Lossless Join ppt

8/12/2019 Lossless Join ppt

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Goal of Decomposition

Eliminate redundancy by decomposing a relation

into several relations in a higher normal form.

It is important to check that a decomposition

does not lead to bad design

There are three desirable properties:

1. Lossless.

2. Dependency preservation.

3. Minimal redundancy.


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Problem with Decomposition

Given instances of the decomposed relations,

we may not be able to reconstruct the

corresponding instance of the original

relationinformation loss


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Example : Problem with Decomposition

R1 U R2Model Name Price Categorya11 100 Canon

a11 150 Canon

s20 200 Nikon

a70 100 Canona70 150 Canon

Model Name Price Category

a11 100 Canon

s20 200 Nikon

a70 150 Canon

R


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Lossy decomposition

In previous example, additional tuples are

obtained along with original tuples

Although there are more tuples, this leads to

less information

Due to the loss of information, decomposition

for previous example is called lossy

decompositionor lossy-join decomposition


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Example of Relation Decomposition


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Lossless Decomposition

A decomposition {R1, R2,, Rn} of a relation R iscalled a losslessdecompositionfor R if thenatural join of R1, R2,, Rn produces exactly therelation R.

Definition:

Let { R1, R2 } be a decomposition of R (R1 U

R2 = R); the decomposition is lossless if forevery legal instance r of R:

r = R1(r) |X| R2(r)


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Lossless Decomposition

A decomposition is lossless if we can recover:

R(A, B, C)

Decompose

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

Recover

R(A, B, C)

Thus, R = R


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Lossless Check Example 1:

Consider five attributes: ABCDE

Three relations: ABC, AD, BDE

FDs: A ->BD, B ->E

A B C D E

ABC a1 a2 a3 b14 b15

AD a1 b22 b23 a4 b25

BDE b21 a2 b33 a4 a5


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Lossless Check Example


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Lossless Check Example


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Example 2:

R1 (A1, A2, A3, A5)

R2 (A1, A3, A4)

R3 (A4, A5)

FD1: A1A3 A5

FD2: A5A1 A4

FD3: A3 A4A2


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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: A1A3 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: A1A3 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: A5A1 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: A5A1 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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Why do we preserve the dependency?

We would like to check easily that updates to

the database do not result in illegal relations

being created.

It would be nice if our design allowed us to

check updates without having to compute

natural joins.


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Dependency Preservation Decomposition

Definition: Each FD specified in F either appears directly in

one of the relations in the decomposition, or be inferred from

FDs that appear in some relation.


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Test of Dependency Preservation

If a decomposition is not dependency-preserving, some

dependency is lost in the decomposition.

One way to verify that a dependency is not lost is to take

joins of two or more relations in the decomposition to geta relation that contains all of the attributes in the

dependency under consideration and then check that the

dependency holds on the result of the joins.


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Test of Dependency Preservation II

Find F- F', the functional dependencies not

checkable in one relation.

See whether this set is obtainable from F' by

using Armstrong's Axioms.

This should take a great deal less work, as we

have (usually) just a few functional

dependencies to work on.


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Dependency Preserving Example

Consider relation ABCD, with FDs :

A ->B, B ->C, C ->D

Decompose into two relations: ABC and CD.

ABC supports the FDs A->B, B->C.

CD supports the FD C->D.

All the original dependencies are preserved.


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Non-Dependency Preserving Example

Consider relation ABCD, with FDs:

A ->B, B ->C, C->D

Decompose into two relations: ACD and BC.

ACD supports the FD B ->C and implied FD A ->C.

BC supports the FD B->C.

However, no relation supports A ->B.

So the dependency is not preserved.


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Minimal Redundancy

In order to achieve the lack of redundancy, we

do some decomposition which is represented

by several normal forms.


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Conclusion

Decompositions should always be lossless.

Decompositions should be dependency

preserving whenever possible.

We have to perform the normal decomposition

to make sure we get rid of the minimal

redundant information.

Lossless Join ppt

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