Lecture No 13Functional Dependencies &

Normalization ( II )Mar 3rd 2011

Database SystemsDatabase Systems

Functional Dependency

Formal Definition: Attribute B is functionally dependant upon attribute A (or a collection of attributes) if a value of A determines a single value of attribute B at any one time.

Formal NotationFormal Notation: : A A B B This should be read as This should be read as ‘A ‘A determines B’determines B’ or or ‘B is functionally dependant on A’‘B is functionally dependant on A’. A . A is called the is called the determinantdeterminant and B is called the and B is called the object of the object of the determinantdeterminant..

staffNo job dept dname SL10 Salesman 10 SalesSA51 Manager 20 AccountsDS40 Clerk 20 AccountsOS45 Clerk 30 Operations

Example:

staffNo jobstaffNo deptstaffNo dnamedept dname

Functional Dependencies

Inference Rules for FDs Given a set of FDs F, we can infer additional FDs that hold whenever the

FDs in F hold

Armstrong's inference rules

1. A1. (Reflexive) If Y subset-of X, then X Ye.g. {Name,Gender}->{Name}

1. A2. (Augmentation) If X Y, then XZ YZ (Notation: XZ stands for X U Z){Name, Gender}->{Name}implies {Name,Gender,Age}->{Name,Age}

1. A3. (Transitive) If X Y and Y Z, then X Z

A1, A2, A3 form a sound and complete set of inference rules

Additional Useful Inference Rules

Decomposition If X YZ, then X Y and X Z

Union If X Y and X Z, then X YZ

Psuedotransitivity If X Y and WY Z, then WX Z

Closure of a set F of FDs is the set F+ of all FDs that can be inferred from F

Full Functional Dependency

Full Functional Dependency: Only of relevance with composite determinants. This is the situation when it is necessary to use all the attributes of the composite determinant to identify its object uniquely.

order# line# qty price A001 001 10 200A002 001 20 400A002 002 20 800A004 001 15 300

(Order#, line#) qty(Order#, line#) price

Full Functional Dependencies

Partial Functional Dependency

Partial Functional Dependency: This is the situation that exists if it is necessary to only use a subset of the attributes of the composite determinant to identify its object uniquely.

student# unit# room grade9900100 A01 TH224 29900010 A01 TH224 149901011 A02 JS075 39900001 A01 TH224 16

Example:

(student#, unit#) grade

Full Functional Dependencies

unit# room

Partial Functional Dependencies

Repetition of data!

Functional Dependency: examples Example1: Suppose we are given relation R with

attributes A, B, C, D, E, F and the FD’sA BCB ECD EFYou can even take the example as :A emp no, B dept no, C manager’s emp no, D proj

no directed by that manager, E dept name, F time allocated to specified proj.

We now show that FD (AD F) holds in R, and so is a member of the closure of the given set:

Example 1

Abbreviations Meaning

A Employee no

B Dept no

C Manager’s emp no

D Proj no directed by that manager

E Dept name

F Time allocated to specified proj

Example 1 (…cont)

Abbreviations Meaning

A BC Employee no Deptno, Manager’s emp no

B E Dept no Dept name

CD EF Manager’s emp no, Proj no directed by that manager Dept name, Time allocated to specified proj

We now show that FD (AD F) holds in R

Example 1: Solution

1. A BC (given) 2. A C (1, decomposition) 3. AD CD (2, augmentation) 4. CD EF (given) 5. AD EF (3 and 4, transitivity) 6. AD F (5, decomposition)

Irreducible Sets of Dependencies

1. The right hand side of every FD in set S involves just one attribute (singleton set)

2. The left hand side of every FD in set S is irreducible, meaning that no attribute can be discarded from the determinant without changing the closure S’ (without converting S into some set not equivalent to S). Such an FD is left-irreducible

3. No FD in S can be discarded from S without changing the closure S’.

Example 2:

Suppose we are given relation R with attributes A, B, C, D, and FD’s

A BC

B C

A B

AB C

AC D Compute an irreducible set of FD’s that is

equivalent to this given set?

Example 2: Solution

1. The first step is to rewrite the FD’s such that each one has a singleton right hand side:

A B A C B C A B AB C AC D

Example 2: Solution

2. Next, attribute C can be eliminated from the left hand side of the FD AC D, because we have A C, so A AC by augmentation, and we are given AC D, so A D by transitivity; thus the C on the left hand side of AC D is redundant.

3. Next, we observe that the FD AB C can be eliminated, because again we have

A C, so AB CB by augmentation, so AB C by decomposition.

4. Finally, the FD A C is implied by the FD’s

A B and B C, so it can also be eliminated. We are left with:

Example 2: Solution

A B

B C

A D

This set is irreducible.

Quiz…

Given R = (A,B,C,D,E,F,G,H, I, J) and

F = {AB → E, AG → J, BE → I, E → G, GI → H}

Does F AB → GH?