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Florida International University
COP 4710 – Database Management
Final Project
Value: 100 points (30% weight of the grading of the course)
Due Date: July 24, 2022 11:59 pm
Instructor: Lenis Hernandez
In this project, you are required to design and implement a database for a real miniworld environment
based on the given database requirements. The deliverable for this assignment must include:
1. The ER diagram, relational schemas and the corresponding relation instance for the database.
2. The SQL scripts for the database creation and data population.
3. Your deliverable for this assignment should include the SQL scripts for each query question and
corresponding answer (you can include the query results as plain text files in a readable format).
This is an individual project.
You are given a set of requirements for a hospital system database. Based on the set of requirements
for the hospital system database, you are asked to do the following:
1. Design the conceptual schema for the hospital system database by using an ER diagram (you can use
UML format). Your conceptual design of the database should include the followings but not limit to:
a. Entities
b. Relationships
c. Keys
d. Structural constraints (Cardinality ratio and participation constraints) (20 points)
2. Transform the ER schema of database you get from step 1 into the corresponding relational
database schema. Example Fig. 5.8 page 172 of the textbook. (10 points)
a. Specify all the key attributes of relations and any referential integrity constraints.
b. Specify the data item format for each attribute in each relation schema.
c. Specify all the functional dependencies you could infer from the requirements.
3. Normalize relation schema in the database design that you get from step 4 into either 3NF or
BCNF if it is necessary. (10 points)
4. Implement the relational database you get in step 5, via PostgreSQL, this includes creating the
database, creating the corresponding relation schemas, data preparation and loading data into the
database. (30 points)
5. Implement the given queries using PostgreSQL. Provide the SQL script for each query (30 points)
The requirements for a hospital system database:
In a hospital we represent data about patients, patient treatments, medicine, doctors, nurses and other
employees. These are the requirements:
1. The database keeps track of each patient’s name, last name, SSN, Date of birth, age, address
(house/apt number, street, city, state, zip code), phone, sex (gender), date admitted, date
discharged. Each patient has a unique identifier.
2. Each patient is assigned to a room.
3. The database will keep track if a patient was admitted through the Emergency Room (ER) for
history record purposes.
4. Each room is described by room number, and room type (single bed, two bed).
5. The patient is billed a treatment and the treatment is formed by the list of medicines used in the
treatment including the quantity of each medicine.
6. Each medicine is described by code, description and price.
7. The database keeps track of three types of employees: doctors, nurses and receptionists. Each
employee can only belong to one of these types. Each employee has a name, last name, SSN, sex
(gender), address (house/apt number, street, city, state, zip code), phone number, salary, birth
date and employee number.
8. Every doctor has a specialty that can be internist, cardiologist, pulmonologist, nephrologist, ENT,
neurologist, neurologist or endocrinologist.
9. A doctor can attend more than one patient.
10. A nurse governs a room.
11. The receptionist maintains a record of the patient and this record contains the record number,
the patient id, the appointment date and observations about the patient.
Hint:
1. For any unspecified requirements, add the appropriate assumptions to make the specification
complete.
2. You may want to identify multi-value attributes, composite attribute, and multi-valued
composite attributes.
Queries (2 points each one):
1. List the last name, name, employee number, type of employee of all employees ordered by last
name.
2. List the last name, name, employee number, type of employee of all employees ordered by last
name grouped by employee type.
3. List the name, last name, employee number, Specialty of doctors. Group by specialty and order
by last name.
4. List the count of doctors per specialty order the list by specialty name.
5. List the name, last name, employee number of all the nurses.
6. List the employees name, last name, employee type, salary with salaries greater than 85K.
7. List the name, last name, sex, patient id and room number of all the patients not discharged yet
and who are older than 65 years old.
8. List the patient name, last name, patient id of patients discharged in one specific month
(specified the month based on the data you used to populate the database).
9. List the name and last name and room number of patients admitted through the ER and already
discharged
10. List the name and last name, assigned room, room type and assigned nurse name and last name
of patients not discharged yet.
11. List the nurse name, nurse last name and average patient they took care per month.
12. List the doctor name and last name and average patient attended per month for all the doctors.
13. List the specialty and number of doctors for the specialty that has more than 3 doctors that
make more than 100K a year.
14. List the room number of any empty room.
15. List the average cost of a treatment.
Notes:
For the generation of the data, you can use: www.mockaroo.com or generatedata.com
Question 1
Question 2
COP 4710
Database Management
Instructor: Lenis Hernandez
Basics of Functional Dependencies and
Normalization for Relational Databases
Topics covered
1 Informal Design Guidelines for Relational Databases
1.1 Semantics of the Relation Attributes
1.2 Redundant Information in Tuples and Update Anomalies
1.3 Null Values in Tuples
1.4 Spurious Tuples
2 Functional Dependencies (FDs)
2.1 Definition of Functional Dependency
3 Normal Forms Based on Primary Keys
3.1 Normalization of Relations
3.2 Practical Use of Normal Forms
3.3 Definitions of Keys and Attributes Participating in Keys
3.4 First Normal Form
3.5 Second Normal Form
3.6 Third Normal Form
4 General Normal Form Definitions for 2NF and 3NF (For Multiple Candidate Keys)
5 BCNF (Boyce-Codd Normal Form)
1. 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?
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)
1.1 Semantics of the Relational Attributes
must be clear
• 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.
Figure 14.1 A simplified COMPANY relational
database schema
Figure 14.1 A
simplified COMPANY
relational database
schema.
1.2 Redundant Information in Tuples and
Update Anomalies
• Information is stored redundantly
• Wastes storage
• Causes problems with update anomalies
• Insertion anomalies
• Deletion anomalies
• Modification anomalies
EXAMPLE OF AN UPDATE ANOMALY
• Consider the relation:
• EMP_PROJ(Emp#, Proj#, Ename, Pname, No_hours)
• Update Anomaly:
• Changing the name of project number P1 from “Billing” to “CustomerAccounting” may cause this update to be made for all 100 employees working
on project P1.
