lecture 17 4 th normal form prof. sin-min lee department of computer science

Lecture 17 4th Normal Form

Prof. Sin-Min LeeDepartment of Computer Science

Functional Dependencies

o Dependencies for this relation:A BA DBC EF

o Do they all hold in this instance of the relation R?

R A B C D E F a1 b1 c1 d1 e1 f1 a1 b1 c2 d1 e2 f3 a2 b1 c2 d3 e2 f3 a3 b2 c3 d4 e3 f2 a2 b1 c3 d3 e4 f4 a4 b1 c1 d5 e1 f1

• Functional dependencies are specified by the database programmer based on the intended meaning of the attributes.

Armstrong’s Axioms

• Armstrong’s Axioms: Let X, Y be sets of attributes from a relation T.[1] Inclusion rule: If Y X, then X Y.[2] Transitivity rule: If X Y, and Y Z, then X Z.[3] Augmentation rule: If X Y, then XZ YZ.

• Other derived rules:[1] Union rule: If X Y and X Z, then X YZ[2] Decomposition rule: If X YZ, then X Y and X Z[3] Pseudotransitivity: If X Y and WY Z, then XW Z[4] Accumulation rule: If X YZ and Z BW,

then X YZB

Closure

• Let F be a set of functional dependencies.• We say F implies a functional dependency g if g can be

derived from the FDs in F using the axioms.• The closure of F, written as F+, is the set of all FDs that

are implied by F.• Example: Given FDs { A BC, C D, AD B }, what

is the closure?• The closure is a potentially exponential set:

Trivial dependencies: {A A, B B,…,ABC ABC,…}, other dependencies obtained by augmentation {AB ABC, BC BD,…}, dependencies obtained by other rules (or multiple rules), {A BC, C D, AD B, A D }

Closure

• Given a set F of functional dependencies. A functional dependency X Y is said to be entailed (implied) by F, if X Y is in F+.

• To sets of functional dependencies F1, F2 are said to be equivalent, if their closures are equivalent, I.e. F1+ = F2+.

Checking Entailment

• To find whether a functional dependency X Y is implied by a set of functional dependencies F– We can apply all the rules in Armstrong’s Axioms to

find whether we can obtain this dependency from the dependencies in F

OR– We can determine the closure of attributes X, denoted

by X+ with respect to F, and check whether Y X+

Closure of a set of attributes• Given a set of F of functional dependencies, the

closure of a set of attributes X, denoted by X+ is the largest set of attributes Y such that X Y.Algorithm Closure(X,F)X[0] = X; I = 0;repeat

I = I + 1; X[I] = X[I-1];

FOR ALL Z W in FIF Z X[I] THEN X[I] = X[I] W

END FORuntil X[I] = X[I-1];RETURN X+ = X[I]

Entailment

• Given F = { C DE, AB CE, EB CF, G A }

• Find: GB+

– Initialize: GB+ = {G,B}– Use G A , add A, GB+ = {A,B,G}– Use AB CE, add C,E, GB+ = {A, B, C, E, G}– Use C DE, add D, GB+ = {A, B, C, D, E, G}– Use EB CF, add F, GB+ = {A, B, C, D, E, F, G}– Incidentally, GB is a superkey. Is it also a key?

Closure of a set of attributes

• Given a set of functional dependencies F for a relation R, X is said to be a superkey, if X+ contains all the attributes in R.– In other words, X implies all other attributes.– Alternatively, if two tuples are the same with

respect to X then they should be the same with respect to all other attributes.

Boyce-Codd Normal Form• A table T is said to be in Boyce-Codd Normal Form

(BCNF) with respect to a given set of functional dependencies F if for all functional dependencies of the form X A entailed by F the following is true:– If A is not a subset of X then X is a superkey, or – If A is not contained in X then, X contains all the

attributes in a key.• Given {AB C, AB D, AE D, C F} with Key:

{A,B,E}– not in BCNF since C is a single attribute not in AB, but AB is not

a superkey.

Boyce-Codd Normal FormGiven head(T)={A,B,C,D,E,F} with functional dependencies

{AC D, AC E, AF B, AD F, BC A, ABC F } and keys: {A, C}, {B, C}, is this relation T in BCNF?

No. It is sufficient to find one violation!

– AF B violates BCNF since B is not in AF and AF is not a superkey.

– AD F violates BCNF since F is not in AD and AD is not a superkey.

Note: ABC F does not violate BCNF since ABC is a superkey.

Decomposition• A decomposition of a relation R with functional

dependency set F is a sequence of pairs of the form– (R1,F1), …, (Rn,Fn) such that– The union of attributes in R1,…,Rn is equivalent to the attributes

in R– All functional dependencies in F1,…,Fn are entailed by F.

