3NF and Boyce-Codd Normal Form

Prof. Sin-Min Lee

Department of Computer Science

San Jose State University

What it’s all about• Given a relation, R, and a set of functional

dependencies, F, on R. • Assume that R is not in a desirable form for enforcing

F.

• Decompose relation R into relations, R1,..., Rk, with associated functional dependencies, F1,..., Fk, such that R1,..., Rk are in a more desirable form, 3NF or BCNF.

• While decomposing R, make sure to preserve the dependencies, and make sure not to lose information.

Primitive DomainsFLT-SCHEDULE

flt# weekday airline dtime from atime to

DL242 MO WE FR DELTA 10:40 ATL 12:30 BOSSK912 SA SU SAS 12:00 CPH 15:30 JFKAA242 MO FR AA 08:00 CHI 10:10 ATL

Attributes must be defined over domains with atomic values

FLT-SCHEDULE

flt# weekday airline dtime from atime to

DL242 MO DELTA 10:40 ATL 12:30 BOS

SK912 SA SAS 12:00 CPH 15:30 JFK

AA242 MO AA 08:00 CHI 10:10 ATL

DL242 WE DELTA 10:40 ATL 12:30 BOSDL242 FR DELTA 10:40 ATL 12:30 BOS

SK912 SU SAS 12:00 CPH 15:30 JFK

AA242 FR AA 08:00 CHI 10:10 ATL

Bad Database Design- redundancy of fact

FLIGHTS

flt# date airline plane#DL242 10/23/00 Delta k-yo-33297

DL242 10/24/00 Delta t-up-73356

DL242 10/25/00 Delta o-ge-98722

AA121 10/24/00 American p-rw-84663

AA121 10/25/00 American q-yg-98237

AA411 10/22/00 American h-fe-65748

• redundancy: airline name repeated for same flight• inconsistency: when airline name for a flight

changes, it must be changed many places

Bad Database Design- fact clutter

• insertion anomalies: how do we represent that SK912 is flown by Scandinavian without there being a date and a plane assigned?

• deletion anomalies: cancelling AA411 on 10/22/00 makes us lose that it is flown by American.

• update anomalies: if DL242 is flown by Sabena, we must change it everywhere.

FLIGHTS

flt# date airline plane#DL242 10/23/00 Delta k-yo-33297

DL242 10/24/00 Delta t-up-73356

DL242 10/25/00 Delta o-ge-98722

AA121 10/24/00 American p-rw-84663

AA121 10/25/00 American q-yg-98237

AA411 10/22/00 American h-fe-65748

Bad Database Design- information loss

FLIGHTS

flt# date airline plane#

DL242 10/23/00 Delta k-yo-33297

DL242 10/24/00 Delta t-up-73356

DL242 10/25/00 Delta o-ge-98722

AA121 10/24/00 American p-rw-84663

AA121 10/25/00 American q-yg-98237

AA411 10/22/00 American h-fe-65748FLIGHTS-AIRLINE

flt# airlineDL242 Delta

AA121 American

AA411 American

DATE-AIRLINE-PLANE

date airline plane#10/23/00 Delta k-yo-33297

10/24/00 Delta t-up-73356

10/25/00 Delta o-ge-98722

10/24/00 American p-rw-84663

10/25/00 American q-yg-98237

10/22/00 American h-fe-65748

Bad Database Design- information loss

FLIGHTS

flt# date airline plane#DL242 10/23/00 Delta k-yo-33297

DL242 10/24/00 Delta t-up-73356

DL242 10/25/00 Delta o-ge-98722

AA121 10/24/00 American p-rw-84663

AA121 10/25/00 American q-yg-98237

AA211 10/22/00 American h-fe-65748

AA411 10/24/00 American p-rw-84663

AA411 10/25/00 American q-yg-98237

AA411 10/22/00 American h-fe-65748

DATE-AIRLINE-PLANE

date airline plane#10/23/00 Delta k-yo-33297

10/24/00 Delta t-up-73356

10/25/00 Delta o-ge-98722

10/24/00 American p-rw-84663

10/25/00 American q-yg-98237

10/22/00 American h-fe-65748

FLIGHTS-AIRLINE

flt# airlineDL242 Delta

AA121 American

AA411 American

• information loss: we polluted the database with false facts; we can’t find the true facts.