EXAMPLE OF AN INSERT ANOMALY
• Consider the relation:
• EMP_PROJ(Emp#, Proj#, Ename, Pname, No_hours)
• Insert Anomaly:
• Cannot insert a project unless an employee is assigned to it.
• Conversely
• Cannot insert an employee unless an he/she is assigned to a project.
EXAMPLE OF A DELETE ANOMALY
• Consider the relation:
• EMP_PROJ(Emp#, Proj#, Ename, Pname, No_hours)
• 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.
Figure 14.3 Two relation schemas suffering
from update anomalies
Figure 14.3
Two relation schemas
suffering from update
anomalies. (a)
EMP_DEPT and (b)
EMP_PROJ.
Figure 14.4 Sample states for EMP_DEPT and
EMP_PROJ
Figure 14.4
Sample states for EMP_DEPT
and EMP_PROJ resulting from
applying NATURAL JOIN to the
relations in Figure 14.2. These
may be stored as base
relations for performance
reasons.
Guideline for 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 anomalies present, then note them so that applications can
be made to take them into account.
1.3 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
1.4 Generation of Spurious Tuples – avoid at
any cost
• 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.
Spurious Tuples (2)
•
There are two important properties of decompositions:
a) Non-additive or losslessness 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 Chapter 15).
2. Functional Dependencies
• Functional dependencies (FDs)
• Are used to specify formal measures of the “goodness” of relational designs
• And keys are used to define normal forms for relations
• 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
2.1 Defining Functional Dependencies
• 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
Examples of FD constraints (1)
• Social security number determines employee name
• SSN  ENAME
• Project number determines project name and location
• PNUMBER  {PNAME, PLOCATION}
• Employee ssn and project number determines the hours per week
that the employee works on the project
• {SSN, PNUMBER}  HOURS
Examples of FD constraints (2)
• An FD is a property of the attributes in the schema R
• The 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])
Defining FDs from instances
• Note that in order to define the FDs, we need to understand the
meaning of the attributes involved and the relationship between
them.
• An FD is a property of the attributes in the schema R
• Given the instance (population) of a relation, all we can conclude is
that an FD may exist between certain attributes.
• What we can definitely conclude is – that certain FDs do not exist
because there are tuples that show a violation of those
dependencies.
Figure 14.7 Ruling Out FDs
Note that given the state of the TEACH relation, we can say that
the FD: Text → Course may exist. However, the FDs Teacher →
Course, Teacher → Text and
Couse → Text are ruled out.
Figure 14.8 What FDs may exist?
• A relation R(A, B, C, D) with its extension.
• Which FDs may exist in this relation?
3 Normal Forms Based on Primary Keys
• 3.1 Normalization of Relations
• 3.2 Practical Use of Normal Forms
• 3.3 Definitions of Keys and Attributes Participating in Keys
• 3.4 First Normal Form
• 3.5 Second Normal Form
• 3.6 Third Normal Form
3.1 Normalization of Relations (1)
• Normalization:
• The process of decomposing unsatisfactory “bad” relations by breaking up
their attributes into smaller relations
• Normal form:
• Condition using keys and FDs of a relation to certify whether a relation
schema is in a particular normal form
Normalization of Relations (2)
• 2NF, 3NF, BCNF
• based on keys and FDs of a relation schema
• 4NF
• based on keys, multi-valued dependencies : MVDs;
• 5NF
• based on keys, join dependencies : JDs
• Additional properties may be needed to ensure a good relational
design (lossless join, dependency preservation; see Chapter 15)
3.2 Practical Use of Normal Forms
• Normalization is carried out in practice so that the resulting designs are of high
quality and meet the desirable properties
• The practical utility of these normal forms becomes questionable when the
constraints on which they are based are hard to understand or to detect
• The database designers need not normalize to the highest possible normal form
• (usually up to 3NF and BCNF. 4NF rarely used in practice.)
• Denormalization:
• The process of storing the join of higher normal form relations as a base relation—
which is in a lower normal form
3.3 Definitions of Keys and Attributes
Participating in Keys (1)
• A superkey of a relation schema R = {A1, A2, …., An} is a set of
attributes S subset-of R with the property that no two tuples t1 and t2
in any legal relation state r of R will have t1[S] = t2[S]
• A key K is a superkey with the additional property that removal of any
attribute from K will cause K not to be a superkey any more.
Definitions of Keys and Attributes
Participating in Keys (2)
• If a relation schema has more than one key, each is called a candidate
key.
• One of the candidate keys is arbitrarily designated to be the primary key, and
the others are called secondary keys.
• A Prime attribute must be a member of some candidate key
• A Nonprime attribute is not a prime attribute—that is, it is not a
member of any candidate key.
3.4 First Normal Form
• Disallows
• composite attributes
• multivalued attributes
• nested relations; attributes whose values for an individual tuple are nonatomic
• Considered to be part of the definition of a relation
• Most RDBMSs allow only those relations to be defined that are in First
Normal Form
Figure 14.9 Normalization into 1NF
Figure 14.9
Normalization into 1NF. (a)
A relation schema that is not
in 1NF. (b) Sample state of
relation DEPARTMENT. (c)
1NF version of the same
relation with redundancy.
Figure 14.10 Normalizing nested relations into
1NF
Figure 14.10
Normalizing nested relations into 1NF. (a) Schema of the EMP_PROJ relation with a
nested relation attribute PROJS. (b) Sample extension of the EMP_PROJ relation
showing nested relations within each tuple. (c) Decomposition of EMP_PROJ into
relations EMP_PROJ1 and EMP_PROJ2 by propagating the primary key.
