Information Security I

By

Fahad Layth MalallahReference 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 ScienceCihan UniversityFirst Semester, 2014-2015.Lecture-W7-D1-D2.

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4-CryptographyA- 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.

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-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).

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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!!!

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Hard Mathematical Problem

• The concept HMP is best understood as amathematical problem which is computationallyinfeasible to solve.

• The HMP is proven mathematically.

• Among the concepts that are HMP that we haveseen are:

– DLP (Discrete Logarithm Problem).

– Integer Factorization.

MCS 1413 - CRYPTOGRAPHY 5

RSA Algorithm:

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Ali:1-Alie encrypts M byusing 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])= 13- 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 , Aliwill encrypt a M and Bob will decrypt the message. Bob shouldcreate a private key to decryption.

.

RSA numerical Example 1:Part A wants to send a message M to Part B. encrypt the messagem=10 and decrypt the cipher c by using asymmetric cryptosystemRSA. 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

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 72d=29.Private key is 29. this should be kept with Part B for decryption.

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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

-Summary-Encryption and Decryption of Asymmetric cryptography of RSAhave been illustrated with an example.

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-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 ande = 23. In the following, we will try to crack the system and todetermine the original message M.(i) What parameters comprises the public key and what parametersthe private key?.(ii) What steps are necessary to determine the private key from thepublic 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 dcorresponding to the RSA system.

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-Exercises:6- Local Area Network uses a public key infrastructure based on RSAwith 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 ofn = 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>, whereN is the known public number and e is her public key. Accordingly,Bob sent her the cipher text 81. Find the corresponding message.

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-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 .

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RSA Al-Gamal

Depend on DLP, Number factorization Depend on DLP

Cipher text size is the same as the messagesize

Cipher text size is the double of messagesize

Public key (n,e), private key= p,q, d. public key g,p,A private key: a

4- In RSA, the cipher-text C = 9. The public key is given by n = 143 ande = 23. In the following, we will try to crack the system and todetermine the original message M.(i) What parameters comprises the public key and what parametersthe private key?.(ii) What steps are necessary to determine the private key from thepublic 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)= 11.9143/3143/7143/11= 13 ok. Now p=11, q= 13

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Now p=11, q= 13d.23 = 1 mod (11-1) (13-1) 23. d= 1 mod 120Now compute the inverse as 1 =ax + by : a= 23, b= 120X= 47, y= -9 , the inverse is 47, so d= 47.

3- Original message is M

http://www.cs.princeton.edu/~dsri/modular-inversion.html5- Given p = 19, q = 29, N = pq and e = 17, compute the private key dcorresponding 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

http://www.cs.princeton.edu/~dsri/modular-inversion.html5- Given p = 19, q = 29, N = pq and e = 17, compute the private key dcorresponding 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 504Now, compute the inverse of d as:1 = ax + by : a= 17, b= 504.1= 17 x + 504 yNow, compute q from gcd (504,17), then compute x(s) and y(s).Finally: x= 89, y= -3.

Accordingly, the inverse d = 89.

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6- Local Area Network uses a public key infrastructure based on RSAwith 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

6- Local Area Network uses a public key infrastructure based on RSAwith 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

7- Consider a RSA public-key system where the public key consists ofn = 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 120to compute the inverse 1= ax + by: a=71, b=120So, x= -49 , y= 29.d= 1 * -49 mod 120 d=71.

B-

C-

19

nCM d mod

73

143mod1271

M

M

8-Alice has published her RSA public keys as <N; e> = <91;5>, whereN 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

205

91mod81

mod

29

M

M

nCM d