Relational Database Design Management Databaseptw/teaching/DBM/normal-forms.pdf · Normalisation...

DatabaseManagement

Peter Wood

RelationalDatabase DesignUpdate Anomalies

Data Redundancy

Normal FormsFD Inference

Boyce-Codd Normal Form

Third Normal Form

NormalisationAlgorithmsLossless Join

BCNF Algorithm

BCNF Examples

Dependency Preservation

3NF Algorithm

Relational Database Design

There are two interconnected problems which are causedby bad database design:

I Redundancy problemsI Update anomalies

Good database design is based on using certain normalforms for relation schemas.

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

Example 1

Let F1 = {E → D,D → M,M → D}.E stands for ENAME, D stands for DNAME and M standsfor MNAME

A relation r1 over EMP1 (whose schema is {ENAME,DNAME, MNAME}):

ENAME DNAME MNAMEMark Computing Peter

Angela Computing PeterGraham Computing Peter

Paul Math DonaldGeorge Math Donald

E is the only key for EMP1 w.r.t. F1.

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

Problems with EMP1 and F1

1. We cannot represent a department and managerwithout any employees (i.e., we cannot insert a tuplewith a null ENAME because of entity integrity);such a problem is called an insertion anomaly.

2. For the same reason as (1), we cannot delete all theemployees in a department and keep just thedepartment information;such a problem is called a deletion anomaly.

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

More problems with EMP1 and F1

3. E.g. in the first tuple, modifying “Peter” to “Philip” or“Computing” to “Math”, does not violate any FDresulting from a key but D → M would be violated(D is not a key for EMP1 w.r.t. F1) ;such a problem is called a modification anomaly.

? In (3) it is not sufficient to check that r1 satisfies the FDsresulting from the keys of EMP1 w.r.t. F1.

? Ideally, we would like all the FDs of a relation schema tobe inferred from key dependencies, i.e. FDs of the formK → schema(R), where K is a key for R w.r.t. F .

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

Final problem with EMP1 and F1

4. There is redundancy in r1, i.e. for every employee ina given department MNAME is repeated.

? “Peter” appears three times for “Computing” and“Donald” twice for “Math”.

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

Example 2

Let F2 = {E → S}.E stands for ENAME, S stands for SAL and C stands forCNAME.

A relation r2 over EMP2 (whose schema is {ENAME,CNAME, SAL}):

ENAME CNAME SALJack Jill 25Jack Jake 25Jack John 25

Donald Dan 30Donald David 30

EC is the only key for EMP2 w.r.t. F2.

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

Problems with EMP2 and F2

1. Insertion anomaly: we cannot insert an employeewithout any children.

2. Deletion anomaly: if there is a mistake and “Donald”does not have any children, we cannot record thisfact by deleting the two tuples for “Donald”.

3. Modification anomaly: if we try to modify the salaryof “Jack” in the first tuple to be 27 instead of 25,since no FD resulting from a key will be violated, butE → S would be violated.

4. Redundancy: the salary of each employee isrepeated for every child.

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

Formalising Redundancy Problems

Let R be a relation schema and F be a set of FDs over R.

Definition. R has a redundancy problem if(1) there exists a relation r over R that satisfies F , and(2) there exists an FD X → A in F and two distinct tuplesin r that have equal XA values.

• It can be shown that redundancy problems, give rise toupdate anomalies and vice versa.

? Verify that the schemas of Examples, 1 and 2 haveredundancy problems.

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

Problem for you to work on

Consider the following relation over schema Films:

Title Year Genre StarNameStar Wars 1977 SciFi Carrie FisherStar Wars 1977 SciFi Harrison Ford

Raiders . . . 1981 Action Harrison FordRaiders . . . 1981 Adventure Harrison Ford

When Harry . . . 1989 Comedy Carrie Fisher

Assume that the only FD that holds on Films isTitle→ Year.

What is the only key for Films?

Give an example of1. an insertion anomaly2. a deletion anomaly3. a modification anomaly4. a redundancy problem

for Films.

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

Normal Forms

I We assume that we are given a (1NF) relationschema R and a set F of functional dependencies(FDs) over R.

I We define two normal forms for relation schemas:I Boyce-Codd Normal Form (BCNF)I Third Normal Form (3NF)

I BCNF guarantees that the relation schema has noredundancy problems

I BCNF is stronger than 3NF: If R is in BCNF, then Ris in 3NF

I 3NF, however, does sometimes have someadvantages (see later)

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

Rules of inference for FDs

Given a set F of FDs, other FDs can be derived fromthose in F .

