multi-valued dependencies and fourth normal form

Multi-valued Dependencies and Fourth Normal Form

COSC 6340

Ch. Eick: 4thNF and MVD's 2

Topics Covered1. Definition of Multivalued

Dependencies2. Reasoning about MVDs3. Fourth Normal Form


Motivation There are schemas that are in BCNF that do

not seem to be sufficiently normalized

name street

Starscity title year

C. Fisher

C. Fisher

C. Fisher

C. Fisher

C. Fisher

123 Maple Str.

5 Locust Ln.

123 Maple Str.

5 Locust Ln.

123 Maple Str.

5 Locust Ln.C. Fisher

Hollywood

Malibu

Hollywood

Malibu

Hollywood

Malibu

Star Wars 1977

Star Wars 1977

Empire Strikes Back 1980


Return of the Jedi 1983



Attribute Independence No reason to associate address with one

movie and not another When we repeat address and movie facts

in all combinations, there is obvious redundancy

However, NO BCNF violation in Stars relation There are no non-trivial FD’s at all, all five

attributes form the only superkey Why?


Multi-valued DependencyDefinition: Multivalued dependency (MVD): A1A2…An B1B2…Bm holds for relation R if: For all tuples t, u in R If t[A1A2...An] = u[A1A2...An], then there exists a v in

R such that: (1) v[A1A2...An] = t[A1A2...An] = u[A1A2...An] (2) v[B1B2…Bm] = t[B1B2…Bm] (3) v[C1C2…Ck] = u[C1C2…Ck], where C1C2…Ck is

all attributes in R except (A1A2...An B1B2…Bm)


Pictorially Speaking...

An MVD guarantees v exists The existence of a fourth tuple w is implied

by interchanging t and u

t

u

A’s B’s Others

v

w


Example: name street city

name street

Starscity title year

C. Fisher

C. Fisher

C. Fisher

123 Maple Str.

123 Maple Str.

5 Locust Ln.

Hollywood

Hollywood

Malibu

Star Wars 1977



C. Fisher

C. Fisher

5 Locust Ln.

123 Maple Str.


Malibu

Hollywood

Malibu

Star Wars 1977



t

u

v


Example: name street city

name street

Stars

city title year

C. Fisher

C. Fisher

C. Fisher

C. Fisher

123 Maple Str.

5 Locust Ln.

123 Maple Str.

5 Locust Ln.

Hollywood

Malibu

Hollywood

Malibu

Star Wars 1977

Star Wars 1977



C. Fisher 123 Maple Str.


Hollywood

Malibu



u

t

wv


More on MVDs Intuitively, A1A2…An B1B2…Bm says that the

relationship between A1A2…An and B1B2…Bm is independent of the relationship between A1A2…An and R -{B1B2…Bm}

MVD's uncover situations where independent facts related to a certain object are being squished together in one relation

Functional dependencies rule out certain tuples from being in a relation

How? Multivalued dependencies require that other tuples

of a certain form be present in the relation a.k.a. tuple-generating dependencies


Let’s Illustrate In Stars, we must repeat the movie (title, year)

once for each address (street, city) a movie star has Alternatively, we must repeat the address for each movie a

star has made Example: Stars with name street city name street city title year

C. Fisher

C. Fisher

C. Fisher

123 Maple Str.

5 Locust Ln.

123 Maple Str.

Hollywood

Malibu

Hollywood

Star Wars 1977



Is an incomplete extent of Stars Infer the existence of a fourth tuple under the given MVD


Trivial MVDs Trivial MVD A1A2…An B1B2…Bm where B1B2…

Bm is a subset of A1A2…An or (A1A2…An B1B2…Bm ) contains all attributes of R


Reasoning About MVDs FD-IS-AN-MVD Rule (Replication) If A1A2…An B1B2…Bm then A1A2…An B1B2…Bm holds


Reasoning About MVDs COMPLEMENTATION Rule If A1A2…An B1B2…Bm then A1A2…An

C1C2…Ck where C1C2…Ck is all attributes in R except (A1A2…An B1B2…Bm )

AUGMENTATION Rule If XY and WZ then WX YZ TRANSITIVITY Rule If XY and YZ then X (ZY)


Coalescence Rule for MVDX Y

W:W Z

Then: X Z

If:

Remark: Y and W have to be disjoint and Z has to be a subset of or equal to Y


Definition 4NF Given: relation R and set of MVD's for R Definition: R is in 4NF with respect to its

MVD's if for every non-trivial MVD A1A2…AnB1B2…Bm , A1A2…An is a superkey

Note: Since every FD is also an MVD, 4NF implies BCNF

Example: Stars is not in 4NF


Decomposition Algorithm(1) apply closure to the user-specified FD's and MVD's**:(2) repeat until no more 4NF violations: if R with AA ->> BB violates 4NF then: (2a) decompose R into R1(AA,BB) and R2(AA,CC), where CC is all attributes in R except (AA BB) (2b) assign FD's and MVD's to the new relations**

** MVD's: hard problem! No simple test analogous to computing the attribute

closure for FD’s exists for MVD’s. You are stuck to have to use the 5 inference rules for MVD’s when computing the closure!


