

F = { CUSTOMER->->STREET,CITY

CUSTOMER->->LOAN_NO }

LOANS

CUSTOMER LOAN

Sue 101

ADDRESS

CUSTOMER STREET CITY

Sue El in Ottawa

X R - - Y

Frank 112Sue 113

Sue Eagleson KanataFrank Bank Ottawa


Note: in the decomposed tables both MVD of F are trivial, hence 4NF


27/30

Given a database design situation with functionaland multi-valued dependencies, it isadvanta eous to find a decom osition that is

4NFLossless JoinDependency Preserving

,would be to relax 4NF and go to BCNF or 3NF andlose some dependency preservation

You never want to relax the lossless-joinconstraint



28/30

Join Dependencies

It is not always possible to decompose a relation

join However it may be possible to decompose R into

more an wo re a ons , , ... , n w c s alossless join decomposition

These cases are rare and difficult to detect inractice



29/30

Join Dependency

A join dependency JD(R1, R2, ... ,Rn) specifies aconstraint on instances r(R) that every legalinstance r R should have a lossless oindecomposition into R1, R2, ... ,Rn.

,

JOIN( (r), (r), ... , (r) ) = r



30/30

Fifth Normal Form (Project Normal Form)

A relation schema R is in 5NF with res ect to a setF of functional, multi-valued, and joindependencies if, for every nontrivial joinde endenc JD R1, R2, ... ,Rn im lied b F, everRi is a superkey of R

Current practical database design does not pay


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Multi Valued Dependencies & 4NF - Good

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