Bad Database Design- dependency loss

DATE-AIRLINE-PLANE

date airline plane#10/23/00 Delta k-yo-33297

10/24/00 Delta t-up-73356

10/25/00 Delta o-ge-98722

10/24/00 American p-rw-84663

10/25/00 American q-yg-98237

10/22/00 American h-fe-65748

FLIGHTS-AIRLINE

flt# airlineDL242 Delta

AA121 American

AA411 American

• dependency loss: we lost the fact that (flt#, date) plane#

Good Database Design

• no redundancy of FACT (!)• no inconsistency• no insertion, deletion or update anomalies• no information loss• no dependency loss

FLIGHTS-DATE-PLANE

flt# date plane#DL242 10/23/00 k-yo-33297

DL242 10/24/00 t-up-73356

DL242 10/25/00 o-ge-98722

AA121 10/24/00 p-rw-84663

AA121 10/25/00 q-yg-98237

AA411 10/22/00 h-fe-65748

FLIGHTS-AIRLINE

flt# airlineDL242 Delta

AA121 American

AA411 American

Let X and Y be sets of attributes in R

• Y is functionally dependent on X in R iff for each x R.X there is precisely one y R.Y

• Y is fully functional dependent on X in R if Y is functional dependent on X and Y is not functional dependent on any proper subset of X

• We use keys to enforce functional dependencies in relations:

X Y

X Y

Functional Dependencies and Keys

FLIGHTS

flt# date airline plane#

FLIGHTS

flt# date airline plane#

FLIGHTS

flt# date airline plane#

Functional Dependencies and Keys

plane# is not determined by flt# alone

airline is not determined by flt# and date

the FLIGHT relation will not allow the FDs to be enforced by keys

Functional Dependencies and Keys

real world database

name

address

cust# name address

cust# name address

Consider the meaning

cust# name address

cust# name address

cust# name address combined

separate

How to Compute Meaning- Armstrong’s inference rules

Rules of the computation:– reflexivity: if YX, then XY– Augmentation: if XY, then WXWY– Transitivity: if XY and YZ, then XZ

Derived rules:– Union: if XY and XZ, the XYZ– Decomposition: if XYZ, then XY and XZ– Pseudotransitivity: if XY and WYZ, then XWZ

Armstrong’s Axioms:– sound– complete

Overview of NFsNF2

1NF

2NF

3NF

BCNF

Normal Forms- definitions

• NF: non-first normal form• 1NF: R is in 1NF. iff all domain values are

atomic2• 2NF: R is in 2. NF. iff R is in 1NF and every

nonkey attribute is fully dependent on the key• 3NF: R is in 3NF iff R is 2NF and every nonkey

attribute is non-transitively dependent on the key• BCNF: R is in BCNF iff every determinant is a

candidate key

• Determinant: an attribute on which some other attribute is fully functionally dependent.

Example of Normalization

flt# date plane# airline from to milesFLT-INSTANCE

flt#date

plane#

airlinefrom

to

miles

flt#date

plane#

airlinefrom

to

miles

flt#date

plane# flt#airlinefrom

to

miles

fromto

miles

flt#airlinefrom

toflt#

dateplane#

Example of Normalization1NF:

3NF & BCNF:

2NF:

3NF that is not BCNF

A

B C

Candidate keys: {A,B} and {A,C} Determinants: {A,B} and {C}

A decomposition:

Lossless, but not dependency preserving!

A B CR

C BR1

A CR2

• When a relation has more than one candidate key, anomalies may result even though the relation is in 3NF.

• 3NF does not deal satisfactorily with the case of a relation with overlapping candidate keys– i.e. composite candidate keys with at least one attribute in

common.

• BCNF is based on the concept of a determinant.– A determinant is any attribute (simple or composite) on which

some other attribute is fully functionally dependent.

• A relation is in BCNF is, and only if, every determinant is a candidate key.

The theory• Consider the following relation and determinants.

Example 1. Given R(a,b,c,d)a,c -> b,da,d -> b

• To be in BCNF, all valid determinants must be a candidate key. In the relation R, a,c->b,d is the determinate used, so the first determinate is fine.

• Example 2. If {a, b} is not a key, a,d->b suggests that a,d can be the primary key, which would determine b. However this would not determine c. This is not a candidate key, and thus R is not in BCNF.

