Answer:

Because ...

• allows the designer to initially concentrate on an abstract view of data
• allows the designer to initially concentrate on an abstract view of relationships
• no need to initially worry about concrete representation details
• no need to initially worry about fine-grained details of constraints
• the structural parts of the mapping are straightforward
• new information (concrete data types, constraints, FKs) must be added
  for the relational model, but this is easier if other design work already done

Answer:

Relational models for the three Teacher-Teaches-Subject scenarios:

1. 
   
2. 
   
3. 
   

In (a), we implement the relationship via a foreign key in the relation that has only one associated entity.

In (c), we place the foreign key in the relation that totally participates in the relationship, so as to minimise wasted space.

Answer:

Relational mappings for a class hierarchy:

Note: the arrows show how the foreign keys in the relations R, S, T reference the primary key id attribute in relation P.

Answer:

Relational mappings for a class hierarchy:

Note: the ER and OO mappings cannot represent the disjoint constraint. There is nothing in the data model to prevent an object occurring in multiple sub-classes. In the single-table mapping, however, we can add an extra attribute which indicates which sub-class each tuple belongs to; this forces each tuple to belong to just a single sub-class. However, this still doesn't prevent a tuple from e.g. containing a value for attribute b when it also has a value of 'S' for the subClass attribute. In SQL, we can write table constraints to prevent such situations. Preventing sub-class overlap in the ER or OO models in SQL requires us to use global constraints, typically implemented as triggers.

Answer:

1. -- preferred approach
   create table R (
       id      integer,
       name    text,
       address text,
       d_o_b   date,
       primary key (id)
   );
   -- possible approach
   create table R (
       id      integer primary key,
       name    text,
       address text,
       d_o_b   date
   );
   

2. -- only possible approach
   create table S (
       name    text,
       address text,
       d_o_b   date,
       primary key (name,address)
   );
   

Answer:

1. any constaint on the value that an attribute can take, e.g.

   x integer check (x > 0)
   d date check (t between '2020-01-01' and '2022-12-31')
   y integer not null
   

2. a constraint requiring consideration of all tuples in the table, e.g.

   id integer primary key
   z integer unique
   

3. foreign keys indicate that a value in attribute must appear as a value of a corresponding attribute in another table, e.g.

   create table R (
      x integer primary key,
      y float
   );
   create table S (
      a integer primary key,
      b integer references R(a)
   );
   

create table R (                    create table R (
   a integer primary key,              a integer,
   b integer,                          b integer,
   ...                                 ...
);                                     primary key (a)
                                    );

Answer:

No difference for a single attribute primary key (as above).

But you cannot define a multi-attribute primary key inline, e.g.

-- this does not work               -- define it like this

create table R (                    create table R (
   a integer primary key,              a integer,
   b integer primary key,              b integer,
   ...                                 ...
);                                     primary key (a,b)
                                    );

create table R (
   id    serial,
   name  text,
   d_o_b date,
   ...
   primary key (id)
);

Answer:

1. It creates an integer attribute and a sequence. The sequence generates, suprise, a sequence of consecutive (unique) integer values. Every time you insert a new tuple and don't give a value for the id attribute, the sequence supplies a new one and increments itself.

2. You need to use the returning clause to capture the generated value, e.g.

   insert into R(name,d_o_b) values ('John','1972-02-28') returning id;
   

3. Since the serial attribute contains an integer value, you would reference it as e.g.

   fk integer references R(id)
   

   Defining the fk attribute as serial actually works, but generates a useless sequence as a side-effect.

Answer:

Convert CompanyListing entity into an SQL CREATE TABLE definition:

CREATE TABLE CompanyListing
(
	name         char(4) PRIMARY KEY,
	sharePrice   numeric(6,2),
	netWorth     numeric(20,2)
);

Stock-market listings typically use 3-4 character abbreviations for company names, so a fixed-length character string is ok. Since share prices are money values, using numeric values with two decimal places (for cents) would be suitable. Individual share prices are rarely more than $1000.00, so we use a field with 6 digits, including 2 decimal places. The nett worth of a listed company is likely to be very large, so we allow for up to 20 digits. Note that both of these allow only integer values of cents. Since the stock market probably requires more precision, then a float value (or allowing more digits after the decimal point) might be more appropriate. PostgreSQL provides a money data type which uses floating point and has the added advantage of displaying the field in a format like $999.99

