1 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. <632895, "John Smith", "101 Main St. Atlanta, GA 30332",

"(404) 894-2000"> 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) = {<0.a> , <0,b> , <1,c> }

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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 t=<v1, v2, ..., vn> 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

t[Ai] = vi (the value of attribute Ai for tuple t).

- Similarly, t[Au, 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), t1[SK] <> t2[SK].

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.

t[PK] <> 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 t1[FK] = t2[PK].

 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 and PROJECT 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 and PROJECT

SELECT operation (denoted bys ): 

- Selects the tuples (rows) from a relation R that satisfy a certain selection condition c

- Form of the operation: 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:

DNO=4 (EMPLOYEE)

SALARY>30000 (EMPLOYEE)

(DNO=4 AND SALARY>25000) OR DNO=5(EMPLOYEE)

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PROJECT operation (denoted by ):  - Keeps only certain attributes (columns) from a relation R

specified in an attribute list L 

- Form of operation: L(R)  - Resulting relation has only those attributes of R specified

in L 

Example: 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: SEX,SALARY(EMPLOYEE)

If several male employees have salary 30000, only a single tuple <M, 30000> is

kept in the resulting relation.

Duplicate tuples are eliminated by the 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:

FNAME,LNAME,SALARY ( DNO=4(EMPLOYEE) )  - Alternatively, we specify explicit intermediate relations for

each step: DEPT4_EMPS <- DNO=4(EMPLOYEE) R <- 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 <- DNO=4(EMPLOYEE)

R(FIRSTNAME,LASTNAME,SALARY) <-

FNAME,LNAME,SALARY(DEPT4_EMPS)

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Figure 7.8 Results of SELECT and PROJECT operations.(a) (DNO=4 AND SALARY>25000) OR (DNO=5 AND SALARY>30000)(EMPLOYEE).

(b) LNAME, FNAME, SALARY(EMPLOYEE). (c) SEX, SALARY(EMPLOYEE).

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Class Number – CS 304Class Number – CS 304

Class Name - DBMSClass Name - DBMS

Instructor – Sanjay Madria

Instructor – Sanjay Madria

Lesson Title – Relational Algebra –3rd July

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 i=1, 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 operation:RESULT 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) <- 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: t[A1, A2, ..., Am]=t1 and t[B1, B2, ..., Bn]=t2  - If R1 has n1 tuples and R2 has n2 tuples, then R will have

n1*n2 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 <-DEPARTMENT X EMPLOYEE

DEPT_MANAGER <-MGRSSN=SSN(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) <-R1(A1, A2, ..., Am) X c R2 (B1, B2, ..., Bn)

Here c can be <, >, =, <=, >=  

EQUIJOIN: The join condition c includes one or more equality comparisons involving attributes from R1 and R2. That is, c is of the form:

(Ai=Bj) AND ... AND (Ah=Bk); 1<i,h<m, 1<j,k<n In the above EQUIJOIN operation:Ai, ..., Ah are called the join attributes of R1Bj, ..., 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 <-DEPARTMENT X MGRSSN=SSN EMPLOYEE

RESULT <-DNAME,FNAME,LNAME(T)

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NATURAL JOIN (*):

In an EQUIJOIN R <- 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 <- 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:

T<- EMPLOYEE *(DNO),(DNUMBER) DEPARTMENT

RESULT <-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 <- R1 * R2.

Example: Retrieve each EMPLOYEE's name and the name of his/her SUPERVISOR:

SUPERVISOR(SUPERSSN,SFN,SLN)<-SSN,FNAME,LNAME(EMPLOYEE)

T<-EMPLOYEE * SUPERVISOR

RESULT <-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.DNUMBERthe 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 { , , 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

<grouping attributes> F<function list> (R)

- The grouping attributes are optional

Example 1: Retrieve the average salary of all employees (no grouping needed):

R(AVGSAL) <- 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) <-

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 Bi’s.  

S(R) – Renaming R as S

(B1, B2,…Bn) (R) – Renaming attributes of R as Bi’s.

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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 Operation1. Insert <‘Cecilia’, ‘F”, “Kolonsky’, null, ‘1960-04-

05’, ‘6357 Windy Lane, Katy, TX’, F, 28000, null, 4> into EMPLOYEEThis insertion violates the entity integrity constraint (null for the primary key SSN), so it is rejected.

2. Insert <‘Alicia’, ‘J’, ‘Zelaya’, 999887777, 1960-04-05’, ‘6357 Windy Lane, Katy, TX’, F, 28000, 987654321’, 4> into EMPLOYEEThis 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 <‘Cecilia’, ‘F’, ‘Kolonsky’, ‘67768989’, 1960-

04-05’, ‘6357 Windswept, katy, TX’, F, 28000, ‘987654321’, 7> into EMPLOYEE

This insertion violates the referential integrity constraint specified on DNO because no DEPARTMENT tuple exists with DNUMBER = 7

4. Insert <‘Cecilia’, ‘F’, ‘Kolonsky’, ‘67768989’, 1960-04-05’, ‘6357 Windy Lane, Katy, TX’, F, 28000, null, 4> into EMPLOYEE.

This insertion satisfies all constraints, so it is acceptable.

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The Delete Operation1. 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 28000Acceptable

2. Update the DNO of the EMPLOYEE tuple with SSN = ‘999887777’ to 1Acceptable

3. Update the DNO of the EMPLOYEE tuple with SSN = ‘999887777’ to 7Unacceptable, 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.