C20.0046: Database Management Systems Lecture #5

M.P. Johnson, DBMS, Stern/NYU, Spring 2005

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C20.0046: Database Management SystemsLecture #5

Matthew P. JohnsonStern School of Business, NYUSpring, 2005


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Converting weak ESs – differences

Atts of Crew Rel are: attributes of Crew key attributes of

supporting ESs

Crew Unit-of Studio

StudioNameCrew_ID address

C2MiramaxC1Disney

C1Miramax

Crew_IDStudioNameCrew

Supporting relships may be omitted (why?)


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Weak entity sets - relationships

Crew Studio

StudioNameCrew_ID address

Insurance

IName

Address 1260 7th Av.NYBlueCross1250 6th Av.NYAetna

AddressINameInsuranceSubscribes

Unit-of


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Weak entity sets - relationships Non-supporting relationships for weak ESs

are converted keys include entire weak ES key

C21C22C21

Crew_ID

AetnaBlueCrossAetna

Insurer

UniversalDisney

Universal

StudioNameSubscribes


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Conversion example Video store rental example, plus some atts

Q: Conversion to relations?

Rental

VideoStore

Customer

Movie

date

yearMNameaddress

Cname

MID


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Conversion example, continued Resulting binary-relationship version

Q: Conversion to relations?

Rental

Customer

Store

Movie

StoreOf

MovieOf

BuyerOf

dateyearMName

address

Cname

MID


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Converting inheritance hierarchies No best way Several non-ideal methods:

E/R-style: each ES relation OO-style: each possible “object” relation nulls-style: each rooted hierarchy relation

non-applicable fields filled in with nulls Pros & cons

for each method, exist situations favoring it


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Converting inheritance hierarchies

Movies

Cartoons Murder-Mysteries

isa isaVoices

Weapon

stars

length title year

Lion King

Component


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Inheritance: E/R-style conversion Each ES relation

Root entity set: Movies(title, year, length)

1301993Lion King198819901980

Year

110115120

length

Roger RabbitScream

Star Wars

Title

Knife1990R. Rabbit1988

YearKnife

murderWeaponScream

TitleSubclass: MurderMysteries(title, year, murderWeapon)

Subclass: Cartoons(title, year)

1993Lion King1990

YearRoger Rabbit

Title


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Subclasses: object-oriented approach Every possible “subtree” (what’s this?):

1. Movies2. Movies + Cartoons3. Movies + Murder-Mysteries4. Movies + Cartoons + Murder-Mysteries

Title Year length

Star Wars 1980 120

Title Year length Murder-Weapon

Scream 1988 110 Knife

Title Year length

Lion King 1990 115


Roger Rabbit 1988 110 Knife

1. 3.

2. 4.


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Subclasses: nulls approach One relation for entire hierarchy Any non-applicable fields are NULL

Q: How do we know if a movie is a MM? Q: How do we know if a movie is a cartoon?


Star Wars 1980 120 NULLLion King 1993 130 NULLScream 1988 110 Knife

Roger Rabbit 1990 115 Knife


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Subclasses methods: considerations1. Query time ~ number of tables accessed in query

nulls: best, since each entity single row multi-node questions: “Find 1999 films with length > 150 mins”

E/R: just Movies, so fast OO: Movies AND cartoons, so slow

single-node questions: “Find weapons in >150-min. cartoons” E/R: Movies, Cartoons AND MMs, so slow OO: just MoviesCMM, so fast

2. Number of relations per entity set nulls: just one, so very few E/R: one per ES, so more OO: exponential in #ESs, so very many (how many?)

3. Number/size of rows per entity nulls: one “long” row per entity

But all non non-applicables become null OO: one (all-relevant) row per entity, so smallest E/R: several (tho all relevant) rows per entity

but keys are duplicated


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E/R-style & quasi-redundancy Name and year of Roger Rabbit were listed in three different

rows (in different tables) Suppose title changes (“Roger” “Roget”)

must change all three places Q: Is this redundancy? A: No!

name and year are independent multiple movies may have same name

Real redundancy reqs. dependency

two rows agree on SSN must agree on rest conflicting hair colors in these rows is an error

two rows agree on movie title may still disagree conflicting years may be correct – or may not be

can forbid first case but not second Better: introduce “movie-id” key att


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Combined isa/weak example Exercise 3.3.1

Convert from E/R to R, by E/R, OO and nulls

courses

Lab-courses

Depts

#machines

room

cnumber

givenBy

dname chair

isaplatform

cname


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Agenda Last time: relational model This time:1. Functional dependencies

