1. 10/3/2017

2. Chapter 6 The Relational Algebra and Calculus
10/3/2017
Chapter 6 The Relational Algebra and Calculus
Copyright © 2004 Ramez Elmasri and Shamkant Navathe

3. Recap of Relational Algebra Operations

4. Database State for COMPANY
All examples discussed below refer to the COMPANY database shown here.

5. Q4: For every project located in ‘stafford’, list the project no, the controlling depart no, and the department manager’s last name, address and birth date.
STAFFORD_PROJS   Plocation=’stafford’ (project)
CONTR_DEPT  (STAFFORD_PROJS DNUM= DNUMBER DEPARTMENT)
PROJ_DEPT_MGR (CONTR_DEPT MGR-SSN= SSN EMPLOYEE)
RESULT   PNUMBER,DNUM, LNAME, ADDRESS,BDATE (PROJ_DEPT_MGR)

6. Examples of in Relational Algebra
Q5: Find the names of employees who work on all the projects controlled by department no 5.
ADEPT5_PROJS(PNO)   PNUMBER( DNUM=’5’ PROJECT)
EMP_PROJ(SSN,PNO)   ESSN,PNO(WORKS_ON)
RESULT_EMP_SSNS  (EMP_PROJ ÷ DEPT5_PROJS)
RESULT   LNAME, FNAME (RESULT_EMP_SSNS * EMPLOYEE)

7. Examples of in Relational Algebra
Q7. List the names of all employee with two or more dependants
Here we use the aggregate function count to count the number of employees to be more than 2
T1(ssn,no_of_dependant) ESSN ʒ Count dependant_name(DEPENDANT)
T2σ NO_of_dependant >= 2 (T1)
RESULTLNAME,FNAME (T2*EMPLOYEE)

8. Examples of in Relational Algebra
Q8. List the names of managers who have at least one dependant
MGRS(ssn) MGR_SSN(DEPARTMENT)
EMPS_WITH_DEPS(SSN) ESSN(DEPENDANT)
MGRS_WITH_DEPS (MGRS∩EMPS_WITH_DEPS)
RESULTLNAME,FNAME (MGRS_WITH_DEPS*EMPLOYEE)

9. Examples of Queries in Relational Algebra
STUDENT Course
STUDENT
Consider the relations student ,course, and studentCourse as follows:
Course

10. Examples of Queries in Relational Algebra
Q1. Get student names for the students who have completed course c1.
Here we can make natural join because both tables have common attribute we need in join condition
RESULT   S_name(Student *C_course=“c1” StudentCourse)
Q2. Get student names for the students who have studied by at least book named “introduction to compilers”
RESULT  s_name( s_code( c_code((σ c_book=‘introduction to compilers’(Course))* StudentCourse)) * Student)

11. Examples of Queries in Relational Algebra
Q3. Get student names for the students who have completed all courses
RESULT  s_name(( s_code,c_code(StudentCourse)÷
 C_code(course))*(student)
Q4. Get student codes for the students who have completed at least all those courses completed by student s2
RESULT  s_code( s_code,c_code(studentCourse)÷  c_code(σ s_code=‘s2’(studentCourse)))

12. Examples of Queries in Relational Algebra
Q5. Get student names for the students who have not completed course c2
Here we firstly get students who have completed the course c2 and make the difference between the result and all student in student realtion
RESULT  s_name(( s_code (Student) –
 s_code (σ c_code=‘c2’(StudentCourse)))* Student)
Q6. get all pairs of student names such that the two students are in the same year.
Here we take acopy from table student and rename it as stud1 And then we make join between both relaions
RESULT stud1.s_name,stud1.s_code(
σ((stud1.s_Code=student.s_code)&(stud1.s_year=student.s_year)) (ρ stud1(student)) * (student))