Ch. Eick: 4thNF and MVD's 2
Topics Covered
1. Definition of Multivalued Dependencies
2. Reasoning about MVDs3. Fourth Normal Form
Ch. Eick: 4thNF and MVD's 3
Motivation There are schemas that are in BCNF that do
not seem to be sufficiently normalized
name street
Stars
city 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
Empire Strikes Back 1980
Return of the Jedi 1983
Return of the Jedi 1983
Ch. Eick: 4thNF and MVD's 4
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?
Ch. Eick: 4thNF and MVD's 5
Multi-valued Dependency
Definition: 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)
Ch. Eick: 4thNF and MVD's 6
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
Ch. Eick: 4thNF and MVD's 7
Example: name street city
name street
Stars
city title year
C. Fisher
C. Fisher
C. Fisher
123 Maple Str.
123 Maple Str.
5 Locust Ln.
Hollywood
Hollywood
Malibu
Star Wars 1977
Empire Strikes Back 1980
Empire Strikes Back 1980
C. Fisher
C. Fisher
5 Locust Ln.
123 Maple Str.
5 Locust Ln.C. Fisher
Malibu
Hollywood
Malibu
Star Wars 1977
Return of the Jedi 1983
Return of the Jedi 1983
t
u
v
Ch. Eick: 4thNF and MVD's 8
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
Empire Strikes Back 1980
Empire Strikes Back 1980
C. Fisher 123 Maple Str.
5 Locust Ln.C. Fisher
Hollywood
Malibu
Return of the Jedi 1983
Return of the Jedi 1983
u
t
w
v
Ch. Eick: 4thNF and MVD's 9
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
Ch. Eick: 4thNF and MVD's 10
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
Empire Strikes Back 1980
Return of the Jedi 1983
Is an incomplete extent of Stars Infer the existence of a fourth tuple under the given MVD
Ch. Eick: 4thNF and MVD's 11
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
Ch. Eick: 4thNF and MVD's 12
Reasoning About MVDs
FD-IS-AN-MVD Rule (Replication) If A1A2…An B1B2…Bm then
A1A2…An B1B2…Bm holds
Ch. Eick: 4thNF and MVD's 13
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)
Ch. Eick: 4thNF and MVD's 14
Coalescence Rule for MVD
X 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
Ch. Eick: 4thNF and MVD's 15
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
Ch. Eick: 4thNF and MVD's 16
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!
Ch. Eick: 4thNF and MVD's 17
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?
Ch. Eick: 4thNF and MVD's 18
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.
Ch. Eick: 4thNF and MVD's 19
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.
Ch. Eick: 4thNF and MVD's 20
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?
Ch. Eick: 4thNF and MVD's 21
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.
Ch. Eick: 4thNF and MVD's 22
‘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
Ch. Eick: 4thNF and MVD's 23
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..
Ch. Eick: 4thNF and MVD's 24
Problem 1; H1-2004
a. Candidate keys are: {a,b}, {a,d}, {a,e}
b. 14 superkeys totalc. All but the first functional
dependency are bad
Ch. Eick: 4thNF and MVD's 25
Problem 2; H2-04
a. 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 CBDEF
d. 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 holds
f. No R is not in BCNF because EADF holds and E is not a candidate key.
Ch. Eick: 4thNF and MVD's 26
Problem 3a-2004
From 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!
Ch. Eick: 4thNF and MVD's 27
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
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