4nf (multivalued dependency), and 5nf (join dependency)

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

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 University2

Review (cont.) 1NF

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.


4NF Definition

A relation R is in 4NF if and only if, for every one of its non-trivial multivalued dependencies XY, X is a superkey—that is, X is either a candidate key or a superset thereof. Nontrivial MVD means that:1. Y is not a subset of X, and2. X and Y are not, together, all the attributes.


Decomposition into 4NF If XY 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. All but Y – X is the other.


Decomposition into 4NF Method Find a 4NF violation in R, say AB, 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.1. R1, whose schema is A’s and B’s.2. 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.


4NF Decomposition ExampleDrinkers(name, addr, phones, beersLiked)FD: nameaddrMVD’s: namephones

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


4NF Decomposition Example (cont.) Decompose using nameaddr:1. Drinkers1(name, addr)

In 4NF; only dependency is nameaddr.

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

namebeersLiked apply. No FD’s, so all three attributes form the key.


4NF Decomposition Example (cont.) Decompose Drinkers2

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


MVD ExampleDrinkers(name, addr, phones, beersLiked) A drinker’s phones are independent of the

beers they like. namephones and namebeersLiked.

Thus, each of a drinker’s phones appears with each of the beers they like in all combinations.


MVD Example (cont.) Tuples Implied by namephones

If we have tuples:

Then these tuples must also be in the relation.

name addr phones beersLikedsue a p1 b1sue a p2 b2

sue a p2 b1sue a p1 b2


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:

name addr phones beersLiked 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.


Example (cont’d)

Violates 4NF because CUSTOMER is not a superkey

(and CUSTOMER->->LOAN_NO is non-trivial)


Solution:


1NF 2NF 3NF BCNF 4NF 5NF

functional dependencies

multivalueddependencies

joindependencies

HIGHER NORMAL FORMS


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 WEATHER@ORIGIN ELIMINATED


UNIVERSITY DISCIPLINE DEGREE

Old Town Computing BSc

Old Town Mathematics PhD

New City Computing PhD

Old Town Computing PhD

AWARD teaches (UNIVERSITY, DISCIPLINE)is_read_for (DISCIPLINE, DEGREE)awards (UNIVERSITY, DEGREE)

teaches (NewCity, Computing) = true awards (NewCity, PhD) = trueis_read_for (Computing, BSc) = true

FROM(NewCity teaches Computing) and (Computing is_read_for BSc)

IT DOES NOT FOLLOWNewCity awards BSc for_reading Computing


UNIVERSITY DISCIPLINE DISCIPLINE DEGREE UNIVERSITY DEGREE

Old Town Computing Computing BSc Old Town BSc

Old Town Mathematics Mathematics PhD Old Town PhD

New City Computing Computing PhD New City PhD





New City Computing BSc







spurious tuple

AWARD


Join Dependency

R

if R1 R2

R

then

JD*( )

holds for R

for

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


R

JD* ( , )

then R is in 5NF

if

JD* ( , )

A relation R is in 5NF iff

for all JD * (R1, R2, R3, ..., Rm) in R,

every Ri is a superkey for R.

preventing illogical conjunction of factsFifth Normal Form

JD* ( , ) holds for R

then R is not in 5NF

if

does not contain key


UNIVERSITY DISCIPLINE

Old Town Computing

Old Town Mathematics

New City Computing

DISCIPLINE DEGREE

Computing BSc

Mathematics PhD

Computing PhD

UNIVERSITY DEGREE

Old Town BSc

Old Town PhD

New City PhD

AWARD


NAME CODE TEACHING RESEARCH

Old Town P05 21 4

New City C01 23 3

RANKING

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)).............................................................................................

candidate keys - NAME or CODE

all projections in JD1 - to JD5 are superkeys for RANKING 5NF


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.


References http://www.cs.sjsu.edu/~lee/cs157b/Spring09P

resentation/4NF_and_Multivalued_Dependency_by_Kristina_Miguel.ppt

http://www.cs.sjsu.edu/faculty/lee/cs157/4NF%20and%205NF.ppt


4nf (multivalued dependency), and 5nf (join dependency)

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