Answer:

Convert Person entity into SQL CREATE TABLE definition:

CREATE TABLE Person
(
	familyName      varchar(30),
	givenName       varchar(30),
	initial         char(1),
	streetNumber    integer,
	streetName      varchar(40),
	suburb          varchar(40),
	birthday        date,
	PRIMARY KEY     (familyName,givenName,initial)
);

The choice of a three-part name is tricky. The family-name and given-name parts are pretty much as described above. However, the initial creates a problem. It is part of the key, and so the above definition requires it to be provided, even though not everyone is going to have a middle initial. It ought to remain part of the key, however, so that we can distinguish between people called John A. Smith and John B. Smith. Since no part of the key is allowed to be NULL, we need to adopt some convention for people with no initials; a plausible approach would to use a single space character (i.e. ' '). If we need to deal with addresses like 1a Smith Street, then we'd need to change the number attribute to a string type. Since all DBMSs have a date type, along with functions for extracting the components, we may as well collapse the components of the birthday attribute into a single field of date type.

Answer:

Relational models for very small University ER model:

Answer:

Supplier/Parts ER design expressed as an SQL schema:

CREATE TABLE Supplier (
	name    varchar(50),
	city    varchar(50),
	PRIMARY KEY (name)
);
create TABLE Part (
	number  integer,
	colour  varchar(20),
	PRIMARY KEY (number)

);
CREATE TABLE Supply (
	supplier varchar(50),
        part     integer,
	quantity integer,
	PRIMARY KEY (supplier,part),
	FOREIGN KEY (supplier) REFERENCES Supplier(name),
	FOREIGN KEY (part) REFERENCES Part(number)
);

In this example, we write all constraints at the table level. In subsequent examples, we write constraints in a more compact form.

Which elements of the E/R design do not appear in the relational version?

All of the elements appear. The translation is a straightforward mapping because we have an N:M relationship. Each entity becomes a table; the relationship becomes a table. Attributes in the ER model become attributes in the relational model. The only information we need to add are specific domain definitions for the attributes; we've chosen reasonable domains.

Answer:

Relational design for Person-Car-Accident

Box-and-arrow version:

SQL Schema:

CREATE TABLE People
(
	licenceNo       integer PRIMARY KEY,
	name            varchar(40),
	address         varchar(60)
);
CREATE TABLE Cars
(
	registrationNo  char(6) PRIMARY KEY, -- e.g. "ABC123"
	model           varchar(20),
	year            integer
);
CREATE TABLE Accidents
(
    reportNo        integer PRIMARY KEY,
	happenedAt      date,
	location        varchar(60)
);
CREATE TABLE Owns
	person          integer REFERENCES People(licenceNo),
	car             char(6) REFERENCES Cars(registrationNo),
	PRIMARY KEY     (person,car)
);
CREATE TABLE Involved
(
	accident        integer REFERENCES Accidents(reportNo),
	person          integer REFERENCES People(licenceNo),
	car             char(6) REFERENCES Cars(registrationNo),
	damage          money,
	PRIMARY KEY     (accident,person,car)
);

Which elements of the E/R design do not appear in the relational version?

At a syntactic level, all of entities, relationships and attributes are explicitly represented in the relational schema. At a semantic level, the total participation constraints on Person and Car in the Owns relation are not represented, so that there could be people in the database who do not own a car, and cars that are not owned by anyone.

Note that it is not possible to express these constraints in standard SQL. They would need to be implemented by e.g. stored trigger procedures.

Answer:

[Based on GUW 2.1.3] Relational design for Teams-Players-Fans

Box-and-arrow version:

SQL schema:

We use plural-ised names for tables

CREATE TABLE Teams
(
	name         varchar(50) PRIMARY KEY,
	captain      varchar(40) NOT NULL REFERENCES Players(name)
);
CREATE TABLE Players
(
	name         varchar(40) PRIMARY KEY,
	team         varchar(50) NOT NULL REFERENCES Teams(name)
);
CREATE TABLE Fans
(
	name         varchar(40) PRIMARY KEY,
);
CREATE TABLE TeamColours
(
	team         varchar(50) REFERENCES Teams(name),
	colour       varchar(30),
	PRIMARY KEY  (team,colour)
);
CREATE TABLE FavTeams
(
	fan          varchar(50) REFERENCES Fans(name),
	team         varchar(50) REFERENCES Teams(name),
	PRIMARY KEY  (fan,team)
);
CREATE TABLE FavPlayers
(
	fan             varchar(50) REFERENCES Fans(name),
	player          varchar(50) REFERENCES Players(name),
	PRIMARY KEY     (fan,player)
);
CREATE TABLE FavColours
(
	fan             varchar(50) REFERENCES Fans(name),
	colour          varchar(30),
	PRIMARY KEY     (fan,colour)
);

Which elements of the E/R design do not appear in the relational version?

