Fundamentals of Database Systems Lecture 5 & 6 PDF

Fundamentals of Database Systems Seventh Edition Chapter 8 The Relational Algebra Slides in this presentation contain...

Fundamentals of Database Systems Seventh Edition Chapter 8 The Relational Algebra Slides in this presentation contain hyperlinks. JAWS users should be able to get a list of links by using INSERT+F7 Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Chapter Outline 8.1 Relational Algebra – Unary Relational Operations – Relational Algebra Operations From Set Theory – Binary Relational Operations – Additional Relational Operations – Examples of Queries in Relational Algebra 8.2 Relational Calculus – Tuple Relational Calculus – Domain Relational Calculus 8.3 Example Database Application (COMPANY) 8.4 Overview of the QBE language (appendix D). Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Relational Algebra Overview (1 of 4) 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) Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Relational Algebra Overview (2 of 4) 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) Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Relational Algebra Overview (3 of 4) Relational Algebra consists of several groups of operations – Unary Relational Operations ▪ SELECT (symbol (  sigma )) ▪ PROJECT (symbol : p ( pi )) ▪ RENAME (symbol : p ( rho )) – Relational Algebra Operations From Set Theory ▪ Union – Intersection ( ) Difference (or Minus, − ) ▪ CARTESIAN PRODUCT ( x ) – Binary Relational Operations ▪ JOIN (several variations of JOIN exist) ▪ DIVISIONs Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Relational Algebra Overview (4 of 4) – Additional Relational Operations ▪ OUTER JOINS, OUTER UNION ▪ AGGREGATE FUNCTIONS (These compute summary of information: for example, SUM, COUNT, AVG, MIN, MAX) Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Database State for COMPANY All examples discussed below refer to the Company database shown here. Figure 5.7 Referential integrity constraints displayed on the COMPANY relational database schema. Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved The Following Query Results Refer to This Database State Figure 5.6 One possible database state for the COMPANY relational database schema. Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Unary Relational Operations: SELECT (1 of 4) The SELECT operation (denoted by σ (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: σDNO= 4 (EMPLOYEE ) – Select the employee tuples whose salary is greater than $30,000: σSalary>30,000 (EMPLOYEE ) Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Unary Relational Operations: SELECT (2 of 4) In general, the select operation is denoted by σ (R ) where – The symbol σ (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 Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Unary Relational Operations: SELECT (3 of 4) SELECT Operation Properties – The SELECT operation σ (R ) produces a relation S that has the same schema (same attributes) as R – SELECT σ is commutative: ▪ σ ( σ (R ) ) = σ ( σ (R ) ) – Because of commutativity property, a cascade (sequence) of SELECT operations may be applied in any order: ▪ σ ( σ ( σ (R ) ) ) = σ ( σ ( σ (R ) ) ) Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Unary Relational Operations: SELECT (4 of 4) – A cascade of SELECT operations may be replaced by a single selection with a conjunction of all the conditions: ▪ σ ( σ ( σ (R ) ) = σ ANDAND (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 Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Unary Relational Operations: Project (1 of 3) PROJECT Operation is denoted by π (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 ) Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Unary Relational Operations: Project (2 of 3) The general form of the project operation is:  ( R ) – π (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. Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Unary Relational Operations: Project (3 of 3) 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 Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Examples of Applying Select and Project Operations Figure 8.1 Results of SELECT and PROJECT operations. (a) σ (Dno=4 AND Salary>25000) OR (Dno=5 AND Salary>30000) (EMPLOYEE ). (b) Lname,Fname, Salary (EMPLOYEE ). (c) Sex, Salary (EMPLOYEE ). Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved 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. Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved 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 ( σDNO=5 (EMPLOYEE) ) OR We can explicitly show the sequence of operations, giving a name to each intermediate relation: – DEP5_EMPS  σDNO=5 (EMPLOYEE) – RESULT  FNAME, LNAME,SALARY (DEP5_EMPS ) Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Unary Relational Operations: RENAME (1 of 3) 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) Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Unary Relational Operations: RENAME (2 of 3) 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 – S ( R ) changes: ▪ the relation name only to S –  (B1, B2, …,Bn ) ( R ) changes: ▪ the column (attribute) names only to Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Unary Relational Operations: RENAME (3 of 3) 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 Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Example of Applying Multiple Operations and RENAME Figure 8.2 Results of a sequence of operations. (a) Fname,Lname, Salary (σDno=5 (EMPLOYEE ) ). (b) Using intermediate relations and renaming of attributes Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Relational Algebra Operations from Set Theory: UNION (1 of 2) 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) Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Relational Algebra Operations from Set Theory: Union (2 of 2) 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  σDNO=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 Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Figure 8.3 Result of the Union Operation RESULT ← RESULT1  RESULT2 union symbol Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved 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) are type compatible if: – they have the same number of attributes, and – the domains of corresponding attributes are type compatible 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) Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved 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” Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved 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” Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Example to Illustrate the Result of UNION, INTERSECT, and DIFFERENCE Figure 8.4 The set operations UNION, INTERSECTION, and MINUS. (a) Two union-compatible relations. (b) STUDENT  INSTRUCTOR. (c) STUDENT  INSTRUCTOR. (d) STUDENT – INSTRUCTOR. (e) INSTRUCTOR – STUDENT. Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved 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 Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved The Following Query Results Refer to This Database State Figure 5.6 One possible database state for the COMPANY relational database schema. Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Relational Algebra Operations from Set Theory: CARTESIAN PRODUCT (1 of 3) CARTESIAN (or CROSS) PRODUCT Operation – This operation is used to combine tuples from two relations in a combinatorial fashion. – Denoted by – Result is a relation Q with degree n + m attributes: ▪ – 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 × S will have nR * nS tuples. – The two operands do NOT have to be "type compatible” Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved  Cartesian product Example ❑ To compute the cross product of these two relations, we need to create a new relation that contains all possible combinations of tuples from both relations. ❑ In this case, we'll end up with a new relation with four attributes: "ID", "Color", "Size", and "Price". Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved  Relational Algebra Operations from Set Theory: CARTESIAN PRODUCT (2 of 3) Generally, CROSS PRODUCT is not a meaningful operation – Can become meaningful when followed by other operations Example (not meaningful) Return (FNAME, LNAME, dependent_name) of the female employees and their dependents : – FEMALE_EMPS  σ 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 Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Relational Algebra Operations from Set Theory: CARTESIAN PRODUCT (3 of 3) To keep only combinations where the DEPENDENT is related to the EMPLOYEE, we add a SELECT operation as follows Example (meaningful): – FEMALE_EMPS  σ SEX=’F’ (EMPLOYEE ) – EMPNAMES  FNAME, LNAME, SSN (FEMALE_EMPS ) – EMP_DEPENDENTS  EMPNAMES x DEPENDENT – ACTUAL_DEPS  σ SSN=ESSN (EMP_DEPENDENTS ) – RESULT  FNAME, LNAME, DEPENDENT_NAME ( ACTUAL_DEPS ) RESULT will now contain the name of female employees and their dependents Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Figure 8.5 The Cartesian Product (CROSS PRODUCT) Operation (1 of 3) Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Figure 8.5 The Cartesian Product (CROSS PRODUCT) Operation (2 of 3) Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Figure 8.5 The Cartesian Product (CROSS PRODUCT) Operation (3 of 3) Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Example 2 Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Example 1 OR Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Example 2 Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Example 2 Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Binary Relational Operations: JOIN (1 of 2) JOIN Operation – 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 – where R and S can be any relations that result from general relational algebra expressions. Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Binary Relational Operations: JOIN (2 of 2) Example: 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 Mgr_ssn=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 Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Figure 8.6 Result of the JOIN Operation D E P T dash M X s s n = S s n end expression EMPLOYEE G R left arrow DEPARTMENT super absolute value of x sub start expression m g r dash ← DEPT_MGR DEPARTMENT Mgr_ssn=Ssn EMPLOYEE Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Some Properties of JOIN (1 of 2) Consider the following JOIN operation: – R(A1, A 2,..., A n ) R.A i =S.B j 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—r from R and s from S, but only if they satisfy the join condition – 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 Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Some Properties of JOIN (2 of 2) The general case of JOIN operation is called a Theta- join: R theta S The join condition is called theta Theta can be any general boolean expression on the attributes of R and S; for example: – Most join conditions involve one or more equality conditions “AND”ed together; for example: – Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved  Theta Join Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Binary Relational Operations: EQUIJOIN EQUIJOIN Operation The most common use of join involves join conditions with equality comparisons only Such a join, where the only comparison operator used is =, is called an EQUIJOIN. – In the result of an EQUIJOIN we always have one or more pairs of attributes (whose names need not be identical) that have identical values in every tuple. – The JOIN seen in the previous example was an EQUIJOIN. Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved The problem of join operation Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Binary Relational Operations: NATURAL JOIN Operation (1 of 2) NATURAL JOIN Operation – Another variation of JOIN called NATURAL JOIN — denoted by * — was created to get rid of the second (superfluous ‫ )زائده‬attribute in an EQUIJOIN condition. ▪ because one of each pair of attributes with identical values is superfluous – The standard definition of natural join requires that the two join attributes, or each pair of corresponding join attributes, have the same name in both relations – If this is not the case, a renaming operation is applied first. Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Binary Relational Operations: NATURAL JOIN Operation (2 of 2) Example: To apply a natural join on the DNUMBER attributes of DEPARTMENT and DEPT_LOCATIONS, it is sufficient to write: – DEPT_LOCS  DEPARTMENT * DEPT_LOCATIONS Only attribute with the same name is DNUMBER An implicit join condition is created based on this attribute: DEPARTMENT.DNUMBER=DEPT_LOCATIONS.DNUMBER Another example: Q  R ( A,B,C,D ) * S ( C,D,E ) – The implicit join condition includes each pair of attributes with the same name, “AND”ed together: ▪ R.C=S.C AND R.D.S.D – Result keeps only one attribute of each such pair: ▪ Q(A,B,C,D,E) Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Example of NATURAL JOIN Operation Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Example of NATURAL JOIN Operation 2 Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Additional Relational Operations (1 of 2) The OUTER JOIN Operation – In NATURAL JOIN and EQUIJOIN, tuples without a matching (or related) tuple are eliminated from the join result ▪ Tuples with null in the join attributes are also eliminated ▪ This amounts to loss of information. – A set of operations, called OUTER joins, can be used when we want to keep all the tuples in R, or all those in S, or all those in both relations in the result of the join, regardless of whether or not they have matching tuples in the other relation. Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Additional Relational Operations (2 of 2) The left outer join operation keeps every tuple in the first or left relation R in ; if no matching tuple is found in S, then the attributes of S in the join result are filled or “padded” with null values. A similar operation, right outer join, keeps every tuple in the second or right relation S in the result of A third operation, full outer join, denoted by keeps all tuples in both the left and the right relations when no matching tuples are found, padding them with null values as needed. Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Example Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Example of types pf OUTER JOIN Operation Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Example of types pf OUTER JOIN Operation Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Additional Relational Operations: Aggregate Functions and Grouping A type of request that cannot be easily expressed in the basic relational algebra is to specify mathematical aggregate functions on collections of values from the database. Examples of such functions include retrieving the average or total salary of all employees or the total number of employee tuples. – These functions are used in simple statistical queries that summarize information from the database tuples. Common functions applied to collections of numeric values include – SUM, AVERAGE, MAXIMUM, and MINIMUM. The COUNT function is used for counting tuples or values. Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Aggregate Function Operation Use of the Aggregate Functional operation Ƒ – F MAX Salary (EMPLOYEE ) retrieves the maximum salary value from the EMPLOYEE relation – F MIN Salary (EMPLOYEE ) retrieves the minimum Salary value from the EMPLOYEE relation – F SUM Salary (EMPLOYEE ) retrieves the sum of the Salary from the EMPLOYEE relation – F COUNT Ssn,AVERAGE Salary (EMPLOYEE ) computes the count (number) of employees and their average salary ▪ Note: count just counts the number of rows, without removing duplicates Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Using Grouping With Aggregation The previous examples all summarized one or more attributes for a set of tuples – Maximum Salary or Count (number of) Ssn Grouping can be combined with Aggregate Functions Example: For each department, retrieve the DNO, COUNT SSN, and AVERAGE SALARY A variation of aggregate operation Ƒ allows this: – Grouping attribute placed to left of symbol – Aggregate functions to right of symbol – Dno F COUNT Ssn, AVERAGE Salary ( EMPLOYEE ) Above operation groups employees by DNO (department number) and computes the count of employees and average salary per department Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Figure 8.10 the Aggregate Function Operation a. b. c. (a) Dno No_of_employees Average_sal (b) Dno Count_ssn Average_salary 5 4 33250 5 4 33250 4 3 31000 4 3 31000 1 1 55000 1 1 55000 (c) Count_ssn Average_salary 8 35125 Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Aggregation function Example Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved SQL and Relation Algebra Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Examples of Queries in Relational Algebra Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Examples of Queries in Relational Algebra: Procedural Form Q1: Retrieve the name and address of all employees who work for the ‘Research’ department. RESEARCH_DEPT   DNAME=’Research’ (DEPARTMENT ) RESEARCH_EMPS  (RESEARCH_DEPT EMPLOYEE) DNUMBER= DNOEMPLOYEE RESULT  FNAME, LNAME, ADDRESS (RESEARCH_EMPS ) Q2: Retrieve the names of employees who have no dependents. ALL_EMPS  SSN (EMPLOYEE) EMPS_WITH_DEPS(SSN)  ESSN (DEPENDENT) EMPS_WITHOUT_DEPS  (ALL_EMPS − EMPS_WITH_DEPS) RESULT  LNAME, FNAME (EMPS _ WITHOUT _ DEPS * EMPLOYEE ) Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Examples of Queries in Relational Algebra – Single Expressions As a single expression, these queries become: Q1: Retrieve the name and address of all employees who work for the ‘Research’ department.  Fname,Lname, Address( Dname= ‘Research’ (DEPARTMENT Dnumber=Dno (EMPLOYEE )) Q2: Retrieve the names of employees who have no dependents. Lname,Fname ((Ssn (EMPLOYEE) − Ssn(Essn (DEPENDENT))) * EMPLOYEE) Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved  Exercise Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved Examples of Queries in Relational Algebra Retrieve the names of all employees who work for the ‘Headquarters’ Department.. For every project located in ‘Stafford’, list the project number, controlling department, and the manager’s name. Find the names of the employees who work on all the projects controlled by DNo = 3. Make a list of project numbers for projects that involve an employee with LName = ‘Smith’, either as a worker or as a manager of the department that controls the project. Copyright © 2016, 2011, 2007 Pearson Education, Inc. All Rights Reserved