3.5 Second Normal Form (1)
• Uses the concepts of FDs, primary key
• Definitions
• Prime attribute: An attribute that is member of the primary key K
• Full functional dependency: a FD Y -> Z where removal of any attribute from Y
means the FD does not hold any more
• Examples:
• {SSN, PNUMBER} -> HOURS is a full FD since neither SSN -> HOURS nor PNUMBER ->
HOURS hold
• {SSN, PNUMBER} -> ENAME is not a full FD (it is called a partial dependency ) since
SSN -> ENAME also holds
Second Normal Form (2)
• A relation schema R is in second normal form (2NF) if every nonprime attribute A in R is fully functionally dependent on the primary
key
• R can be decomposed into 2NF relations via the process of 2NF
normalization or “second normalization”
Figure 14.11 Normalizing into 2NF and 3NF
Figure 14.11
Normalizing into 2NF and 3NF.
(a) Normalizing EMP_PROJ into
2NF relations. (b) Normalizing
EMP_DEPT into 3NF relations.
Figure 14.12 Normalization into 2NF and 3NF
Figure 14.12
Normalization into 2NF
and 3NF. (a) The LOTS
relation with its
functional dependencies
FD1 through FD4.
(b) Decomposing into
the 2NF relations LOTS1
and LOTS2. (c)
Decomposing LOTS1
into the 3NF relations
LOTS1A and LOTS1B.
(d) Progressive
normalization of LOTS
into a 3NF design.
3.6 Third Normal Form (1)
• Definition:
• Transitive functional dependency: a FD X -> Z that can be derived from two
FDs X -> Y and Y -> Z
• Examples:
• SSN -> DMGRSSN is a transitive FD
• Since SSN -> DNUMBER and DNUMBER -> DMGRSSN hold
• SSN -> ENAME is non-transitive
• Since there is no set of attributes X where SSN -> X and X -> ENAME
Third Normal Form (2)
• A relation schema R is in third normal form (3NF) if it is in 2NF and no non-prime
attribute A in R is transitively dependent on the primary key
• R can be decomposed into 3NF relations via the process of 3NF normalization
• NOTE:
• In X -> Y and Y -> Z, with X as the primary key, we consider this a problem only if Y is
not a candidate key.
• When Y is a candidate key, there is no problem with the transitive dependency .
• E.g., Consider EMP (SSN, Emp#, Salary ).
• Here, SSN -> Emp# -> Salary and Emp# is a candidate key.
Normal Forms Defined Informally
• 1st normal form
• All attributes depend on the key
• 2nd normal form
• All attributes depend on the whole key
• 3rd normal form
• All attributes depend on nothing but the key
4. General Normal Form Definitions (For
Multiple Keys) (1)
• The above definitions consider the primary key only
• The following more general definitions take into account relations
with multiple candidate keys
• Any attribute involved in a candidate key is a prime attribute
• All other attributes are called non-prime attributes.
4.1 General Definition of 2NF (For Multiple
Candidate Keys)
• A relation schema R is in second normal form (2NF) if every nonprime attribute A in R is fully functionally dependent on every key of
R
• In Figure 14.12 the FD
County_name → Tax_rate violates 2NF.
So second normalization converts LOTS into
LOTS1 (Property_id#, County_name, Lot#, Area, Price)
LOTS2 ( County_name, Tax_rate)
4.2 General Definition of Third Normal Form
• Definition:
• Superkey of relation schema R – a set of attributes S of R that contains a key
of R
• A relation schema R is in third normal form (3NF) if whenever a FD X → A
holds in R, then either:
• (a) X is a superkey of R, or
• (b) A is a prime attribute of R
• LOTS1 relation violates 3NF because
Area → Price ; and Area is not a superkey in LOTS1. (see Figure 14.12).
4.3 Interpreting the General Definition of Third
Normal Form
• Consider the 2 conditions in the Definition of 3NF:
A relation schema R is in third normal form (3NF) if whenever
a FD X → A holds in R, then either:
• (a) X is a superkey of R, or
• (b) A is a prime attribute of R
• Condition (a) catches two types of violations :
– one where a prime attribute functionally determines a
non-prime attribute. This catches 2NF violations due to non-full
functional dependencies.
-second, where a non-prime attribute functionally
determines a non-prime attribute. This catches 3NF violations
due to a transitive dependency.
4.3 Interpreting the General Definition of Third
Normal
Form
(2)
• ALTERNATIVE DEFINITION of 3NF: We can restate the definition as:
A relation schema R is in third normal form (3NF) if every
non-prime attribute in R meets both of these conditions:
• It is fully functionally dependent on every key of R
• It is non-transitively dependent on every key of R
Note that stated this way, a relation in 3NF also meets the
requirements for 2NF.
• The condition (b) from the last slide takes care of the
dependencies that “slip through” (are allowable to) 3NF but are
“caught by” BCNF which we discuss next.
5. BCNF (Boyce-Codd Normal Form)
• A relation schema R is in Boyce-Codd Normal Form (BCNF) if whenever an FD X
→ A holds in R, then X is a superkey of R
• Each normal form is strictly stronger than the previous one
• Every 2NF relation is in 1NF
• Every 3NF relation is in 2NF
• Every BCNF relation is in 3NF
• There exist relations that are in 3NF but not in BCNF
• Hence BCNF is considered a stronger form of 3NF
• The goal is to have each relation in BCNF (or 3NF)
Figure 14.13 Boyce-Codd normal form
Figure 14.13
Boyce-Codd normal form. (a) BCNF normalization of
LOTS1A with the functional dependency FD2 being lost in
the decomposition. (b) A schematic relation with FDs; it is
in 3NF, but not in BCNF due to the f.d. C → B.
Figure 14.14 A relation TEACH that is in 3NF
but not in BCNF
Figure 14.14
A relation TEACH that is in 3NF
but not BCNF.
Achieving the BCNF by Decomposition (1)
• Two FDs exist in the relation TEACH:
• fd1: { student, course} -> instructor
• fd2: instructor -> course
• {student, course} is a candidate key for this relation and that the dependencies
shown follow the pattern in Figure 14.13 (b).
• So this relation is in 3NF but not in BCNF
• A relation NOT in BCNF should be decomposed so as to meet this property, while
possibly forgoing the preservation of all functional dependencies in the
decomposed relations.
