HW #1 Suggested Solutions
1. Find the last name, first name and major of all students taking any section of the course “4380”
offered by the “Computer Science” department in “Fall 99”.
CD := (COURSE  DEPARTMENT) WHERE (course_number = 4380) AND
(department_name = “Computer Science”)
CS := (CD  ( SECTION WHERE semester=”Fall 99” ) )[section_id]
TS := ((CS  TRANSCRIPT)  STUDENT)[lastname, firstname, major]
Note: CD is all courses numbered 4380 offered by the Computer Science Department.
CS is the identifier of all sections of this course offered in Fall99.
TS is all students taking any of the sections in CS.
2. Find the first name and last name of all instructors who taught the course “4380” offered by the
“Computer Science” department.
IN := (CD  SECTION)[instructor_id]
IN2 := (IN  INSTRUCTOR)[firstname, lastname]
Note. CD is the same as above. IN is the id of instructor who taught any section of
course 4380 from Computer Science Department. IN2 is the result which has the
name of the instructors.
3. Find the number and department identifier of all courses in which no student ever got an “F”.
FSections := (TRANSCRIPT where grade = ‘F’)[section_id]
FC := (Fsections  Section)[course_number, dept_id]
NoF := (COURSE[course_number, dept_id]) – FC
Note. Fsections are the identifiers of all section in which a student got an F.
FC is all courses (given by course number and department id) in which a student
got an F.
Finally, NoF is the result. We take all courses and subtract from this the courses
in which a student got an F.
4. Find the identifier of all instructors who are teaching at least three sections of a course in the
same semester.
S1(section_id1, section_number1, course_number1,
dept_id1, semester1, instructor_id1) :=
SECTION(section_id, section_number, course_number,
dept_id, semester, instructor_id)
S2(section_id2, section_number2, course_number2,
dept_id2, semester2, instructor_id2) :=
SECTION(section_id, section_number, course_number,
dept_id, semester, instructor_id)
Temp1 := (S1 x S2 x SECTION) WHERE (course_number = course_number1) AND
(course_number = course_number2) AND
(course_number1 = course_number2)
Temp2 := Temp1 WHERE (instructor_id = instructor_id1) AND
(instructor_id = instructor_id2) AND (instructor_id1 = instructor_id2)
Temp3 := Temp2 WHERE (dept_id = dept_id1) AND
(dept_id = dept_id2) AND (dept_id1 = dept_id2)
Temp4 := Temp3 WHERE (semester = semester1) AND
(semester = semester2) AND (semester1 = semester2)
Temp5 := Temp4 WHERE (section <> section1) AND
(section1 <> section2) AND (section <> section2)
Note. We first create three different copies of the SECTION relation. Then, we take the
cartesian product. The cartesian product contains combinations of three tuples, all
coming from the section relation. We are now looking for a tuple in the cartesian
product that is composed of three different tuples from the SECTION relation, each
tuple is for the same professor, same course number, same department id and the
same semester, the three sections are different. The result is then what we are looking
for.
5. Find the identifier of all students who took all the required courses for their majors –even if they
do not have a grade for some of the courses yet.
SR := (STUDENT  REQUIREMENT)[sid, course_number, department_id]
SC := (SECTION  TRANSCRIPT)[sid, course_number, department_id]
ReqsNotTaken := (SR – SC)[sid]
TakenAll := STUDENT[sid] – ReqsNotTaken
Note. SR is the requirements for each student. SC is the all courses students have
taken. ReqsNotTaken is the identifier of all students who have not taken a required
course. If we subtract this from all students, we are left with student who have taken all
their required courses.