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Normalization 02

CSE3421 notes

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Closure of a set of attributes

Given a set of FDs F, find the set of attributes functionally determined by a given set of attributes X. This is called the closure of X, and denoted as X+.

F, X � ? (or, X+ = ? ) Example: F = { A � B, B � C} Closure of A: A+ = ABC

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Example

F = { C � A,

BC � D,

ACD � B,

D � EG,

AB � C}

AB � ?C

BC

ACD

D

AB

GEDCBA

AB � ABC � ABC(A)D � ABCD(B)EG

Redundant from C

From BC

Redundant from ACDFrom D

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Closure of a set of FDs

• Given a set of FDs F, find all FDs that can be produced by F. This is called the closure of F, and denoted as F+.

• F+ can be found by applying the Armstrong Axioms.

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Armstrong Axioms

• (A1) Reflexivity (this produces all trivial FDs)

X Y X Y⊇ ⇒ →• (A2) Augmentation

,X Y XZ YZ Z→ ⇒ → ∀

• (A3) Transitivity

,X Y Y Z X Z→ → ⇒ →

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Additional rules• Union

,X Y X Z X YZ→ → ⇒ →• Decomposition

X YZ X Y and X Z→ ⇒ → →

By applying (A1), (A2), (A3) repeatedly, we get F+.

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Example

• Let R: (A, B, C) and F = { A � B, B � C}. F+ = ?

• Apply Armstrong Axioms

• (A1) (reflexivity) produces all trivial FDs, i.e., those whose left-hand-side (LHS) is a superset of the right-hand-side (RHS). For example, AB � A, AB � B, etc).

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• Non-trivial FDs:– Apply (A3), transitivity:

• A � B with B � C generate A � C.

• (A3) cannot be further applied.

– Apply (A2), augmentation, in all possible permutations.• For A � B: remaining attribute is C. Therefore, AC�BC is in

F+.

• For B�C: add A and get BA�CA. Therefore, AB�AC is in F+.

• For A�C: add B and get AB�BC. Therefore, AB�BC is in F+.

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So, F+ = { A � B, B � C, all trivial FDs, A � C, AC � BC, AB � AC, AB � BC}

From (A3)

From (A2)

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Where are we …X+: Closure of set of attributes

F+: closure of set of FDs

MC: minimal cover of F

Lossless join property

Algorithm to compute 3NF

Preservation of dependencies

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Lossless join

• If a relation R is decomposed into R1, R2, …and the natural join R1 join R2 join R3 …is exactly equal to R, then the decomposition is said to have the lossless join property.

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Example

c2b2a2

c1b1a1

CBA

R

b2a2

b1a1

BA

R1:(A,B)

c2b2

c1b1

CB

R2:(B,C)

c1b1a1

c2b2a2

CBA

R1 join R2

Exactly equal to R. so have lossless join

property.

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Example

c2b1a2

c1b1a1

CBA

R

b1a2

b1a1

BA

R1:(A,B)

c2b1

c1b1

CB

R2:(B,C)

c2b1a2

c1b1a2

c1b1a1

c2b1a1

CBA

R1 join R2

Not the same as R. So do

not have lossless join

property. (have lossy

join)

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Proposition

A decomposition

( )1 2,R Rρ =Has a lossless join with respect to F, if and only if, either

[ ]1 2 1 2R R R R F+∩ → − ∈

or

[ ]1 2 2 1R R R R F+∩ → − ∈

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Example Assume R1 = {A, B}, R2 = {B, C} is a decomposition of R. Then,

{ }1 2R R B∩ =

{ }1 2R R A− =

{ }2 1R R C− =

For the decomposition to be lossless, it should be either B�A in F+ or B�C in F+. But to figure if this is the case, we should have F first (which we don’t in this example) … stay tuned.

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Proposition

If X�Y is in F of R and X Y∩ = ∅ then

The decomposition of R into R1: R-Y; R2 : XY, is lossless.

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Proof

( )( )

1 2

1 2

R R R Y XY

XYW Y XY

XW XY

X

R R X

∩ = − ∩

= − ∩= ∩=⇒ ∩ =

For some W

Now need to show that

X�R1 (i.e., X� R-Y), or

X�R2 (i.e., X�XY)

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Proof …/ Since

X�Y (assumption), and

X�X (trivial),

we have that X�XY, which is R2.

Therefore, X�R2. (done)

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Preservation of dependencies

Definition: A decomposition

( )1 2,...,, NR R Rρ =

Preserves a set of FDs F, if

( )1

i

N

Ri

F Fπ+

+ +

=

= ∪

where ( )iR Fπ +

is the set of all dependencies from F+ that are

comprised of attributes in Ri.

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Note: for simplicity, we may denote

( )iR Fπ +

as

iF +

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Example

R:(A, B, C)

F= { A � C,

B � C,

A � B}

Assume R1 = (A, B) and R2 = (B, C).

Does the decomposition of R into R1, R2 preserve dependencies?

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Example …/

Check 1 2F F∪ first (and if not succeed then have to try

1 2F F+ +∪

{ }1 2 ,F F A B B C∪ = → →

Need only show that the 3rd dependency of F, A � C can be generated by

1 2F F∪

Since A�B and B�C, we have that A�C. (done).

Therefore the decomposition preserves dependencies.

