Chapter 7 Functional Dependencies Copyright © 2004 Pearson Education, Inc.

Chapter 7

Functional Dependencies

Copyright © 2004 Pearson Education, Inc.

Copyright © 2004 Ramez Elmasri and Shamkant NavatheElmasri/Navathe, Fundamentals of Database Systems, Fourth Edition

Outline

Informal Design Guidelines for Relational Databases– Semantics of the Relation Attributes– Redundant Information in Tuples and Update Anomalies– Null Values in Tuples– Spurious Tuples

Functional Dependencies (FDs)– Definition of FD – Inference Rules for FDs– Equivalence of Sets of FDs– Minimal Sets of FDs

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Informal Design Guidelines for

Relational Databases (1)

What is relational database design?

The grouping of attributes to form "good" relation schemas

Two levels of relation schemas– The logical "user view" level

– The storage "base relation" level

Design is concerned mainly with base relations What are the criteria for "good" base relations?

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Informal Design Guidelines for Relational Databases (2)

We first discuss informal guidelines for good relational design

Then we discuss formal concepts of functional dependencies and normal forms- 1NF (First Normal Form)- 2NF (Second Normal Form)- 3NF (Third Normal Form)- BCNF (Boyce-Codd Normal Form)

Additional types of dependencies, further normal forms, relational design algorithms by synthesis are discussed in Chapter 16

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Semantics of the Relation

Attributes GUIDELINE 1: Informally, each tuple in a relation

should represent one entity or relationship instance. (Applies to individual relations and their attributes).

Attributes of different entities (EMPLOYEEs, DEPARTMENTs, PROJECTs) should not be mixed in the same relation

Only foreign keys should be used to refer to other entities Entity and relationship attributes should be kept apart as much as

possible.

Bottom Line: Design a schema that can be explained easily relation by relation. The semantics of attributes should be easy to interpret.

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A simplified COMPANY relational database schema

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Redundant Information in Tuples

and Update Anomalies Mixing attributes of multiple entities may cause

problems– Information is stored redundantly wasting storage

– Data inconsistency

Problems with update anomalies– Insertion anomalies– Deletion anomalies– Modification anomalies

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EXAMPLE OF AN UPDATE

ANOMALY (1) Consider the relation:EMP_PROJ ( Emp#, Proj#, Ename, Pname, No_hours)

Update Anomaly: Changing the name of project

number P1 from “Billing” to “Customer-Accounting” may cause this update to be made for all 100 employees working on project P1.

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EXAMPLE OF AN UPDATE ANOMALY (2)

Insert Anomaly: Cannot insert a project unless an employee is assigned to .

Inversely - Cannot insert an employee unless an he/she is assigned to a project.

Delete Anomaly: When a project is deleted, it will result in deleting all the employees who work on that project. Alternately, if an employee is the sole employee on a project, deleting that employee would result in deleting the corresponding project.

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Guideline to Redundant Information in Tuples and Update Anomalies

GUIDELINE 2: Design a schema that does not suffer from the insertion, deletion and update anomalies. If there are any present, then note them so that applications can be made to take them into account

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Null Values in Tuples

GUIDELINE 3: Relations should be designed such that their tuples will have as few NULL values as possible

Attributes that are NULL frequently could be placed in separate relations (with the primary key)

Reasons for nulls:– attribute not applicable or invalid– attribute value unknown (may exist)– value known to exist, but unavailable

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Spurious Tuples

Bad designs for a relational database may result in erroneous results for certain JOIN operations

The "lossless join" property is used to guarantee meaningful results for join operations

GUIDELINE 4: The relations should be designed to satisfy the lossless join condition. No spurious tuples should be generated by doing a natural-join of any relations. Avoid relations that contain matching attributes that are not (foreign key, primary key) combinations because joining on such attributes may produce spurious tuples

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Spurious Tuples (2)

There are two important possleroperties of decompositions:

(a) non-additive or lssness of the corresponding join

(b) preservation of the functional dependencies.

Note that property (a) is extremely important and cannot be sacrificed. Property (b) is less stringent and may be sacrificed. (See more in chapter 16 [1]).

