Tutorial 5

The Hong Kong University of Scienceand Technology

COMP231 Tutorial 5

Functional Dependencies, 3NF

HKUST 2 Database Management Systems

• SuperkeyK R

• Candidate KeyK R

no K’ K, s.t. K’ R (minimal)

• Primary KeyThe candidate key chosen to uniquely identify tuples in a

relation

superkey

candidate key

primary key

Review: Key


• For a set of functional dependencies F, we can get the closure, F+, by applying Armstrong’s Axioms.

• Armstrong’s Axioms:• Reflexivity

If X Y, then X Y

• Augmentation

If X Y, then XZ YZ

• Transitivity

If X Y, Y Z, then X Z

• Derived rules:• Decomposition

If X YZ, then X Y and X Z

• Union

If X Y and X Z, then X YZ

• Pseudo-transitivity

If X Y and WY Z, then WX Z

Review: The Closure of FD


Definition:

X, Y are attributes of a relation R:

X Y is in F+ Y X+

Example:

Given R = (loan_no, amount, branch_name, customer_name)• If loan_no amount

then loan_no+ = {loan_no, amount}

• If we also have loan_no branch_namethen loan_no+ = {loan_no, amount, branch_name}

• If we also have loan_no customer_name

then loan_no+ = {loan_no, amount, branch_name, customer_name}

Algorithm:

•X(0) := X•Repeat

X(i+1) := X(i) Z,where Z is the set of attributes such that there exists YZ in F, and Y X(i)

•Until X(i+1) := X(i)

•Return X(i+1)

Review: The Closure of Attributes


Review: Canonical Cover of FD

Definition:

A canonical cover for F is a set of dependencies Fc such that

• F and Fc are equivalent

• Fc contains no redundancy

• Each left hand side of functional dependency in Fc is unique


Review: Normalization

• Decomposition of a relation R with the following goals• Lossless (necessary)

Information lost?

• Dependency preservation (desirable)

(i Fi)+ = F+ ?

• Good form

1NF, 2NF, 3NF, BCNF

2NF:

R is in 2NF if and only iffor each FD: X {A} in F+

ThenA X (the FD is trivial), ORX is not a proper subset of a candidate key for R, ORA is a prime attribute

3NF:R is in 3NF if and only if

for each FD: X {A} in F+

ThenA X (trivial FD), ORX is a superkey for R, ORA is prime attribute for R

•A primary attribute is an attribute that is part of a candidate key


R = (A, B, C, D, E)

F = {ABC, CDE, BD, EA}

Compute A+ and B+:

A+ := {A}

:= {A, B, C} ABC and {A} A+

:= {A, B, C, D} BD and {B} A+

:= {A, B, C, D, E} CDE and {C, D} A+

ends because A+ stops changing

B+ := {B}

:= {B, D} BD and {B} B+

ends because B+ stops changing

Exercise 1: The Closure of Attributes


R = (A, B, C, D, E)

F = {ABC, CDE, BD, EA}

List all candidate keys of R.

• We have A+ = {A, B, C, D, E} in Exercise 1,

then AABCDE, it is a candidate key of R.

• Since EA,

then EABCDE. (transitivity)

• Since CDE,

then CDABCDE. (transitivity)

• Since BD,

then BCCD, then BCABCDE. (augmentation, transitivity)

So A, E, CD, BC are candidate keys of R.

Exercise 2: Candidate Keys


R = (A, B, C, D, E)

F = {ACE, ACDB, CED, BE}

Find the canonical cover of F.

Algorithm:

Repeat

UnionX1Y1 and X1Y2 replaced with X1Y1Y2

Find an extraneous attribute

If an extraneous attribute is found in XY,

delete it from XY

Until F does not change

Exercise 3: Compute Canonical Cover


R = (A, B, C, D, E)

F = {ACE, ACDB, CED, BE}

Find the canonical cover of F.

First loop:

Union

Fc(1) = {ACE, ACDB, CED, BE}


Consider ACDB:

D is extraneous because ACE and CED

Remove D in ACDB

Fc(1) = {ACE, ACB, CED, BE}

Exercise 3: Compute Canonical Cover (cont)


R = (A, B, C, D, E)

Fc(1) = {ACE, ACB, CED, BE}

Second loop:

Union

Fc(2) = {ACBE, CED, BE}


Consider ACBE:

E is extraneous because BE

Remove E in ACBE

Fc(2) = {ACB, CED, BE}



R = (A, B, C, D, E)


Third loop:

Union



No extraneous attributes found

Ends because Fc stops changing

Fc = {ACB, CED, BE}




• Different order of removing the extraneous attributes may result in different FC

• Example:

R=(A, B, C, D)

FD = {AC, BCA, ABCD}

• In ABCD, A is extraneous or C is extraneous

• If we remove A first, we get Fc = {AC, BCAD}

• If we remove C first, we get Fc = {AC, BCA, ABD}


Exercise 4: Normal forms

• R=(A, B, C, D, E)

• FD = {ABC, CDE, BD, EA}

• Is R in 1NF?• Yes. Relational tables are always in 1NF.

