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Boyce-Codd Normal Form

By: Thanh Truong

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Boyce-Codd Normal Form

Eliminates all redundancy that can bediscovered by functional dependencies

But, we can create a normal form morerestrictive called 4NF

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Definition

A relation schema Ris in BCNF with respectto a set Fif:

For all functional dependencies ofFof theform , where R and R is a trivial functional dependency()

is a superkey for schemaR

A database design is in BCNF if each memberof the set of relational schemas thatconstitute the design is in BCNF

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Rule for schema not in BCNF

Let Rbe a schema not in BCNF, thenthere is at least one nontrivial functional

dependency such that is not asuperkey

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Example of not BCNF

bor_loan = (customer_id, loan_number,amount)

loan_numberamountbut loan_number is not a superkey

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BCNF Decomposition

The definition of BCNF can be used todirectly test if a relationship is in BCNF

If a relation is not in BCNF it can bedecomposed to create relations that arein BCNF

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Example

borrower = (customer_id, load_number)

Is BCNF because no nontrivial functional

dependency hold onto itloan = (loan_number, amount)

Has one nontrivial functional dependency thatholds, loan_numberamount, but

loan_number is a superkey so loan is in BCNF

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Testing for BCNF

1) To check if a nontrivial dependency causes a violation of BCNF, compute

a

+

(attribute closure of), and verify that itincludes all attributes ofR; that is, is is thesuperkey ofR

2) If we can show that none of the

dependencies in Fcauses a violation ofBCNF, then none of the dependencies in F+will cause a violation of BCNF either

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Alternate test for 2)

For every subset of attributes in Richeck that a+(the attribute closer of

under F) either includes no attribute ofRi-, or includes all attributes ofRi

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BCNF DecompositionAlgorithm

IfRis not in BCNF, we can decompose Rintoa collection of BCNF schemas R1 , R2, , RnResult := {R};

done := false;computer F+

while(not done) do

if(there is a schema Ri in result that is not in BCNF)

then begin

be a nontrivial functional dependency that holds

on Ri such that -> Ri is not in F+, and = ;

result :=(resultRi) (Ri - ) (,);

endelse done := true;

Pg 289 figure 7.12

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Example

lending = (branch_name, branch_city, assets,customer_name, loan_number, amount)

branch_nameassets branch_cityloan_numberamount branch_name

candidate key is {loan_number,customer_name}

branch_name is not superkey so not in BCNFso lending is not BCNF

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Example (cont)

So we replace lendingby:

branch = (branch_name, branch_city, assets)

loan_info = (branch_name, customer_name,loan_number, amount)

The only nontrivial functional dependenciesthat hold on branchinclude branch_nameon

the left side of the arrow. Sincebranch_nameis a key for branch, the relationbranchis in BCNF

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Example (cont)

For loan_info

The functional dependency

loan_number

amount branch_nameholds on loan_infobut loan_numberis not akey for loan_info, so we replace loan_infoby

loanb = (loan_number, branch_name,

amount)borrower = (customer_name, loan_number)

loanband borrowerare in BCNF

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3NF pro/con

Advantage of 3NF: it is always possible toobtain a 3NF design without sacrificinglosslessness or dependency preservation

Disadvantage of 3NF: we may have to usenull values to represent some of the possiblemeaningful relationships among data items,

and there is the problem of repetition ofinformation

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Conclusion

Our goals of database design withfunctional dependencies are

1) BCNF2) Losslessness

3) Dependency preservation

Not possible to get all 3, we have tochoose between BCNF or dependencypreservation

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Source

Silberschatz, Korth, Sudarshan (2006).Database System Concepts. New York:

McGraw-Hill