Functional Dependencies and Normalization - WPIweb.cs.wpi.edu/~cs3431/c15/lectures/Week...

transcript

Functional Dependencies and Normalization

Instructor: Mohamed Eltabakh

meltabakh@cs.wpi.edu

Goal… l  Given a database schema, how do you judge whether or not

the design is good?

l  How do you ensure it does not have redundancy or anomaly problems?

l  To ensure your database schema is in a good form we use: l  Functional Dependencies l  Normalization Rules

What is Normalization l  Normalization is a set of rules to systematically

achieve a good design

l  If these rules are followed, then the DB design is guarantee to avoid several problems: l  Inconsistent data l  Anomalies: insert, delete and update l  Redundancy: which wastes storage, and often slows down

query processing

Problem I: Insert Anomaly

sNumber sName pNumber pName s1 Dave p1 MM s2 Greg p2 ER

Student

Question: Could we insert a professor without student? Note: We cannot insert a professor who has no students.

Insert Anomaly: We are not able to insert “valid” value/(s)

Student Info Professor Info

Problem II: Delete Anomaly

Question: Can we delete a student and keep a professor info ? Note: We cannot delete a student that is the only student of a professor.

Delete Anomaly: We are not able to perform a delete without losing some “valid” information.

sNumber sName pNumber pName s1 Dave p1 MM s2 Greg p2 ER

Student

Problem III: Update Anomaly

Question: Can we simply update a professor’s name ? Note: To update the name of a professor, we have to update in multiple tuples.

Update Anomaly: To update a value, we have to update multiple rows. Update anomalies are due to redundancy.

sNumber sName pNumber pName s1 Dave p1 MM s2 Greg p1 MM

Student

Problem IV: Inconsistency

What if the name of professor p1 is updated in one place and not the other!!!

Inconsistent Data: The same object has multiple values. Inconsistency is due to redundancy.

sNumber sName pNumber pName s1 Dave p1 MM s2 Greg p1 MM

Student

Schema Normalization

l  Following the normalization rules, we avoid l  Insert anomaly l  Delete anomaly l  Update anomaly l  Inconsistency

What to Cover

l  Functional Dependencies (FDs)

l  Closure of Functional Dependencies

l  Lossy & Lossless Decomposition

l  Normalization 9

Usage of Functional Dependencies

l  Discover all dependencies between attributes

l  Identify the keys of relations

l  Enable good (Lossless) decomposition of a given relation

Functional Dependencies (FDs)

sNumber sName address

1 Dave 144FL

2 Greg 320FL

Student

Suppose we have the FD: sNumber → address That is, there is a functional dependency from sNumber to address

Meaning: A student number determines the student address

Or: For any two rows in the Student relation with the same value for sNumber, the value for address must be same.

l  Require that the value for a certain set of attributes determines uniquely the value for another set of attributes

l  A functional dependency is a generalization of the notion of a key

l  FD: A1,A2,…An → B1, B2,…Bm

L.H.S R.H.S

l  The basic form of a FDs A1,A2,…An → B1, B2,…Bm

L.H.S R.H.S

>> The values in the L.H.S uniquely determine the values in the R.H.S attributes (when you lookup the DB) >> It does not mean that L.H.S values compute the R.H.S values

Examples: SSN à personName, personDoB, personAddress DepartmentID, CourseNum à CourseTitle, NumCredits personName personAddress X

FD and Keys

sNumber sName address

1 Dave 144FL

2 Greg 320FL

Student

Questions : • Does a primary key implies functional dependencies? Which ones ? • Does unique keys imply functional dependencies? Which ones ? • Does a functional dependency imply keys ? Which ones ?

Observation : Any key (primary or candidate) or superkey of a relation R functionally determines all attributes of R.

Primary Key : <sNumber>

Functional Dependencies (FDs) l  Let R be a relation schema where

l  α⊆R and β⊆R -- α and β are subsets of R’s attributes

l  The functional dependency α→β holds on R if and only if: l  For any legal instance of R, whenever any two tuples t1 and t2 agree

on the attributes α, they also agree on the attributes β. That is, l  t1[α]=t2[α] ⇒ t1[β] =t2[β]

©Silberschatz, Korth and Sudarshan7.12Database System Concepts - 5th Edition, July 28, 2005.

