l02: logical database design

H.Lu/HKUST

L02: Logical Database Design

Introduction Functional Dependencies Normal Forms Normalization by decomposition

H.Lu/HKUST L02: Database Design -- 2

Life Cycle of Database Applications

Database initial study

Database design

Implement. & loading

Operation

Maint. & evolution

Analyze the company situation Define problems & constraints Define objectives; scope and

boundaries

Conceptual design; DBMS selection Logical & physical design

Introducing changes; make enhancements

Produce the required information flow

Install the DBMS; create the DB; load & convert the data

Testing & evaluation Test, fine-tune, evaluate the DB and its application programs


Database Design Procedures

Data & requirement analysis

Conceptual modeling

Data model verification

Logical design

Physical design

Determine end user views, outputs, and transaction-processing requirements

Make the semantics clear

Define storage structures and access paths for optimum performance

Translate the conceptual schema into definitions of tables, views, and so on

Identifying main processes, insert, update, and delete rules, validate reports, queries, views, integrity, sharing, and security

DBMS software selection


Overview of Relational Database Design Requirement Analysis Conceptual design: make the semantics clear

What are the entities and relationships in the enterprise? What information about these entities and relationships

should we store in the database? What are the integrity constraints or business rules that

hold? A database `schema’ in the ER Model can be represented

pictorially (ER diagrams). Can map an ER diagram into a relational schema.

Logical design: (Normalization) Check relational schema for redundancies and related anomalies.

Physical Database Design and Tuning: Consider typical workloads and

further refine the database design.


Logical Design of Relational Databases

Logical design of relational database requires that we find a “good” collection of relation schemas. A bad design may lead to

Repetition of information.

Inability to represent certain information.

Design goals:

Avoid redundant data

Ensure that relationships among attributes are represented

Facilitate the checking of updates for violation of database integrity constraints


The Evils of Redundancy

Redundancy is at the root of several problems associated with relational schemas: redundant storage, insert/delete/update anomalies

Input of our design phase: An initial schema. Consider relation obtained from Hourly_Emps:

• Hourly_Emps (eid, name, office, rating, hrly_wages, hrs_worked)

Notation: We will denote this relation schema by listing the attributes: ENORWH

• This is really the set of attributes {E,N,O,R,W,H}.• Sometimes, we will refer to all attributes of a relation by using

the relation name. (e.g., Hourly_Emps for ENORWH)


Anomalies

E N O R W H

123-22-3666 Attishoo 48 8 10 40

231-31-5368 Smiley 22 8 10 30

131-24-3650 Smethurst 35 5 7 30

434-26-3751 Guldu 35 5 7 32

612-67-4134 Madayan 35 8 10 40

IC: the hourly_wage is determined by rating. Update anomaly: Can we change W in just the

1st tuple of ENORWH? Insertion anomaly: What if we want to insert an

employee and don’t know the hourly wage for his rating?

Deletion anomaly: If we delete all employees with rating 5, we lose the information about the wage for rating 5!


Decomposition – The Basic Approach Main refinement technique:

decomposition (replacing ENORWH with ENORH and RW)

Decomposition should be used judiciously: Is there reason to decompose

a relation? What problems (if any) does

the decomposition cause? Integrity constraints, in particular

functional dependencies, can be used to identify schemas with such problems and to suggest refinements.

E N O R H

123-22-3666 Attishoo 48 8 40

231-31-5368 Smiley 22 8 30

131-24-3650 Smethurst 35 5 30

434-26-3751 Guldu 35 5 32

612-67-4134 Madayan 35 8 40

Hourly_Emps2

R W

8 10

5 7

Wages


Database Design Theory

Decide whether a particular relation R is in “good” form. In the case that a relation R is not in “good” form, decompose

it into a set of relations {R1, R2, ..., Rn} such that each relation is in good form the decomposition is a lossless-join decomposition

Our theory is based on: functional dependencies multivalued dependencies (not going to cover in this

course)

H.Lu/HKUST




Functional Dependencies A functional dependency X Y holds over relation R if, for

every allowable instance r of R:

t1 r, t2 r, X(t1) = X (t2) implies Y(t1) = Y (t2)

i.e., given two tuples in r, if the X values agree, then the Y values must also agree. (X and Y are sets of attributes.)

A functional dependency is a generalization of the notion of a key. Constraints on the set of legal relations. Require that the value for a certain set of attributes

determines uniquely the value for another set of attributes


Functional Dependencies (Cont.)

