1. 1. Explain the following concepts: (a) superkey (b) key
(c) candidate key
(d) primary key
(e) foreign key
key constraints
a superkey is any combination of attributes that uniquely
identify a tuple: t1[superkey]  t2[superkey].
a key is superkey that has a minimal set of attributes
Sept. 2004
Yangjun Chen

2. one candidate key is chosen as the primary key (PK)
If a relation schema has more than one key, each of them is called a candidate key.
one candidate key is chosen as the primary key (PK)
foreign key (FK) is defined as follows:
i) Consider two relation schemas R1 and R2;
ii) The attributes in FK in R1 have the same domain(s) as the primary key attributes PK in R2; the attributes FK are said to reference or refer to the relation R2;
iii) A value of FK in a tuple t1 of the current state r(R1) either occurs as a value of PK for some tuple t2 in the current state r(R2) or is null. In the former case, we have t1[FK] = t2[PK], and we say that the tuple t1 references or refers to the tuple t2.
Sept. 2004
Yangjun Chen

3. 2. Draw an ER-diagram to describe the following real world problem.
(a) A university is organized into faculties.
(b) Each faculty has a unique name, ID and number of professors and a specific professor is chosen as the faculty head.
(c) Each faculty provides a number of courses.
(d) Each course has a unique name and courseID.
(e) Each professor has a name, SIN, address, salary, sex and courses taught by him/her.
(f) Each professor belongs to a faculty and can teach several sections of a course.
(g) Each student has a name, ID, SIN, address, GPA, sex, and major.
(h) Each student can choose one faculty as his/her major faculty and take several courses with certain credit hours. Some of the courses are mandatory and some are optional.
Sept. 2004
Yangjun Chen

4. ER-model: professor faculty course student belong teach head provide
salary
name ID
addr.
SIN
belong N 1
professor
NoProf
F-Id
sex
F-name M
startdate 1
teach
section-IDs
faculty
head 1 M N
couresId 1
provide N
course
name
choose N ID N
addr.
SIN
take M
student
major
mandatory-optional
creditHours
name
sex
birthdate
Sept. 2004
Yangjun Chen

5. - collision resolution strategy: chaining
3. Linear Hashing
- collision resolution strategy: chaining
- split rule: load factor > 0.7
- initially M = 4 (M: size of the primary area)
- hash functions: hi(key) = key mod 2i  M (i = 0, 1, 2, …)
- bucket capacity = 2
Trace the insertion process of the following keys into a linear
hashing file:
24, 7, 10, 5, 14, 4, 1, 8, 2, 3, 17, 13, 15.
Sept. 2004
Yangjun Chen

6. when inserting the sixth record we would have
The first phase – phase0
when inserting the sixth record we would have
but the load factor 6/8= 0.75 > 0.70 and so bucket 0 must be split (using h1 = Key mod 2M): 24 4 5 10 14 7
n=0 before the split
(n is the point to the bucket to be split.) 24 5 10 14 7 4
n=1 after the split
load factor: 6/10=0.6
no split
Sept. 2004
Yangjun Chen

7. 24 5 10 14 7 4 0 1 2 3 4 n=1 load factor: 7/10=0.7 no split 24 5 1 10
insert(1) 24 5 10 14 7 4
n=1
load factor: 7/10=0.7
no split 24 5 1 10 14 7 4
Sept. 2004
Yangjun Chen

8. 24 5 1 10 14 7 4 0 1 2 3 4 n=1 load factor: 8/10=0.8 split using h1.
insert(8) 24 5 1 10 14 7 4
n=1
load factor: 8/10=0.8
split using h1. 24 8 5 1 10 14 7 4
Sept. 2004
Yangjun Chen

9. 24 8 1 10 14 7 4 5 0 1 2 3 4 5 n=2 load factor: 8/12=0.66 no split 24
n=2
load factor: 8/12=0.66
no split
insert(2) 24 8 1 10 14 7 4 5
Sept. 2004
Yangjun Chen

10. 24 8 1 10 14 7 4 5 n=2 load factor: 9/12=0.75 split using h1. 2
overflow 2 24 8 1 10 2 7 4 5 14
Sept. 2004
Yangjun Chen

