1. Multi-valued Dependencies and Fourth Normal Form
COSC 6340

2. Topics Covered Definition of Multivalued Dependencies
Reasoning about MVDs
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
Stars
name
street
city
title
year
C. Fisher
123 Maple Str.
Hollywood
Star Wars
1977
C. Fisher
5 Locust Ln.
Malibu
Star Wars
1977
C. Fisher
123 Maple Str.
Hollywood
Empire Strikes Back
1980
C. Fisher
5 Locust Ln.
Malibu
Empire Strikes Back
1980
C. Fisher
123 Maple Str.
Hollywood
Return of the Jedi
1983
C. Fisher
5 Locust Ln.
Malibu
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 A’s B’s Others t vw
An MVD guarantees v exists
The existence of a fourth tuple w is implied by interchanging t and u
Ch. Eick: 4thNF and MVD's

7. Example: name  street city
Stars
name
street
city
title
year
C. Fisher
123 Maple Str.
Hollywood
Star Wars
1977 t
C. Fisher
5 Locust Ln.
123 Maple Str.
Malibu
Hollywood
Star Wars
1977
Return of the Jedi
1983 v
C. Fisher
123 Maple Str.
Hollywood
Empire Strikes Back
1980 u
C. Fisher
5 Locust Ln.
Malibu
Empire Strikes Back
1980
Ch. Eick: 4thNF and MVD's

8. Example: name  street city
Stars
name
street
city
title
year
C. Fisher
123 Maple Str.
Hollywood
Star Wars
1977 u
C. Fisher
5 Locust Ln.
Malibu
Star Wars
1977 w v
C. Fisher
123 Maple Str.
Hollywood
Empire Strikes Back
1980 t
C. Fisher
5 Locust Ln.
Malibu
Empire Strikes Back
1980
C. Fisher
123 Maple Str.
5 Locust Ln.
Hollywood
Malibu
Return of the Jedi
1983
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
123 Maple Str.
Hollywood
Star Wars
1977
C. Fisher
5 Locust Ln.
Malibu
Empire Strikes Back
1980
C. Fisher
123 Maple Str.
Hollywood
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 XY and WZ then WX YZ
TRANSITIVITY Rule
If XY and YZ then X  (Z-Y)
Ch. Eick: 4thNF and MVD's

14. Coalescence Rule for MVD
X  Y
Then: X  Z
If:  
W:W  Z
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…AnB1B2…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.
namestreet city is a 4NF violation
Apply decomposition:
R(name, street, city)
S(name, title, year)
What about namestreet city in R and nametitle year in S?
Ch. Eick: 4thNF and MVD's

18. Exercise
For the relation R(A,B,C,D) with only MVD’s AB and AC 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 AB. 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. Another View of 4NF True MVD’s MVD’s that are also ‘s FD’s Trivial
True MVD XY:= non-trivial &
XY does not hold ‘s
True MVD’s
MVD’s that are also
FD’s
Remark: If XY is a
true MVD then X cannot
be a superkey (because
XY does not hold);
Therefore, true MVD’s
always violate 4NF (“true
MVD’s are always bad)
Trivial
MVD’s
4NF:= Relation is in
BCNF and there are
no true MVD’s (yellow
part is empty)
XY and XY
XY and not XY
XY
Ch. Eick: 4thNF and MVD's

23. H1-2005-Problem8 8) Normalization [6] graded
R(A,B,C,D,E,F) is given with: (1) ABCD (2)CDAB (3)ABF (4) FE
What are the candidate keys of relation R? [1]
b) Transform R into a relational schema that is in BCNF and does not have any lost functional dependencies! [5]
Correct Solution:
Candidate keys: AB and CD
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 ABCD and CDAB, R2(A,B,F) with local functional dependencies ABF and R3(F,E) with local functional dependencies FA..
Ch. Eick: 4thNF and MVD's

24. Problem 1; H1-2004 Candidate keys are: {a,b}, {a,d}, {a,e}
14 superkeys total
All but the first functional dependency are bad
Ch. Eick: 4thNF and MVD's

25. Problem 2; H2-04
No; EBC 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)
No (but just mentioning neither E ABC nor E CF holds is not sufficient (e.g. if EABC holds then the decomposition is lossless!) ) --- a counter examples should be given to show that the statement is false!
Yes C is candidate key; therefore CBDEF; therefore CBDEF
Yes E BC and BC BCD implies ED due to MVD-transitivity (CCDBCBCD  BCBCD)
Yes EBC; therefore EADF; moreover, CADF and using the Coalescence Rule we obtain EADF; therefore, EA holds
No R is not in BCNF because EADF holds and E is not a candidate key.
Ch. Eick: 4thNF and MVD's

26. Problem 3a-2004 From AB and AC we can infer: ABC??
(1) AC AAC
(2) AB  ACABC
(3) ACABC  ACDE
(4) AAC, ACDE ADE
(5) ADE ABC
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 DB
Answer 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? yes
c)      Does CD  B hold for R? yes
d)      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 ED)? No
Ch. Eick: 4thNF and MVD's