Example: Consider r(A,B) with the instance of r. On this instance, A B does NOT hold, but B A does

hold. An FD is a statement about all allowable relations.

Must be identified based on semantics of application. Given some allowable instance r1 of R, we can check if it violates some

FD f, but we cannot tell if f holds over R! A functional dependency is trivial if it is satisfied by all

instances of a relation E.g.

• Sid, Sname Sname• Sname Sname

In general, is trivial if

1 41 53 7


Use of Functional Dependencies

We use functional dependencies to: test relations to see if they are legal under a given set of

functional dependencies. • If a relation r is legal under a set F of functional dependencies,

we say that r satisfies F.

specify constraints on the set of legal relations• We say that F holds on R if all legal relations on R satisfy the set

of functional dependencies F.

Note: A specific instance of a relation schema may satisfy a functional dependency even if the functional dependency does not hold on all legal instances. For example, a specific instance of Sailors may, by chance, satisfy Sname Sid.


Armstrong’s Axioms Given some FDs, we can usually infer additional FDs:

ssn did, did lot implies ssn lot An FD f is implied by a set of FDs F if f holds whenever all

FDs in F hold. F+ = closure of F is the set of all FDs that are implied by F.

Armstrong’s Axioms (X, Y, Z are sets of attributes): Reflexivity: If X Y, then X Y Augmentation: If X Y, then XZ YZ for any Z Transitivity: If X Y and Y Z, then X Z

These are sound and complete inference rules for FDs!


Reasoning About FDs

Couple of additional rules (that follow from AA): Union: If X Y and X Z, then X YZ Decomposition: If X YZ, then X Y and X Z Pseudo-transitivity: If X Y and ZY T, then X Z T

Example: Contracts(cid, sid, jid, did, pid, qty, value), and: C is the key: C CSJDPQV Project purchases each part using single contract: JP C Dept purchases at most one part from a supplier: SD P

JP C, C CSJDPQV imply JP CSJDPQVSD P implies SDJ JPSDJ JP, JP CSJDPQV imply SDJ CSJDPQV


Closure of Attribute Sets

Computing the closure of a set of FDs can be expensive. (Size of closure is exponential in # attrs!)

Typically, we just want to check if a given FD X Y is in the closure of a set of FDs F. An efficient check: Compute attribute closure of X (denoted X+ ) wrt F:

• Set of all attributes A such that X A is in F+

• There is a linear time algorithm to compute this.

Check if Y is in X+ Does F = {A B, B C, C D E } imply A E?

i.e, is A E in the closure F+ ? Equivalently, is E in A+ ?


Algorithm to Compute Closure

Algorithm to compute +, the closure of under Fresult:= ;while (changes to result) do

for each in F dobegin

if result then result :=result ;end

R = (A, B, C, G, H, I)F = ( A B

A CCG HCG IB H }

(AG+)1. Result= AG2. Result= ABCG (A C; A B and A AGB)3. Result= ABCGH (CG H and CG AGBC)4. Result=ABCGHI (CG I and CG AGBCH)Is AG a candidate key?


Uses of Attribute Closure There are several uses of the attribute closure algorithm: Testing for superkey:

To test if is a superkey, we compute +, and check if + contains all attributes of R.

Testing functional dependencies To check if a functional dependency holds (or, in

other words, is in F+), just check if +. That is, we compute + by using attribute closure, and then

check if it contains . Is a simple and cheap test, and very useful

Computing closure of F For each R, we find the closure +, and for each S

+, we output a functional dependency S.


Closure of a Set of Functional Dependencies

Given a set F set of functional dependencies, there are certain other functional dependencies that are logically implied by F. E.g. If A B and B C, then we can infer that

A C The set of all functional dependencies logically

implied by F is the closure of F. We denote the closure of F by F+.


Procedure for Computing F+

F+ = Frepeat for each FD f in F+ apply reflexivity and augmentation rules on f add the resulting functional dependencies to F+

for each pair of FD f1 and f2 in F+ if f1 and f2 can be combined using

transitivity then add the resulting FD to F+

until F+ does not change any further


Example R = (A, B, C, G, H, I)

F = { A B, A C, CG H, CG I, B H} some members of F+

A H • by transitivity from A B and B H

AG I • by augmenting A C with G, to get AG CG

and then transitivity with CG I CG HI

• from CG H and CG I : “union rule” can be inferred from– definition of functional dependencies, or – Augmentation of CG I to infer CG CGI, augmentation of

