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CS 772/872
Network and Systems Security
Fall 2006
Midterm Exam
Time 2 & 1/2 hours
Open Book & Notes
Name:
Dr. wahab
Login: wahab@cs.odu.edu
Each Question is 10 points
Question 1:
Dr. Wahab uses the crypto program to encrypt all his
files. Not only that but he use the same key for encrypting all his files.
Assume that you got hold of one of his plain files and the
corresponding cipher file.
Describe how you can find his key?
% crypto “`cat plainfile`”
cipherfile keyfile
and look at keyfile for the key “repeated”.
Question 2:
Briefly describe how UNIX implements:
For
each user N unix saves <n, p>,
where
n is the user login name & p is the message digest of the user password P
(<= 8 char) .
At login, user N types n and P.
The user is allowed to login if n and MD(P) matches
<n,p>.
Unix access control is based on: user, group IDs where every
user, file, and process is assigned a user ID, group ID. A user can access a
file that belongs to him or his group. In addition Unix support the concept of
“others” where if a file may be given the permission to be accessed by everyone
who can login to the system.
Question 3:
The traditional
combination lock (T) requires three turns between 1 and 40. Assume that you
have the option of buying new combination locks (N) and (C) such that:
(N) Requires seven turns between numbers 0 and
9.
(C) Requires five turns between charters A and
Z.
Which lock (T, N or C) is more secure?
Explain?
Maximum number of trials:
T = 40**3 = 64,000
N = 10**7= 10,000,000
C = 26**5 = 11,881,376
This C is more secure.
Question 4:
Assume we have the
following encryption scheme:
(C) Caesar-based cipher: Divide the string into 8 bits long blocks
and replace each block with another block K positions away with wrap around.
(M) Monoalphabetic-based cipher: Arbitrary map one 8 bits block with another
block.
Which encryption
scheme (C or M) is more secure? Explain?
For C:
Number of different blocks = 2**8= 256.
Thus number of possible keys is 256
keys.
For M:
Number of possible maps: 256! = Huge (
10! = 3,628,800)
No doubt, M is more secure.
Question 5:
In this question, we will use
the following notations:
·
P and V are the Public and its corresponding priVate keys.
·
K is a
symmetric Key.
·
M is a Message
and C is the corresponding cipher
·
K {M} : message
M is encrypted with K.
·
K [M] : message M is decrypted with K.
·
{M}P :
message M is encrypted/verified with P.
·
[M]V :
message M is decrypted/signed with V.
·
H(M): the hash
of message M.
Assume that Bob and Alice agree on a shared secret K and eA
/ dA are the public/private key pair of
Bob may authenticate himself to
1.
Bob sends
2.
Bob sends
3.
Bob sends
In each method, describe what
Since Alice knows the shared key K with
1.
if
K[C] is equal to K
2.
if [C]
dA is equal to K
3.
if H(K) is equal to C.
Question 6:
Assume that Bob and Alice agree on a shared
secret K, explain how to use a message digest algorithm to encrypt a message M
sent from Alice to
Both
Alice and Bob knows the shared secret K
and generates:
b1= MD(K)
bi
= MD(K|bi-1), i=2,3, ....
Alice
Bob
send ci = mi ® bi
>> recv ci and
compute mi= ci ® bi
Question 7:
A. In the following:
%
openssl genrsa -out
rsaprivatekey.pem -des3 1024
1. What is 1024 represents?
The
private key length.
2. What
–des3 is used for ?
Used
to encrypt the private key.
B. In
the following:
% openssl rsautl -encrypt -pubin -inkey rsapublickey.pem -in
file1 -out file2
1. What is maximum length of file1?
1024
2. How to convert file2 to
base64?
openssl
enc
-base64 –e -out file2.base64 -in
file2
Question 8:
A. Show the result, in HEX, of
the first 4 octets of the 1st round of
DES to:
Encrypt one data block of all 1s using a key of all 1s.
Just look at
http://www.cs.odu.edu/~cs772/fall06/lectures/secret_key_cryptography.html#fig3-6
The answer is clearly, without
any computations is:
FF FF FF FF
B. Show the result, in HEX, of the first 4 octets of 1st round of IDEA to:
Encrypt one data block of all 1s with a key of all 1s.
According
to
http://www.cs.odu.edu/~cs772/fall06/lectures/secret_key_cryptography.html#fig._3-21
Both Xa and Ka are FFFF= 2**16
The answer is (see explanation
page 76):
Xa x Ka mod 2*16 + 1 =
65,536 x 65,536 mod 65,537 =
00 00 00 01
Question 9:
Show the result, in HEX, of the
first four octets of the 1st round of AES-128 to:
Encrypt one data block of all 1s with a
key of all 1s.
8B 74 8B 74
Question 10:
.Let M be a 15 octet message of all 1s.
1. What is the value, in HEX, of the first octet of the message checksum of MD2?
14
2. What is the value, in HEX, of
the first octet of the 0th pass of the MD2 digest?
29