For example, if F contains E → D and D → M, thenE → M can be derived from F (transitivity).

An FD X → Y is trivial if Y ⊆ X ; otherwise it isnontrivial.

There are 3 rules of inference for FDs, known asArmstrong’s Axioms:

1. Reflexivity. If Y ⊆ X , then X → Y (trivial FDs).2. Augmentation. If X → Y , then XA→ YA for any

attribute A not in X or Y .3. Transitivity. If X → Y and Y → Z , then X → Z .

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

Closure of a set of FDs

If X → Y can be derived for a set of FDs F , we write thisas F ` X → Y .

Given F , the closure of F , denoted by F+, is the set of allFDs that can be derived (or proven) from F . That is,

F+ = {X → Y | F ` X → Y}.

The closure of a set of attributes, CLOSURE(X ,F ),effectively uses Armstrong’s Axioms to find all attributesdetermined by X .

From CLOSURE(X ,F ) one can find all FDs in F+ thathave X on the lefthand side.

For example, if CLOSURE(HR,F ) = HRCT , then F+

contains HR → C, HR → T , . . .

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm


Assume we are given a set F of FDs, along with itsclosure F+.

Definition. R is in Boyce-Codd Normal Form (BCNF)w.r.t. F if for every non-trivial FD X → Y in F+, X is asuperkey for R w.r.t. F .

Example 1

Let schema(R1) = {STUDENT, POSITION, SUBJECT};S stands for STUDENT, J stands for SUBJECT and Pstands for POSITION.

Let F1 = {SJ → P,PJ → S}.

• Is R1 in BCNF w.r.t. F1 ?

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

Example 2

Let schema(R2) = {STREET, CITY, POSTCODE};S stands for STREET,C stands for CITY andP stands for POSTCODE.

Let F2 = {SC → P,P → C}.

• Is R2 in BCNF w.r.t. F2 ?

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

Third Normal Form

Definition. An attribute A in schema(R) is said to beprime w.r.t. F if A is a member of one of the keys of Rw.r.t. F .

Definition. R is in Third Normal Form (3NF) w.r.t. F if forevery non-trivial FD X → A in F+ either X is a superkeyfor R w.r.t. F or A is prime.

So 3NF is weaker than BCNF

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

Examples

I Example 1, with R1 = {S, J,P} andF1 = {SJ → P,PJ → S}, is in BCNF

I Therefore R is in 3NF

I What about Example 2, with R2 = {S,C,P} andF2 = {SC → P,P → C}, which was not in BCNF?

I What are the keys and prime attributes of R2 w.r.t.F2?

I Is R2 in 3NF w.r.t. F2 ?

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

Examples

I Example 1, with R1 = {S, J,P} andF1 = {SJ → P,PJ → S}, is in BCNF

I Therefore R is in 3NFI What about Example 2, with R2 = {S,C,P} and

F2 = {SC → P,P → C}, which was not in BCNF?I What are the keys and prime attributes of R2 w.r.t.

F2?I Is R2 in 3NF w.r.t. F2 ?

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

BCNF/3NF Example 3

Let R3 be a relation schema, with schema(R3) ={ENAME, DNAME, MNAME};E stands for ENAME, D stands for DNAME and M standsfor MNAME.

Let F3 = {E → D,D → M}.

I Is R3 in BCNF w.r.t. F3?I What are the keys and prime attributes of R3 w.r.t.

F3?I Is R3 in 3NF w.r.t. F3?

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

BCNF/3NF Example 4

Let R4 be a relation schema, with schema(R4) ={ENAME, CNAME, SAL};E stands for ENAME, C stands for CNAME and S standsfor SAL.

Let F4 = {E → S}.

I Is R4 in BCNF w.r.t. F4?I What are the keys and prime attributes of R4 w.r.t.

F4?I Is R4 in 3NF w.r.t. F4?

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

Problem for you to work on

Consider relation schema R where schema(R) = ABCD.

Let the set F of FDs which hold on R be{AB → C,C → D,D → A}.

1. What are all the keys of R? (done earlier)2. Which FDs violate BCNF?3. Which FDs violate 3NF?

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

BCNF Normalisation Algorithm

In practice, we are given:I an entity-relationship diagram (ERD) andI a set of functional dependencies (FDs) F .