Keys and superkeys in terms of FDs Finding keys for relations

2. Rules for combining FDs Next time: anomalies & normalization


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Next topic: Functional dependencies FDs are constraints

part of the schema can’t tell from particular relation instances FD may hold for some instances “accidentally”

Finding all FDs is part of DB design Used in normalization


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Functional dependencies Definition:

Notation: Read: Ai functionally determines Bj

If two tuples agree on the attributes

A1, A2, …, An

then they must also agree on the attributes

B1, B2, …, Bm

A1, A2, …, An B1, B2, …, Bm


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Typical Examples of FDs Product

name price, manufacturer

Person ssn name, age father’s/husband’s-name last-name zipcode state phone state (notwithstanding inter-state area codes)

Company name stockprice, president symbol name name symbol


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To check A B, erase all other columns; for each rows t1, t2

i.e., check if remaining relation is many-one no “divergences” i.e., if AB is a well-defined function thus, functional dependency

Functional dependenciesBm...B1Am...A1

t1

t2

if t 1, t 2 agree here then t 1, t 2 agree here


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FDs ExampleProduct(name, category, color, department, price)

name colorcategory departmentcolor, category price

Consider these FDs:

What do they say ?


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FDs ExampleFDs are constraints:• On some instances they hold• On others they don’t

name category color department price

Gizmo Gadget Green Toys 49

Tweaker Gadget Green Toys 99

Does this instance satisfy all the FDs ?



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FDs Example

name category color department price

Gizmo Gadget Green Toys 49

Tweaker Gadget Black Toys 99

Gizmo Stationary Green Office-supp. 59

What about this one ?



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Q: Is PositionPhone an FD here?

A: It is for this instance, but no, presumably not in general

Others FDs? EmpID Name, Phone, Position but Phone Position

Recognizing FDs

EmpID Name Phone PositionE0045 Smith 1234 ClerkE1847 John 9876 SalesrepE1111 Smith 9876 SalesrepE9999 Mary 1234 Lawyer


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Keys of relations {A1A2A3…An} is a key for relation R if

1. A1A2A3…An functionally determine all other attributes

Usual notation: A1A2A3…An B1B2…Bk

rels = sets distinct rows can’t agree on all Ai

2. A1A2A3…An is minimal No proper subset of A1A2A3…An functionally determines

all other attributes of R Primary key: chosen if there are several

possible keys


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Keys example Relation: Student(Name, Address, DoB,

Email, Credits) Which (/why) of the following are keys?

SSN Name, Address (on reasonable assumptions) Name, SSN Email, SSN Email

NB: minimal != smallest


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Superkeys A set of attributes that contains a key Satisfies first condition:

functionally determines every other attribute in the relation

Might not satisfy the second condition: minimality may be possible to peel away some attributes

from the superkey keys are superkeys

key are special case of superkey superkey set is superset of key set

name;ssn is a superkey but not a key


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Discovering keys for relations Relation entity set

Key of relation = (minimized) key of entity set

Relation binary relationship Many-many: union of keys of both entity sets Many(M)-one(O): only key of M (Why?) One-one: key of either entity set (but not both!)


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Example – entity sets Key of entity set = (minimized) key of relation

Student(Name, Address, DoB, SSN, Email, Credits)

Student

Name

Address

DoB

SSN

Email

Credits


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Example – many-many Many-many key: union of both ES keys

Student Enrolls Course

SSN Credits CourseID Name

Enrolls(SSN,CourseID)


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Example – many-one Key of the many ES but not of the one ES

keys from both would be non-minimal

Course MeetsIn Room

CourseID Name RoomNo Capacity

MeetsIn(CourseID,RoomNo)


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Example – one-one Keys of both ESs included in relation Key is key of either ES (but not both!)

Husbands Married Wives

SSN Name SSN Name

Married(HSSN, WSSN) or

Married(HSSN, WSSN)


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Discovering keys: multiway Multiway relationships:

Multiple ways – may not be obvious R:F,G,HE is many-one E’s key is included

but not part of key Recall that relship atts are implicitly many-one

Course Enrolls Student

CourseID Name SSN NameSection

RoomNo CapacityEnrolls(CourseID,SSN,RoomNo)


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Example Exercise 3.4.2 Relation relating particles in a box to

locations and velocitiesInPosition(id,x,y,z,vx,vy,vz)

Q: What FDs hold? Q: What are the keys?


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Next time Do reading on web

Finish project part 1

Office hours now

C20.0046: Database Management Systems Lecture #5

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