1. Information Security I
By
Fahad Layth Malallah
Reference Books:
1. Introduction to Computer Security, by Matt Bishop.
2. Security in Computing, 4th Edition, by Charls P. Pfleeger.
3. Principle of Computer Security. 2nd edition, by Arthur.
4th grade, Computer Science
Cihan University
First Semester, 2014-2015.
Lecture-W7-D1-D2.
1

2. 4-Cryptography
A- Definitions.
B-Encryption and Decryption Definition(Symmetric & Asymmetric).
C-Classical Cryptosystems.
1- Transposition Ciphers (Permutation):
- Shuffling Scheme.
2- Substitution Ciphers :
- Caesar’s Cipher.
-Vigenère Cipher.
-One-Time Pad.
D-Symmetric Cryptography :
-Data Encryption Standard (DES) Algorithm.
E- Asymmetric Cryptography (Public-Key Systems):
1-Basic on modular arithmetic, Number Theory.
2-Modular arithmetic inverse computation.
3- Al-Gamal Algorithm (ciphering & de-ciphering).
4-RSA Algorithm (ciphering & de-ciphering).
Summary- Exercises.
2

3. -Aim of this lecture:
Students will be familiar and able to secure information by using:
E- Asymmetric Cryptography (Public-Key Systems):
1-Basic on modular arithmetic, Number Theory.
2-Modular arithmetic inverse computation.
3- Al-Gamal Algorithm (ciphering & de-ciphering).
4-RSA Algorithm (ciphering & de-ciphering).
3

4. 4- RSA Algorithm (ciphering & de-ciphering).
• Three scientist have invented a security algorithm named it by
first character of their names:
• Adleman the mathematician.
• Rivest and Shamir the computer scientists.
– Alice must create a Public Key, which she can publish so that Bob (and
everyone else) can use it to encrypt messages to her. Because the
public key is a one way function, it must be virtually impossible for
anybody to reverse it and decrypt Alice’s message.
– However, Alice needs to decrypt the messages being sent to her. She
must therefore have a Private Key, which allows her to reverse the
effect of the Public Key.
– There is a mathematical relation between the Public Key and Private
Key, but if the Public Key is known the ability to find the Private Key is
zero, even if the mathematical relation is known!!!
4

5. Hard Mathematical Problem
• The concept HMP is best understood as a
mathematical problem which is computationally
infeasible to solve.
• The HMP is proven mathematically.
• Among the concepts that are HMP that we have
seen are:
– DLP (Discrete Logarithm Problem).
– Integer Factorization.
MCS 1413 - CRYPTOGRAPHY 5

6. RSA Algorithm:
6
Ali:
1-Alie encrypts M by
using public keys (e, n)as:
Bob:
1- chooses secret primes p and q and
computes n=pq .
2- chooses an exponent e as:
gcd( e, [p-1 ]. [q-1])= 1
3- then, computes d as :
de= 1 mod (p-1)(q-1)
4- Bob makes (p,q,d) public and keeps
(e,n) secret keys, then send only the
public to Ali….
5-Bob decrypts by computing .
Procedures is : Ali want to send a Secret message M to Bob. So , Ali
will encrypt a M and Bob will decrypt the message. Bob should
create a private key to decryption.
.

7. RSA numerical Example 1:
Part A wants to send a message M to Part B. encrypt the message
m=10 and decrypt the cipher c by using asymmetric cryptosystem
RSA. Let p = 7 and q = 13 be the two primes.
Solution:
1- Part B must select n= pq. and e where: gcd(e, [p-1][q-1])
n = pq = 91 and (p − 1)(q − 1) = 72.
To find e : gcd(e,72)=1 :
Choose e. Let’s look among the primes.
• Try e = 2. gcd(2, 72) = 2 (does not work)
• Try e = 3. gcd(3, 72) = 3 (does not work)
• Try e = 5. gcd(5, 72) = 1 (it works)
We choose e = 5. (e,n) is the public key
2- Part B also must find d (private key) next slides… 7

8. RSA numerical Example 1: Continue…
2- Part B also must find d (private key) by :
d.e = 1 mod (p-1) (q-1)  d.e=1 mod (7-1) (13-1)
d. 5 = 1 mod (6 ) (12)  d.5 = 1 mod 72
Now, we find multiplication inverse for 5 mod 72.
inverse equation: 1= ax + by a=5, b=72  1= 5x + 72 y.
1= (5*29) + (-2 * 72) correct.
Inverse(5)= 29.
29 = 1 mod 72
d=29.
Private key is 29. this should be kept with Part B for decryption. 8

9. RSA numerical Example 1: Continue…
3- Now, Part B sends the public key (e,n) and keeps the private key
(p,q,d).
4- Now, Part A encrypt the message m=10 as:
9
82
91mod10
mod
5



C
C
nMC e
5- Now , Part B will decrypt the C by using the private key 29
10
91mod82
mod
29



m
m
nCm d

10. -Summary
-Encryption and Decryption of Asymmetric cryptography of RSA
have been illustrated with an example.
10

11. -Exercises:
1-On which hard mathematical problem does RSA base its security?
2- Explain the ciphering and deciphering operations of RSA.
3-Compare between Al-Gamal and RSA .
4- In RSA, the cipher-text C = 9. The public key is given by n = 143 and
e = 23. In the following, we will try to crack the system and to
determine the original message M.
(i) What parameters comprises the public key and what parameters
the private key?.
(ii) What steps are necessary to determine the private key from the
public key?.
(iii) Determine the private key for the given system.
(iv) What is the original message M?.
5- Given p = 19, q = 29, N = pq and e = 17, compute the private key d
corresponding to the RSA system.
11

