Name: _
Quiz 2
CSE 565
October 4, 2016
P. No _
Time: 20 minutes
Problem 1. Determine (23) and (21) where (n) is Eulers totient function. (4 points)
Ans:
Since 23 is prime, (23) = 22. Since 21 = 7 x 3,
Name: _
Quiz 2
CSE 565
October 4, 2016
P. No _
Time: 20 minutes
Problem 1. Determine (19) and (33) where (n) is Eulers totient function.
Ans:
(4 points)
Since 19 is prime, (19) = 18. Since 33 = 11 x 3
CSE 565 Computer Security
Solutions to Homework 4 Problems
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Question 1:
The four approaches are i) secret key distribution using symmetri
CSE 565 Computer Security Solution Key for Midterm 1
Question 1: 18 (3 + 7 + 8 points)
a) If you are an adversary (cryptanalyst), how would you determine if
a given ciphertext is encrypted using eithe
Name: _
CSE 435/535 Information Retrieval Fall 2015
Midterm (90 minutes, 100 points)
Question 1: State whether the following statements are True or False (10 points)
You do not have to give reasons.
CSE565: Computer Security
Shambhu Upadhyaya Computer Science & Eng. University at Buffalo
UB Fall 2010
Buffalo, New York, 14260
UB Fall 2010 CSE565: S. Upadhyaya Lec 1.1
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Name: _ P. No _
Quiz 1
CSE 565
September 15, 2016
Time: 20 minutes
Question 1: (8 + 2 points)
a) Vernam cipher is a type of one-time pad where an arbitrarily long nonrepeating sequence of numbers is
c
CSE 565 Computer Security Solution Key for Midterm 1
Question 1: (4 + 6 + 5 points) a) A simple frequency analysis will determine the difference. If a monoalphabetic substitution is used, then the sta
9/15/2009
CSE 565: Computer Security
Due: Oct. 1, 2009
Project 1 – Vulnerability Analysis
1. Background
Vulnerability is a weakness in a system that makes it possible for a threat to precipitate. A th
9/3/2013
CSE 565: Computer Security Due: September 17, 2013
Homework 1
No questions will be graded for correctness. However, the questions will be graded for completeness.
Problem 1. (Optional only, w
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11/15/2013
CSE 565: Computer Security Due: 11/26/2013 in class
Homework 5
Question 1:
Problem 19.2 from Stallings (6th Edition).
In the PGP scheme, what is the expected number of session keys generate
CSE 565 Computer Security
Solutions to Homework 5 Problems
Question 1: (optional)
You need to refer to RFC 2401 (http:/rfc.net/rfc2401.html) Section 5.
IPv4 Header Fields
version
header length
TOS
tot
9/17/2013
CSE 565: Computer Security Due: 10/01/13 in class
Homework 2
There are four optional problems which are not counted for grade. You may do them if you feel that
you are not challenged enough.
CSE 565 Computer Security, Fall 2011
Solution Key for Homework 1
Problem 1. (5 points)
a) Main threats are (not necessarily in any particular order) 1) state-sponsored attacks on national security, 2)
CSE 565 Computer Security
Solutions to Homework 4 Problems
Question 1:
The four approaches are i) secret key distribution using symmetric encryption and a KDC, ii) secret key
distribution using symmet
Name: _ P. No _
Quiz 1
CSE 565
September 14, 2017
Time: 20 minutes
Question 1: (10 + 2 points)
a) Vernam cipher uses the formula ci = (ri + pi) mod n where ri is ith random number, pi is the ith messa
Name: _
Quiz 3
CSE 565
Sample
P. No _
Time: 20 minutes
Problem 1: (8 + 6 points)
a) One local area network vendor provides a key distribution facility, as illustrated in the following figure.
Describe
10/09/2017
CSE 565: Computer Security Due: 10/26/17 in class
Homework 3
Problem 1: (optional)
Problem 9.6 from Stallings, 7e.
Suppose we have a set of blocks encrypted with the RSA algorithm and we do
11/10/2016
CSE 565: Computer Security Due: 11/22/2016 in class
Homework 5
Question 1:
Problem 19.6 from Stallings, 7e.
What is the basic difference between X.509 and PGP in tems of key hierarchies and
CSE 565 Computer Security
Solution Key for Midterm 1
Question 1: 12 (3 + 3 + 6) points
a) If a monoalphabetic substitution is used, then the statistical properties of the ciphertext should be the
same
Name: _
Quiz 1
CSE 565
September 16, 2011
Time: 15 minutes
Question: (15 points)
Let X' be the bit-wise complement of X and K be the key. Prove that if the complement of the plaintext
block (X) is tak
CSE 565, Quiz 3, 10/24/11
20 minutes
Name:
Person Number:
1) List two disputes that can arise in the context of message authentication where a secret key is used to
generate a small fixed-size block o
Quiz 2 Solutions
CSE 565
October 3, 2011
Time: 15 minutes
Problem 1. Evaluate 117 mod 13
(3 points)
Ans:
112 mod 13 = 4; 117 = 112 x 112 x 112 x 11
116 mod 13 = 4 x 4 x 4 mod 13 = 12
117 mod 13 = 12 x
CSE 565, Quiz 4, 11/14/11 (15 minutes)
Name:
Question 1 (5 points)
In Kerberos, what is the purpose of Authenticator (also called a session key) that is issued along with
the ticket? Answer briefly in
Name: _ P. No _
Quiz 5
CSE 565
November 28, 2017
Time: 20 minutes
Question 1: (4 points)
List two disadvantages to using physical separation in a computing system. List two disadvantages to using
temp
CSE 565 Computer Security
Solutions to Homework 5 Problems
(All questions carry equal points. Maximum you can earn is 60 points.)
Question 1: (Optional)
SMTP encapsulates an email message in an envelo