Functional Dependencies (Cont.)Functional Dependencies (Cont.)

� Let R be a relation schema� � R and � � R

� The functional dependency� � �

holds on R if and only if for any legal relations r(R), whenever any two tuples t1 and t2 of r agree on the attributes �, they also agree on the attributes �. That is,

t1[�] = t2 [�] � t1[� ] = t2 [� ] � Example: Consider r(A,B ) with the following instance of r.

� On this instance, A � B does NOT hold, but B � A does hold.

1 41 53 7

A à B (Does not hold) B à A (holds)

Functional Dependencies & Keys

l  K is a superkey for relation schema R if and only if l  K → R -- K determines all attributes of R

l  K is a candidate key for R if and only if l  K→R, and l  No α⊂K, α→R

Keys imply FDs, and FDs imply keys

Example I

Student(SSN, Fname, Mname, Lname, DoB, address, age, admissionDate)

l  If you know that SSN is a key, Then l  SSN à Fname, Mname, Lname, DoB, address, age, admissionDate

l  If you know that (Fname, Mname, Lname) is a key, Then l  Fname, Mname, Lname à SSN, DoB, address, age, admissionDate

Need to know all of L.H.S to determine any of the R.H.S

Example II

Student(SSN, Fname, Mname, Lname, DoB, address, age, admissionDate)

l  If you know that SSN à Fname, Mname, Lname, DoB, address, age, admissionDate l  Then, we infer that SSN is a candidate key

l  If you know that Fname, Mname, Lname à SSN, DoB, address, age, admissionDate l  Then, we infer that (Fname, Mname, Lname) is a key. Is it Candidate or super key??? l  Does any pair of attributes together form a key??

l  If no à (Fname, Mname, Lname) is a candidate key (minimal) l  If yes à (Fname, Mname, Lname) is a super key

Example III

l  Does this FD hold? l  Title, year à length, genre, studioName

l  Does this FD hold? l  Title, year à starName

l  What is a key of this relation instance? l  {title, year, starName} l  Is it candidate key?

>> For this instance à not a candidate key (title, starName) can be a key

Properties of FDs

l  Consider A, B, C, Z are sets of attributes l  Reflexive (trivial):

l  A → B is trivial if B ⊆ A

©Silberschatz, Korth and Sudarshan7.15Database System Concepts - 5th Edition, July 28, 2005.

Functional Dependencies (Cont.)Functional Dependencies (Cont.)

� A functional dependency is trivial if it is satisfied by all instances of a relation

� Example:

� customer_name, loan_number � customer_name

� customer_name � customer_name

� In general, � � � is trivial if � � �

Properties of FDs (Cont’d) l  Consider A, B, C, Z are sets of attributes l  Transitive:

l  if A → B, and B → C, then A → C

l  Augmentation: l  if A → B, then AZ → BZ

l  Union: l  if A → B, A → C, then A → BC

l  Decomposition: l  if A → BC, then A → B, A → C

Use these properties to derive more FDs

Example l  Given R( A, B, C, D, E)

l  F = {A à BC, DE à C, B à D}

l  Is A a key for R or not? Does A determine all other attributes? l  A à A B C D

l  Is BE a key for R? l  BE à B E D C

l  Is ABE a candidate or super key for R? l  ABE à A B E D C l  AE à A E B C D

>> ABE is a super key >> AE is a candidate key

Use the FD properties to derive more FDs

What to Cover

Closure of a Set of Functional Dependencies l  Given a set F set of functional dependencies, there

are other FDs that can be inferred based on F l  For example: If A → B and B → C, then we can infer that

A → C

l  Closure set F è F+

l  The set of all FDs that can be inferred from F l  We denote the closure of F by F+ l  F+ is a superset of F

l  Computing the closure F+ of a set of FDs can be expensive

Inferring FDs l  Suppose we have:

l  a relation R (A, B, C, D) and l  functional dependencies A → B, C → D, A à C

l  Question: l  What is a key for R?

l  We can infer A → ABC, and since C → D, then l  A → ABCD

l  Hence A is a key in R 25

Is it is the only candidate key ???

