The Relational Data Model (Based on Chapter 5)

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The Relational Data Model(Based on Chapter 5)
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1. Relational Model Concepts

• BASIS OF THE MODEL
•  
•  
• The relational Model of Data is based on the
  concept of a Relation.
•  
• A Relation is a mathematical concept based on
  the ideas of sets.
•  
• The strength of the relational approach to
  data management comes from the formal foundation
  provided by the theory of relations.
•  
•  

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INFORMAL DEFINITIONS

• RELATION A table of values
•  
• A relation may be thought of as a set of rows.
•  
• A relation may alternately be though of as a
  set of columns.
•  
• Each row of the relation may be given an
  identifier.
•  
• Each column typically is called by its column
  name or column header or attribute name.

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FORMAL DEFINITIONS

• A Relation may be defined in multiple ways.
•  
• The Schema of a Relation R (A1, A2, .....An)
• Relation R is defined over attributes A1, A2,
  .....An
•  
• For Example -
• CUSTOMER (Cust-id, Cust-name, Address, Phone)
•  
• Here, CUSTOMER is a relation defined over the
  four attributes Cust-id, Cust-name, Address,
  Phone, each of which has a domain or a set of
  valid values.

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• For example, the domain of Cust-id is 6 digit
  numbers.
• A tuple is an ordered set of values
•  
• Each value is derived from an appropriate
  domain.
•  
• Each row in the CUSTOMER table may be called
  as a tuple in the table and would consist of four
  values.
•  
• lt632895, "John Smith", "101 Main St. Atlanta, GA
  30332", "[phone]"gt is a triple belonging
  to the CUSTOMER relation.
•  
• A relation may be regarded as a set of tuples
  (rows).
• Columns in a table are also called as
  attributes of the relation.

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FORMAL DEFINITIONS (contd..)

• The relation is formed over the cartesian
  product of the sets each set has values from a
  domain that domain is used in a specific role
  which is conveyed by the attribute name.
• For example, attribute Cust-name is defined
  over the domain of strings of 25 characters. The
  role these strings play in the CUSTOMER relation
  is that of the name of customers.
• Formally, Given R(A1, A2, .........., An)
• r(R) subset-of dom (A1) X dom (A2) X ....X
  dom(An)
•  
• R schema of the relation
• r of R a specific "value" or population of R.
•  

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• R is also called the intension of a relation
• r is also called the extension of a relation
• Let S1 0,1
• Let S2 a,b,c
•   Let R be a subset-of S1 X S2
•  
• for example r(R) lt0.agt , lt0,bgt , lt1,cgt

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DEFINITION SUMMARY

• Informal Terms Formal Terms
• Table Relation
• Column Attribute/Domain
• Row Tuple
• Values in a column Domain
• Table Definition Schema of Relation
• Populated Table Extension

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Figure 7.1 The attributes and tuples of a
relation STUDENT.
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2 Characteristics of Relations

• Ordering of tuples in a relation r(R) The tuples
  are not considered to be ordered, even though
  they appear to be in the tabular form.
•  
• Ordering of attributes in a relation schema R
  (and of values within each tuple) We will
  consider the attributes in R(A1, A2, ..., An) and
  the values in tltv1, v2, ..., vngt to be ordered .
•  
•  

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• Values in a tuple All values are considered
  atomic (indivisible). A special null value is
  used to represent values that are unknown or
  inapplicable to certain tuples.
• Notation
• - We refer to component values of a tuple t by
  
• tAi vi (the value of attribute Ai for tuple
  t).
• - Similarly, tAu, Av, ..., Aw refers to the
  subtuple of t containing the values of attributes
  Au, Av, ..., Aw, respectively.

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Figure 7.2 The relation STUDENT from Figure 7.1,
with a different order of tuples
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3 Relational Integrity Constraints

• Constraints are conditions that must hold on all
  valid relation instances. There are three main
  types of constraints
• Key constraints
• Entity integrity constraints,
• Referential integrity constraints

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3.1 Key Constraints

• Superkey of R A set of attributes SK of R such
  that no two tuples in any valid relation instance
  r(R) will have the same value for SK. That is,
  for any distinct tuples t1 and t2 in r(R), t1SK
  ltgt t2SK.
• Key of R A "minimal" superkey that is, a
  superkey K such that removal of any attribute
  from K results in a set of attributes that is not
  a superkey.