• (See Algorithm 15.3)
Achieving the BCNF by Decomposition (2)
• Three possible decompositions for relation TEACH
• D1: {student, instructor} and {student, course}
• D2: {course, instructor } and {course, student}
• D3: {instructor, course } and {instructor, student} ✓
• All three decompositions will lose fd1.
• We have to settle for sacrificing the functional dependency preservation. But we cannot
sacrifice the non-additivity property after decomposition.
• Out of the above three, only the 3rd decomposition will not generate spurious tuples after
join.(and hence has the non-additivity property).
• A test to determine whether a binary decomposition (decomposition into two relations) is nonadditive (lossless) is discussed under Property NJB on the next slide. We then show how the third
decomposition above meets the property.
Test for checking non-additivity of Binary
Relational Decompositions
• Testing Binary Decompositions for Lossless Join (Non-additive Join)
Property
• Binary Decomposition: Decomposition of a relation R into two relations.
• PROPERTY NJB (non-additive join test for binary decompositions): A
decomposition D = {R1, R2} of R has the lossless join property with respect to
a set of functional dependencies F on R if and only if either
• The f.d. ((R1 ∩ R2)  (R1- R2)) is in F+, or
• The f.d. ((R1 ∩ R2)  (R2 – R1)) is in F+.
Test for checking non-additivity of Binary
Relational Decompositions – cont
If you apply the NJB test to the 3 decompositions of the TEACH
relation:
• D1 gives Student  Instructor or Student  Course, none of which is
true.
• D2 gives Course  Instructor or Course  Student, none of which is
true.
• However, in D3 we get Instructor  Course or Instructor  Student.
Since Instructor  Course is indeed true, the NJB property is satisfied
and D3 is determined as a non-additive (good) decomposition.
General Procedure for achieving BCNF when a
relation fails BCNF
Here we make use the algorithm from Chapter 15 (Algorithm 15.5):
• Let R be the relation not in BCNF, let X be a subset-of R, and let X → A be the FD
that causes a violation of BCNF. Then R may be decomposed into two relations:
• (i) R –A and (ii) X υ A.
• If either R –A or X υ A. is not in BCNF, repeat the process.
Note that the f.d. that violated BCNF in TEACH was Instructor →Course. Hence its BCNF
decomposition would be :
(TEACH – COURSE) and (Instructor υ Course), which gives
the relations: (Instructor, Student) and (Instructor, Course) that we obtained before in
decomposition D3.
5. Multivalued Dependencies and Fourth
Normal Form (1)
Definition:
•
•
A multivalued dependency (MVD) X —>> Y specified on relation schema
R, where X and Y are both subsets of R, specifies the following constraint
on any relation state r of R: If two tuples t1 and t2 exist in r such that t1[X]
= t2[X], then two tuples t3 and t4 should also exist in r with the following
properties, where we use Z to denote (R 2 (X Ï… Y)):
•
t3[X] = t4[X] = t1[X] = t2[X].
•
t3[Y] = t1[Y] and t4[Y] = t2[Y].
•
t3[Z] = t2[Z] and t4[Z] = t1[Z].
An MVD X —>> Y in R is called a trivial MVD if (a) Y is a subset of X, or (b)
X Ï… Y = R.
Multivalued Dependencies and Fourth
Normal Form (3)
Definition:
•
A relation schema R is in 4NF with respect to a set of
dependencies F (that includes functional dependencies and
multivalued dependencies) if, for every nontrivial
multivalued dependency X —>> Y in F+, X is a superkey for R.
•
Note: F+ is the (complete) set of all dependencies (functional
or multivalued) that will hold in every relation state r of R that
satisfies F. It is also called the closure of F.
Figure 14.15 Fourth and fifth normal forms.
Figure 14.15
Fourth and fifth normal forms. (a) The EMP relation with two MVDs: Ename –>> Pname and Ename –>>
Dname. (b) Decomposing the EMP relation into two 4NF relations EMP_PROJECTS and EMP_DEPENDENTS.
(c) The relation SUPPLY with no MVDs is in 4NF but not in 5NF if it has the JD(R1, R2, R3). (d)
Decomposing the relation SUPPLY into the 5NF relations R1, R2, R3.
6. Join Dependencies and Fifth Normal
Form (1)
Definition:
•
A join dependency (JD), denoted by JD(R1, R2, …, Rn),
specified on relation schema R, specifies a constraint on the
states r of R.
•
•
•
The constraint states that every legal state r of R should have a
non-additive join decomposition into R1, R2, …, Rn; that is, for
every such r we have
* (R1(r), R2(r), …, Rn(r)) = r
Note: an MVD is a special case of a JD where n = 2.
A join dependency JD(R1, R2, …, Rn), specified on relation
schema R, is a trivial JD if one of the relation schemas Ri in
JD(R1, R2, …, Rn) is equal to R.
Join Dependencies and Fifth Normal Form
(2)
Definition:
• A relation schema R is in fifth normal form (5NF) (or
Project-Join Normal Form (PJNF)) with respect to a
set F of functional, multivalued, and join
dependencies if,
•
for every nontrivial join dependency JD(R1, R2, …, Rn) in F+
(that is, implied by F),
•
•
every Ri is a superkey of R.
Discovering join dependencies in practical databases with
hundreds of relations is next to impossible. Therefore, 5NF is
rarely used in practice.
COP 4710
Database Management
Instructor: Lenis Hernandez
Relational Calculus
Topics covered:
• Relational Calculus
• Tuple Relational Calculus
• Domain Relational Calculus
• Example Database Application (COMPANY)
• Overview of the QBE language (appendix D)
Relational Calculus
• A relational calculus expression creates a new relation, which is
specified in terms of variables that range over rows of the stored
database relations (in tuple calculus) or over columns of the stored
relations (in domain calculus).
• In a calculus expression, there is no order of operations to specify
how to retrieve the query result—a calculus expression specifies only
what information the result should contain.
• This is the main distinguishing feature between relational algebra and
relational calculus.
Relational Calculus (continued)
• Relational calculus is considered to be a nonprocedural or declarative
language.