CG H to infer CGI HI, and then transitivity


Minimal Cover

Sets of functional dependencies may have redundant dependencies that can be inferred from the others Eg: A C is redundant in: {A B, B C, A C} Parts of a functional dependency may be redundant

• E.g. on RHS: {A B, B C, A CD} can be simplified to {A B, B C, A D}

• E.g. on LHS: {A B, B C, AC D} can be simplified to {A B, B C, A D}

Intuitively, a minimal cover of F is a “minimal” set of functional dependencies equivalent to F, with no redundant dependencies or having redundant parts of dependencies


Minimal Cover for a Set of FDs

Minimal cover G for a set of FDs F: Closure of F = closure of G. Right hand side of each FD in G is a single attribute. If we modify G by deleting an FD or by deleting attributes

from an FD in G, the closure changes. Intuitively, every FD in G is needed, and as small as possible

in order to get the same closure as F. Example:

F: {A B, ABCD E, EF GH, ACDF EG}

has minimal cover G: {A B, ACD E, EF G, EF H}


Extraneous Attributes Consider a set F of functional dependencies and the functional

dependency in F. Attribute A is extraneous in if A

and F logically implies (F – { }) {( – A) }. Attribute A is extraneous in if A

and the set of functional dependencies (F – { }) { ( – A)} logically implies F.

Note: implication in the opposite direction is trivial in each of the cases above, since a “stronger” functional dependency always implies a weaker one Example: Given F = {A C, AB C }

• B is extraneous in AB C because A C logically implies AB C.

Example: Given F = {A C, AB CD}• C is extraneous in AB CD since A C can be inferred even after

deleting C


Testing if an Attribute is Extraneous

Consider a set F of functional dependencies and the functional dependency in F. To test if attribute A is extraneous in

• compute (a – {A})+ using the dependencies in F

• if (a – {A})+ contains A, A is extraneous

To test if attribute A is extraneous in • compute + using only the dependencies in

F’ = (F – { }) { ( – A)},

• check that + contains A; if it does, A is extraneous


Computing a Minimal Cover

1. Rewrite FDs in F into G so that all FDs in G have a single attribute on the right side

2. For each FD in G, check each attribute in the left side to see if it can be deleted while preserving equivalence to F+.

3. Check each remaining FD in G to see if it can be deleted while preserving equivalence to F+.


Minimal Cover – An Example

F: {A B, ABCD E, EF G, EF H, ACDF EG} Rewrite F into G

G: {A B, ABCD E, EF G, EF H, ACDF E, ACDF G} Since A B, B in ABCD E can be deleted

G: {A B, ACD E, EF G, EF H, ACDF E, ACDF G} Since A B, ACD E, EF G,

ACDF G is redundant Since ACD E,

ACDF E is redundant No more extraneous attributes at the left side and no more redundant

FDSMinimal cover: {A B, ACD E, EF G, EF H}


Minimal Cover – Another Example

F = {A BC, B C, A B, AB C}

G = {A B, A C, B C, A B, AB C}

G = {A B, A C, B C, AB C}

AB C is redundant,

G = {A B, A C, B C}

A C is logically implied by A B and B C, A C is redundant

The minimal cover is: {A B, B C}


Summary

Functional dependency (FD) theory forms the base of database design theory. We will see how they are used in database design.

There are other forms dependencies, such as multivalued dependency (MVD), inclusion dependency, etc.

In practical applications, FD seems sufficient to derive good schema

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Normal Forms Returning to the issue of schema refinement, the first

question to ask is whether any refinement is needed! If a relation is in a certain normal form (BCNF, 3NF

etc.), it is known that certain kinds of problems are avoided/minimized. This can be used to help us decide whether decomposing the relation will help.

Role of FDs in detecting redundancy: Consider a relation R with 3 attributes, ABC.

• No FDs hold: There is no redundancy here.• Given A B: Several tuples could have the same A

value, and if so, they all have the same B value!


First Normal Form

Domain is atomic if its elements are considered to be indivisible units Examples of non-atomic domains:

• Set of names, composite attributes

• Identification numbers like CS101 that can be broken up into parts

A relational schema R is in first normal form if the domains of all attributes of R are atomic

Non-atomic values complicate storage and encourage redundant (repeated) storage of data We assume all relations are in first normal form


Second Normal Form

R: Sales ( dept, item, price)FD: dept, item price; item price R is in 1NF

dept

item

price

KEY

Anomalies?Yes. Caused by non-full functional

dependency

A 2NF relation is a 1NF relation whose nonprime attributes are fully and functionally dependent on the keys

Sales (dept, item)

Iteminfo (item, price)


Third Normal Form

Iteminfo (item, price,discount)

item price

price discount

discount

item price

KEY

Anomalies?

