Computer and Network Security Assignment 1

Computer and Network Security
Assignment 1
Name: Ho Ka Kay Ivan
Class No.: 754-01
Student ID: 10454353

Q1. While in the school, a naughty student notices the password used by the teacher to access the school’s
database system(compromise of confidentiality). Later, he uses the password to change the data values of the
database system (compromise of integrity).

Q2a)Policy: Passwords need to be as unpredictable and safe as possible.
Mechanism: (1)Passwords with containing words found in Oxford dictionary are rejected. (2) Passwords
with less than 6 letters are rejected.

b) Policy: Prevent non-users from guessing their passwords into the system.
Mechanism: If the user inputs the wrong password for 3 consecutive times, the login system is locked.

c) Policy: Only students in computer science stream will be given accounts on computers in KEC909.
Mechanism: (1) If the student is studying computer science, he will be given an account in the KEC909
computers. (2) If the login account is not a computer science student, the system is locked.

Q3. Confidentiality prevents unauthorized disclosure of a system, while integrity prevent both unauthorized
access and modification of a system, thus preventing deception of a system.

Q4 a) Public key of Lily: 17
5
mod 53 = 40
Public key of Kenneth: 17
7
mod 53 = 6

b) Shared Secret Key: 40
7
mod 53 = 38

c) For two users A and B, if A wants to send b a message, the shared secret key is:
S
A.B
= (g
kb
mod p)
ka
mod p

where kb = B’s private key, ka = A’s private key.
= (g
kb x ka
) mod p
On the other hand, if B wants to send A a message, the shared secret key is:
S
B.A
= (g
ka
mod p)
kb
mod p
= (g
ka x kb
) mod p
= (g
kb x ka
) mod p
= S
A.B
Therefore, two users using the same Diffie-Hellman system to communitcate will have the same shared
secret key.


Q5a) All prime numbers p have only 2 factors: 1 and p.
Therefore, all integers n with no common factor with p are: 1, 2, 3, 4, 5, 6, … , p-1. Therefore ϕ(p) = p-1

b) The numbers with common factors with pq are:
p, 2p, 3p, 4p, … , (q-1)p, q, 2q, 3q, 4q, … , (p-1)q, pq
Therefore, no. of integers with no common factor with pq are:
pq – (q-1) – (p-1) – 1 = p – q + 1
= (p – 1)(q – 1)

Q6. Creating a table of the first 4 letters, we have:
W Y X Y J L K L
X Z Y Z K M L M
Y A Z A L N M N
Z B A B M O N O
A C B C N P O P
B D C D O Q P Q
C E D E P R Q R
D F E F Q S R S
E G F G R T S T
F H G H S U T U
G I H I T V U V
H J I J U W V W
I K J K V X W X

We can see a commonly used word part, so we guess that the to decipher the code, we move the message
back 10 letters, and the result is:
MONOALPHABETIC

Q7. Observing that after jumping every five letters, we see the word “THIS A”, so we guess that there are
six columns, then we have:
T S M S I U
H S E N F L
I I N O F T
S G T T I X
A N I D C X
So when read from top to bottom, we have:
“THIS ASSIGNMENT IS NOT DIFFICULT XX”.