• This differs from relational algebra, where we must write a sequence
of operations to specify a retrieval request; hence relational algebra
can be considered as a procedural way of stating a query.
Tuple Relational Calculus
• The tuple relational calculus is based on specifying a number of tuple variables.
• Each tuple variable usually ranges over a particular database relation, meaning
that the variable may take as its value any individual tuple from that relation.
• A simple tuple relational calculus query is of the form
{t | COND(t)}
• where t is a tuple variable and COND (t) is a conditional expression involving t.
• The result of such a query is the set of all tuples t that satisfy COND (t).
Tuple Relational Calculus (continued)
• Example: To find the first and last names of all employees whose salary is above
$50,000, we can write the following tuple calculus expression:
{t.FNAME, t.LNAME | EMPLOYEE(t) AND t.SALARY>50000}
• The condition EMPLOYEE(t) specifies that the range relation of tuple variable t is
EMPLOYEE.
• The first and last name (PROJECTION FNAME, LNAME) of each EMPLOYEE tuple t that
satisfies the condition t.SALARY>50000 (SELECTION  SALARY >50000) will be
retrieved.
The Existential and Universal Quantifiers
• Two special symbols called quantifiers can appear in formulas; these are the
universal quantifier () and the existential quantifier ().
• Informally, a tuple variable t is bound if it is quantified, meaning that it appears in
an ( t) or ( t) clause; otherwise, it is free.
• If F is a formula, then so are ( t)(F) and ( t)(F), where t is a tuple variable.
• The formula ( t)(F) is true if the formula F evaluates to true for some (at least one)
tuple assigned to free occurrences of t in F; otherwise ( t)(F) is false.
• The formula ( t)(F) is true if the formula F evaluates to true for every tuple (in the
universe) assigned to free occurrences of t in F; otherwise ( t)(F) is false.
The Existential and Universal Quantifiers (continued)
•  is called the universal or “for all” quantifier because every tuple in
“the universe of” tuples must make F true to make the quantified
formula true.
•  is called the existential or “there exists” quantifier because any
tuple that exists in “the universe of” tuples may make F true to make
the quantified formula true.
Example Query Using Existential Quantifier
• Retrieve the name and address of all employees who work for the ‘Research’ department. The
query can be expressed as :
{t.FNAME, t.LNAME, t.ADDRESS | EMPLOYEE(t) and ( d)
(DEPARTMENT(d) and d.DNAME=‘Research’ and d.DNUMBER=t.DNO) }
• The only free tuple variables in a relational calculus expression should be those that appear to the
left of the bar ( | ).
• In above query, t is the only free variable; it is then bound successively to each tuple.
• If a tuple satisfies the conditions specified in the query, the attributes FNAME, LNAME, and
ADDRESS are retrieved for each such tuple.
• The conditions EMPLOYEE (t) and DEPARTMENT(d) specify the range relations for t and d.
• The condition d.DNAME = ‘Research’ is a selection condition and corresponds to a SELECT
operation in the relational algebra, whereas the condition d.DNUMBER = t.DNO is a JOIN
condition.
Example Query Using Universal Quantifier
• Find the names of employees who work on all the projects controlled by department number 5. The query
can be:
{e.LNAME, e.FNAME | EMPLOYEE(e) and ( ( x)(not(PROJECT(x)) or not(x.DNUM=5)
OR ( ( w)(WORKS_ON(w) and w.ESSN=e.SSN and x.PNUMBER=w.PNO))))}
• Exclude from the universal quantification all tuples that we are not interested in by making the condition
true for all such tuples.
• The first tuples to exclude (by making them evaluate automatically to true) are those that are not in the
relation R of interest.
• In query above, using the expression not(PROJECT(x)) inside the universally quantified formula evaluates to
true all tuples x that are not in the PROJECT relation.
• Then we exclude the tuples we are not interested in from R itself. The expression not(x.DNUM=5) evaluates to
true all tuples x that are in the project relation but are not controlled by department 5.
• Finally, we specify a condition that must hold on all the remaining tuples in R.
( ( w)(WORKS_ON(w) and w.ESSN=e.SSN and x.PNUMBER=w.PNO)
Languages Based on Tuple Relational Calculus
• The language SQL is based on tuple calculus. It uses the basic block structure to
express the queries in tuple calculus:
• SELECT
• FROM
• WHERE
• SELECT clause mentions the attributes being projected, the FROM clause
mentions the relations needed in the query, and the WHERE clause mentions the
selection as well as the join conditions.
• SQL syntax is expanded further to accommodate other operations.
Languages Based on Tuple Relational Calculus (continued)
• Another language which is based on tuple calculus is QUEL which actually uses
the range variables as in tuple calculus. Its syntax includes:
• RANGE OF IS
• Then it uses
• RETRIEVE
• WHERE
• This language was proposed in the relational DBMS INGRES. (system is currently
still supported by Computer Associates – but the QUEL language is no longer
there).
The Domain Relational Calculus
• Another variation of relational calculus called the domain relational calculus, or simply, domain
calculus is equivalent to tuple calculus and to relational algebra.
• The language called QBE (Query-By-Example) that is related to domain calculus was developed
almost concurrently to SQL at IBM Research, Yorktown Heights, New York.
• Domain calculus was thought of as a way to explain what QBE does.
• Domain calculus differs from tuple calculus in the type of variables used in formulas:
• Rather than having variables range over tuples, the variables range over single values from
domains of attributes.
• To form a relation of degree n for a query result, we must have n of these domain variables— one
for each attribute.
The Domain Relational Calculus (continued)
• An expression of the domain calculus is of the form
{ x1, x2, . . ., xn |
COND(x1, x2, . . ., xn, xn+1, xn+2, . . ., xn+m)}
• where x1, x2, . . ., xn, xn+1, xn+2, . . ., xn+m are domain variables that range over
domains (of attributes)
• and COND is a condition or formula of the domain relational calculus.
Example Query Using Domain Calculus
Retrieve the birthdate and address of the employee whose name is ‘John B. Smith’.