Attribute Closure

l  Attribute Closure of A l  Given a set of FDs, compute all attributes X that A

determines l  A à X

l  Attribute closure is easy to compute l  Just recursively apply the transitive property

l  A can be a single attribute or set of attributes

Algorithm for Computing Attribute Closures l  Computing the closure of set of attributes {A1, A2, …, An}:

1.  Let X = {A1, A2, …, An} 2.  If there exists a FD: B1, B2, …, Bm → C, such that

every Bi ∈ X, then X = X ∪ C 3.  Repeat step 2 until no more attributes can be added.

l  X is the closure of the {A1, A2, …, An} attributes l  X = {A1, A2, …, An} +

Example 1: Inferring FDs l  Assume relation R (A, B, C) l  Given FDs : A → B, B → C, C → A

l  What are the possible keys for R ? l  Compute the closure of each attribute X, i.e., X+

l  X+ contains all attributes, then X is a key

l  For example: l  {A}+ = {A, B, C} l  {B}+

= {A, B, C} l  {C}+

= {A, B, C}

l  So keys for R are <A>, <B>, <C>

Example 2: Attribute Closure

l  Given R( A, B, C, D, E) l  F = {A à BC, DE à C, B à D}

l  What is the attribute closure {AB}+ ? l  {AB}+ = {A B} l  {AB}+ = {A B C} l  {AB}+ = {A B C D}

l  What is the attribute closure {BE}+ ? l  {BE}+ = {B E} l  {BE}+ = {B E D} l  {BE}+ = {B E D C}

Set of attributes α is a key if α+ contains all

attributes

Example 3: Inferring FDs

l  Assume relation R (A, B, C, D, E)

l  Given F = {A à B, B à C, C D à E } l  Does A à E?

l  The above question is the same as l  Is E in the attribute closure of A (A+)? l  Is A à E in the function closure F+ ?

A à E does not hold

A D à ABCDE does hold A D is a key for R

Summary of FDs l  They capture the dependencies between attributes

l  How to infer more FDs using properties such as transitivity, augmentation, and union

l  Functional closure F+

l  Attribute closure A+

l  Relationship between FDs and keys

What to Cover

Decomposing Relations

Greg Dave

pNumber

MM s2 MM s1

pName sNumber StudentProf

FDs: pNumber → pName

Greg Dave

pNumber

sNumber

Student

pNumber

Professor

Greg Dave

sNumber

Student

pNumber

Professor

Lossless

Lossless vs. Lossy Decomposition

l  Assume R is divided into R1 and R2

l  Lossless Decomposition l  R1 natural join R2 should create exactly R

l  Lossy Decomposition l  R1 natural join R2 adds more records (or deletes

records) from R

Lossless Decomposition

Greg Dave

pNumber

MM s2 MM s1

Greg Dave

pNumber

sNumber

Student

pNumber

Professor Lossless

Student & Professor are lossless decomposition of StudentProf (Student ⋈ Professor = StudentProf)

Lossy Decomposition

Greg Dave

pNumber

MM s2 MM s1

Greg Dave

sNumber

Student

pNumber

Professor Lossy

Student & Professor are lossy decomposition of StudentProf (Student ⋈ Professor != StudentProf)

Goal: Ensure Lossless Decomposition

l  How to ensure lossless decomposition?

l  Answer: l  The common columns must be candidate key in

one of the two relations

Back to our example

Greg Dave

pNumber

MM s2 MM s1

Greg Dave

pNumber

sNumber

Student

pNumber

Professor

Greg Dave

sNumber

Student

pNumber

Professor

Lossless

pNumber is candidate key

pName is not candidate key

What to Cover

Functional Dependencies and Normalization - WPIweb.cs.wpi.edu/~cs3431/c15/lectures/Week...

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