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Functional Dependencies函数依赖Meaning of FD’s

Keys and SuperkeysFunctional Dependencies

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Functional Dependencies X → A is an assertion about a relation R that

whenever two tuples of R agree on all the attributes of X, then they must also agree on the attribute A. Say “X → A holds in R.” Convention: …, X, Y, Z represent sets of attributes; A,

B, C,… represent single attributes. Convention: no set formers in sets of attributes, just

ABC, rather than {A,B,C }.

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Example

Movies(title, year,length,filmType,studioname)

Reasonable FD’s to assert:1. tile year → length2. tile year →filmType3. tile year → studioName

title year length

film type

studioName

starName

Star WarsStar WarsStar WarsMighty DucksWayne’s WorldWayne’s World

197719771977199119921992

1241241241049595

colorcolorcolorcolorcolorcolor

FoxFoxFoxDisneyParamountParamount

Carrie FisherMark HamillHarrison FordEmilio EstevezDana CarveyMike Meyers

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FD’s With Multiple Attributes

No need for FD’s with > 1 attribute on right. But sometimes convenient to combine

FD’s as a shorthand. Example: tile year → length , tile

year →filmType and tile year → studioName become tile year → length filmType studioName

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Keys of Relations K is a superkey for relation R if K

functionally determines all of R. K is a key for R if K is a superkey,

but no proper subset of K is a superkey.

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ExampleMovies(title,

year,length,filmType,studioname) {tile, year,starName} is a superkey

because together these attributes determine all the other attributes.

{tile, year,starName,length,filmType} is a superkey too.

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E/R and Relational Keys Keys in E/R concern entities. Keys in relations concern tuples. Usually, one tuple corresponds to one

entity, so the ideas are the same. But --- in poor relational designs, one

entity can become several tuples, so E/R keys and Relational keys are different.

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Example

Drinkers addrname

Beers

manfname

Bars

name

license

addr

Note:license =beer, full,none

Sells Bars sell somebeers.

LikesDrinkers likesome beers.Frequents

Drinkers frequentsome bars.

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Example Data (Cont.)

name addr beersLiked manf favBeerJaneway Voyager Bud A.B. WickedAleJaneway Voyager WickedAle Pete’s WickedAleSpock Enterprise Bud A.B. Bud

Relational key = {name beersLiked}But in E/R, name is a key for Drinkers, and beersLiked is a key

for Beers.Note: 2 tuples for Janeway entity and 2 tuples for Bud entity.

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Where Do Keys Come From?

1. Just assert a key K. The only FD’s are K → A for all attributes A.

2. Assert FD’s and deduce the keys by systematic exploration.

E/R model gives us FD’s from entity-set keys and from many-one relationships.

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More FD’s From “Physics” Example: “no two courses can meet in

the same room at the same time” tells us: hour room → course.

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Rules About Functional Dependencies

The Splitting/Combining Rule We can replace a FD A1A2…An→ B1B2…Bm by a set

of FD’s A1A2…An→ Bi,for i=1,2,…,m. This transformation we call splitting rules.

We can replace a set of FD’s A1A2…An→ Bi for i=1,2,…,m by a set A1A2…An→ B1B2…Bm,. This transformation we call combining rules.

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Examplestitle year → length title year → studioName

title year → length studioName

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MoreConsider FD:

title year → length if we try to split the left side into

title → length year → length

Then we get two false FD’s.

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A functional dependency A1A2…An→B is said to be trivial( 平凡 ) if B is one of the A’s, otherwise is said to be nontrivial( 非平凡 ). Example: Suppose Functional Dependencies

title, year → title is a trivial dependency.

Trivial Dependencies

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The Closure of Attributes

Suppose {A1,A2,…,An} is a set of attributes and S is a set of FD’s. The closure of {A1,A2,…,An} under the FD’s in S is the set of attributes B such that every relation that satisfies all the FD’s in the set S also satisfies A1A2…An→ B.That is, A1A2…An→ B follows from the FD’s of S.

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Computing the closure

An easier way to test is to compute the closure of Y, denoted Y +.

Basis: Y + = Y. Induction: Look for an FD’s left side X

that is a subset of the current Y +. If the FD is X -> A, add A to Y +.

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Y+new Y+

X A

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Example R = (A, B, C, G, H, I) F = {A B

A C CG HCG IB H}

(AG)+

1. result = AG2. result = ABCG (A C and A B)3. result = ABCGH (CG H and CG AGBC)4. result = ABCGHI (CG I and CG AGBCH)

Is AG a candidate key? 1. Is AG a super key?

1. Does AG R? == Is (AG)+ R2. Is any subset of AG a superkey?

1. Does A R? == Is (A)+ R2. Does G R? == Is (G)+ R

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例 设关系模式 R （ U,F ） , 其中 ,U={A,B,C,D,E,I},F={A→D,AB→C,BI→C

,ED→I,C→E}, 求（ AC ） F+ 。

解： (1) 令 X={AC}, 则 X(0)=AC 。 (2) 在 F 中找出左边是 AC 子集的函数依赖： A→D,C→E 。 (3) X(1)=X(0) D E=ACDE∪ ∪ 。

Another Examples

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(4) 很明显 X(1)≠X(0), 所以 X(i)=X(1), 并转向算法中的步骤 (2) 。

(5) 在 F 中找出左边是 ACDE 子集的函数依赖： ED→I 。(6) X(2)=X(1)∪I=ACDEI 。(7) 虽然 X(2)≠X(1), 但是 F 中未用过的函数依赖的左边

属性已没有 X(2) 的子集 , 所以 , 可停止计算 , 输出（ AC） F

+ = X(2)=ACDEI 。

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Transitive( 传递 ) Functional Dependencies Suppose we have a relation R with three attributes A,

B, and C, the FD’s A→B and B→C both hold for R. Then it is easy to see that the FD A→C also holds for R, So C is said to depend on A transitively, via B.

The transitive rule

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Examples

Two of the FD’s title year → studioName studioName →studioAddr

Combine the two FD’s above to get title year → studioAddr

title year length

film type

studioName

studioAddr

Star WarsMighty DucksWayne’s World

197719911992

12410495

colorcolorcolor

FoxDisneyParamount

HolleywoodBuena VistaHolleywood

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Closing Sets of Functional Dependencies

Any set of given FD’s from which we can infer all the FD’s for a relation will be called a basis for that relation. If no proper subset of the FD’s in the basis can also derive the complete set of FD’s, then we say the basis is minimal, that is the closing set of the functional dependencies of the relation

Consider a relation R(A,B,C), One of its FD’s is{A→B,B →A, B→C,C →B}

Another is{A→B,B→C,C →A}