12. -Exercises:
6- Local Area Network uses a public key infrastructure based on RSA
with public key n =pq=55 and e=7.
(i) Find the private key d. For RSA we have de= 1 mod (p-1)(q-1)
(ii) Find the corresponding message M for a cipher C = 3.
7- Consider a RSA public-key system where the public key consists of
n = pq = 143 and e = 71.
A: Find a number d such that ed = 1 modulo (p-1)(q -1).
B: Give the decryption function for RSA.
C: Decrypt the cipher C = 12.
8-Alice has published her RSA public keys as <N; e> = <91;5>, where
N is the known public number and e is her public key. Accordingly,
Bob sent her the cipher text 81. Find the corresponding message.
12

13. -Exercises:
1-On which hard mathematical problem does RSA base its security?
1-discrete Logarithm Problem.
2- Number factorization.
2- Explain the ciphering and deciphering operations of RSA.
It is available in the lecture documents (slide 6).
3-Compare between Al-Gamal and RSA .
13
RSA Al-Gamal
Depend on DLP, Number factorization Depend on DLP
Cipher text size is the same as the message
size
Cipher text size is the double of message
size
Public key (n,e), private key= p,q, d. public key g,p,A private key: a

14. 4- In RSA, the cipher-text C = 9. The public key is given by n = 143 and
e = 23. In the following, we will try to crack the system and to
determine the original message M.
(i) What parameters comprises the public key and what parameters
the private key?.
(ii) What steps are necessary to determine the private key from the
public key?.
(iii) Determine the private key for the given system.
(iv) What is the original message M?.
Sol:
1-Public key : n=143, e= 23. private key is d. ( d.e= 1 mod (p-1)(q-1))
2- d.e= 1 mod (p-1) (q-1), how do we find p & q.
Divide n by sqrt(n). Sqrt(143)= 11.9
143/3
143/7
143/11= 13 ok. Now p=11, q= 13 14

15. Now p=11, q= 13
d.23 = 1 mod (11-1) (13-1)  23. d= 1 mod 120
Now compute the inverse as 1 =ax + by : a= 23, b= 120
X= 47, y= -9 , the inverse is 47, so d= 47.
3- Original message is M
http://www.cs.princeton.edu/~dsri/modular-inversion.html
5- Given p = 19, q = 29, N = pq and e = 17, compute the private key d
corresponding to the RSA system.
Sol:
d.e = 1 mod (p-1)( q-1)  d. 17= 1 mod (19-1) (29-1)
15
46
143mod9
mod
47



M
M
nCM d

16. http://www.cs.princeton.edu/~dsri/modular-inversion.html
5- Given p = 19, q = 29, N = pq and e = 17, compute the private key d
corresponding to the RSA system.
Sol:
d.e = 1 mod (p-1)( q-1)  d. 17= 1 mod (19-1) (29-1)
17. d = 1 mod 504
Now, compute the inverse of d as:
1 = ax + by : a= 17, b= 504.
1= 17 x + 504 y
Now, compute q from gcd (504,17), then compute x(s) and y(s).
Finally: x= 89, y= -3.
Accordingly, the inverse d = 89.
16

17. 6- Local Area Network uses a public key infrastructure based on RSA
with public key n =pq=55 and e=7.
(i) Find the private key d. For RSA we have de= 1 mod (p-1)(q-1)
(ii) Find the corresponding message M for a cipher C = 3.
Sol:
1- d.e= 1 mod (p-1) (q-1)  we have to find p & q.
So p= 11, q=5.
-To compute d: d. 7 = 1 mod (11-1) (5-1)  7.d =1 mod 40
-to compute inverse : 1= ax + by as a = 7, b= 40
- Compute x & y , x=-17 ,y= 3 d= (-17*1 + 40) mod 40 d= 23 17
11555
3.18355
4.755




18. 6- Local Area Network uses a public key infrastructure based on RSA
with public key n =pq=55 and e=7.
(i) Find the private key d. For RSA we have de= 1 mod (p-1)(q-1)
(ii) Find the corresponding message M for a cipher C = 3.
Sol:
-Compute x & y , x=-17 ,y= 3 d= (-17*1 + 40) mod 40 d= 23
2-
18
27
55mod3
mod
23



M
M
nCM d

19. 7- Consider a RSA public-key system where the public key consists of
n = pq = 143 and e = 71.
A: Find a number d such that ed = 1 modulo (p-1)(q -1).
B: Give the decryption function for RSA.
C: Decrypt the cipher C = 12.
Sol:
A- n=143=pq=11.13  d ed=1 mod (p-1)(q-1)
71. d = 1 mod (11-1)(13-1).  71 d = 1 mod 120
to compute the inverse 1= ax + by: a=71, b=120
So, x= -49 , y= 29.
d= 1 * -49 mod 120  d=71.
B-
C-
19
nCM d
mod
73
143mod1271


M
M

20. 8-Alice has published her RSA public keys as <N; e> = <91;5>, where
N is the known public number and e is her public key. Accordingly,
Bob sent her the cipher text 81. Find the corresponding message.
Sol:
In order to find the message , we have to firstly find the private key
which is d.
d.e =1 mod (p-1) (q-1)  now we have to find q & p from n where
n=pq.
91= 7 . 13= p.q.
d.5 = 1 mod (7-1) (13-1)  5.d = 1 mod 72  by finding the inverse
so: d=29.
Now, apply the decryption rule 
20
5
91mod81
mod
29



M
M
nCM d