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• Example The CAR relation schema
• CAR(State, Reg, SerialNo, Make, Model, Year)
• has two keys Key1 State, Reg, Key2
  SerialNo,
• which are also superkeys. SerialNo, Make is a
  superkey but not a key.
• If a relation has several candidate keys, one is
  chosen arbitrarily to be the primary key.
• The primary key attributes are underlined.

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Figure 7.4 The CAR relation with two candidate
keys LicenseNumber and EngineSerialNumber.
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Figure 7.5 Schema diagram for the COMPANY
relational database schema the primary keys are
underlined.
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Figure 7.5 Schema diagram for the COMPANY
relational database schema the primary keys are
underlined.
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Figure 7.6 (continued)
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3.2 Entity Integrity

• Relational Database Schema A set S of relation
  schemas that belong to the same database. S is
  the name of the database.
• S R1, R2, ..., Rn
•  
• Entity Integrity The primary key attributes PK
  of each relation schema R in S cannot have null
  values in any tuple of r(R). This is because
  primary key values are used to identify the
  individual tuples.
• tPK ltgt null for any tuple t in r(R)
•  
• Note Other attributes of R may be similarly
  constrained to disallow null values, even though
  they are not members of the primary key.

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3.3 Referential Integrity

• A constraint involving two relations (the
  previous constraints involve a single relation).
•  
• Used to specify a relationship among tuples in
  two relations the referencing relation and the
  referenced relation.
•  
• Tuples in the referencing relation R1 have
  attributes FK (called foreign key attributes)
  that reference the primary key attributes PK of
  the referenced relation
• R2. A tuple t1 in R1 is said to reference a tuple
  t2 in R2 if t1FK t2PK.
•  
• A referential integrity constraint can be
  displayed in a relational database schema as a
  directed arc from R1.FK to R2.

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Figure 7.7 Referential integrity constraints
displayed on the COMPANY relational database
schema diagram.
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Figure 7.6 One possible relational database
state corresponding to the company schema.
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4 Update Operations on Relations

• - INSERT a tuple.
• - DELETE a tuple.
• - MODIFY a tuple.
•  
• - Integrity constraints should not be
  violated by the update operations.
•  
• - Several update operations may have to be
  grouped together.
•  
• - Updates may propagate to cause other
  updates automatically. This may be necessary to
  maintain integrity constraints.
•  
• -

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• In case of integrity violation, several actions
  can be taken
• - cancel the operation that causes the violation
  (REJECT optiom)
• - perform the operation but inform the user of
  the violation
• - trigger additional updates so the violation is
  corrected (CASCADE option, SET NULL option)
• - execute a user-specified error-correction
  routine

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5 The Relational Algebra

• - Operations to manipulate relations.
• - Used to specify retrieval requests
  (queries).
• - Query result is in the form of a relation.
•  
• Relational Operations
• 5.1 SELECT s and PROJECT P
  operations.
• 5.2 Set operations These include
  UNION U, INTERSECTION ,
• DIFFERENCE -, CARTESIAN
  PRODUCT X.
• 5.3 JOIN operations X.
• 5.4 Other relational operations DIVISION,
  OUTER JOIN, AGGREGATE FUNCTIONS.

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5.1 SELECT s and PROJECT P

• SELECT operation (denoted bys )
•  
• - Selects the tuples (rows) from a relation
  R that satisfy a certain selection condition c
• - Form of the operation s c(R)
•  
• - The condition c is an arbitrary Boolean
  expression on the attributes of R
•  
• - Resulting relation has the same attributes
  as R
•  
• - Resulting relation includes each tuple in
  r(R) whose attribute values satisfy the condition
  c
•  

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• Examples
• s DNO4 (EMPLOYEE)
• s SALARYgt30000 (EMPLOYEE)
• s(DNO4 AND SALARYgt25000) OR DNO5(EMPLOYEE)

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• PROJECT operation (denoted byP )
•  
• - Keeps only certain attributes (columns)
  from a relation R specified in an attribute list
  L
•  
• - Form of operation P L(R)
•  
• - Resulting relation has only those
  attributes of R specified in L
•  
• Example P FNAME,LNAME,SALARY(EMPLOYEE)
•  
• - The PROJECT operation eliminates duplicate
  tuples in the resulting relation so that it
  remains a mathematical set (no duplicate
  elements)

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• Example P SEX,SALARY(EMPLOYEE)
• If several male employees have salary 30000, only
  a single tuple ltM, 30000gt is
• kept in the resulting relation.
• Duplicate tuples are eliminated by the P
  operation.