Example 1

Patient No

Patient Name Appointment Id Time Doctor

1 John 0 09:00 Zorro

2 Kerr 0 09:00 Killer

3 Adam 1 10:00 Zorro

4 Robert 0 13:00 Killer

5 Zane 1 14:00 Zorro

Two possible keys

• DB(Patno,PatName,appNo,time,doctor)• Determinants:

– Patno -> PatName– Patno,appNo -> Time,doctor– Time -> appNo

• Two options for 1NF primary key selection:– DB(Patno,PatName,appNo,time,doctor) (example 1a)– DB(Patno,PatName,appNo,time,doctor) (example 1b)

Example 1a

• DB(Patno,PatName,appNo,time,doctor)

• No repeating groups, so in 1NF

• 2NF – eliminate partial key dependencies:– DB(Patno,appNo,time,doctor)– R1(Patno,PatName)

• 3NF – no transient dependences so in 3NF

• Now try BCNF.

BCNF Every determinant is a candidate key

DB(Patno,appNo,time,doctor)R1(Patno,PatName)

• Is determinant a candidate key?– Patno -> PatName

Patno is present in DB, but not PatName, so irrelevant.

Continued…

DB(Patno,appNo,time,doctor)R1(Patno,PatName)

– Patno,appNo -> Time,doctorAll LHS and RHS present so relevant. Is this a candidate key? Patno,appNo IS the key, so this is a candidate key.

– Time -> appNoTime is present, and so is appNo, so relevant. Is this a candidate key? If it was then we could rewrite DB as: DB(Patno,appNo,time,doctor)This will not work, so not BCNF.

Rewrite to BCNF

• DB(Patno,appNo,time,doctor)R1(Patno,PatName)

• BCNF: rewrite to DB(Patno,time,doctor) R1(Patno,PatName) R2(time,appNo)

• time is enough to work out the appointment number of a patient. Now BCNF is satisfied, and the final relations shown are in BCNF

Example 1b

• DB(Patno,PatName,appNo,time,doctor) • No repeating groups, so in 1NF• 2NF – eliminate partial key dependencies:

– DB(Patno,time,doctor)– R1(Patno,PatName) – R2(time,appNo)

• 3NF – no transient dependences so in 3NF• Now try BCNF.

BCNF Every determinant is a candidate key

DB(Patno,time,doctor)R1(Patno,PatName) R2(time,appNo)

• Is determinant a candidate key?– Patno -> PatName

Patno is present in DB, but not PatName, irrelevant.– Patno,appNo -> Time,doctor

Not all LHS present so not relevant– Time -> appNo

Time is present, but not appNo, so not relevant. – Relations are in BCNF.

Summary - Example 1

This example has demonstrated three things:

• BCNF is stronger than 3NF, relations that are in 3NF are not necessarily inBCNF

• BCNF is needed in certain situations to obtain full understanding of the data model

• there are several routes to take to arrive at the same set of relations in BCNF.– Unfortunately there are no rules as to which route will

be the easiest one to take.

Example 2

Grade_report(StudNo,StudName,(Major,Adviser,(CourseNo,Ctitle,InstrucName,InstructLocn,Grade)))

• Functional dependencies– StudNo -> StudName– CourseNo -> Ctitle,InstrucName– InstrucName -> InstrucLocn– StudNo,CourseNo,Major -> Grade– StudNo,Major -> Advisor– Advisor -> Major

Example 2 cont...

• UnnormalisedGrade_report(StudNo,StudName,(Major,Advisor, (CourseNo,Ctitle,InstrucName,InstructLocn,Grade)))

• 1NF Remove repeating groups– Student(StudNo,StudName)– StudMajor(StudNo,Major,Advisor)– StudCourse(StudNo,Major,CourseNo,

Ctitle,InstrucName,InstructLocn,Grade)

Example 2 cont...

