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Analysis of Midterm-Examination
1.(5) Describe the working process with a database system.
a) database definition
By the database definition, we create an entity-relationship diagram for a
realistic world problem, which is then transformed into a database schema.
b) populating tables with data
The second phase is the database construction, by which a database is
initially populated with data.
c) database manipulation
Once a database is constructed, one can conduct some operation on it, such
as data insertion, data deletion and data updating, which changes the
database from a state to another state. It is the responsibility of a database
system to guarantee that the database goes from one correct state to another.
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Analysis of Midterm-Examination
2.(10) Describe the three-schema architecture of a database
system and the main functionality of each level in this
architecture.
a) internal level
The first level is the internal level, which possesses an internal schema. It
describes the physical storage structure of the database. The internal
schema uses a physical data model and describes the complete details of
data storage and access paths for the database.
b) conceptual level
The second level is the conceptual level, which has a conceptual schema.
It describes the structure of the whole database for a community of users.
The conceptual level hides the details of physical structures and
concentrates on entity types, data types, relationships, user operations, and
constraints. A high level data model or an implementation data model can
be used at this level.
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Analysis of Midterm-Examination
c) external level
The third level is the so-called external level or the view level, which includes a
number of external schemas or user views. It describes the part of the database that a
particular user group is interested in and hides the rest of the database from that user
group. A high level data model or the implementation data model can be used at this
level.
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Analysis of Midterm-Examination
3.(10) 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
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Analysis of Midterm-Examination
•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.
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Analysis of Midterm-Examination
4.(15) 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.
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Analysis of Midterm-Examination
ER-model:
F-name
F-Id
salary
N
belong
1
NoProf
startdate
faculty
name
1
addr.
SIN
professor
M
1
M
N
N
sections
provide
N
has
1
ID
course
SIN
N
M
student
name
sex
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birthdate
sectionId
M
choose
addr.
sex
teach
head
1
ID
major
mandatory-optional
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take
name
couresId
creditHours
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Analysis of Midterm-Examination
5.(30) 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.
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Analysis of Midterm-Examination
The first phase – phase0
•when inserting the sixth record we would have
24
4
5
0
1
10
14
7
2
3
n=0 before the split
(n is the point to the
bucket to be split.)
•but the load factor 6/8= 0.75 > 0.70 and so bucket 0 must be
split (using h1 = Key mod 2M):
24
10
14
5
0
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1
2
7
n=1 after the split
load factor: 6/10=0.6
no split
4
3
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Analysis of Midterm-Examination
insert(1)
24
5
10
14
0
1
2
24
0
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5
1
1
7
3
10
14
7
2
3
4
4
4
n=1
load factor: 7/10=0.7
no split
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Analysis of Midterm-Examination
insert(8)
24
5
1
10
14
7
4
0
1
2
3
4
24
8
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1
10
14
7
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n=1
load factor: 8/10=0.8
split using h1.
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Analysis of Midterm-Examination
24
8
0
1
1
10
14
7
2
3
4
5
4
5
n=2
load factor: 8/12=0.66
no split
insert(2)
24
8
0
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1
1
10
14
7
2
3
4
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5
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Analysis of Midterm-Examination
24
8
1
10
14
7
4
n=2
load factor: 9/12=0.75
split using h1.
overflow
2
24
8
0
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1
5
10
2
7
2
3
4
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14
5
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Analysis of Midterm-Examination
insert(3)
24
8
24
8
1
10
2
7
4
5
14
1
10
2
7
3
4
5
14
n=3
load factor: 10/14=0.714
split using h1.
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Analysis of Midterm-Examination
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
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1
10
2
3
4
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14
7
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Analysis of Midterm-Examination
8
24
1
17
10
2
3
4
5
14
7
n=0
load factor: 11/16=0.687
no split.
insert(13)
8
24
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1
17
10
2
3
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14
7
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Analysis of Midterm-Examination
8
24
1
17
10
2
3
5
13
4
14
7
n=0
load factor: 12/16=0.75
split bucket 0, using h2.
1
17
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10
2
3
4
5
13
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8
24
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Analysis of Midterm-Examination
insert(15)
1
17
10
2
1
17
10
2
3
3
4
5
13
4
5
13
14
7
8
24
14
7
15
8
24
n=1
load factor: 13/18=0.722
split bucket 1, using h2.
1
17
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2
3
4
5
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15
8
24
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Analysis of Midterm-Examination
6.(15) 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
a
b
f
c
e
f
h
Fig. 1
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Analysis of Midterm-Examination
remove h:
c
b
a
b
e
c
e
f
remove c:
c
a
a
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e
b
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Analysis of Midterm-Examination
remove e:
c
a
a
e
b
f
remove f:
e
a
a
b
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Analysis of Midterm-Examination
7.(15) Given the relation schemas shown in Fig. 2, 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 ‘Business Computing’.
EMPLOYEE
fname, minit, lname, ssn, bdate, address, sex, salary, superssn, dno
DEPARTMENT
Fig. 2
Dname, dnumber, mgrssn, mgrstartdate
PROJECT
WORKS_ON
Pname, pnumber, plocation, dnum
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Analysis of Midterm-Examination
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)
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