# rsa-w7(rsa) d1-d2

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Information Security IByFahad 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 ScienceCihan UniversityFirst Semester, 2014-2015.Lecture-W7-D1-D2.11

4-CryptographyA- Definitions.B-Encryption and Decryption Definition(Symmetric & Asymmetric).C-Classical Cryptosystems. 1- Transposition Ciphers (Permutation): - Shuffling Scheme. 2- Substitution Ciphers : - Caesars Cipher. -Vigenre 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).

34- 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 Alices 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

Hard Mathematical ProblemThe 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 - CRYPTOGRAPHY5

RSA Algorithm:6Ali: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])= 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 , Ali will encrypt a M and Bob will decrypt the message. Bob should create a private key to decryption.

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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. Lets 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 slides7Model of security: policies of securities

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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.

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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

5- Now , Part B will decrypt the C by using the private key 29

Model of security: policies of securities

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-SummaryEncryption and Decryption of Asymmetric cryptography of RSA have 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 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.

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Factorization of big numbers Finding big prime numbers Multiplication of big prime numbers Exponentiation of big numbers Computing discrete logarithms.11

-Exercises:6- Local Area Network uses a public key infrastructure based on RSA with public key n =pq=55 and e=7.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 = , 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

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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 .

13RSAAl-GamalDepend on DLP, Number factorizationDepend on DLPCipher text size is the same as the message sizeCipher text size is the double of message sizePublic key (n,e), private key= p,q, d.public key g,p,A private key: a13

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)= 11.9143/3143/7143/11= 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 d corresponding to the RSA system.Sol: d.e = 1 mod (p-1)( q-1) d. 17= 1 mod (19-1) (29-1)15

Factorization of big numbers Finding big prime numbers Multiplication of big prime numbers Exponentiation of big numbers Computing discrete logarithms.15

http://www.cs.princeton.edu/~dsri/modular-inversion.html5- 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 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 = 89.1616

6- Local Area Network uses a public key infrastructure based on RSA with public key n =pq=55 and e=7.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

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6- Local Area Network uses a public key infrastructure based on RSA with public key n =pq=55 and e=7.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

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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=120So, x= -49 , y= 29.d= 1 * -49 mod 120 d=71.

B-

C-

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8-Alice has published her RSA public keys as = , where N is the known public number and e is her public key. Accordingly, Bob sent her the cipher text 81. Find the correspon