1

Loss-Less Joins

2

Decompositions Dependency-preservation

property: enforce constraints on original relation by enforcing some constraints on resulting relations

Lossless-join property: get original relation by joining the resulting relations

Boyce-Codd normal form (BCNF): lossless join

Third normal form (3NF): lossless join and dependency preservation

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Testing for a Dependency Preservation

If we project R with FD set F, onto R1 with FD F1, R2 with FD F2,…, Rk with FD Fk

We say that dependencies are preserved if and ony if

F+ = (F1+ U F2

+ U … U Fk+)

4

Testing for a Lossless Join

If we project R onto R1, R2,…, Rk , can we recover R by rejoining?

Any tuple in R can be recovered from its projected fragments.

So the only question is: when we rejoin, do we ever get back something we didn’t have originally?

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Lossy Decomposition

R R1 R2 ... Rn

R R1

SSN Name Address SSN Name NameAddress1111 Joe 1 Pine 1111 Joe Joe 1 Pine2222 Alice 2 Oak 2222 Alice Alice 2 Oak3333 Alice 3 Pine 3333 Alice Alice 3 Pine

R2 Rn...

R1 R2R

Problem: Name is not a key

Always true!Always true!

Have to CheckHave to Check?

6

The Chase Test

Suppose tuple t comes back in the join.

Then t is the join of projections of some tuples of R, one for each Ri of the decomposition.

Can we use the given FD’s to show that one of these tuples in R must be t ?

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The Chase – (2)

Start by assuming t = abc… . For each i, there is a tuple si of R that

has a, b, c,… in the attributes of Ri.

si can have any values in other attributes.

We’ll use the same letter as in t, but with a subscript, for these components.

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Example: The Chase

Let R = ABCD, and the decomposition be AB, BC, and CD.

Let the given FD’s be C->D and B ->A.

Suppose the tuple t = abcd is the join of tuples projected onto AB, BC, CD.

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The Tableau

A B C Da b c1 d1

a2 b c d2

a3 b3 c d

d

Use C->D

a

Use B ->AWe’ve proved thesecond tuple must be t.

The tuplesof R pro-jected ontoAB, BC, CD.

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Summary of the Chase

1. If two rows agree in the left side of a FD, make their right sides agree too.

2. Always replace a subscripted symbol by the corresponding unsubscripted one, if possible.

3. If we ever get an unsubscripted row, we know any tuple in the project-join is in the original (the join is lossless).

4. Otherwise, the final tableau is a counterexample.

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Example: Lossy Join

Same relation R = ABCD and same decomposition.

But with only the FD C->D.

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The Tableau

A B C Da b c1 d1

a2 b c d2

a3 b3 c d

d

Use C->DThese three tuples are an exampleR that shows the join lossy. abcdis not in R, but we can project andrejoin to get abcd.

These projectionsrejoin to formabcd.

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Attribute Closure as Chase

A B C D E F

a b c1 d1 e1 f1

a b c2 d2 e2 f2

R = ABCDE, AB ->C, BC ->AD, D ->E, CF ->B

Compute (AB)+

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Multivalued Dependencies

Fourth Normal FormReasoning About FD’s +

MVD’s

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Definition of MVD

A multivalued dependency (MVD) on R, X ->->Y

says that if two tuples of R agree on all the attributes of X, then their components in Y may be swapped, and the result will be two tuples that are also in the relation.

Let Z = R - (X+Y), then for each value of X, values of Y are independent of values of Z.

16

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•Example: The MVD Course →→ Prof holds

•MVD: X →→ Y holds over R, then for any value of attribute set X = x, the following holds true (let Z = R-XY):∏YZ (σX=x (R)) = ∏Y (σ X=x (R)) x ∏Z (σX = x(R))

That is Y and Z=R-XY are independent of each other given X

That is Y and Z=R-XY are independent of each other given X

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Example: MVD

Drinkers(name, addr, phones, beersLiked) A drinker’s phones are independent of

the beers they like. name->->phones and name ->-

>beersLiked. Thus, each of a drinker’s phones

appears with each of the beers they like in all combinations.

This repetition is unlike FD redundancy. name->addr is the only FD.

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Tuples Implied by name->->phones

If we have tuples:

name addr phones beersLikedsue a p1 b1sue a p2 b2sue a p2 b1sue a p1 b2

Then these tuples must also be in the relation.

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Picture of MVD X ->->Y

X Y others

equal

exchange

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MVD Rules

Every FD is an MVD (promotion ). If X ->Y, then swapping Y ’s between two

tuples that agree on X doesn’t change the tuples (same X values imply same Y values!).

Therefore, the “new” tuples are surely in the relation, and we know X ->->Y.

Complementation : If X ->->Y, and Z is all the other attributes, then X ->->Z.

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Splitting Doesn’t Hold

Unlike FD’s, we cannot split the right side --- sometimes you have to leave several attributes on the right side.

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Example: Multiattribute Right Sides

Drinkers(name, areaCode, phone, beersLiked, manf)

A drinker can have several phones, with the number divided between areaCode and phone (last 7 digits).

A drinker can like several beers, each with its own manufacturer.

