database management cop4540, scs, fiu functional dependencies (chapter 14)

Database ManagementCOP4540, SCS, FIU

Functional Dependencies(Chapter 14)


What are functional dependencies

• A FD is a constraint between two sets of attributes from the database..

• A FD, denoted by X Y, between two sets of attributes X and Y forms a constraint that states:– For any two tuples t1 and t2 such that t1[X]

=t2[X], we much also have t1[Y] =t2[Y].

– The values of the Y component of a tuple depend on, or are determined by, the values of X component.


Example

Employee(SSN, EName)

Project(ProjectNo, ProjectName, DeptName)

WorksOn(SSN, ProjectNo, Hours)

Functional dependencies:

SSN Ename

ProjectNo ProjectName DeptName

SSN ProjectNo Hour


Additional Notes

• A functional dependency is a property of the semantics or meaning of the attributes.

• A functional dependency is a property of the relation schema R, not of a particular legal relation state r of R.– FD cannot be inferred from a given relation

state.– It must be defined explicitly by someone who

knows the semantics of the attributes of R.


Keys and FDs• A key is a determinant, while a determinant may or may

not be a key.

• A key is always unique to the relation, while a determinant may or may not be unique to the relation.

• The most significant difference between keys and FDs is UNIQUENESS and MINIMALITY.

• Several terms:

– Superkey

– Candidate Key

– Primary Key


Example

Student(SSN, Name, Phone, City, Street, Zip)

SSN Phone Name City Street Zip

SSN Name City Street Zip

City Street Zip


Example

ENAME SSN BDATE ADDRESS DNUMBER DNAME DMGRSSN

EMP_DEPT

EMP_PROJSSN PNUMBER HOURS ENAME PNAME PLOCATION


Computing the Closure of Attributes• What is the closure of attributes?

– Given a set of attributes A = {A1, A2, …, An} and a set of FDs S, the closure of A under the FDs in S is the set of attributes B such that every relation that satisfies all the FDs in S also satisfies A1, A2, …, An B.

– B is denoted by A+ = {A1, A2, …, An}+

• How to compute a closure?Algorithm: Determining X+, the closure of X under F

X+ = X;repeat oldX+ = X+;

for each FD Y Z in F doif Y X+ then X+ = X+ Z;

until ( oldX+ = X+ );


Let us consider a relation with attribute A, B, C, D,E and F. Suppose that this relation has the functional dependencies AB C, BC AD. D E, andCF B. 1. What is the closure of {A, B}, that is {A,B}+ ?2. How to test whether AB D follows from these dependencies? How about D A?3. What can be the primary key of this relation?

Example


Splitting/Combing Rule

• Splitting rule A FD A1 A2 … An B1 B2 … Bm can be replaced by

a set of FDs A1 A2 … An Bi for i = 1, 2, …, m.

• Combining rule A set of FDs A1 A2 … An Bi for i = 1, 2, …, m can

be replaced by a single FD A1 A2 … An B1 B2 … Bm


Trivial-dependency Rule

A FD A1 A2 … An B1 B2 … Bm is

– Trivial if the B’s are a subset of the A’s.– Nontrivial if at least one of the B’s is not among the

A’s.– Completely nontrivial if none of the B’s is also one of

the A’s.

Trivial-dependency rule The FD A1 A2 … An B1 B2 … Bm is equivalent to

A1 A2 … An C1 C2 … Ck, where the C’s are all those B’s that are not also A’s


The Transition Rule

• If two FDs A1, A2, …, An B1, B2, …, Bm and B1, B2, …, Bm C1, C2, …, Ck hold in relation R, then A1, A2, …, An C1, C2, …, Ck also holds in R.

• This rule can be proved by computing the closure of {A1, A2, …, An} under {A1, A2, …, An B1, B2, …, Bm B1, B2, …, Bm C1, C2, …, Ck }.


A complete Set of Inference Rules

• Armstrong’s axioms (1974)– Reflexivity

• if X Y then X Y

– Augmentation• {X Y} XZ YZ

– Transitivity• {X Y, Y Z} X Z


Other rules

• Decomposition ( or projective) rule– { X YZ} X Y

• Union ( or additive) rule– {X Y, X Z} X YZ

• Pseudotransitive rule– {X Y, WY Z} WX Z


Proof1. X--->YZ (given)

YZ ---> Y ( using Reflexivity)

X ---> Y ( applying transitivity to the above two FDs)

2. X ---> Y (given)

X----> Z (given)

X ----> XY (applying augmentation to rule 1)

XY --> YZ (applying augmentation to rule 2)

X ---> YZ ( applying transitivity to the above two rules)

3. X ---> Y (given)

WY ---> Z (given)

WX ---> WY ( using augmentation to rule 1)

WX ---> Z (applying transitivity to rule 2 and 3)


Equivalence of Sets Dependencies

Given two sets of dependencies E and F, the closures of these two sets are E+ and F+ respectively, We say– E is covered by F, if E F+.– E and F are equivalent if E+ = F+.– How to determine the equivalence?

• calculating X+ under F for each FD X Y in E, and then checking whether this X+ includes the attributes in Y. (This is to determine whether E is covered by F)

• determining whether F is covered by E by the same approach above.


Example

• E = {A B, B A, B C, C A}

• F = {A B, B C, C A }


Minimal Sets of FDs• A set of FDs F is minimal if it satisfies the

following condition:– Every FD in F has a single attribute for its right-hand

side.– We cannot replace any FD X A with a FD Y A,

where Y is a proper subset of X, and still have a set of dependencies that is equivalent to F.

– We cannot remove any dependency from F and still have a set of FDs that is equivalent to F.

• A minimal basis of F is denoted by Fmin, and there may be several minimal bases for one particular F.

database management cop4540, scs, fiu functional dependencies (chapter 14)

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