Lt17 Decomposition

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Database Systems and

Applications

Lecture 17

Decomposition


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What is Decomposition?

Decomposition the process of breaking downin parts or elements.

Decomposition in database means breakingtables down into multiple tables

From Database perspective means going to ahigher normal form


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Decomposition

Important that decompositions are good,

Two Characteristics of Good Decompositions

1) Lossless

2) Preserve dependencies


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Problems with Decomposition

Some queries become more expensive

Given instances of the decomposed relations,we may not be able to reconstruct thecorresponding instance of the original relation information loss.


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What is lossless?

Lossless means functioning without a loss.

In other words, retain everything.

Important for databases to have this feature.


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A decomposition is lossless if we can recover:R(A,B,C)

R1(A,B) R2(A,C)

R(A,B,C) should be the same asR(A,B,C)

Must ensure R = R

Decompose

Recover

Lossless Decomposition


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Lossless Decomposition

Sometimes the same set of data is reproduced:

(Word, 100) + (Word, WP) (Word, 100, WP) (Oracle, 1000) + (Oracle, DB) (Oracle, 1000, DB) (Access, 100) + (Access, DB) (Access, 100, DB)

Name Price Category

Word 100 WP

Oracle 1000 DB

Access 100 DB

Name Price

Word 100

Oracle 1000

Access 100

Name Category

Word WP

Oracle DB

Access DB


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Lossy Decomposition Sometimes its not:

(Word, WP) + (100, WP) = (Word, 100, WP) (Oracle, DB) + (1000, DB) = (Oracle, 1000, DB) (Oracle, DB) + (100, DB) = (Oracle, 100, DB) (Access, DB) + (1000, DB) = (Access, 1000, DB) (Access, DB) + (100, DB) = (Access, 100, DB)

Name Price Category

Word 100 WP

Oracle 1000 DB

Access 100 DB

Category Name

WP Word

DB Oracle

DB Access

Category Price

WP 100

DB 1000

DB 100

Whats

wrong?


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Lossy Decomposition

Employee Project Branch

Samu Mars Warangal

Soumini Jupiter Hyderabad

Soumini Venus Hyderabad

Sekhar Saturn HyderabadSekhar Venus Hyderabad

Functional dependencies:

Employee Branch, Proje ct Branch

T


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Lossy Decomposition

Decomposition of the previous relation

Employee Branch

Samu Warangal

Soumini Hyderabad

Sekhar Hyderabad

Project Branch

Mars Warangal

Jupiter Hyderabad

Saturn Hyderabad

Venus Hyderabad

T1 T2


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Lossy Decomposition

After Natural Join


Samu Mars Warangal



Sekhar Saturn Hyderabad

Sekhar Venus Hyderabad

Soumini Saturn Hyderabad

Sekhar Jupiter Hyderabad

Original Relation

The result is different from the original relation:the information can not be reconstructed.


Samu Mars Warangal



Sekhar Saturn Hyderabad

Sekhar Venus Hyderabad


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Lossless Decomposition Property

The decomposition of R into R1 and R2 islossless withrespect to F if and only if the closure of F containseither:

R1 R2 (R1 intersect R2) R1, that is: allattributes common to both R1 and R2 functionallydetermine ALL the attributes in R1 OR

R1 R2 (R1 intersect R2) R2, that is: allattributes common to both R1 and R2 functionallydetermine ALL the attributes in R2


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Ensuring lossless decomposition

R(A1, ..., An, B1, ..., Bm, C1, ..., Cp)R(A1, ..., An, B1, ..., Bm, C1, ..., Cp)

If A1, ..., AnB1, ..., Bmor A1, ..., AnC1, ..., CpThen the decomposition is lossless

R1(A1, ..., An, B1, ..., Bm)R1(A1, ..., An, B1, ..., Bm) R2(A1, ..., An, C1, ..., Cp)

R2(A1, ..., An, C1, ..., Cp)

Note: dont need both


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In Simpler Terms

R1 R2 R1

R1 R2 R2

If R is split into R1 and R2, for the decomposition to belossless then at leastone of the two should hold true.

Projecting on R1 and R2, and joining back, results in the

relation you started with


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Why lossless?

Ensures that attributes involved in the natural join(R1 R2) are a candidate key for at least one ofthe two relations.

This ensures we can never get the situation wherefalse tuples are generated, as for any value on thejoin attributes there will be a unique tuple in one

of the relations.