At a syntactic level, the multi-valued attributes from the E/R design do not appear directly in the relational model, but are replaced by tuples in the TeamColours and FavColours tables.

At a semantic level, it doesn't capture the total participation of the Team entity in the PlaysFor relationship. While all players have to play for a team, the diagram does not enforce that each team must have at least one player who plays for it (except indirectly via the fact that it has to have a captain).

It also doesn't require that a team has at least one colour or that a fan has any favourite colours. Of course, the E/R diagram doesn't imply this either (non-key attributes are not required to have a value), but if it did state this, the relational model as given could not capture it.

The above SQL schema is simple, but doesn't actually load because of the mutual interdependence of Player and Team. To fix this, you would need something like the following:

-- create Team without the foreign key and then add it once Player exists

CREATE TABLE Teams
(
	name      varchar(50) PRIMARY KEY,
	captain   varchar(40) NOT NULL
);
CREATE TABLE Players
(
	name      varchar(40) PRIMARY KEY,
	team      varchar(50) NOT NULL REFERENCES Teams(name)
);
ALTER TABLE Teams ADD FOREIGN KEY (captain) REFERENCES Players(name);

-- alternatively, move the captain foreign key to the Player table
--   which is allowed because it's a 1:1 mapping
-- this isn't as efficient because players who are not captain will
--   have a null value for the captain foreign key

CREATE TABLE Team
(
	name      varchar(50) PRIMARY KEY
);
CREATE TABLE Player
(
	name      varchar(40) PRIMARY KEY,
	team      varchar(50) NOT NULL REFERENCES Teams(name)
	captain   varchar(50) REFERENCES Teams(name)
);

Answer:

Relational design for Trucking Company

Box-and-arrow version:

SQL schema:

CREATE TABLE Truck
(
	truckNo         integer PRIMARY KEY,
	maxVolume   float,
	maxWeight   float
);
CREATE TABLE Trip
(
	tripNo      integer PRIMARY KEY,
	tripDate    date,
	truck       integer REFERENCES Truck(truckNo)
);
CREATE TABLE Store
(
	address     varchar(60) PRIMARY KEY,
	storeName   varchar(50)
);
CREATE TABLE Warehouse
(
	location    varchar(60) PRIMARY KEY
);
CREATE TABLE Shipment
(
	shipmentNo  integer PRIMARY KEY,
	volume      float,
	weight      float,
	trip        integer REFERENCES Trip(tripNo),
	store       varchar(60) REFERENCES Store(address)
);
CREATE TABLE Source
(
	trip        integer REFERENCES Trip(tripNo),
	warehouse   varchar(60) REFERENCES Warehouse(location),
	PRIMARY KEY (trip,warehouse)
);

Which elements of the E/R design do not appear in the relational version?

At a syntactic level, the 1:n relationships (Includes, Uses, Destination) do not appear as tables in the relational model. They are implemented by foreign keys in the table which has only one associated entity.

Answer:

Relational models for company ER model:

Box-and-arrows schema:

SQL schema:

create table Employee (
	ssn         integer,
	birthdate   date,
	name        varchar(50),
	worksFor    varchar(50) not null,
	primary key (ssn)
--	foreign key (worksFor) is added later
);

create table Department (
	name        varchar(50),
	phone       varchar(20),
	location    varchar(30),
	manager     integer not null unique,
	mdate       date,
	primary key (name),
	foreign key (manager) references Employee(ssn)
);

alter table Employee add
	foreign key (worksFor)
	references Department(name);

create table Project (
	pnum        integer,
	title       varchar(100),
	primary key (pnum)
);

create table Dependent (
	ssn         integer not null,
	name        varchar(50),
	birthdate   date,
	relation    varchar(10) check
	              (relation in ('spouse','child')),
	primary key (ssn,name),
	foreign key (ssn) references Employee(ssn)
);

create table Participation (
	ssn         integer,
	pnum        integer,
	"time"      integer, -- number of hours on project
	primary key (ssn,pnum),
	foreign key (pnum) references Project(pnum),
	foreign key (ssn) references Employee(ssn)
);