• A decomposition is obtained by a simple projection of R onto the attributes in the decomposed relations. – For example, given R1 with schema R1(A1) , then

R1 = A1 (R)

Lossless Decompositions

• A decomposition of R to (R1, F1) and (R2,F2) is said to be lossless, if we join R1 and R2 on the common attributes, we are guaranteed to get R.

• In other words, given R1(A1) and R2(A2), we need to make sure that it is always the case that

R=R1 join(A1A2) R2

where join(A1A2) means equi-join on the common attributes in A1 and A2.

R A B C D a1 b1 c1 d1 a1 b1 c2 d1 a2 b1 c2 d3 a3 b2 c3 d4 a4 b1 c1 d1

R1 A B a1 b1 a2 b1 a3 b2 a4 b1

R2 B C D b1 c1 d1 b1 c2 d1 b1 c2 d3 b2 c3 d4

R1 join R2 A B C Da1 b1 c1 d1a1 b1 c2 d1a1 b1 c2 d3a2 b1 c1 d1a2 b1 c2 d1a2 b1 c2 d3a3 b2 c3 d4a4 b1 c1 d1a4 b1 c2 d1a4 b1 c2 d3

{R1, R2} is not a lossless decompostion

of relation R.

The join with respect toR1.B=R2.B is not equal

to R.

Lossless Decomposition

Given a relation R with a set F of functional dependencies, and a decomposition of to (R1, F1) and (R2,F2) such that R1(A1) and R2(A2) is said to be lossless iff either

A1A2 A1 or A1A2 A2

is entailed by F.

Lossless Decompositions

Let F={AB C, CD E, AB E, DE AF}and

R1(A,B,C,D) F1={AB C}R2(C,D,E,F) F2={CD E,DE F}

is this lossless?ABCD CDEF = CDis CD CDEF entailed by F?

Dependency preservation

• A decomposition of relation (R,F) into (R1(A1), F1) and (R2(A2), F2) is said to be dependency preserving iff F1F2 is equivalent to F.

• To check whether F1F2 is equivalent to F, we need to check

1. Whether all functional dependencies in F1F2 are entailed by F, and

2. Whether all functional dependencies in F are entailed by F1F2 .


• Given a decomposition R1(A1) of a relation R with functional dependency set F, the only dependencies that can be preserved in R1 are all dependencies in F+ of the form B C such that BC A1.

Let F={AB C, CD E, AB E, DE AF}Find all the functional dependencies that can be

preserved in:R1(A,B,C,D) R2(C,D,E,F)


• When a decomposition is lossy, then information connecting tuples is lost. We get errorneous information.

• When dependencies are lost in a decomposition, they cannot be enforced as table constraints. They have to be enforced as additional constraints.

• It is vital that decompositions are lossless. It is important but not vital that decompositions are dependency preserving.

Normal Forms

• If a relation is not in BNCF normal form, then it can be decomposed by lossless decompositions into smaller relations that are in BCNF.

• BCNF decomposition:– Until all relations are in BCNF

• Find a dependency X Y in R(A) that violated BCNF• Replace R, with R1(XY), and R2(A-Y)X

• This algorithm may cause many dependencies to be lost.

3NF Conversion

• Given R(A,B,C,D,E,F) and F= { ABCD, ABCE, BDE, EC, EF}• Keys: ABC and ABE. • Not in BNCF (violations: BDE, EC, EF), • Not in 3NF (violations: EF).• Convert to 3NF using the algorithm:

– First compact functional dependencies with common left side, to get F= { ABCDE, BDE, ECF}

– Create relations R1(A,B,C,D,E), R2(B,D,E), R3(E,C,F)– Since there exists relations that contain ABC, and ABE, we are done.– Incidentally, all relations are also in BCNF.

Multivalued Dependencies (MVDs)

• Let R be a relation schema and let R and R. The multivalued dependency holds on R if in any legal relation r(R), for all pairs for tuples t1

and t2 in r such that t1[] = t2 [], there exist tuples t3 and t4 in r such that: t1[] = t2 [] = t3 [] = t4 [] t3[] = t1 [] t3[R – ] = t2[R – ] t4 [] = t2[] t4[R – ] = t1[R – ]

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 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)

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 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

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?

MVD (Cont.)

• Tabular representation of

X ->> Y is trivial if

(a) Y X or

(b) Y U X = R

Multivalued Dependencies

• There are database schemas in BCNF that do not seem to be sufficiently normalized

• Consider a database classes(course, teacher, book)such that (c,t,b) classes means that t is qualified to teach c, and b is a required textbook for c

• The database is supposed to list for each course the set of teachers any one of which can be the course’s instructor, and the set of books, all of which are required for the course (no matter who teaches it).