Relation R with FDs F is in 3NF if, for all X A in F +

A X, or X contains a key for R, or A is part of some key for

R.

Iteminfo (item, price)

Discnt(price,discount)Yes. Caused by transitive dependency


What Does 3NF Achieve?

If 3NF violated by X A, one of the following holds: X is a subset of some key K

• We store (X, A) pairs redundantly.

X is not a proper subset of any key.• There is a chain of FDs K X A, which means that we cannot

associate an X value with a K value unless we also associate an A value with an X value.

But: even if a relation is in 3NF, these problems could arise. e.g., Reserves SBDC, SBD is a key, if C S, SDBC is

in 3NF, but for each reservation of sailor S, same (S, C) pair is stored.


Third Normal Form – Another Example

SalaryInfo(salary-bracket, status, bonus)

Salary-bracket, status bonus;

bonus status

Salary-bracket

status

price

KEY

40K Engi neer 7K50K Engi neer 9K40K Manager 10K45K Manager 10K60K Manager 10K


Boyce-Codd Normal Form (BCNF)

Relationn R with FDs F is in BCNF if, for all X A in F +

A X (called a trivial FD), or X contains a key for R.

In other words, R is in BCNF if the only non-trivial FDs that hold over R are key constraints. No dependency in R that can be predicted using FDs alone. If we are shown two tuples that agree upon the X

value, we cannot infer the A value in one tuple

different from the A value in the other. If example relation is in BCNF, the 2 tuples

must be identical (since X is a key).

X Y A

x y1 a

x y2 ?


BCNF – Examples

R = (A, B, C)F = { A B, B C }Is R in BCNF?

Property

F = { P CAL, CL AP, A C }

Property_id Country_name Area Lot#

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Decomposition of a Relation Scheme

Suppose that relation R contains attributes A1 ... An. A decomposition of R consists of replacing R by two or more relations such that: Each new relation scheme contains a subset of the attributes

of R (and no attributes that do not appear in R), and Every attribute of R appears as an attribute of one of the

new relations. Intuitively, decomposing R means we will store instances of

the relation schemes produced by the decomposition, instead of instances of R.

E.g., Can decompose ENORWH into ENORH and RW.


Example Decomposition

Decompositions should be used only when needed. SNLRWH has FDs E ENORWH and R W Second FD causes violation of 3NF; W values repeatedly

associated with R values. Easiest way to fix this is to create a relation RW to store these associations, and to remove W from the main schema:

• i.e., we decompose ENORWH into ENORH and RW

The information to be stored consists of ENORWH tuples. If we just store the projections of these tuples onto ENORH and RW, are there any potential problems that we should be aware of?


Problems with Decompositions There are three potential problems to consider:

Some queries become more expensive. • e.g., How much did sailor Joe earn? (salary = W*H)

Given instances of the decomposed relations, we may not be able to reconstruct the corresponding instance of the original relation!

• Fortunately, not in the ENORWH example. Checking some dependencies may require joining the

instances of the decomposed relations.• Fortunately, not in the ENORWH example.

Tradeoff: Must consider these issues vs. redundancy.


Lossless Join Decompositions

Decomposition of R into X and Y is lossless-join w.r.t. a set of FDs F if, for every instance r that satisfies F: x(r) y(r) = r

It is always true that r x(r) y(r)

In general, the other direction does not hold! If it does, the decomposition is lossless-join.

Definition extended to decomposition into 3 or more relations in a straightforward way.

It is essential that all decompositions used to deal with redundancy be lossless! (Avoids Problem (2).)


More on Lossless Join

The decomposition of R into X and Y is lossless-join wrt F if and only if the closure of F contains: X Y X, or X Y Y

In particular, the decomposition of R into UV and R - V is lossless-join if UV holds over R.

A B C1 2 34 5 67 2 81 2 87 2 3

A B C1 2 34 5 67 2 8

A B1 24 57 2

B C2 35 62 8


Losslessness Test – Case 1

Case 1: R is decomposed into R1 and R2. The decomposition is lossless if at least one of the following FDs is contained in F+.

R1 R2 R1 – R2 or R1 R2 R2 – R1Example: R (A, B, C) FD: A B

Decomposition #1: R1 (A, B), R2 (A, C)Decomposition #2: R1 (A, B), R2 (B, C)


Losslessness Test – Case 2

1. Given a set of relations and FD2. Construct a table: 3. columns – attributes, rows – relations4. (i, j) = ‘a’ if Aj is an attribute of Ri5. For each FD XY, if column X contains an ‘a’, mark

column Y as ‘a’.6. Repeat the above until no change7. If there is at least one row with all ‘a’s, the

decomposition is lossless, not otherwise.