11. insert(3) 24 8 1 10 2 7 4 5 14 24 8 1 10 2 7 3 4 5 14
n=3
load factor: 10/14=0.714
split.using h1.
Sept. 2004
Yangjun Chen

12. The second phase – phase1 n = 0; using h1 = Key mod 2M to insert and 24 8 1 10 2 3 4 5 14 7
n=4
The second phase – phase1
n = 0; using h1 = Key mod 2M to insert and
h2 = Key mod 4M to split.
insert(17) 8 24 1 10 2 3 4 5 14 7
Sept. 2004
Yangjun Chen

13. 8 24 1 17 10 2 3 4 5 14 7 n=0 load factor: 11/16=0.687 no split. 8 24
insert(13) 8 24 1 17 10 2 3 4 5 14 7
Sept. 2004
Yangjun Chen

14. 8 24 1 17 10 2 3 4 5 13 14 7
n=0
load factor: 12/16=0.75
split bucket 0, using h2. 1 17 10 2 3 4 5 13 14 7 8 24
Sept. 2004
Yangjun Chen

15. insert(15) 1 17 10 2 3 4 5 13 14 7 8 24 1 17 10 2 3 4 5 13 14 7 15 8 24
n=1
load factor: 13/18=0.722
split bucket 1, using h2. 1 17 10 2 3 4 5 13 14 7 15 8 24
Sept. 2004
Yangjun Chen

16. Here, we assume that each internal node can contain at most two
4. Given the following B+-tree, trace the deletion sequence: h, c, e, f.
Here, we assume that each internal node can contain at most two
keys and each leaf node can contain at most two value/point pairs. c b e a b c e f h
Sept. 2004
Yangjun Chen

17. f e a b c f e a c b
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Yangjun Chen

18. c a e a b f a e a b f a b a
Sept. 2004
Yangjun Chen

19. 5. Given the relation schemas shown in Fig. 1, construct expressions
(using relational algebraic operations) to evaluate the
following query:
Find the names of employees who works on all the projects
controlled by department Number ‘Business Computing’.
EMPLOYEE
fname, minit, lname, ssn, bdate, address, sex, salary, superssn, dno
DEPARTMENT
Dname, dnumber, mgrssn, mgrstartdate
PROJECT
WORKS_ON
Pname, pnumber, plocation, dnum
Essn pno, hours
Sept. 2004
Yangjun Chen

20. : DN dnumber(Dname = ‘Business Computing’(Department))
DEPT_P PNUMBER(DNDN.dnumber = PROJECT.dnumber(PROJECT))
EMP_PNOS  ESSN,PNO(WORK_ON)
SSNS EMP_PNOS :
DEPT_P
RESULT  FNAME, LNAME(SSNS * EMPLOYEE)
Sept. 2004
Yangjun Chen

21. Pname, pnumber, plocation, dnum
6. Given the relation schemas as shown in Fig. 2, construct a SQL
clause to evaluate the following query:
For each department, retrieve the department number, the department
name, the depatment manager, and the number of employees who
work in that department.
PROJECT
WORKS ON
Pname, pnumber, plocation, dnum
Essn pno, hours
Sept. 2004
Yangjun Chen

22. SELECT x.Dname, x.dnumber, z.fname, z.lname, count(y.ssn)
FROM department x, employee y, employee z
WHERE x.Dnumber=y.dno
and x.mgrssn=z.ssn
Group by x.dnumber
The following SQL statement is wrong.
SELECT Dname, dnumber, fname, lname, count(ssn)
FROM department , employee
WHERE Dnumber=dno and mgrssn=ssn
Group by dnumber
Sept. 2004
Yangjun Chen

23. 4. Trace the insertion process of the following keys into an empty
b+-tree: 7, 6, 5, 3, 1, 2.
Here, we assume that each internal node can contain at most two
keys and each leaf node can contain at most two value/point pairs. 7 6 7
Sept. 2004
Yangjun Chen

24. 5 6 7 5 5 6 3 5 7 6
Sept. 2004
Yangjun Chen

25. 5 3 6 1 3 5 6 7 3 6 2 3 3 5 1 2 6 7
Sept. 2004
Yangjun Chen