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Example of Lossless Decomposition

GIVEN:LENDINGSCHEME=(BRANCHNAME, ASSETS, BRANCHCITY,LOANNUMBER, CUSTOMERNAME, AMOUNT)

FD'S:

BRANCHNAME ASSETS BRANCHCITYLOANNUMBER AMOUNT BRANCHNAME

DECOMPOSE LENDINGSCHEME INTO:

1. BRANCHSCHEME=(BRANCHNAME, ASSETS, BRANCHCITY)

2. BORROWSCHEME=(BRANCHNAME, LOANNUMBER,CUSTOMERNAME, AMOUNT)


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Example of Lossless Decomposition

SHOW THAT THE DECOMPOSITION IS A LOSSLESSDECOMPOSITION

USE AUGMENTATION RULE ON FIRST FD TO OBTAIN:BRANCHNAME BRANCHNAME ASSETSBRANCHCITY

INTERSECTION OF BRANCHSCHEME ANDBORROWSCHEME IS BRANCHNAME

BRANCHNAME BRANCHSCHEMESO, INITIAL DECOMPOSITION IS A LOSSLESS


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Example 2

GIVEN:BORROWSCHEME=(BRANCHNAME, LOANNUMBER,

CUSTOMERNAME, AMOUNT)FD'S:

LOANNUMBER AMOUNT BRANCHNAME

DECOMPOSE BORROWSCHEME INTO:

1. LOAN-INFO-SCHEME=(BRANCHNAME, LOANNUMBER,AMOUNT)

2. CUSTOMER-LOAN-SCHEME=(LOANNUMBER,CUSTOMERNAME)


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Example 2 (cont)

SHOW THAT THE DECOMPOSITION IS A LOSSLESSDECOMPOSITION

1. USE AUGMENTATION RULE ON FD TO OBTAIN:LOANNUMBER LOANNUMBER AMOUNT

BRANCHNAME

1. INTERSECTION OF LOAN-INFO-SCHEME AND CUSTOMER-LOAN-SCHEME IS LOANNUMBER

1. LOANNUMBER LOAN-INFO-SCHEME1. SO, INITIAL DECOMPOSITION IS A LOSSLESS


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Example

R1 (A1, A2, A3, A5)

R2 (A1, A3, A4)

R3 (A4, A5)

FD1: A1 A3 A5

FD2: A5 A1 A4

FD3: A3 A4 A2


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A1 A2 A3 A4 A5

R1 a(1) a(2) a(3) b(1,4) a(5)

R2 a(1) b(2,2) a(3) a(4) b(2,5)

R3 b(3,1) b(3,2) b(3,3) a(4) a(5)

Example (cont)


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By FD1: A1 A3 A5

A1 A2 A3 A4 A5

R1 a(1) a(2) a(3) b(1,4) a(5)

R2 a(1) b(2,2) a(3) a(4) b(2,5)

R3 b(3,1) b(3,2) b(3,3) a(4) a(5)

Example (cont)


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By FD1: A1 A3 A5

we have a new result table

A1 A2 A3 A4 A5

R1 a(1) a(2) a(3) b(1,4) a(5)

R2 a(1) b(2,2) a(3) a(4) a(5)

R3 b(3,1) b(3,2) b(3,3) a(4) a(5)

Example (cont)


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By FD2: A5 A1 A4

A1 A2 A3 A4 A5

R1 a(1) a(2) a(3) b(1,4) a(5)

R2 a(1) b(2,2) a(3) a(4) a(5)

R3 b(3,1) b(3,2) b(3,3) a(4) a(5)

Example (cont)


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By FD2: A5 A1 A4

we have a new result table

A1 A2 A3 A4 A5

R1 a(1) a(2) a(3) a(4) a(5)

R2 a(1) b(2,2) a(3) a(4) a(5)

R3 a(1) b(3,2) b(3,3) a(4) a(5)

Example (cont)


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Example 1

R(A B C D E)

FD1 = (A B)

FD2 = (BC E)

FD3 = (ED

A)R1=(AB);

R2=(ACDE);


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Example 2

Is this decomposition lossless?

R (A B C D E)

FD1 AB CFD2 C E

FD3 BD

FD4 EA

R1=(BCD);

R2=(ACE);


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Example 3

R(A B C D E)

FD1: A BC

FD2: BD CE

FD3: E AD

FD4: CE A

R1(ABC) =

R2 (BCDE) =


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Conclusion

Decomposing is the act of breaking tablesdown in order to achieve higher normal form.

Decompositions should always be lossless.This confirms that information in the originalrelation can be accurately reconstructedbased on the decomposed relations.

Remember that for a decomposition to beconsidered GOOD it must also preservefunctional dependencies.

Date post:	03-Apr-2018
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