• Is R in 2NF?• We found candidate keys: A, E, CD, BC.

ABC BC are prime attribute

CDE E is a prime attribute

BD D is a prime attribute

EA A is a prime attribute

So R is in 2NF

What is 2NF?

For each FD: X->{A}

1. A X (FD is trivial)

2. X is NOT subset of candidate key

3. A is prime attribute


Exercise 4: Normal forms (cont)

• R=(A, B, C, D, E)

• FD = {ABC, CDE, BD, EA}

• Is R in 3NF?• We found candidate keys: A, E, CD, BC.

ABC A is a candidate key

CDE CD is a candidate key

BD D is a prime attribute

EA E is a candidate key

So R is in 3NF What is 3NF?

For each FD: X->{A}

1. A X (FD is trivial)

2. X is super key

3. A is prime attribute

Different with 2NF



• R=(A, B, C, D, E, F)

• FD = {AB, BCD, CE, BF}

• Is R in 2NF?• Candidate key: AC.

AB A is a proper subset of candidate key AND B is not a prime attribute

BCD BC is not a proper subset of candidate key

CE C is a proper subset of candidate key AND E is not a prime attribute

BF B is not a proper subset of candidate key

AB or CE makes R not in 2NF

What is 2NF?

For each FD: X->{A}

A X (FD is trivial)

X is NOT subset of candidate key

A is prime attribute



• R=(A, B, C, D, E, F)

• FD = {AB, BCD, CE, BF}

• Is R in 3NF?• 3NF 2NF 1NF, R is not in 2NF, so R is not in 3NF either.

• Candidate key: AC.

AB A is not a super-key AND B is not a prime attribute

BCD BC is not a super-key AND D is not a prime attribute

CE C is not a super-key AND E is not a prime attribute

BF B is not a super-key AND F is not a prime attribute

Either one of the FD makes R not in 3NF

What is 3NF?

For each FD: X->{A}

1.

A X (FD is trivial)

2.

X is super key

3.

A is prime attribute


Exercise 5: Decomposition

R = (A, B, C, D, E, F, G, H)

F = {ACG, DEG, BCD, CGBD, ACDB, CEAG}

• A decomposition of R:• Table1: (A, B, C, D)

• Table2: (D, E, G)

• Table3: (A, C, D, F, H)

• Is it lossless?Yes

A decomposition of R into R1 and R2 is lossless if and only if the common attributes of R1 and R2 is a candidate key for R1 or R2

• (Table1 Table3) Table2 = D (candidate key of Table2)

• Table1 Table3 = ACD (candidate key of Table1)

R

Table1Table3 Table2

Table1 Table3


Exercise 5: Decomposition (cont)

R = (A, B, C, D, E, F, G, H)


• A decomposition of R:• Table1: (A, B, C, D)

• Table2: (D, E, G)

• Table3: (A, C, D, F, H)

• Is it dependency preserving?

• No (CGBD is lost)


Exercise 6: 3NF

R = (A, B, C, D, E, F, G, H)


Decompose R into 3NF

• Algorithm:

Compute Fc of F

S :=

For each FD XY in Fc:

S := S (X,Y)

If no scheme contains a candidate key for R:

Choose any candidate key CN

S := S table with attributes in CN


Exercise 6: 3NF (cont)

R = (A, B, C, D, E, F, G, H)


• In Exercise 3, we get

Fc = {ACB, DEG, BCD, CGD, CEA}

• For each FD, we generate a table:

Table1: (A, B, C)

Table2: (D, E, G)

Table3: (B, C, D)

Table4: (C, D, G)

Table5: (A, C, E)


R = (A, B, C, D, E, F, G, H)

Fc = {ACB, DEG, BCD, CGD, CEA}

• Any table in S contains a candidate key for R?

No

• ACFH is a candidate key:

Generate Table6: (A, C, F, H)

• So a possible 3NF decomposition of R is:

(A, B, C), (D, E, G), (B, C, D), (C, D, G), (A, C, E), (A, C, F, H)

• Lossless?

Yes

• Dependency preserving?

Yes

Exercise 6: 3NF (cont)

R

ACBDEG

ACFH

CEBDG

ACE

BCDG

DGE

BCDCDG

Tutorial 5

Documents

x p y augmentation

x p z union

trivial x

prime attribute d

x trivial fd

p branch

p customer

candidate keys of r