To produce a database design, we1. Convert the ERD into a database schema S.2. If any of the relation schemas in S are not in BCNF

with respect to F , we decompose them.

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

Decomposing a relation schema

Given a relation schema R which is not in BCNF, wedecompose it by

I replacing it by two (smaller) relation schemas R1 andR2

I such that R = R1 ∪ R2.

For example, given R = {E ,C,S} (employee, child,salary) and F = {E → S}, we might decompose R into

I R1 = {E ,C} andI R2 = {E ,S}

How do we decide which attributes go into whichdecomposed relation schemas?

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

Lossless join

What if we choose a different decomposition for ourexample?schema(R) = {ENAME, CNAME, SAL} andsingle FD: ENAME→ SAL

(Modified) relation r over R is given by

ENAME CNAME SALJack Diane 25Jack John 25

Donald Diane 30Donald David 30

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

If we decompose R into {ENAME,CNAME} and{CNAME,SAL} as follows:

ENAME CNAMEJack DianeJack John

Donald DianeDonald David

CNAME SALDiane 25John 25Diane 30David 30

and then perform the natural join, we get

ENAME CNAME SALJack Diane 25Jack Diane 30Jack John 25

Donald Diane 25Donald Diane 30Donald David 30

⇒ with two tuples that were not in the original relation

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

I A decomposition such as that into {ENAME,CNAME}and {CNAME,SAL} is called lossy

I We started knowing Jack’s salary was 25I After decomposing, if we query Jack’s salary we get

both 25 and 30I The decomposition does not faithfully represent the

original information we had

I A decomposition which does faithfully represent theoriginal information is called lossless

I Losslessness is guaranteed if we ensure that thecommon attributes between a pair of decomposedrelation schemas is a key for one of them

I The BCNF algorithm ensures losslessdecompositions

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

I A decomposition such as that into {ENAME,CNAME}and {CNAME,SAL} is called lossy

I We started knowing Jack’s salary was 25I After decomposing, if we query Jack’s salary we get

both 25 and 30I The decomposition does not faithfully represent the

original information we hadI A decomposition which does faithfully represent the

original information is called losslessI Losslessness is guaranteed if we ensure that the

common attributes between a pair of decomposedrelation schemas is a key for one of them

I The BCNF algorithm ensures losslessdecompositions

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

Decomposition Condition used by Algorithm

The FD ENAME→ SAL violates BCNF in schema R.

I ENAME is the left-hand side of the violating FDI SAL is the right-hand side (RHS) of the violating FD

Split {ENAME, CNAME, SAL} into two relation schemas:

1. R1 = EMPLOYEE, containing all the attributes in theviolating FD, i.e.,schema(EMPLOYEE) = { ENAME, SAL }, andF1 = { ENAME→ SAL }.

2. R2 = DEPENDENT, containing all attributes in Rexcept those on the RHS of the violating FD, i.e.,schema(DEPENDENT) = { ENAME, CNAME }, andF2 = ∅ (excluding trivial FDs).

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

Algorithm DECOMPOSE(R, F )

let the output database schema Out be empty;if IS-BCNF(R, F) then

add R to Out;else

let X → A in F+ be nontrivial (i.e. A is not in X)such that X is not a superkey with respect to F;

let R1 have schema(R1) = X ∪ {A};merge DECOMPOSE(R1, F ) and Out;let R2 have schema(R2) = schema(R) − {A};merge DECOMPOSE(R2, F ) and Out;

end ifreturn Out;

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

Algorithm properties

The natural join can be applied to all of the relations inDECOMPOSE(R, F) to recover precisely the informationstored in any relation over schema(R); this is known asthe lossless join property.

Note that, in general, we need to consider F+, theclosure of F , to check whether there are any FDs whichviolate BCNF.

But we can start trying to find violations in F , and onlyconsider F+ once we find no violations in F .

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

Another Example of BCNF DecompositionLet STUD be a relation schema, with schema(STUD) ={SNUM, POSTCODE, CITY, COUNTRY}, with FDs{SNUM→ POSTCODE, POSTCODE→ CITY,CITY→ COUNTRY}

I CITY→ COUNTRY violates BCNF in STUD, sodecompose STUD intoCC, with schema(CC) = {CITY, COUNTRY}, andSTUD1, with schema(STUD1) = {SNUM,POSTCODE, CITY}

I CC is in BCNF while POSTCODE→ CITY violatesBCNF in STUD1, so decompose STUD1 intoPC, with schema(PC) = {POSTCODE, CITY}, andSINFO = {SNUM, POSTCODE}.