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Functional Dependencies (FDs)

Definition of FDDirect, indirect, partial dependenciesInference Rules for FDsEquivalence of Sets of FDsMinimal Sets of FDs

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Functional Dependencies (1)

Functional dependencies (FDs) are used to specify formal measures of the "goodness" of relational designs

FDs and keys are used to define normal forms for relations

FDs are constraints that are derived from the meaning and interrelationships of the data attributes

A set of attributes X functionally determines a set of attributes Y if the value of X determines a unique value for Y

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X -> Y holds if whenever two tuples have the same value for X, they must have the same value for Y

For any two tuples t1 and t2 in any relation instance r(R): If t1[X]=t2[X], then t1[Y]=t2[Y]

X -> Y in R specifies a constraint on all relation instances r(R)Written as X -> Y; can be displayed graphically on a relation

schema as in Figures. ( denoted by the arrow).

FDs are derived from the real-world constraints on the attributes

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Examples of FD constraints (1)

social security number determines employee nameSSN -> ENAME

project number determines project name and locationPNUMBER -> {PNAME, PLOCATION}

employee ssn and project number determines the hours per week that the employee works on the project{SSN, PNUMBER} -> HOURS

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Examples of FD constraints (2)

An FD is a property of the attributes in the schema RThe constraint must hold on every relation instance

r(R)If K is a key of R, then K functionally determines all

attributes in R (since we never have two distinct tuples with t1[K]=t2[K])

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Direct dependency (fully functional dependency): All attributes in a R must be fully functionally dependent on the primary key (or the PK is a determinant of all attributes in R)

TicketID TicketName

TicketType

TicketLocation

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Indirect dependency (transitive dependency): Value of an attribute is not determined directly by the primary key

TicketID TicketName

TicketType

TicketLocation

Price

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Partial dependency– Composite determinant - more than one value is

required to determine the value of another attribute, the combination of values is called a composite determinantEMP_PROJ(SSN, PNUMBER, HOURS, ENAME, PNAME,

PLOCATION){SSN, PNUMBER} -> HOURS

– Partial dependency - if the value of an attribute does not depend on an entire composite determinant, but only part of it, the relationship is known as the partial dependency

SSN -> ENAME PNUMBER -> {PNAME, PLOCATION}


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Partial dependency

TicketID TicketName

TicketType

TicketLocation

Price

Agent-id AgentName

AgentLocation

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Inference Rules for FDs (1)

An FD X → Y is inferred from a set of dependencies F specified on R if whenever r satisfies all the dependencies in F, X → Y also holds in r

F |=X → Y to denote that the functional dependency X→Y is inferred from the set of functional dependencies F

Exp: U = {ABC}, F = {AB, BC},

We can say F A⊨ C

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Armstrong's inference rules:IR1. (Reflexive) If Y subset-of X, then X -> YIR2. (Augmentation) If X -> Y, then XZ -> YZ

(Notation: XZ stands for X U Z)IR3. (Transitive) If X -> Y and Y -> Z, then X -> Z

Lemma 1: IR1, IR2, IR3 form a sound and complete set of

inference rules

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Lemma 2: Some additional inference rules that are useful:

(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

The last three inference rules, as well as any other inference rules, can be deduced from IR1, IR2, and IR3 (completeness property)

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Sample Exercises

F = {AB, BC} – AC is inferred from F? (transitive)

F = {ABC} – AB , AC are inferred from F?

Ans: ABC và BCB (reflexive)

=> AB (transitive)F = {AB, BC}, ABC ?F = {AB}, ACB?

A AC ACA & AB

ACB is right

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

Closure of a set of attributes X with respect to F is the set X + of all attributes that are functionally determined by X

X + can be calculated by repeatedly applying IR1, IR2, IR3 using the FDs in F

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Determining X+

Example: Emp_Proj(Ssn, Ename,Pnumber, Pname, Plocation, Hours)

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Sample Exercises

R(ABCDEG) and FDs as follows:F = {AB→C, C →A, BC →D,

ACD→B, D→EG, BE→C,

CG→BD, CE→AG}X = {BD}, calculate X+

Result:X+ = R

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Lemma 3:X Y is inferred from F based on Armstrong’s

rules if and only if Y is a subset of X+ with respect to F

F Arm (X ⊢ Y) Y XF+Note:

– We can check whether X is a key of R by calculating X+. If X+ = R then X is a key

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Checking if an FD Holds on FUsing the Closure

Let R(ABCDEFGH) satisfy the following functional dependencies: {A->B, CH->A, B->E, BD->C, EG->H, DE->F}

Which of the following FD is also guaranteed to be satisfied by R?

1. BFG AE

2. ACG DH

3. CEG AB

Hint: Compute the closure of the LHS of each FD that you get as a choice. If the RHS of the candidate FD is contained in the closure, then the candidate follows from the given FDs, otherwise not.