The reason why the foreign key constraint is added later is because there is a mutually recursive pair of foreign key references between Employee and Department. We can't add the foreign key reference until the relevant table exists, so we need to create one table first, without the foreign key, add the other table (which refers to the first table), and then add the foreign key reference from the first table to the second table.

The not null constraints on Department.manager, Employee.worksFor and Depdendent.ssn implement the total participation requirements from the ER model.

Answer:

SQL schemas for the Person class hierarchy in the medical scenario:

-- NOTE that these schemas include representations of
-- the relationships between various sub-classes of Person

-- Using ER-style mapping for subclasses of Person

create table Person (
	ssn         integer,
	name        varchar(50) not null,
	address     varchar(100),
	primary key (ssn)
);

-- subclasses are overlapping; a Person could thus be
-- in any combination of the Doctor, Patient or Pharmacist tables

create table Doctor (
	ssn         integer,
	yearsExp    integer,
	primary key (ssn),
	foreign key (ssn) references Person(ssn)
);

create table Specialties (
	doctor      integer,
	specialty   varchar(20) check
	              (specialty in ('Feet','Ears','Throat')),
	primary key (doctor,specialty),
	foreign key (doctor) references Doctor(ssn)
);

create table Patient (
	ssn         integer,
	birthdate   date,
	primaryPhys integer not null, -- total participation
	primary key (ssn),
	foreign key (ssn) references Person(ssn),
	foreign key (primaryPhys) references Doctor(ssn)
);

create table Pharmacist (
	ssn         integer,
	phName      varchar(30),
	phAddress   varchar(100),
	qual        varchar(30),
	primary key (ssn),
	foreign key (ssn) references Person(ssn)
--	foreign key (phName,phAddress) is added later
);


-- Using single-table-style mapping for subclasses of Person

create table Person (
	ssn         integer,
	name        varchar(50) not null,
	address     varchar(100),
	-- an Person can belong to any combination of subclasses
	isPatient   boolean,
	isDoctor    boolean,
	isPharmacist boolean,
	-- patient-specific attributes
	primaryPhys integer, -- total participation handled below
	birthdate   date,
	-- doctor-specific attributes
	-- ... none ... Specialities are in separate table
	-- pharmacist-specific attributes
	phName      varchar(30),
	phAddress   varchar(100),
	qual        varchar(30),
	primary key (ssn),
	foreign key (primaryPhys) references Person(ssn),
--	foreign key (phName,phAddress) is added later
--	constraint  ClassAttributeCheck ...
--	-- trying to write a boolean expression that determines
--	-- that there is an appropriate combination of subclass
--	-- flags (isX) and attribute values is extremely tedious,
--	-- so we don't even bother to try
);

Answer:

SQL schema for the rest of the medical scenario:

Box-and-arrow schema:

SQL Schema:

-- Assume the definitions of Person/Patient/Doctor/Pharmacist
-- as given above in the ER-style

create table Pharmacy (
	name        varchar(30),
	address     varchar(100),
	manager     integer not null,  -- total participation
	primary key (name,address),
	foreign key (manager) references Pharmacist(ssn)
);

alter table Pharmacist add
	foreign key (phName,phAddress)
	references Pharmacy(name,address);

create table Drug (
	tradename   varchar(40),
	formula     varchar(100),
	primary key (tradename)
);

-- if "treatDate" is date only, and is part of primary key,
-- then a doctor cannot treat a patient more than once/day.
-- if this is required, make "treatDate" as a timestamp
create table Treats (
	doctor      integer,
	patient     integer,
	treatDate   date,
	primary key (doctor,patient,treatDate),
	foreign key (doctor) references Doctor(ssn),
	foreign key (patient) references Patient(ssn)
);

-- if "prescDate" is date only, then cannot prescibe
-- the same drug more than once on the same day
create table Prescribes (
	doctor      integer,
	patient     integer,
	drug        varchar(40),
	prescDate   date,
	quantity    integer, -- float if mass/volume/...
	primary key (doctor,patient,drug,prescDate),
	foreign key (doctor) references Doctor(ssn),
	foreign key (patient) references Patient(ssn),
	foreign key (drug) references Drug(tradename)
);

create table SoldIn (
	drug        varchar(40),
	phName      varchar(30),
	phAddress   varchar(100),
	price       money,
	primary key (drug,phName,phAddress),
	foreign key (phName,phAddress)
	              references Pharmacy(name,address),
	foreign key (drug) references Drug(tradename)
);

Which elements of the ER design do not appear in the relational version?

