Solutions to Midterm Examination

Prof. Li-Yan Yuan
CMPUT 291: File and Database Management Systems
Solutions to Midterm Examination
Section B1, February 24, 2004
1. Construct an E-R diagram for a car-insurance company that has a set of customers, each of
whom owns one or more cars. Each car has associated with it zero to any number of recorded
accidents. The company also maintains the statistical information for each type of vehicle,
such as the accident and car theft ratio.
If you have any doubt about the specification, make a reasonable assumption. Document all
your assumptions. Please use the textbook notations to specify the mapping constraints. [18]
2. The following ER diagram characterizes information about an on-line book store. Customer,
Store, Book store information about customers, stores, and books respectively. Order represents
each order placed by a customer. Relationship set supply indicates which store supplies
which book to which order; and ordered indicates who placed the order.
Customer
Store
✻
❅
ordered
❅
❅❅
❅
✗
Date
✖
✔
✕
❅
supply
❅
❅❅
❅
✲
Order
Book
The attributes for all entity sets are Customer(c id, c name, c address), Order(o id, o data),
Store(s id, s address), Book(isbn, title, price), with c id, o id, s id, isbn as respective
keys. Note that the arrow indicates the one side of the corresponding relationship set.
Use the SQL’s CREATE TABLE statement to create the database schema from the ER
diagram, specifying all the essential constraints. State any reasonable assumptions you may
have. [12]
CREATE TABLE store (
s_id int,
s_address VARCHAR(200),
PRIMARY KEY (s_id)
)
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CREATE TABLE book (
isbn VARCHAR(24),
title VARCHAR(48),
price NUMERIC(10,2),
PRIMARY KEY (isbn)
)
CREATE TABLE customer (
c_id int,
c_name VARCHAR(128),
c_address VARCHAR(200),
PRIMARY KEY (c_id)
)
CREATE TABLE order (
o_id int,
o_data VARCHAR(500),
c_id int,
o_date DATE,
PRIMARY KEY (o_id),
FOREIGN KEY c_id REFERENCES customer
)
CREATE TABLE supply (
s_id int,
isbn VARCHAR(24),
o_id int,
PRIMARY KEY ( s_id, isbn, o_id),
FOREIGH KEY s_id REFERENCES store,
FOREIGN KEY isbn REFERENCES book,
FOREIGH KEY o_id REFERENCES order
)
3. Consider a university registration database with the following tables.
student(s id, s name, status, phone, major)
professor(p id, p name, dept)
teach(p id, c id, term)
transcript(s id, c id, term, grade)
The first two tables store information about students and professors; the third table indicates
who teaches which course at which term; and the last table stores information about course
registration.
Use SQL to express the following queries. (You results may contain duplicate tuples.)
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(a) Find the names (s name) of all students who have taken a course taught by a professor
in the ’CS’ department.
SELECT s_name
FROM student s, transcript t, teach c, professor p
WHERE s.s_id = t.s_id AND t.c_id = c.c_id AND
c.p_id = p.p_id AND p.dept = ’CS’ AND
t.term = c.term
(b) Find the names of all students who have taken more than one course taught by a professor
from the ’CS’ department. You cannot not use aggregate functions with or without the
group by clause for this question.
SELECT s_name
FROM student s, transcript t1, transcript t2, teach c1, teach c2,
professor p1, professor p2
WHERE s.s_id = t1.s_id AND t1.c_id = c1.c_id AND t1.term = c1.term AND
c1.p_id = p1.p_id AND p1.dept = ’CS’ AND
s.s_id = t2.s_id AND t2.c_id = c2.c_id AND t2.term = c2.term AND
c2.p_id = p2.p_id AND p2.dept = ’CS’ AND
(t1.c_id t2.c_id OR t1.term t2.term)
(c) List the names of all students who have taken a course from every professor in the
’MUSIC’ department.
SELECT s_name
FROM student s, transcript t, teach c, professor p
WHERE s.s_id = t.s_id AND t.c_id = c.c_id AND t.term = c.term AND
c.p_id = p.p_id AND p.dept = ’MUSIC’
GROUP BY s_name
HAVING COUNT( DISTINCT p_id) >= ANY ( SELECT count(*)
FROPM professor
WHERE dept = ’MUSIC’
)
SELECT s_name
FROM student s
WHERE NOT EXIST ( SELECT *
FROM professor p
WHERE p.dept = ’MUSIC’ AND
NOT EXIST ( SELECT *
FROM transacript t, teach c
WHERE s.s_id = t.s_id AND
t.c_id = c.c_id AND
p.c_id = c.c_id AND
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)
)
t.term = c.term
(d) List the names of students whose average grade is higher than that of all junior students.
A junior student is a student whose status is ’junior’.
SELECT s_name
FROM student s, transcript t
WHERE s.s_id = t.s_id
GROUP BY s_name, s.s_id
HAVING AVG(grade) > ALL ( SELECT AVG(grade)
FROM student s2, transacript t2
WHERE s2.s_id = t2.s_id AND s2.status = ’junior’
GROUP BY s2.s_id
)
(e) List the course id and the average grade taught by each professor in the ’CS’ department
in the ’winter03’ term. That is, you are asked to list p id, c id, and the average grade
for each course c id taught by p id in ’winter03’ term.
SELECT p.p_id, c.c_id, AVG(t.grade)
FROM teach c, transcript t, professor p
WHERE c.c_id = t.c_id AND c.p_id = p.p_id AND p.dept = ’CS’ AND
c.term = ’winter03’
GROUP BY p.p_id, c.c_id
(f) List p id and p name of all professors who have never taught a course taken by a junior
student.
SELECT p_id, p_name
FROM professor
EXCEPT
SELECT p.p_id, p.p_name
FROM professor p, teach c, transacript t, student s
WHERE p.p_id = c.p_id AND c.c_id = t.c_id AND c.term = t.term AND
t.s_id = s.s_id AND s.status = ’junior’
(g) Assume that a major always coincides with a department name. list, for each department,
the number of professors who are working for the department, the number of
students who major in the department, and the number of courses taught by professors
from the department.
Note that a course taught in two different terms is counted as one, not two, course taught
by the department.
SELECT p.dept, count(distinct p.p_id), count(distinct s.s_id),
count(distinct c.c_id)
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FROM professor p, teach c, student s
WHERE p.dept = s.major AND p.p_id = c.p_id
GROUP BY p.dept
4. Consider a relational database about a university with the following three relations
teach(Prof, Course)
take(Student, Course, Grade)
advise (Prof, Student)
The first relation indicates the courses a prof teaches; the second tells what courses each
student takes and the corresponding grades; and the last indicates advisers of a student.
Write a relational algebra expression to retrieve
(a) all students who do not take a course from their own advisers.
Π Student (take) − Π Student (teach ✶ take ✶ advise)
(b) A tough course is a course in which no one receives a grade higher than ’6’. List the
names of all students who take only tough courses. [14]
Π Student (take) − Π Student1 (σ C1=C2∧G2>6 (take[S1, C1, G] × take[S2, C2, G2]))
Or, alternatively,
Π Student (take)−Π take1.Student (σ take1.Course=take2.Course∧take2.Course.grade>6 (take1×take2))
Note that both take1 and take2 are take.
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