Figure 1: The er diagram for the terrorist database ... - ByteLABS

Name
Nationality
Name
City
Terrorists
Belong_To
Groups
Wanted_For
Stolen
Date
Location
Crimes
Used_In
Weapons
Date
Location
Type
Serial
Figure 1: The er diagram for the terrorist database.
Module Title: Inf1B
Exam Diet (Dec/April/Aug): April 2005
Brief notes on answers:
Part A DATA AND ANALYSIS ANSWERS
Part A, answers to questions 1 - 3
1. (a) A possible er diagram is shown in Figure 1.
(b) For the er diagram of Figure 1, the sql statements are the following:
create table Groups (name : char(30), city : char(30),
primary key (name));
create table Terrorists (name : char(30),
nationality : char(30),
gname : char(30),
primary key (name, gname),
foreign key (gname)
references groups
on delete cascade);
create table Crimes (date : char(20), location : char(30),
primary key (date, location));
create table Weapons (serial : char(20), type : char(30),
primary key (serial));
create table Wanted_For (name : char(30), gname(30),
date : char(20), location : char(30),
primary key (name, gname,
date, location),
foreign key (name, gname)
references Terrorists,
foreign key (date, location)
references Crimes);
create table Used_In (date : char(20), location : char(30),
i

serial : char(20),
primary key (date, location, serial),
foreign key (date, location)
references Crimes,
foreign key (serial),
references Weapons);
create table Stolen (name : char(30), serial : char(20),
date : char(20), location : char(30),
primary key (name, serial),
foreign key (name)
referencesGroups,
foreign key (serial)
references Weapons);
(c) The query in tuple-relational calculus is:
{P | T ∈ Terrorists ∧ W ∈ Wanted For∧
U ∈ Used In ∧ V ∈ Weapons ∧
T.name = W.name ∧
W.date = U.date ∧ W.location = U.location ∧
U.serial = V.serial ∧
P.name = T.name ∧ P.type = V.type}
2. (a) (i). The cosine between two vectors ⃗x and ⃗y is defined as:
cos(⃗x, ⃗y) =
⃗x · ⃗y
|⃗x||⃗y| =
∑ ni=1
x i y i
√ ∑ni=1
x 2 √ ∑ni=1
y 2 (1)
(ii).
cos(Q, D 1 ) =
0 · 2 + 1 · 4 + 1 · 2 + 0 · 1 + 0 · 0
√
02 + 1 2 + 1 2 + 0 2 + 0 2√ 2 2 + 4 2 + 2 2 + 1 2 + 0 = 6
2 5 √ 2 (2)
cos(Q, D 2 ) =
0 · 0 + 1 · 0 + 1 · 2 + 0 · 2 + 0 · 1
√
02 + 1 2 + 1 2 + 0 2 + 0 2√ 0 2 + 0 2 + 2 2 + 2 2 + 1 = √ 2 (3)
2 2
This means that D 1 is closer to Q than D 2 and will be ranked higher by the
information retrieval engine.
(b) The basic pattern for noun phrases is: DT? JJ* NN. Instead of determiners,
we can have numerals, and instead of simple nouns, we can have compounds
(such as Apogee common stock). This yields the following regular expression:
(DT|CD)? JJ* (NN|JJ)* NN. If we allow pre-nominal determiner phrases such
as each of, we get: (DT IN)? (DT|CD)? JJ* (NN|JJ)* NN.
In CQP notation (as required in the question), this regular expression is:
([pos="DT"] [pos="IN"])? [pos="DT|CD"]? [pos="JJ"]*
[pos="NN|JJ"]* [pos="NN"]
ii

Other solutions are possible. For example, two properties of NPs have not been
taken into account: coordination (e.g., the cat and the dog) and genitives (e.g.,
Tony’s wife). However, students should receive full credit if their regular expression
covers all the NPs given in the question.
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Part B OBJECT ORIENTED PROGRAMMING ANSWERS
Part B, answers to questions 4 - 6
3. Examples of appropriate definitions are as follows.
(a) An iterative version is:
public static int fibonacci( int n )
{ int a = 0;
int b = 1;
int f = 0;
if ( n == 0 || n == 1 )
return n;
else
return fib( a, b, f, n );
}
public static int fib( int a, int b, int f, int n)
{ for ( int i = 2; i

Marking guide: 3 marks for correct handling of the array. 2 marks for the for
loop (or similar).
(e) A printing definition is:
public class fibbo {
public static void main( String args[] )
{ int number = 7;
int array[] = new int[number];
fibonacciArray( array );
for ( int counter = 0; counter < array.length; counter++ )
System.out.println(counter + " has fibonacci " + array[ counter ]);
}
}
Marking guide: 2 marks for correct handling of the array. 2 marks for the for
loop (or similar). 1 mark for an appropriate print statement.
4. (a) public abstract class Animal {
private String name;
public Animal () { this(""); }
public Animal (String n) { setName(n); }
public void setName (String n) { name = n; }
public String getName () { return name; }
public abstract boolean canWalk ();
public abstract boolean canFly ();
}
public abstract boolean canSwim ();
public class Bird {
public Bird () { super(); }
public Bird (String n) { super(n); }
public boolean canWalk () { return false; }
public boolean canFly () { return true; }
}
public boolean canSwim () { return false; }
public class Mammal {
public Mammal () { super(); }
v

public Mammal (String n) { super(n); }
public boolean canWalk () { return true; }
public boolean canFly () { return false; }
}
public boolean canSwim () { return false; }
public class Fish {
public Fish () { super(); }
public Fish (String n) { super(n); }
public boolean canWalk () { return false; }
public boolean canFly () { return false; }
}
public boolean canSwim () { return true; }
(b) (i). False. One cannot instantiate an abstract superclass.
(ii). True. It is perfectly acceptable for an abstract superclass reference to refer
to a subclass instance.
(iii). True. Same reasons as above.
(iv). False. Birds and Mammals are at the same level of the hierarchy; one cannot
refer to the other.
5. (a) The implementation of levelOrder() is the following:
public void printLevelOrder () {
Queue q = new Queue();
q.enqueue(root);
while (! q.isEmpty()) {
TreeNode head = (TreeNode) q.dequeue();
System.out.print(head.getValue() + " ");
if (head.getLeft() != null)
q.enqueue(head.getLeft());
if (head.getRight() != null)
q.enqueue(head.getRight());
}
}
It generates [11, 7, 17, 1, 8, 16, 22, 25] on the given tree.
(b) It is the equivalent of a preorder traversal.
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