The relational model cannot represent (in standard SQL), the total participation constraints for patients (i.e. every patient must be treated by at least one doctor). This would need to be enforced via e.g. a stored trigger procedure. It also cannot represent the total participation constraint on the relationship WorksFor between pharmacists and pharmacies (i.e. every pharmacy must have at least one pharmacist working in it).

Answer:

SQL schema for book publishing scenario:

SQL Schema: using ER-style mapping of subclasses

create domain TaxFileNum as char(11)
		check (value ~ '^[0-9]{3}-[0-9]{3}-[0-9]{3}$');
create domain ISBNumber as char(15)
		check (value ~ '^[A-Z][0-9]{3}-[0-9]{4}-[0-9]{5}$');
create domain ABNumber as integer check (value > 100000);

create table Publisher (
	abn         ABNumber,
	name        varchar(60),
	address     varchar(100),
	primary key (abn)
);

create table Person (
	tfn         TaxFileNum,
	name        varchar(50),
	address     varchar(100),
	primary key (tfn)
);

create table Author (
	person      TaxFileNum,
	penname     varchar(50),
	primary key (person),
	foreign key (person) references Person(tfn)
);

create table Editor (
	person      TaxFileNum,
	publisher   ABNumber not null,
	primary key (person),
	foreign key (person) references Person(tfn),
	foreign key (publisher) references Publisher(abn)
);

create table Book (
	isbn        ISBNumber,
	title       varchar(100),
	edition     integer check (edition > 0),
	editor      TaxFileNum not null,
	publisher   ABNumber not null,
	primary key (isbn),
	foreign key (editor) references Editor(person),
	foreign key (publisher) references Publisher(abn)
);

create table Writes (
	author      TaxFileNum,
	book        ISBNumber,
	primary key (author,book),
	foreign key (author) references Author(person),
	foreign key (book) references Book(isbn)
);

SQL Schema: using single-table-style mapping of subclasses

-- Uses single-table-style mapping for subclasses of Person

create domain TaxFileNum as char(11)
		check (value ~ '^[0-9]{3}-[0-9]{3}-[0-9]{3}$');
create domain ISBNumber as char(15)
		check (value ~ '^[A-Z][0-9]{3}-[0-9]{4}-[0-9]{5}$');
create domain ABNumber as integer check (value > 100000);

create table Publisher (
	abn         ABNumber,
	name        varchar(60),
	address     varchar(100),
	primary key (abn)
);

create table Person (
	tfn         TaxFileNum,
	name        varchar(50),
	address     varchar(100),
	kind        varchar(10) check (kind in ('author','editor')),
	-- attributes for Authors
	penname     varchar(50),
	-- attributes for Editors
	publisher   ABNumber not null,
	primary key (tfn),
	foreign key (publisher) references Publisher(abn),
	constraint  NoPenNameIfEditor check
			((kind = 'author' and publisher is null) or
			 (kind = 'editor' and penname is null))
);
-- Problem with the above:
-- * publisher attribute defined to be not null
-- * if author type, publisher is required to be null
-- * to resolve this, we have to lose one of the constraints
--   - either lose total participation of Editor with Publisher
--   - or lose check on null poublisher for authors

create table Book (
	isbn        ISBNumber,
	title       varchar(100),
	edition     integer check (edition > 0),
	editor      TaxFileNum not null,
	publisher   ABNumber not null,
	primary key (isbn),
	foreign key (editor) references Person(tfn),
	foreign key (publisher) references Publisher(abn)
);

create table Writes (
	author      TaxFileNum,
	book        ISBNumber,
	primary key (author,book),
	foreign key (author) references Person(tfn),
	foreign key (book) references Book(isbn)
);