Example (Losslessness Test , Case 2)

A B C DR1 a a a aR2 a a aR3 a a a aR4 a a a

A C

B CC D

A B C DR1 a a R2 a aR3 a a a R4 a a a

R (A, B, C, D) R1 (A, B), R2 (B, D), R3 (A, B,C), R4 (B, C, D)

FD: A C, B C, CD B, C D


Dependency Preserving Decomposition

Consider CSJDPQV, C is key, JP C and SD P. BCNF decomposition: CSJDQV and SDP Problem: Checking JP C requires a join!

Dependency preserving decomposition (Intuitive): If R is decomposed into X, Y and Z, and we enforce the

FDs that hold on X, on Y and on Z, then all FDs that were given to hold on R must also hold. (Avoids Problem (3).)

Projection of set of FDs F: If R is decomposed into X, ... projection of F onto X (denoted FX ) is the set of FDs U V in F+ (closure of F ) such that U, V are in X.


Dependency Preserving Decompositions (Contd.)

Decomposition of R into X and Y is dependency preserving if (FX union FY )+ = F +

i.e., if we consider only dependencies in the closure F + that can be checked in X without considering Y, and in Y without considering X, these imply all dependencies in F +.

Important to consider F +, not F, in this definition: ABC, A B, B C, C A, decomposed into AB and

BC. Is this dependency preserving? Is C A preserved?????

Dependency preserving does not imply lossless join: ABC, A B, decomposed into AB and BC.

And vice-versa! (Example?)


Dependency Preserving -- Example

R (A, B, C)

FD: AB C, C B

R1 (A, C), R2 (B, C)

Dependency preserving?

lossless?

R1 { (a1, c1), (a1, c2)}

R2 { (b1, c1), (b1, c2) }R1

R2 ?


Decomposition into BCNF Consider relation R with FDs F. If X Y violates BCNF,

decompose R into R - Y and XY. Repeated application of this idea will give us a collection of

relations that are in BCNF; lossless join decomposition, and guaranteed to terminate.

e.g., CSJDPQV, key C, JP C, SD P, J S To deal with SD P, decompose into SDP, CSJDQV. To deal with J S, decompose CSJDQV into JS and

CJDQV In general, several dependencies may cause violation of

BCNF. The order in which we ``deal with? them could lead to very different sets of relations!


BCNF and Dependency Preservation

In general, there may not be a dependency preserving decomposition into BCNF. e.g., CSZ, CS Z, Z C Can’t decompose while preserving 1st FD; not in BCNF.

Similarly, decomposition of CSJDQV into SDP, JS and CJDQV is not dependency preserving (w.r.t. the FDs JP C, SD P and J S). However, it is a lossless join decomposition. In this case, adding JPC to the collection of relations

gives us a dependency preserving decomposition.• JPC tuples stored only for checking FD! (Redundancy!)


Decomposition into 3NF

Obviously, the algorithm for lossless join decomposition into BCNF can be used to obtain a lossless join decomposition into 3NF (typically, can stop earlier).

To ensure dependency preservation, one idea: If X Y is not preserved, add relation XY. Problem is that XY may violate 3NF! e.g., consider the

addition of CJP to `preserve? JP C. What if we also have J C ?

Refinement: Instead of the given set of FDs F, use a minimal cover for F.


Decomposition to 3NF

1 begin2 S = 3 Fc = minimal cover of F;4 for each X Y Fc do5 if not exists Ri S and XY Ri then6 S = S {Rj(XY)}7 endif8 endfor9 if not exists Ri S and Ri contains a key of R then10 S = S {Rj}11 endif12 remove from S the redundant relations ；13 return(S);14 end.

A relation for each FD

A relation for each key


Summary of Normalization

If a relation is in BCNF, it is free of redundancies that can be detected using FDs. Thus, trying to ensure that all relations are in BCNF is a good heuristic.

If a relation is not in BCNF, we can try to decompose it into a collection of BCNF relations. Must consider whether all FDs are preserved. If a

lossless-join, dependency preserving decomposition into BCNF is not possible (or unsuitable, given typical queries), should consider decomposition into 3NF.

Decompositions should be carried out and/or re-examined while keeping performance requirements in mind.

l02: logical database design

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