Checking for Keys Using the Closure

Which of the following could be a key for R(A,B,C,D,E,F,G) with functional dependencies {ABC, CDE, EFG, FGE, DEC, and BCA}

1. BDF

2. ACDF

3. ABDFG

4. BDFG


Algorithm for Finding a Key

Note: the algorithm determines only one key out of the possible candidate keys for R;

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Finding Keys using FDs

Tricks for finding the key:If an attribute never appears on the RHS of any FD, it

must be part of the keyIf an attribute never appears on the LHS of any FD, but

appears on the RHS of any FD, it must not be part of any key



We have:UL and UR are the set of LHS and RHS attributes

N = U – UR is the set of independent attributes and those which only appear on LHS N must be a part of keys

If N+ = R, then N is a minimal key Stop here!Otherwise:D = UR – UL is the set of attributes which only appears in

RHS D cannot be a part of keyL = U – (N D) is the set of attributes which may or may

not be a part of keysFor each combination X in L, we calculate {N X}+. If

{N X}+ = R so it is a keySlide 7 -41



Consider R = {A, B, C, D, E, F, G, H} with a set of FDs

F = {CD→A, EC→H, GHB→AB, C→D, EG→A, H→B, BE→CD, EC→B}

Find all the candidate keys of R



F = {CD→A, EC→H, GHB→AB, C→D, EG→A, H→B, BE→CD, EC→B}

UR = {AHBDC} = {ABCDH}N = U – UR = {EFG} but EFG+ = EFGA ≠ RUL = {CDEGHB} = {BCDEGH}D = UR – UL = {A}L = U – (N D) = {BCDH}We have combinations such as {B, C, D, H, BC, BD,

BH, CD, CH, DH, BCD, BCH, CDH, BCDH}



For each combination X, calculate {X N}+BEFG+ = ABCDEFGH = R; it’s a key [BE→CD,

EG→A, EC→H] CEFG+ = ABCDEFGH = R; it’s a key [EG→A,

EC→H, H→B, BE→CD] DEFG+ = ADEFG ≠ R; it’s not a key [EG→A] EFGH+ = ABCDEFGH = R; it’s a key [EG→A,

H→B, BE→CD] If we add any further attribute(s), they will form the

superkey. Therefore, we can stop here searching for candidate key(s).

So, candidate keys are: {BEFG, CEFG, EFGH}


Exercises

Consider R = {A, B, C, D, E, F} with a set of FDs

F = {A BC, BD, AD E, CDA}


Consider R = {A, B, C, D, E, F, G} with a set of FDs

F = {ABC→DE, AB→D, DE→ABCF, E→C}


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Equivalence of Sets of FDs

Two sets of FDs F and G are equivalent if:- every FD in F can be inferred from G, and- every FD in G can be inferred from F

Hence, F and G are equivalent if F + =G +

Definition: F covers G if every FD in G can be inferred from F (i.e., if G + subset-of F +)

F and G are equivalent if F covers G and G covers FThere is an algorithm for checking equivalence of

sets of FDsHome Ex:

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Minimal Sets of FDs (1) A minimal cover of a set of functional dependencies E

is a minimal set of dependencies (in the standard canonical form and without redundancy) that is equivalent to E.

A set of FDs is minimal if it satisfies the following conditions:

(1) Every dependency in F has a single attribute for its RHS

(2) We cannot remove any dependency from F and have a set of dependencies that is equivalent to F.

(3) We cannot replace any dependency X -> A in F with a dependency Y -> A, where Y is a subset-of X and still have a set of dependencies that is equivalent to F. (X A also is called a complete functional dependency)

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Minimal Sets of FDs (2)

Every set of FDs has an equivalent minimal setThere can be several equivalent minimal setsThere is no simple algorithm for computing a

minimal set of FDs that is equivalent to a set F of FDs

To synthesize a set of relations, we assume that we start with a set of dependencies that is a minimal set

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Finding a Minimal Cover F for a Set of Functional Dependencies E

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Finding a Minimal Cover F for a Set of Functional Dependencies E

Let the given set of FDs be E :{B→A, D→A, AB→D}. Please find the minimal cover of E.

Step 1: Set F = EStep 2: All FDs in the canonical formStep 3: Determine if AB D has any redundant attribute on

LHS

– Since B A, we have BB AB (IR2) => B AB (However, AB D), so we have B D

– So replace AB D by B D. We have E’ = {B A, D A, B D)

Step 4: We also derive from E’ B A is redundant

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Chapter 7 Functional Dependencies Copyright © 2004 Pearson Education, Inc.

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Transcript of Chapter 7 Functional Dependencies Copyright © 2004 Pearson Education, Inc.