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Example Continued

Since the areaCode-phone combinations for a drinker are independent of the beersLiked-manf combinations, we expect that the following MVD’s hold:

name ->-> areaCode phonename ->-> beersLiked manf

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Example Data

Here is possible data satisfying these MVD’s:

name areaCode phone beersLiked manfSue 650 555-1111 Bud A.B.Sue 650 555-1111 WickedAle Pete’sSue 415 555-9999 Bud A.B.Sue 415 555-9999 WickedAle Pete’s

But we cannot swap area codes or phones by themselves.That is, neither name->->areaCode nor name->->phoneholds for this relation.

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Fourth Normal Form

The redundancy that comes from MVD’s is not removable by putting the database schema in BCNF.

There is a stronger normal form, called 4NF, that (intuitively) treats MVD’s as FD’s when it comes to decomposition, but not when determining keys of the relation.

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4NF Definition

A relation R is in 4NF if: whenever X ->->Y is a nontrivial MVD, then X is a superkey.

Nontrivial MVD means that:• Y is not a subset of X, and• X and Y are not, together, all the

attributes.

Note that the definition of “superkey” still depends on FD’s only.

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BCNF Versus 4NF

Remember that every FD X ->Y is also an MVD, X ->->Y.

Thus, if R is in 4NF, it is certainly in BCNF. Because any BCNF violation is a 4NF

violation (after conversion to an MVD). But R could be in BCNF and not 4NF,

because MVD’s are “invisible” to BCNF.

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Decomposition and 4NF

If X ->->Y is a 4NF violation for relation R, we can decompose R using the same technique as for BCNF.

1. XY is one of the decomposed relations.

2. (R – Y) U X is the other.

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Example: 4NF Decomposition

Drinkers(name, addr, phones, beersLiked)

FD: name -> addrMVD’s: name ->-> phones

name ->-> beersLiked Key is {name, phones, beersLiked}. All dependencies violate 4NF.

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Example Continued

Decompose using name -> addr:1. Drinkers1(name, addr)

In 4NF; only dependency is name -> addr.

2. Drinkers2(name, phones, beersLiked)1. Not in 4NF. MVD’s name ->-> phones and

name ->-> beersLiked apply. No FD’s, so all three attributes form the key.

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Example: Decompose Drinkers2

Either MVD name ->-> phones or name ->-> beersLiked tells us to decompose to: Drinkers3(name, phones) Drinkers4(name, beersLiked)

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Reasoning About MVD’s + FD’s

Problem: given a set of MVD’s and/or FD’s that hold for a relation R, does a certain FD or MVD also hold in R ?

Solution: Use a tableau to explore all inferences from the given set, to see if you can prove the target dependency.

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Why Do We Care?

1. 4NF technically requires an MVD violation.

Need to infer MVD’s from given FD’s and MVD’s that may not be violations themselves.

2. When we decompose, we need to project FD’s + MVD’s.

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Example: Chasing a Tableau With MVD’s and

FD’s To apply a FD, equate symbols, as

before. To apply an MVD, generate one or

both of the tuples we know must also be in the relation represented by the tableau.

We’ll prove: if A->->BC and D->C, then A->C.

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The Tableau for A->C

A B C Da b1 c1 d1

a b2 c2 d2

Goal: prove that c1 = c2. A->->BC and D->C

a b2 c2 d1

Use A->->BC (first row’sD with second row’s BC ).

c2

Use D->C (first andthird row agree on D,therefore agree on C ).

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Example: Transitive Law for MVD’s

If A->->B and B->->C, then A->->C. Obvious from the complementation

rule if the Schema is ABC. But it holds no matter what the

schema; we’ll assume ABCD.

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The Tableau for A->->C

A B C Da b1 c1 d1

a b2 c2 d2

Goal: derive tuple (a,b1,c2,d1). A->->B and B->->C

a b1 c2 d2

Use A->->B to swap B fromthe first row into the second.

a b1 c2

d1 Use B->->C to swap C fromthe third row into the first.

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Rules for Inferring MVD’s + FD’s

Start with a tableau of two rows. These rows agree on the attributes of

the left side of the dependency to be inferred.

And they disagree on all other attributes.

Use unsubscripted variables where they agree, subscripts where they disagree.

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Inference: Applying a FD

Apply a FD X->Y by finding rows that agree on all attributes of X. Force the rows to agree on all attributes of Y. Replace one variable by the other. If the replaced variable is part of the

goal tuple, replace it there too.

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Inference: Applying a MVD

Apply a MVD X->->Y by finding two rows that agree in X. Add to the tableau one or both rows

that are formed by swapping the Y-components of these two rows.

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Inference: Goals To test whether U->V holds, we

succeed by inferring that the two variables in each column of V are actually the same.

If we are testing U->->V, we succeed if we infer in the tableau a row that is the original two rows with the components of V swapped.

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Inference: Endgame

Apply all the given FD’s and MVD’s until we cannot change the tableau.

If we meet the goal, then the dependency is inferred.

If not, then the final tableau is a counterexample relation. Satisfies all given dependencies. Original two rows violate target

dependency.