I All the relation schemas in the database schema{CC, PC, SINFO} are now in BCNF

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

A Third Example

I Consider a modified relation schema EMP, withattributes ENAME, CNAME (child name), DNAME(department name) and MNAME (manager name).

I The set of FDs is F = {E→ D, D→ M, M→ D}, whereE stands for ENAME, D stands for DNAME and Mstands for MNAME (and C stands for child name).

I All three FDs violate BCNF since EC is the only key.I We can choose any one of them as the basis for the

first decomposition step.I We will consider all three decompositions in turn.

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

Third Example: Decomposition 1

I If we first decompose using D→ M, we get twoschemas with attributes {D, M} and {E, C, D}.

I FDs D→ M and M→ D are applicable to {D, M}, butboth D and M are keys.

I FD E→ D is applicable to {E, C, D} and E is not asuperkey.

I So we decompose {E, C, D} into {E, D} and {E, C}.I E is a key for {E, D} and EC is the key for {E, C}.I So the final database schema comprises

{D, M}, {E, D} and {E, C}.

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm


I If we first decompose using E→ D, we get twoschemas with attributes {E, D} and {E, C, M}.

I E→ D is applicable to {E, D}, but E is a key.I What FDs are applicable to {E, C, M}?I None of E→ D, D→ M or M→ D apply because D is

not in {E, C, M}.I We have to consider all FDs in F+.I Recall that E→ M follows from E→ D and D→ M.I E→ M violates BCNF in {E, C, M} because E is not a

key.I So we decompose {E, C, M} into {E, M} and {E, C}.I So the final database schema comprises

{E, D}, {E, M} and {E, C}.

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm


I If we first decompose using M→ D, we get twoschemas with attributes {M, D} and {E, C, M}.

I FDs D→ M and M→ D are applicable to {M, D}, butboth D and M are keys.

I Once again we have {E, C, M}, so it is decomposedas before into {E, M} and {E, C}.

I So the final database schema comprises{M, D}, {E, M} and {E, C}.

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

An example for you to tryLet R be a relation schema, with schema(R) ={C,T,H,R,S,G}.

I C stands for a course,I T stands for a teacher,I H stands for hour,I R stands for room,I S stands for student andI G stands for grade.

An example set of FDs F over R :

1. C→ T,2. HR→ C,3. HT→ R,4. CS→ G and5. HS→ R.

Decompose R into BCNF.

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm


Recall example: F3 = {SC→ P, P→ C}.S stands for Street, C stands for City and P stands forPostcode.

{S,C,P} is not in BCNF

Decompose {S,C,P} into {P,C} and {P,S}

P Cp1 cp2 c

P Sp1 sp2 s

Only FD that can be tested in the decomposition is P→ C

When we join the two relations, we see that SC→ P isviolated.

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm


A decomposition is dependency preserving if the FDswhich hold on the original relation schema can be testedon the decomposed schemas, without using joins.

We cannot always find a BCNF decomposition that isdependency preserving.

To test that no FDs are violated, we may need to joinrelations (expensive).

We can always find a 3NF dependency-preservingdecomposition.

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm


For a starting set of attributes and FDs, some BCNFdecompositions may be dependency preserving andsome not.

Consider the example with attributes { E, C, D, M } andFDs F = {E→ D, D→ M, M→ D}.

We had three possible decompositions1. {D, M}, {E, D} and {E, C}.2. {E, D}, {E, M} and {E, C}.3. {M, D}, {E, M} and {E, C}.

Which of them is dependency-preserving?

DatabaseManagement

Peter Wood


Data Redundancy



Third Normal Form


BCNF Algorithm

BCNF Examples


3NF Algorithm

3NF Algorithm

Given a relation schema R and a set of FDs F , thefollowing steps produce a 3NF decomposition of R thatsatisfies the lossless join condition and is dependencypreserving:

1. Remove all redundancies from F (we haven’tcovered this).

2. For each FD X → A in F , use X ∪ {A} as theschema of one of the relations in the decomposition.

3. If none of the schemas from Step 2 includes asuperkey for R, add another relation schema that is akey for R.

4. Delete any of the schemas from Step 2 that iscontained in another.

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