• Query :
{uv | ( q) ( r) ( s) ( t) ( w) ( x) ( y) ( z)
(EMPLOYEE(qrstuvwxyz) and q=’John’ and r=’B’ and s=’Smith’)}
• Abbreviated notation EMPLOYEE(qrstuvwxyz) uses the
variables without the separating commas: EMPLOYEE(q,r,s,t,u,v,w,x,y,z)
• Ten variables for the employee relation are needed, one to range over the domain of each
attribute in order.
• Of the ten variables q, r, s, . . ., z, only u and v are free.
• Specify the requested attributes, BDATE and ADDRESS, by the free domain variables u for
BDATE and v for ADDRESS.
• Specify the condition for selecting a tuple following the bar ( | )—
• namely, that the sequence of values assigned to the variables qrstuvwxyz be a tuple of
the employee relation and that the values for q (FNAME), r (MINIT), and s (LNAME) be
‘John’, ‘B’, and ‘Smith’, respectively.
QBE: A Query Language Based on Domain Calculus (Appendix C)
• This language is based on the idea of giving an example of a query using “example
elements” which are nothing but domain variables.
• Notation: An example element stands for a domain variable and is specified as an
example value preceded by the underscore character.
• P. (called P dot) operator (for “print”) is placed in those columns which are
requested for the result of the query.
• A user may initially start giving actual values as examples, but later can get used
to providing a minimum number of variables as example elements.
QBE: A Query Language Based on Domain Calculus (Appendix C)
• The language is very user-friendly, because it uses minimal syntax.
• QBE was fully developed further with facilities for grouping,
aggregation, updating etc. and is shown to be equivalent to SQL.
• The language is available under QMF (Query Management Facility) of
DB2 of IBM and has been used in various ways by other products like
ACCESS of Microsoft, and PARADOX.
• For details, see Appendix C in the text.
QBE Examples
• QBE initially presents a relational schema as a “blank schema” in
which the user fills in the query as an example:
Example Schema as a QBE Query Interface
QBE Examples
• The following domain calculus query can be successively minimized by the user as
shown:
• Query :
{uv | ( q) ( r) ( s) ( t) ( w) ( x) ( y) ( z)
(EMPLOYEE(qrstuvwxyz) and q=‘John’ and r=‘B’ and s=‘Smith’)}
Four Successive Ways to Specify a QBE Query
QBE Examples
• Specifying complex conditions in QBE:
• A technique called the “condition box” is used in QBE to state more
involved Boolean expressions as conditions.
• The C.4(a) gives employees who work on either project 1 or 2,
whereas the query in C.4(b) gives those who work on both the
projects.
Complex Conditions with and without a condition
box as a part of QBE Query
Handling AND conditions in a QBE Query
JOIN in QBE : Examples
• The join is simply accomplished by using the same example element
(variable with underscore) in the columns being joined from different
(or same as in C.5 (b)) relation.
• Note that the Result is set us as an independent table to show
variables from multiple relations placed in the result.
Performing Join with common example elements
and use of a RESULT relation
AGGREGATION in QBE
• Aggregation is accomplished by using .CNT for count,.MAX, .MIN,
.AVG for the corresponding aggregation functions
• Grouping is accomplished by .G operator.
• Condition Box may use conditions on groups (similar to HAVING
clause in SQL – see Section 8.5.8)
AGGREGATION in QBE : Examples
NEGATION in QBE : Example
UPDATING in QBE : Examples
COP 4710
Database Management
Instructor: Lenis Hernandez
Relational Algebra
Topics covered:
• Relational Algebra
• Unary Relational Operations
• Relational Algebra Operations From Set Theory
• Binary Relational Operations
• Additional Relational Operations
• Examples of Queries in Relational Algebra
Relational Algebra Overview
• Relational algebra is the basic set of operations for the relational
model
• These operations enable a user to specify basic retrieval requests (or
queries)
• The result of an operation is a new relation, which may have been
formed from one or more input relations
• This property makes the algebra “closed” (all objects in relational algebra are
relations)
Relational Algebra Overview (continued)
• The algebra operations thus produce new relations
• These can be further manipulated using operations of the same algebra
• A sequence of relational algebra operations forms a relational
algebra expression
• The result of a relational algebra expression is also a relation that
represents the result of a database query (or retrieval request)
Brief History of Origins of Algebra
• Muhammad ibn Musa al-Khwarizmi (800-847 CE) – from Morocco wrote a book
titled al-jabr about arithmetic of variables
• Book was translated into Latin.
• Its title (al-jabr) gave Algebra its name.
• Al-Khwarizmi called variables “shay”
• “Shay” is Arabic for “thing”.
• Spanish transliterated “shay” as “xay” (“x” was “sh” in Spain).
• In time this word was abbreviated as x.
• Where does the word Algorithm come from?
• Algorithm originates from “al-Khwarizmi”
• Reference: PBS (http://www.pbs.org/empires/islam/innoalgebra.html)
Relational Algebra Overview
• Relational Algebra consists of several groups of operations
• Unary Relational Operations
• SELECT (symbol:  (sigma))
• PROJECT (symbol:  (pi))
• RENAME (symbol:  (rho))
• Relational Algebra Operations From Set Theory
• UNION (  ), INTERSECTION (  ), DIFFERENCE (or MINUS, – )
• CARTESIAN PRODUCT ( x )
• Binary Relational Operations
• JOIN (several variations of JOIN exist)
• DIVISION
• Additional Relational Operations
• OUTER JOINS, OUTER UNION
• AGGREGATE FUNCTIONS (These compute summary of information: for example, SUM, COUNT,
AVG, MIN, MAX)
Database State for COMPANY
• All examples discussed below refer to the COMPANY database shown here.
Unary Relational Operations: SELECT
• The SELECT operation (denoted by  (sigma)) is used to select a subset of the tuples from a
relation based on a selection condition.