• There are no non-trivial functional dependencies and therefore the relation is in BCNF

• Insertion anomalies – i.e., if Sara is a new teacher that can teach database, two tuples need to be inserted

(database, Sara, DB Concepts)(database, Sara, Ullman)

course teacher book

databasedatabasedatabasedatabasedatabasedatabaseoperating systemsoperating systemsoperating systemsoperating systems

AviAviHankHankSudarshanSudarshanAviAvi Jim Jim

DB ConceptsUllmanDB ConceptsUllmanDB ConceptsUllmanOS ConceptsShawOS ConceptsShaw

classes


• Therefore, it is better to decompose classes into:

course teacher

databasedatabasedatabaseoperating systemsoperating systems

AviHankSudarshanAvi Jim

teaches

course book

databasedatabaseoperating systemsoperating systems

DB ConceptsUllmanOS ConceptsShaw

text

We shall see that these two relations are in Fourth Normal Form (4NF)


Multivalued Dependencies (MVDs)

• Let R be a relation schema and let R and R. The multivalued dependency holds on R if in any legal relation r(R), for all pairs for tuples t1

and t2 in r such that t1[] = t2 [], there exist tuples t3 and t4 in r such that: t1[] = t2 [] = t3 [] = t4 [] t3[] = t1 [] t3[R – ] = t2[R – ] t4 [] = t2[] t4[R – ] = t1[R – ]

MVD (Cont.)

• Tabular representation of

4th Normal Form

No multi-valued dependencies

4th Normal FormNote: 4th Normal Form violations occur

when a triple (or higher) concatenated key represents a pair of double keys

4th Normal Form

4th Normal FormMultuvalued dependencies

Instructor Book Class

Price Inro Comp MIS 2003

Parker Intro Comp MIS 2003

Kemp Data in Action MIS 4533

Kemp ORACLE Tricks MIS 4533

Warner Data in Action MIS 4533

Warner ORACLE Tricks MIS 4533

4th Normal FormINSTR-BOOK-COURSE(InstrID, Book,

CourseID)

COURSE-BOOK(CourseID, Book)COURSE-INSTR(CourseID, InstrID)

4NF(No multivalued dependencies)

TABLE TABLE

TABLETABLE TABLE

TABLE

Independent repeating groups have been treated as a complex relationship.

Example• Let R be a relation schema with a set of attributes that

are partitioned into 3 nonempty subsets.Y, Z, W

• We say that Y Z (Y multidetermines Z)if and only if for all possible relations r(R)< y1, z1, w1 > r and < y2, z2, w2 > rthen< y1, z1, w2 > r and < y2, z2, w1 > r

• Note that since the behavior of Z and W are identical it follows that Y Z if Y W

Theory of MVDs• From the definition of multivalued dependency, we can derive the following rule:– If , then

That is, every functional dependency is also a multivalued dependency

• The closure D+ of D is the set of all functional and multivalued dependencies logically implied by D. – We can compute D+ from D, using the formal definitions of functional

dependencies and multivalued dependencies.– We can manage with such reasoning for very simple multivalued

dependencies, which seem to be most common in practice– For complex dependencies, it is better to reason about sets of

dependencies using a system of inference rules

Fourth Normal Form• A relation schema R is in 4NF with respect

to a set D of functional and multivalued dependencies if for all multivalued dependencies in D+ of the form , where R and R, at least one of the following hold: is trivial (i.e., or = R) is a superkey for schema R

• If a relation is in 4NF it is in BCNF

Restriction of Multivalued Dependencies

• The restriction of D to Ri is the set Di consisting of– All functional dependencies in D+ that include

only attributes of Ri

– All multivalued dependencies of the form ( Ri)

where Ri and is in D+

4NF Decomposition Algorithm result: = {R};

done := false;compute D+;Let Di denote the restriction of D+ to Ri

while (not done) if (there is a schema Ri in result that is not in 4NF) then begin let be a nontrivial multivalued dependency that holds on Ri such that Ri is not in Di, and ; result := (result - Ri) (Ri - ) (, ); end else done:= true;

Note: each Ri is in 4NF, and decomposition is lossless-join

Example• R =(A, B, C, G, H, I)

F ={ A BB HICG H }

• R is not in 4NF since A B and A is not a superkey for R• Decomposition

a) R1 = (A, B) (R1 is in 4NF)b) R2 = (A, C, G, H, I) (R2 is not in 4NF)c) R3 = (C, G, H) (R3 is in 4NF)d) R4 = (A, C, G, I) (R4 is not in 4NF)

• Since A B and B HI, A HI, A Ie) R5 = (A, I) (R5 is in 4NF)f)R6 = (A, C, G) (R6 is in 4NF)

lecture 17 4 th normal form prof. sin-min lee department of computer science

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