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• Sequences of operations
•  
• - Several operations can be combined to form
  a relational algebra expression (query)
• Example Retrieve the names and salaries of
  employees who work in department 4
• P FNAME,LNAME,SALARY (s DNO4(EMPLOYEE) )
•  
• - Alternatively, we specify explicit
  intermediate relations for each step
• DEPT4_EMPS lt-s DNO4(EMPLOYEE)
• R lt-P FNAME,LNAME,SALARY(DEPT4_EMPS)
•  
• -

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• Attributes can optionally be renamed in the
  resulting left-hand-side relation (this may be
  required for some operations that will be
  presented later)
• DEPT4_EMPS lt-s DNO4(EMPLOYEE)
• R(FIRSTNAME,LASTNAME,SALARY) lt-
• P FNAME,LNAME,SALARY(DEPT4_EMPS)

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Figure 7.8 Results of SELECT and PROJECT
operations.(a) ?(DNO4 AND SALARYgt25000) OR
(DNO5 AND SALARYgt30000)(EMPLOYEE).(b) ?LNAME,
FNAME, SALARY(EMPLOYEE). (c) ?SEX,
SALARY(EMPLOYEE).
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Class Number CS 304
Class Name - DBMS
Instructor Sanjay Madria
Lesson Title Relational Algebra 3rd July
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5.2 Set Operations

• - Binary operations from mathematical set
  theory
• UNION R1 U R2,
• INTERSECTION R1 R2,
• SET DIFFERENCE R1 - R2,
• CARTESIAN PRODUCT R1 X R2.
•  
• - For U, , -, the operand relations R1(A1,
  A2, ..., An) and R2(B1, B2, ..., Bn) must have
  the same number of attributes, and the domains of
  corresponding attributes must be compatible that
  is, dom(Ai)dom(Bi) for i1, 2, ..., n. This
  condition is called union compatibility.
•  
• The resulting relation for U, , or -
  has the same attribute names as the first
  operand relation R1 (by convention).

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Figure 7.10 Query result after the UNION
operationRESULT ? RESULT1 ? RESULT2
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Figure 7.11 Illustrating the set operations
union, intersection and difference.(a) Two union
compatible relations.(b) STUDENT ? INSTRUCTOR.
(c) STUDENT ? INSTRUCTOR(d) STUDENT - INSTRUCTOR
(e) INSTRUCTOR - STUDENT
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CARTESIAN PRODUCT

• R(A1, A2, ..., Am, B1, B2, ..., Bn) lt-
• R1(A1, A2, ..., Am) X R2 (B1, B2, ...,
  Bn)
•  
• - A tuple t exists in R for each combination
  of tuples t1 from R1 and t2 from R2 such that
• tA1, A2, ..., Amt1 and tB1, B2, ...,
  Bnt2
•  
• - If R1 has n1 tuples and R2 has n2 tuples,
  then R will have n1n2 tuples.
•  
• - CARTESIAN PRODUCT can combine related
  tuples from two relations if followed by the
  appropriate SELECT operation .
•  

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• Example Combine each DEPARTMENT tuple with the
  EMPLOYEE tuple of the manager.
• DEP_EMP lt-DEPARTMENT X EMPLOYEE
• DEPT_MANAGER lt-s MGRSSNSSN(DEP_EMP)

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5.3 JOIN Operations

• THETA JOIN Similar to a CARTESIAN PRODUCT
  followed by a SELECT.
• The condition c is called a join condition.
• R(A1, A2, ..., Am, B1, B2, ..., Bn) lt-R1(A1, A2,
  ..., Am) X c R2 (B1, B2, ..., Bn)
• Here c can be lt, gt, , lt, gt
•  
• EQUIJOIN The join condition c includes one or
  more equality comparisons involving attributes
  from R1 and R2. That is, c is of the form
• (AiBj) AND ... AND (AhBk) 1lti,hltm, 1ltj,kltn
•  
• In the above EQUIJOIN operation
• Ai, ..., Ah are called the join attributes of R1
• Bj, ..., Bk are called the join attributes of R2
•  

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• Example of using EQUIJOIN
• Retrieve each DEPARTMENT's name and its manager's
  name
• T lt-DEPARTMENT X MGRSSNSSN EMPLOYEE
• RESULT lt-P DNAME,FNAME,LNAME(T)

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• NATURAL JOIN ()
• In an EQUIJOIN R lt- R1 X c R2, the join
  attribute of R2 appear redundantly in the result
  relation R. In a NATURAL JOIN, the redundant join
  attributes of R2 are eliminated from R. The
  equality condition is implied and need not be
  specified.
• R lt- R1 (join attributes of R1),(join attributes
  of R2) R2
• Example Retrieve each EMPLOYEE's name and the
  name of the DEPARTMENT he/she
• works for
• Tlt- EMPLOYEE (DNO),(DNUMBER) DEPARTMENT
• RESULT lt-P FNAME,LNAME,DNAME(T)
•  