Exercise Decompose Stars into a set of

relations that are in 4NF. namestreet city is a 4NF violation Apply decomposition:R(name, street, city)S(name, title, year)

What about namestreet city in R and nametitle year in S?


Exercise For the relation R(A,B,C,D) with only

MVD’s AB and AC find all 4NF violations and decompose R into a collection of relation schemas in 4NF.


Solution Since there are no functional dependencies,

the only key is all four attributes, ABCD. Thus, each of the nontrivial multivalued

dependencies A->->B and A->->C violate 4NF. Separate out the attributes of these

dependencies, first decomposing into AB and ACD

Then decompose the latter into AC and AD because A->->C is still a 4NF violation for ACD.

The final set of relations are AB, AC, and AD.


Exercise Suppose we have relation R(A,B,C)

with MVD AB. If we know that the tuples (a,b1,c1), (a,b2,c2), and (a,b3,c3) are in the current instance of R, what other tuples do we know must also be in R?


Solution Since A->->B, and all the tuples have the

same value for attribute A, we can pair the B-value from any tuple with the value of the remaining attribute C from any other tuple.

Thus, we know that R must have at least the nine tuples of the form (a,b,c), where b is any of b1, b2, or b3, and c is any of c1, c2, or c3. That is, we can derive, using the definition of a multivalued dependency, that each of the tuples (a,b1,c2), (a,b1,c3), (a,b2,c1), (a,b2,c3), (a,b3,c1), and (a,b3,c2) are also in R.


‘s

Another View of 4NF

MVD’s that are also FD’s

True MVD’s

TrivialMVD’s

4NF:= Relation is inBCNF and there are no true MVD’s (yellowpart is empty)

True MVD XY:= non-trivial &XY does not hold

Remark: If XY is a true MVD then X cannotbe a superkey (becauseXY does not hold); Therefore, true MVD’s always violate 4NF (“trueMVD’s are always bad)

XY

XY and XY

XY and not XY


H1-2005-Problem88) Normalization [6] gradedR(A,B,C,D,E,F) is given with: (1) ABCD (2)CDAB (3)ABF (4) FEa) What are the candidate keys of relation R? [1] b) b) Transform R into a relational schema that is in BCNF and

does not have any lost functional dependencies! [5]

Correct Solution:a) Candidate keys: AB and CDb) Decompose R into R1(A,B,C,D,F) with local FD’s (1), (2), (3) and

R2(E,F) with local FD’s (4) Due to the fact that all four dependencies are still present no functional dependency has been lost. Moreover, all functional depencies are good

A non-optimal (“too many relations”) solution I also saw was: Decompose R into R1(A,B,C,D) with local functional dependencies ABCD and CDAB, R2(A,B,F) with local functional dependencies ABF and R3(F,E) with local functional dependencies FA..


Problem 1; H1-2004a. Candidate keys are: {a,b}, {a,d},

{a,e}b. 14 superkeys totalc. All but the first functional

dependency are bad


Problem 2; H2-04a. No; EBC is a “true” multi-valued dependency and E is not a

candidate key (as a matter of fact {E}+={A,D,E,F} see below)b. No (but just mentioning neither E ABC nor E CF holds is not

sufficient (e.g. if EABC holds then the decomposition is lossless!) ) --- a counter examples should be given to show that the statement is false!

c. Yes C is candidate key; therefore CBDEF; therefore CBDEFd. Yes E BC and BC BCD implies ED due to MVD-

transitivity (CCDBCBCD BCBCD)e. Yes EBC; therefore EADF; moreover, CADF and using

the Coalescence Rule we obtain EADF; therefore, EA holdsf. No R is not in BCNF because EADF holds and E is not a

candidate key.


Problem 3a-2004From AB and AC we can infer:

ABC??(1) AC AAC (2) AB ACABC (3) ACABC ACDE (4) AAC, ACDE ADE (5) ADE ABC Using: 1. Augmentation, 2.Augmentation, 3.Complementation,

4.Transitivity, 5. Complementation

Wrong!!

Remark: This problem will be revised in Homework3-2005; it is too complicated to worry about it for the midterm exam!


MDV’s and FD’s --- Ungraded Homework

Assume we have a relation R(A,B,C,D,E) with the following dependencies:

(1) AB CDE (2) CD ABE(3) E DBAnswer the following questions giving reasons for your answers:a) Is R in BCNF? ????? (answer after Spring break) Warning: The

presence of the MVD might imply other functional dependencies (see textbook page 637)

b) Does ABE D hold for R? yesc) Does CD B hold for R? yesd) Does E D always hold for R (either show that this

dependency can be inferred from the given 3 dependencies, or give a counter example of a relation that satisfies (1), (2), (3) but violates ED)? No

multi-valued dependencies and fourth normal form

Documents

fisher123 maple

subset of a1a2an

maple str

locust ln

address street

movie facts

certain tuples

relation r