• The selection condition acts as a filter
• Keeps only those tuples that satisfy the qualifying condition
• Tuples satisfying the condition are selected whereas the other tuples are discarded
(filtered out)
• Examples:
• Select the EMPLOYEE tuples whose department number is 4:
 DNO = 4 (EMPLOYEE)
• Select the employee tuples whose salary is greater than $30,000:
 SALARY > 30,000 (EMPLOYEE)
Unary Relational Operations: SELECT
• In general, the select operation is denoted by  (R)
where
• the symbol  (sigma) is used to denote the select operator
• the selection condition is a Boolean (conditional) expression specified on the attributes
of relation R
• tuples that make the condition true are selected
• appear in the result of the operation
• tuples that make the condition false are filtered out
• discarded from the result of the operation
Unary Relational Operations: SELECT (continued)
• SELECT Operation Properties
• The SELECT operation  (R) produces a relation S that has the same
schema (same attributes) as R
• SELECT  is commutative:
•  ( < condition2> (R)) =  ( < condition1> (R))
• Because of commutativity property, a cascade (sequence) of SELECT operations may be
applied in any order:
• ( ( (R)) =  ( ( ( R)))
• A cascade of SELECT operations may be replaced by a single selection with a conjunction
of all the conditions:
• (< cond2> ((R)) =  AND < cond2> AND < cond3>(R)))
• The number of tuples in the result of a SELECT is less than (or equal to) the number
of tuples in the input relation R
The following query results refer to this database state
Unary Relational Operations: PROJECT
• PROJECT Operation is denoted by  (pi)
• This operation keeps certain columns (attributes) from a relation and
discards the other columns.
• PROJECT creates a vertical partitioning
• The list of specified columns (attributes) is kept in each tuple
• The other attributes in each tuple are discarded
• Example: To list each employee’s first and last name and salary, the
following is used:
LNAME, FNAME,SALARY(EMPLOYEE)
Unary Relational Operations: PROJECT (cont.)
• The general form of the project operation is:
(R)
•  (pi) is the symbol used to represent the project operation
• is the desired list of attributes from relation R.
• The project operation removes any duplicate tuples
• This is because the result of the project operation must be a set of tuples
• Mathematical sets do not allow duplicate elements.
Unary Relational Operations: PROJECT (contd.)
• PROJECT Operation Properties
• The number of tuples in the result of projection (R) is always less or
equal to the number of tuples in R
• If the list of attributes includes a key of R, then the number of tuples in the result of
PROJECT is equal to the number of tuples in R
• PROJECT is not commutative
•  ( (R) ) =  (R) as long as contains the attributes in
Examples of applying SELECT and PROJECT operations
Relational Algebra Expressions
• We may want to apply several relational algebra operations one after
the other
• Either we can write the operations as a single relational algebra expression
by nesting the operations, or
• We can apply one operation at a time and create intermediate result
relations.
• In the latter case, we must give names to the relations that hold
the intermediate results.
Single expression versus sequence of relational operations
(Example)
• To retrieve the first name, last name, and salary of all employees who work in
department number 5, we must apply a select and a project operation
• We can write a single relational algebra expression as follows:
• FNAME, LNAME, SALARY( DNO=5(EMPLOYEE))
• OR We can explicitly show the sequence of operations, giving a name to each
intermediate relation:
• DEP5_EMPS   DNO=5(EMPLOYEE)
• RESULT   FNAME, LNAME, SALARY (DEP5_EMPS)
Unary Relational Operations: RENAME
• The RENAME operator is denoted by  (rho)
• In some cases, we may want to rename the attributes of a relation or
the relation name or both
• Useful when a query requires multiple operations
• Necessary in some cases (see JOIN operation later)
Unary Relational Operations: RENAME (continued)
• The general RENAME operation  can be expressed by any of the
following forms:
• S (B1, B2, …, Bn )(R) changes both:
• the relation name to S, and
• the column (attribute) names to B1, B1, …..Bn
• S(R) changes:
• the relation name only to S
• (B1, B2, …, Bn )(R) changes:
• the column (attribute) names only to B1, B1, …..Bn
Unary Relational Operations: RENAME (continued)
• For convenience, we also use a shorthand for renaming attributes in
an intermediate relation:
• If we write:
• RESULT   FNAME, LNAME, SALARY (DEP5_EMPS)
• RESULT will have the same attribute names as DEP5_EMPS (same
attributes as EMPLOYEE)
• If we write:
• RESULT (F, M, L, S, B, A, SX, SAL, SU, DNO)
 RESULT (F.M.L.S.B,A,SX,SAL,SU,
DNO)(DEP5_EMPS)
• The 10 attributes of DEP5_EMPS are renamed to F, M, L, S, B, A, SX,
SAL, SU, DNO, respectively
Note: the  symbol is an assignment operator
Example of applying multiple operations and RENAME
Relational Algebra Operations from Set Theory: UNION
• UNION Operation
• Binary operation, denoted by 
• The result of R  S, is a relation that includes all tuples that are either in R or
in S or in both R and S
• Duplicate tuples are eliminated
• The two operand relations R and S must be “type compatible” (or UNION
compatible)
• R and S must have same number of attributes
• Each pair of corresponding attributes must be type compatible (have same or
compatible domains)
Relational Algebra Operations from Set Theory: UNION
• Example:
• To retrieve the social security numbers of all employees who either work in department
5 (RESULT1 below) or directly supervise an employee who works in department 5
(RESULT2 below)
• We can use the UNION operation as follows:
DEP5_EMPS  DNO=5 (EMPLOYEE)
RESULT1   SSN(DEP5_EMPS)
RESULT2(SSN)  SUPERSSN(DEP5_EMPS)
RESULT  RESULT1  RESULT2
• The union operation produces the tuples that are in either RESULT1 or RESULT2 or both
Figure 8.3 Result of the UNION
operation RESULT ← RESULT1 ∪ RESULT2.