• 1NFStudent(StudNo,StudName)StudMajor(StudNo,Major,Advisor)StudCourse(StudNo,Major,CourseNo, Ctitle,InstrucName,InstructLocn,Grade)

• 2NF Remove partial key dependenciesStudent(StudNo,StudName)StudMajor(StudNo,Major,Advisor)StudCourse(StudNo,Major,CourseNo,Grade)Course(CourseNo,Ctitle,InstrucName,InstructLocn)

Example 2 cont...• 2NF

Student(StudNo,StudName)StudMajor(StudNo,Major,Advisor)StudCourse(StudNo,Major,CourseNo,Grade)Course(CourseNo,Ctitle,InstrucName,InstructLocn)

• 3NF Remove transitive dependenciesStudent(StudNo,StudName)StudMajor(StudNo,Major,Advisor)StudCourse(StudNo,Major,CourseNo,Grade)Course(CourseNo,Ctitle,InstrucName)Instructor(InstructName,InstructLocn)

Example 2 cont...

• BCNF Every determinant is a candidate key– Student : only determinant is StudNo– StudCourse: only determinant is StudNo,Major– Course: only determinant is CourseNo– Instructor: only determinant is InstrucName– StudMajor: the determinants are

• StudNo,Major, or• AdvisorOnly StudNo,Major is a candidate key.

Example 2: BCNF

• BCNF

Student(StudNo,StudName)StudCourse(StudNo,Major,CourseNo,Grade)Course(CourseNo,Ctitle,InstrucName)Instructor(InstructName,InstructLocn)StudMajor(StudNo,Advisor)Adviser(Adviser,Major)

Problems BCNF overcomes

• If the record for student 456 is deleted we lose not only information on student 456 but also the fact that DARWIN advises in BIOLOGY

• we cannot record the fact that WATSON can advise on COMPUTING until we have a student majoring in COMPUTING to whom we can assign WATSON as an advisor.

STUDENT MAJOR ADVISOR

123 PHYSICS EINSTEIN

123 MUSIC MOZART

456 BIOLOGY DARWIN

789 PHYSICS BOHR

999 PHYSICS EINSTEIN

Split into two tables

In BCNF we have two tables

STUDENT ADVISOR

123 EINSTEIN

123 MOZART

456 DARWIN

789 BOHR

999 EINSTEIN

ADVISOR MAJOR

EINSTEIN PHYSICS

MOZART MUSIC

DARWIN BIOLOGY

BOHR PHYSICS

Returning to the ER Model

• Now that we have reached the end of the normalisation process, you must go back and compare the resulting relations with the original ER model

– You may need to alter it to take account of the changes that have occurred during the normalisation process Your ER diagram should always be a prefect reflection of the model you are going to implement in the database, so keep it up to date!

– The changes required depends on how good the ER model was at first!

Video Library Example

• A video library allows customers to borrow videos.• Assume that there is only 1 of each video.• We are told that:

video(title,director,serial)customer(name,addr,memberno)hire(memberno,serial,date)

title->director,serialserial->titleserial->directorname,addr -> membernomemberno -> name,addrserial,date -> memberno

What NF is this?

• No repeating groups therefore at least 1NF

• 2NF – A Composite key exists:hire(memberno,serial,date)

– Can memberno be found with just serial or date?

– NO, therefore the relations are already in 2NF.

• 3NF?

Test for 3NF

• video(title,director,serial)– title->director,serial– serial->director

• Director can be derived using serial, and serial and director are both non keys, so therefore this is a transitive or non-key dependency.

• Rewrite video…

Rewrite for 3NF

• video(title,director,serial)– title->director,serial– serial->director

• Becomes:• video(title,serial)• serial(serial,director)

Check BCNF

• Is every determinant a candidate key?• video(title,serial) - Determinants are:

– title->director,serial Candidate key– serial->title Candidate key – video in BCNF

• serial(serial,director) Determinants are: – serial->director Candidate key– serial in BCNF

• customer(name,addr,memberno) Determinants are: – name,addr -> memberno Candidate key– memberno -> name,addr Candidate key – customer in BCNF

• hire(memberno,serial,date) Determinants are:– serial,date -> memberno Candidate key– hire in BCNF

• Therefore the relations are also now in BCNF.

R( A B C D) 1 2 3 4 1 2 4 3 1 3 4 1 2 3 2 4 1 2 3 5

A B C D

*

*

*

Q1. For which keys R is 2NF? key {AD} R is 2NF key {BD} R is 2NF key {CD} R is not 2NF

Q2. For which keys R is 3NF? Since prime-attributes are {A, B, C, D} R with key {AD} is 3NF R with key {BD} is 3NF