1. 4NF (Multivalued Dependency), and 5NF (Join Dependency)

2. Review
Superkey – a set of attributes which will uniquely identify each tuple in a relation
Candidate key – a minimal superkey
Primary key – a chosen candidate key
Secondary key – all the rest of candidate keys
Prime attribute – an attribute that is a part of a candidate key (key column)
Non-prime attribute – a non-key column
Narotama University

3. Review (cont.) 1NF 2NF 3NF BCNF
Eliminate repeating groups.
Make a separate table for each set of related attributes, and give each table a primary key.
2NF
Eliminate redundant data.
Each attribute must be functionally dependent on the primary key.
If an attribute depends on only part of a multi-valued key, remove it to a separate table.
3NF
Eliminate columns not dependent on key.
If attributes do not contribute to a description of the key, remove them to a separate table.
Any transitive dependencies are moved into a smaller table.
BCNF
Every determinant in the table is a candidate key.
If there are non-trivial dependencies between candidate key attributes, separate them out into distinct tables.
All normal forms are additive, in that if a model is in 3NF, it is by definition also in 2NF and 1NF.
Narotama University

4. 4NF
Definition
A relation R is in 4NF if and only if, for every one of its non- trivial multivalued dependencies XY, X is a superkey— that is, X is either a candidate key or a superset thereof.
Nontrivial MVD means that:
Y is not a subset of X, and
X and Y are not, together, all the attributes.
Narotama University

5. Decomposition into 4NF
If XY is a 4NF violation for relation R, we can decompose R using the same technique as for BCNF.
XY is one of the decomposed relations.
All but Y – X is the other.
Narotama University

6. Decomposition into 4NF Method
Find a 4NF violation in R, say AB, where A is not a superkey.
If there is such a 4NF violation, break the schema for the relation R that has the 4NF violation into two schemas.
R1, whose schema is A’s and B’s.
R2, whose schema is the A’s and all attributes of R that are not among the A’s or B’s.
Find the FD’s and MVD’s that hold in R1 and R2. Recursively decompose R1 and R2 with respect to their projected dependencies.
Narotama University

7. 4NF Decomposition Example
Drinkers(name, addr, phones, beersLiked)
FD: nameaddr
MVD’s: namephones
namebeersLiked
Key is {name, phones, beersLiked}.
All dependencies violate 4NF.
Narotama University

8. 4NF Decomposition Example (cont.)
Decompose using nameaddr:
Drinkers1(name, addr)
In 4NF; only dependency is nameaddr.
Drinkers2(name, phones, beersLiked)
Not in 4NF. MVD’s namephones and namebeersLiked apply. No FD’s, so all three attributes form the key.
Narotama University

9. 4NF Decomposition Example (cont.)
Decompose Drinkers2
Either MVD name ->-> phones or name ->-> beersLiked tells us to decompose to:
Drinkers3(name, phones)
Drinkers4(name, beersLiked)
Narotama University

10. MVD Example 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.
Narotama University

11. Tuples Implied by namephones
MVD Example (cont.)
Tuples Implied by namephones
If we have tuples:
Then these tuples must also be in the relation.
name addr phones beersLiked
sue a p1 b1
sue a p2 b2
sue a p2 b1
sue a p1 b2
Narotama University

12. Example
Drinkers(name, addr, phones, beersLiked) with MVD Name  phones. If Drinkers has the two tuples:
name addr phones beersLiked
sue a p1 b1
sue a p2 b2
it must also have the same tuples with phones components swapped:
sue a p2 b1
sue a p1 b2
Note: we must check this condition for all pairs of tuples that agree on name, not just one pair.
Narotama University

13. Example (cont’d) Violates 4NF because CUSTOMER is not a superkey
(and CUSTOMER->->LOAN_NO is non-trivial)
Narotama University

14. Solution:
Narotama University

15. HIGHER NORMAL FORMS 1NF 2NF 3NF BCNF 4NF 5NF functional dependencies
multivalued
dependencies
join
dependencies
Narotama University

16. 24-HOUR FLIGHT-TIMETABLE, ALL FLIGHTS EVERY DAY
ASSUMPTIONS
24-HOUR FLIGHT-TIMETABLE, ALL FLIGHTS EVERY DAY
ALL PLANES TAKE-OFF AND LAND (BUT DO NOT CRASH)
NO AIRPORT IS LANDING-ONLY & NO AIRPORT IS TAKE-OFF-ONLY
  TTAB (ORG) =  TTAB (DST)
THERE IS AT LEAST ONE TIME DELAY ENTRY FOR EVERY FLIGHT
SIMILARLY IN WEATHER REPORT HISTORY
IF NO TWO FLIGHTS CAN TAKE OFF AT THE SAME TIME IN THE SAME AIRPORT
WES CAN BE POSTED TO FLIGHT AND ELIMINATED
Narotama University

17. teaches (UNIVERSITY, DISCIPLINE) is_read_for (DISCIPLINE, DEGREE)
Old Town
Computing
BSc
Mathematics
PhD
New City
AWARD
teaches (UNIVERSITY, DISCIPLINE)
is_read_for (DISCIPLINE, DEGREE)
awards (UNIVERSITY, DEGREE)
teaches (NewCity, Computing) = true
awards (NewCity, PhD) = true
is_read_for (Computing, BSc) = true
FROM
(NewCity teaches Computing) and (Computing is_read_for BSc)
IT DOES NOT FOLLOW
NewCity awards BSc for_reading Computing
Narotama University

18. Narotama University

19. Join Dependency
JD * (R1, R2, R3, ..., Rm) holds in R iff R = join (R1, R2, R3, ..., Rm ), Ri - a projection of R
Narotama University

20. Fifth Normal Form preventing illogical conjunction of facts R
A relation R is in 5NF iff
for all JD * (R1, R2, R3, ..., Rm) in R,
every Ri is a superkey for R.
JD* ( ,
) holds for R
then R is not in 5NF if
does not contain key
JD* ( , )
then R is in 5NF if
Narotama University

21. AWARD UNIVERSITY DISCIPLINE Old Town Computing Mathematics New City
DEGREE
Computing
BSc
Mathematics
PhD
UNIVERSITY
DEGREE
Old Town
BSc
PhD
New City
Narotama University

22. candidate keys - NAME or CODE
join dependencies
JD1 * ((NAME, CODE, TEACHING), (NAME, RESEARCH))
JD2 * ((NAME, CODE, RESEARCH), (NAME, TEACHING))
JD3 * ((NAME, CODE, TEACHING), (CODE, RESEARCH))
JD4 * ((NAME, CODE, RESEARCH), (CODE, TEACHING))
JD5 * ((NAME, CODE), (NAME, TEACHING), (CODE,RESEARCH))
all projections in JD1 - to JD5 are superkeys for RANKING  5NF
Narotama University

23. Summary
A multivalued dependency is a statement that two sets of attributes in a relation have sets of values that appear in all possible combinations.
If a relation is in 4NF, then every nontrivial MVD is really an FD with a superkey on the left.
Narotama University

24. References
4NF_and_Multivalued_Dependency_by_Kristina_Miguel. ppt
NF.ppt
Narotama University