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Automate assignment 5., Assignments of Computer Security

Allama Iqbal Open University (AIOU)Computer Security

: Prove that regular languages are closed under intersection and set difference. (05 marks) Must prove using DeMorgan’s Law. Solution:

Typology: Assignments

2019/2020

Uploaded on 04/25/2020

the-knowledge-daily
the-knowledge-daily 🇵🇰

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Normalization example sample solution
Enrollment (StudentID , courseID , course_Instructor , Student_Name ,
Student_Degree , Student_ADD , course_Name , Instructor_Name ,
Instructor_Office , Grade)
StudentSupervisor (StudentID* , specialization , supervisor)
------------------------------------------------------------------------------------------------
•First Normal Form:
Enrollment relation is not in the first Normal Form because the Attribute
Student_Degree is not Atomic.
So we decompose as following:
Enrollment (StudentID , courseID , course_Instructor , course_Name
Student_Name , Student_ADD, Instructor_Name , Instructor_Office , Grade)
Student_Degree (StudentID* , courseID*, StudentDegree)
--
The resulted relations will be:
Enrollment (StudentID , courseID , course_Instructor , course_Name
Student_Name , Student_ADD, Instructor_Name , Instructor_Office , Grade)
Student_Degree (StudentID* , courseID*, StudentDegree)
StudentSupervisor (StudentID* , specialization , supervisor)
1
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Normalization example sample solution

Enrollment ( StudentID , courseID , course_Instructor , Student_Name , Student_Degree , Student_ADD , course_Name , Instructor_Name , Instructor_Office , Grade) StudentSupervisor ( StudentID* , specialization , supervisor)

• First Normal Form:

Enrollment relation is not in the first Normal Form because the Attribute Student_Degree is not Atomic. So we decompose as following: Enrollment (StudentID , courseID , course_Instructor , course_Name Student_Name , Student_ADD, Instructor_Name , Instructor_Office , Grade) Student_Degree (StudentID* , courseID, StudentDegree) -- The resulted relations will be: Enrollment (StudentID , courseID , course_Instructor , course_Name Student_Name , Student_ADD, Instructor_Name , Instructor_Office , Grade) Student_Degree (StudentID , courseID, StudentDegree) StudentSupervisor (StudentID , specialization , supervisor)

- Second Normal Form: 1- First we state the FDs for the Enrollment relation StudentID , courseID , Grade √ StudentID  Student_Name , Student_ADD. X courseID  course_Name , course_Instructor , Instructor_Name , Instructor_Office , X -------------- Enrollment relation is not in the second normal form because only grade attribute is fully functionally dependent on the primary key. However, Student_Degree and StudentSupervisor relations are both in the second normal form. 2- second we decompose: Course_grade (StudentID , CourseID , Grade ) Student (StudentID , Student_Name , Student_ADD ) course (courseID , course_Name , course_Instructor , Instructor_Name , Instructor_Office ) the resulted relations will be : Course_grade (StudentID , CourseID , Grade ) Student (StudentID , Student_Name , Student_ADD ) course (courseID , course_Name , course_Instructor , Instructor_Name , Instructor_Office ,.) StudentSupervisor (StudentID* , specialization , supervisor) Student_Degree (StudentID* , courseID*, Student_Degree)

  • BCNF Course_garde (StudentID , CourseID , Grade ) √ Student (StudentID , Student Name , Student ADD ) √ Course_ Instructor (courseID , course_Name , course_Instructor) √ Instructor (course_Instructor, Instructor_Name , Instructor_Office) √ Student_Degree (StudentID , courseID, Student_Degree) √ StudentSupervisor (StudentID , specialization , supervisor) X Because: supervisor specialization and specialization is a part of the primary key we decompose the relation into the following: Student_ specialization (StudentID* , specialization) Supervisor_ specialization (supervisor ,specialization* ) The resulted relations : Course_garde (StudentID , CourseID , Grade ) Student (StudentID , Student Name , Student ADD ) Course_ Instructor (courseID , course_Name , course_Instructor) Instructor (course_Instructor, Instructor_Name , Instructor_Office) Student_Degree (StudentID* , courseID, Student_Degree) Student_ specialization (StudentID , specialization) Supervisor_ specialization (supervisor ,specialization* )