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• If the join attributes have the same names in
  both relations, they need not be specified and
  we can write R lt- R1 R2.
• Example Retrieve each EMPLOYEE's name and the
  name of his/her SUPERVISOR
• SUPERVISOR(SUPERSSN,SFN,SLN)lt-P
  SSN,FNAME,LNAME(EMPLOYEE)
• Tlt-EMPLOYEE SUPERVISOR
• RESULT lt-P FNAME,LNAME,SFN,SLN(T)

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Figure 7.13 Illustrating the JOIN operation.
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Figure 7.14 An illustration of the NATURAL JOIN
operation. (a) PROJ_DEPT ? PROJECT DEPT.(b)
DEPT_LOCS ? DEPARTMENT DEPT_LOCATIONS.
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• Note In the original definition of NATURAL
  JOIN, the join attributes were required to have
  the same names in both relations.
•  
• There can be a more than one set of join
  attributes with a different meaning between the
  same two relations. For example
•  
• JOIN ATTRIBUTES RELATIONSHIP
• EMPLOYEE.SSN EMPLOYEE manages
• DEPARTMENT.MGRSSN the DEPARTMENT
• EMPLOYEE.DNO EMPLOYEE works for
• DEPARTMENT.DNUMBER the DEPARTMENT
•  

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• A relation can have a set of join attributes to
  join it with itself
•  
• JOIN ATTRIBUTES RELATIONSHIP
• EMPLOYEE(1).SUPERSSN EMPLOYEE(2) supervises
• EMPLOYEE(2).SSN EMPLOYEE(1)
•  
• - One can think of this as joining two
  distinct copies of the relation, although only
  one relation actually exists
•  
• - In this case, renaming can be useful
•  

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Figure 7.15 Illustrating the division
operation.(a) Dividing SSN_PNOS by SMITH_PNOS.
(b) T ? R ? S.
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• Complete Set of Relational Algebra Operations
•  
• - All the operations discussed so far can
  be described as a sequence of only the
  operations SELECT, PROJECT, UNION, SET
  DIFFERENCE, and CARTESIAN PRODUCT.
•  
• - Hence, the set s ,P , U, - , X is
  called a complete set of relational algebra
  operations. Any query language equivalent to
  these operations is called relationally complete.
•  
• - For database applications, additional
  operations are needed that were not part of the
  original relational algebra. These include
• 1. Aggregate functions and grouping.
• 2. OUTER JOIN.

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5.4 Additional Relational Operations

• AGGREGATE FUNCTIONS
• - Functions such as SUM, COUNT, AVERAGE,
  MIN, MAX are often applied to sets of values or
  sets of tuples in database applications
• ltgrouping attributesgt Fltfunction listgt (R)
•  
• - The grouping attributes are optional
•  
• Example 1 Retrieve the average salary of all
  employees (no grouping needed)
• R(AVGSAL) lt- F AVERAGE SALARY (EMPLOYEE)
•  

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• Example 2 For each department, retrieve the
  department number, the number of employees, and
  the average salary (in the department)
• R(DNO,NUMEMPS,AVGSAL) lt-
• DNO F COUNT SSN, AVERAGE SALARY (EMPLOYEE)
•  
• DNO is called the grouping attribute in the
  above example

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Figure 7.16 An illustration of the AGGREGATE
FUNCTION operation.(a) R(DNO, NO_OF_EMPLOYEES,
AVERAGE_SAL) ? DNO ? COUNT SSN, AVERAGE
SALARY(EMPLOYEE). (b) DNO ? COUNT SSN, AVERAGE
SALARY(EMPLOYEE).(C) ? COUNT SSN, AVERAGE
SALARY(EMPLOYEE).
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• OUTER JOIN
•  
• - In a regular EQUIJOIN or NATURAL JOIN
  operation, tuples in R1 or R2 that do not have
  matching tuples in the other relation do not
  appear in the result
•  
• - Some queries require all tuples in R1 (or
  R2 or both) to appear in the result
•  
• - When no matching tuples are found, nulls
  are placed for the missing attributes
•  
• -

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• LEFT OUTER JOIN R1 X R2 lets every tuple in
  R1 appear in the result
•  
• - RIGHT OUTER JOIN R1 X R2 lets every
  tuple in R2 appear in the result
•  
• - FULL OUTER JOIN R1 X R2 lets every
  tuple in R1 or R2 appear in the result