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Example 2

R:(A, B, C)

F= { A � B,

B � C,

C � A}

( )1 2,R Rρ = Where R1 = (A, B), R2 = (B, C)

Therefore, F1 = {A � B} and F2 = { B � C}

Still need C � A. (note, A�B and B�C imply A�C but not C�A.

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Example 2 …

Have to find 1 2 .F and F+ +

1 :F +

Have to calculate F+ and then project on the appropriate attributes.

To calculate F+, apply Armstrong axioms on F.

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Example 2 …

F:

A � B (1)

B � C (2)

C � A (3)

(1), (2) � A � C, in F+.

(2), (3) � B � A, in F+ (and also in F1+)

(3), (1) � C � B, in F+ (and also in F2+)

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Example 2 …/So far …

F1+: A � B (i) B � A (ii)

F2+: B � C (iii)C � B (iv)

Still need C � A. Notice, (iv), (ii) � C � A. i.e. C � A is in � done. Therefore, the decomposition preserves dependencies.

( )1 2F F++ +∪

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Where are we …X+: Closure of set of attributes

F+: closure of set of FDs

MC: minimal cover of F

Lossless join property

Algorithm to compute 3NF

Preservation of dependencies

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Minimal cover for a set of FDs

• A minimal cover G for a set of FDs F, is an equivalent set of FDs that is minimal in the following sense:

1. Every dependency in G is as small as possible (i.e., each LHS of each FD has as few attributes as possible, and the RHS has only one attribute).

2. Every FD in G is required for the closure G+ to be equal to the closure F+.

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How to calculate G?

• Given F, how do we obtain the minimal cover G?

1. Put F in standard form(i.e., all FDs in F have RHS with one attribute).

2. Eliminate extraneous attributes from LHS.

3. Eliminate redundant FDs.

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Important !!!

1. The above steps should be performed in order (1)-(2)-(3), or else the result may not be a minimal cover.

2. The order in which we process the FDs may result in different minimal covers (… which is ok). i.e., the minimal cover is not unique for a set F.

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Example

F:A � B (1)ABCD � E (2)EF � G (3)EF � H (4)ACDF � EG (5)

Calculate the minimal cover of F.

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Step 1: Put F in standard form

• FDs (1)…(4) are already in standard form.

• For FD (5): – ACDF � E (5.1)

– ACDF � G (5.2)

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Step 2: eliminate extraneous attiributes from LHS(minimize LHSs)

(1) A � B : nothing to eliminate.

(2) ABCD � E. • If delete A, will have BCD � E.

• Is this LHS good enough?

• It is, if either • BCD � E, or

• BCD � W, such that W contains ABCD (the original LHS).

Note, (BCD)+ = BCD. Therefore, cannot delete A.

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Can we delete B?If so, then will have

ACD � EIs this LHS good enough?It is, if either

ACD � E, orACD � W that contains ABCD.

Note, ACD � ACD � ABCD. Therefore, ACD � W = ABCD, and thus B can be eliminated!

So, ABCD � E becomes ACD � E.

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Can I delete any more? i.e., delete C or D?If delete C, then ACD � E becomes AD � E. Test:

AD � AD � ABD, does not contain ACD. Therefore, cannot delete C.

Similarly, cannot delete D.

Since we finished scanning the entire LHS of this FD, step 2 is finished for this FD, and the resulting FD is

ACD � E (2.1) – replaces (2) of original F.

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Repeat the above process for FDs (3), (4), (5.1), (5.2).

For (3): EF � G.

1. Can I delete E? ... If so, will have F � G. – not possible … check.

2. Can I delete F? … if so, will have E � G – not possible .. Check.

Therefore, there is no change in (3).

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For (4) : EF � HAgain, cannot delete anything from LHS.

For (5.1) [ ACDF � E ].

Delete A?

�CDF � E ? Or,

�CDF � W that contains ACDF ?

(CDF)+ = CDF, which is neither E nor W.

� cannot delete A.

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For (5.1) [ ACDF � E ] …

Delete C?

�ADF � E, or

�ADF � W that contains ACDF.

ADF � ADF �from (1)� ABDF, which does contain E or ACDF.

� Cannot delete C.

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For (5.1) [ ACDF � E ] …

Delete D?

�ACF � E, or

�ACF � W that contains ACDF.

ACF � ACF �from (1)� ABCF, which does not contain E or ACDF.

� Cannot delete D.

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For (5.1) [ ACDF � E ] …Delete F?�ACD � E, or�ACD � W that contains ACDF.

ACD � ACD �from (1)� ABCD�from (2)� ABCDE, which contains E !!

Therefore, F can be deleted from the LHS of (5.1), and (5.1) becomes

ACD ���� E (5.1.1)

Finished step 2 of (5.1)!! .. On to (5.2) …

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ACDF � G (5.2)• Repeat the above process and find that

nothing can be eliminated from the LHS of (5.2).

• So step 2 of the minimal cover computation is finished (we minimized all LHSs of all FDs).

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The resulting FDs from step 2, are:

• A � B (1) ---------- (1)

• ACD � E (2.1) ------- (2)

• EF � G (3) --------- (3)

• EF � H (4) --------- (4)

• ACD � E (5.1.1)

• ACDF � G (5.2) ------- (5)Same as (2) !!

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End of Normalization 02