Relational Algebra Operations from Set Theory
• Type Compatibility of operands is required for the binary set operation UNION ,
(also for INTERSECTION , and SET DIFFERENCE –, see next slides)
• R1(A1, A2, …, An) and R2(B1, B2, …, Bn) are type compatible if:
• they have the same number of attributes, and
• the domains of corresponding attributes are type compatible (i.e. dom(Ai)=dom(Bi)
for i=1, 2, …, n).
• The resulting relation for R1R2 (also for R1R2, or R1–R2, see next slides) has
the same attribute names as the first operand relation R1 (by convention)
Relational Algebra Operations from Set Theory: INTERSECTION
• INTERSECTION is denoted by 
• The result of the operation R  S, is a relation
that includes all tuples that are in both R and S
• The attribute names in the result will be the
same as the attribute names in R
• The two operand relations R and S must be
“type compatible”
Relational Algebra Operations from Set Theory: SET DIFFERENCE
• SET DIFFERENCE (also called MINUS or EXCEPT) is
denoted by –
• The result of R – S, is a relation that includes all tuples
that are in R but not in S
• The attribute names in the result will be the
same as the attribute names in R
• The two operand relations R and S must be
“type compatible”
Example to illustrate the result of UNION, INTERSECT, and
DIFFERENCE
Some properties of UNION, INTERSECT, and DIFFERENCE
• Notice that both union and intersection are commutative operations; that is
• R  S = S  R, and R  S = S  R
• Both union and intersection can be treated as n-ary operations applicable to any
number of relations as both are associative operations; that is
• R  (S  T) = (R  S)  T
• (R  S)  T = R  (S  T)
• The minus operation is not commutative; that is, in general
• R–S≠S–R
Relational Algebra Operations from Set Theory: CARTESIAN
PRODUCT
• CARTESIAN (or CROSS) PRODUCT Operation
• This operation is used to combine tuples from two relations in a combinatorial
fashion.
• Denoted by R(A1, A2, . . ., An) x S(B1, B2, . . ., Bm)
• Result is a relation Q with degree n + m attributes:
• Q(A1, A2, . . ., An, B1, B2, . . ., Bm), in that order.
• The resulting relation state has one tuple for each combination of tuples—one from
R and one from S.
• Hence, if R has nR tuples (denoted as |R| = nR ), and S has nS tuples, then R x S will
have nR * nS tuples.
• The two operands do NOT have to be “type compatible”
Relational Algebra Operations from Set Theory: CARTESIAN
PRODUCT (cont.)
• Generally, CROSS PRODUCT is not a meaningful operation
• Can become meaningful when followed by other operations
• Example (not meaningful):
• FEMALE_EMPS   SEX=’F’(EMPLOYEE)
• EMPNAMES   FNAME, LNAME, SSN (FEMALE_EMPS)
• EMP_DEPENDENTS  EMPNAMES x DEPENDENT
• EMP_DEPENDENTS will contain every combination of EMPNAMES and
DEPENDENT
• whether or not they are actually related
Relational Algebra Operations from Set Theory: CARTESIAN
PRODUCT (cont.)
• To keep only combinations where the DEPENDENT is related to the
EMPLOYEE, we add a SELECT operation as follows
• Example (meaningful):
• FEMALE_EMPS   SEX=’F’(EMPLOYEE)
• EMPNAMES   FNAME, LNAME, SSN (FEMALE_EMPS)
• EMP_DEPENDENTS  EMPNAMES x DEPENDENT
• ACTUAL_DEPS   SSN=ESSN(EMP_DEPENDENTS)
• RESULT   FNAME, LNAME, DEPENDENT_NAME (ACTUAL_DEPS)
• RESULT will now contain the name of female employees and their dependents
Figure 8.5 The CARTESIAN PRODUCT (CROSS
PRODUCT) operation.
continued on next slide
Figure 8.5 (continued)
.
The CARTESIAN PRODUCT (CROSS
PRODUCT) operation
continued on next slide
Figure 8.5 (continued)
PRODUCT) operation.
The CARTESIAN PRODUCT (CROSS
Binary Relational Operations: JOIN
• JOIN Operation (denoted by
)
• The sequence of CARTESIAN PRODECT followed by SELECT is used quite commonly to
identify and select related tuples from two relations
• A special operation, called JOIN combines this sequence into a single operation
• This operation is very important for any relational database with more than a single
relation, because it allows us combine related tuples from various relations
• The general form of a join operation on two relations R(A1, A2, . . ., An) and S(B1, B2,
. . ., Bm) is:
R S
• where R and S can be any relations that result from general relational algebra
expressions.
Binary Relational Operations: JOIN (cont.)
• Example: Suppose that we want to retrieve the name of the
manager of each department.
• To get the manager’s name, we need to combine each
DEPARTMENT tuple with the EMPLOYEE tuple whose SSN value
matches the MGRSSN value in the department tuple.
• We do this by using the join
operation.
• DEPT_MGR  DEPARTMENT MGRSSN=SSN EMPLOYEE
• MGRSSN=SSN is the join condition
• Combines each department record with the employee who
manages the department
• The join condition can also be specified as DEPARTMENT.MGRSSN=
EMPLOYEE.SSN
Figure 8.6 Result of the JOIN operation
DEPT_MGR ← DEPARTMENT|X| Mgr_ssn=SsnEMPLOYEE.
Some properties of JOIN
• Consider the following JOIN operation:
• R(A1, A2, . . ., An)
S(B1, B2, . . ., Bm)
R.Ai=S.Bj
• Result is a relation Q with degree n + m attributes:
• Q(A1, A2, . . ., An, B1, B2, . . ., Bm), in that order.
• The resulting relation state has one tuple for each combination of tuples—r from R
and s from S, but only if they satisfy the join condition r[Ai]=s[Bj]
• Hence, if R has nR tuples, and S has nS tuples, then the join result will generally have
less than nR * nS tuples.
• Only related tuples (based on the join condition) will appear in the result
Some properties of JOIN
• The general case of JOIN operation is called a Theta-join: R
S
theta
• The join condition is called theta
• Theta can be any general boolean expression on the attributes of R
and S; for example:
• R.Ai
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