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Figure 7.18 The LEFT OUTER JOIN operation.
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RENAME Operator

• ?S(B1, B2,Bn) (R) Renaming relation R as S
  and renaming attributes of R as Bis.
•  
• ?S(R) Renaming R as S
•  
• ?(B1, B2,Bn) (R) Renaming attributes of R as
  Bis.
•  

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• Some Queries
•  
• Q. Retrieve the SSNs of all the employees who
  either work in dept. 5 or supervise an employee
  who works in dept 5.
•  
• Dept-Emps ??DNO 5 (Employee)
• Result1 ? ?SSN (Dept-Emps)
• Result 2(SSN) ? ?SuperSSN (Dept-Emps)
•  
• Result Result 1 ? Result 2

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• Q. Find for each female employee, a list of names
  of her dependents.
•  
• Female-Emp ??SEX F (Employee)
• Empname ? ?FNAME, LNAME, SSN (Female-Emp)
• Emp-dep ? Empname ? Dependent
• Actual-dep ??SSN ESSN (Emp-dep)
• Result ? ?FNAME, LNAME, Dependent-name
  (Actual-dep)
•  
• Q. Find the name of the manager of each
  department.
•  
• Dept-mgr ? Department ?MGRSSN SSN (Employee)
• Result ? ?DNAME, LNAME, FNAME (Dept-mgr)

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• List names of managers who have atleast one
  dependent
• Find names of employees who have no dependents
• List names of all employees with two or more
  dependents

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Figure 7.12 An illustration of the CARTESIAN
PRODUCT operation.
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Figure 7.12 (continued)
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The Insert Operation

• 1. Insert ltCecilia, F, Kolonsky, null,
  1960-04-05, 6357 Windy Lane, Katy, TX, F,
  28000, null, 4gt into EMPLOYEE
• This insertion violates the entity integrity
  constraint (null for the primary key SSN), so it
  is rejected.
• 2. Insert ltAlicia, J, Zelaya, 999887777,
  1960-04-05, 6357 Windy Lane, Katy, TX, F,
  28000, 987654321, 4gt into EMPLOYEE
• This insertion violates the key constraint
  because another tuple with the same SSN Value
  already exists in the EMPLOYEE relation, and so
  it is rejected.

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The Insert Operation (contd.)

• 3. Insert ltCecilia, F, Kolonsky,
  67768989, 1960-04-05, 6357 Windswept, katy,
  TX, F, 28000, 987654321, 7gt into EMPLOYEE
• This insertion violates the referential
  integrity constraint specified on DNO because no
  DEPARTMENT tuple exists with DNUMBER 7
• 4. Insert ltCecilia, F, Kolonsky,
  67768989, 1960-04-05, 6357 Windy Lane, Katy,
  TX, F, 28000, null, 4gt into EMPLOYEE.
• This insertion satisfies all constraints, so it
  is acceptable.

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The Delete Operation

• 1. Delete the WORKS_ON tuple with ESSN
  999887777 and PNO 10.
• This deletion is acceptable
• 2. Delete the EMPLOYEE tuple with SSN
  999887777
• This deletion is not acceptable, because tuples
  in WORKS_ON refer to this tuple. Hence, if the
  tuple is deleted, referential integrity
  violations will result.

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The Delete Operation (contd.)

• 3. Delete the EMPLOYEE tuple with SSN
  333445555
• This deletion will result in even worse
  referential integrity violations, because the
  tuple involved is referenced by tuples from the
  EMPLOYEE, DEPARTMENT, WORKS_ON and DEPENDENT
  relations.
• Options
• Reject deletion
• Propagate deletion by deleting other tuples
• Modify the referencing attributes that cause the
  violations
• Or combination of three above

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The Update Operation

• 1. Update the SALARY of the EMPLOYEE tuple with
  SSN 999887777 to 28000
• Acceptable
• 2. Update the DNO of the EMPLOYEE tuple with SSN
  999887777 to 1
• Acceptable
• 3. Update the DNO of the EMPLOYEE tuple with SSN
  999887777 to 7
• Unacceptable, because it violates referential
  integrity.

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The Update Operation (contd.)

• 4. Update the SSN of the EMPLOYEE tuple with SSN
  999887777 to 987654321
• Unacceptable, because it violates primary key
  and referential integrity constraints.
• Updating an attribute which is neither primary
  key or foreign key usually causes no problems,
  new value should be of correct data type and
  domain.