pert.12.normalization cont (tugas teori)
TRANSCRIPT
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Slide 11-2Elmasri and Navathe, Fundamentals of Database Systems, Fourth EditionRevised by IB & SAM, Fasilkom UI, 2005
Exercise Multiple Choices
Which of the following is known to be true from thefunctional dependency shown as (A, B) (C, D)?a. A is the determinant of Cb. A and B together are determined by C and D togetherc. A and B together determine D
d. C and D together determine Ae. A determines B
Which of the following is not a requirement for 1NF?a. cells must contain single values
b. all entries in a column must be of the same kindc. no two rows may be identicald. rows must be ordered by the value of the primary keye. the order of the columns is insignificant
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Further Normalization
The main reference of this preElmasri &Navathe, Fundamentalthedition,2004, Chapter 11Additional resources: presentDemurjian, Sr (http://www.engr.
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Slide 11-14Elmasri and Navathe, Fundamentals of Database Systems, Fourth EditionRevised by IB & SAM, Fasilkom UI, 2005
What are Possible Decompositions?
S#
S1
S2
S3S4
D1
D1
D2D3
DHead
John
John
Smith
Black
DName
G Information Based
R = ( U, F ) U = { S#, DName, DHead }
F = { S#o DName, DName o DHead }
G = { R1(S#, ), R2(DName, R3(DHead, )}G is Neither Lossless nor FD-Preserving
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Slide 11-15Elmasri and Navathe, Fundamentals of Database Systems, Fourth EditionRevised by IB & SAM, Fasilkom UI, 2005
What are Possible Decompositions?
S# DName
S1
S2
S3
S4
D1
D1D2
D3
S# DHead
S1
S2S3
S4
John
JohnSmith
Black
G Lossless Decomposition but
not Dependency-Preserving
DNameo DHead is lost inthe decomposition
R = ( U, F ) U = { S#, DName, DHead }
F = { S#o DName, DName o DHead }
G = { R1({S# ,DName}, {S#o DName}),R2({S#, DHead}, {S#o DHead})}
Gis Lossless but not FD-Preserving
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Slide 11-16Elmasri and Navathe, Fundamentals of Database Systems, Fourth EditionRevised by IB & SAM, Fasilkom UI, 2005
What are Possible Decompositions?
S# DName
S1
S2
S3
S4
D1
D1
D2D3
DName DHead
John
JohnD1
D1
D2D3
G Lossless & dependency-
preserving decomposition
R = ( U, F ) U = { S#, DName, DHead }
F = { S#o DName, DName o DHead }
= { R1({S# ,DName}, {S# o DName})R3({DName, DHead}, {Dname o DHead})}
G is both Lossless and FD-Preserving
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Slide 11-17Elmasri and Navathe, Fundamentals of Database Systems, Fourth EditionRevised by IB & SAM, Fasilkom UI, 2005BCNF
Eliminate the Partial Functional
Dependencies of Non-primeAttributes to Key Attributes
Eliminate the Transitive
Functional Dependencies ofNon-prime Attributes to KeyAttributes
Eliminate the Partial andTransitive FunctionalDependencies of Prime (Key)Attributes to Key
Lossless Decomposition
but not DependencyPreserving
Lossless Decomposition
and Dependency Preserving2NF
3NF
1NF
Summary of Normalization
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Slide 11-18Elmasri and Navathe, Fundamentals of Database Systems, Fourth EditionRevised by IB & SAM, Fasilkom UI, 2005
The Entire Normalization Picture1NF
2NF
3NF
BCNF
Eliminate Partial FDs of
Non-prime Attributes to Key
Eliminate Transitive FDs of Non-prime
Attributes to Key
Eliminate Partial and Transitive FDsof Prime Attributes to Key
4NF
Eliminate Non-trivial and Non-
functional Multi-Valued Dependencies
5NF
Eliminate Join Dependencies that are
Not Implied by Candidate Key
Wh t M lti V l d
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Slide 11-19Elmasri and Navathe, Fundamentals of Database Systems, Fourth EditionRevised by IB & SAM, Fasilkom UI, 2005
What are Multi-ValuedDependencies?
Focused on the Concept of Multi-Valued Dependencies
A MVD X oo Y Indicates that a Value of X Correspondsto Multiple Values of Y
Consider EMP with MVDs:
ENAME ooPNAME (E works on many Project)
ENAME oo DNAME (E has many Dependents)
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Slide 11-20Elmasri and Navathe, Fundamentals of Database Systems, Fourth EditionRevised by IB & SAM, Fasilkom UI, 2005
What is Fourth Normal Form (4NF)? A Relation Schema R is in Fourth Normal Form (4NF)
w.r.t Dependencies F (FD and MVD) if for every Non-Trivial MVD X oo Y in F+, X is a Superkey for R
MVD X oo Y in R is called trivial ifY is subset of X, or
X U Y = R Reconsider EMP with MVDs:
ENAME oo PNAME (E works on many P)ENAME oo DNAME (E has many Dependents)
ENAME is Not a Superkey of R since Need Triple ofENAME, PNAME, and DNAME to Distinguish
We need to Decompose EMP!
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Slide 11-21Elmasri and Navathe, Fundamentals of Database Systems, Fourth EditionRevised by IB & SAM, Fasilkom UI, 2005
Notes on FD
A functional dependency is trivial if it issatisfied by all instances of a relationE.g.
customer-name, loan-number o customer-name customer-name o customer-name
In general, D o Eis trivial ifE D
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Slide 11-22Elmasri and Navathe, Fundamentals of Database Systems, Fourth EditionRevised by IB & SAM, Fasilkom UI, 2005
Decomposition into 4NF
ENAMEooPNAME is Trivial MVD: EPNAMEqual to EMP_PROJECTS (samENAMEooDNAM
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Slide 11-23Elmasri and Navathe, Fundamentals of Database Systems, Fourth EditionRevised by IB & SAM, Fasilkom UI, 2005
Multivalued Dependencies and 4NF
Decomposing a relation state of EMP that is not in 4NF. (a) EMP relation with
additional tuples. (b) Two corresponding 4NF relations EMP_PROJECTS andEMP_DEPENDENTS.
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Slide 11-24Elmasri and Navathe, Fundamentals of Database Systems, Fourth EditionRevised by IB & SAM, Fasilkom UI, 2005
Other Normalizations
You can read by yourself
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Slide 11-27Elmasri and Navathe, Fundamentals of Database Systems, Fourth Edition
Exercise 10.33
Consider the following relation for published books: BOOK (Book_title, Authorname, Book_type, Listprice,
Author_affil, Publisher) Author_affil referes to the affiliation of the author. Suppose the
following dependencies exist: Book_title -> Publisher, Book_type Book_type -> Listprice Author_name -> Author-affil
(a) What normal form is the relation in? Explain your
answer. (b) Apply normalization until you cannot decompose the
relations further. State the reasons behind eachdecomposition.