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# Cryptography - key sharing - RSA

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Cryptography - Public Key Distribution and RSA basics

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### Cryptography - key sharing - RSA

1. 1. PUBLIC KEY CRYPTOGRAPHY Prime - totient - coprime – relative prime - RSA prasaugus Friday, 28 March 2014 1
2. 2. Prime numbers Divisible by 1 and itself Friday, 28 March 20142
3. 3.  Any number other than prime are broken to prime.  A group of smaller prime number  Factorization involves set of prime numbers to bring back the number  Prime factors are unique Prime Number What is it made of ? Factorization Friday, 28 March 20143
4. 4. Discrete Logarithm Problem  46 mod 12 Ξ 10  3n mod 17 Ξ Equally likely with in 17  Modulus of any given number with any exponent is equally likely Friday, 28 March 20144
5. 5. Friday, 28 March 20145  32 mod 17 Ξ 9  33 mod 17 Ξ 10  34 mod 17 Ξ 13  35 mod 17 Ξ 5  36 mod 17 Ξ 15  37 mod 17 Ξ 11  38 mod 17 Ξ 16  39 mod 17 Ξ 14  310 mod 17 Ξ 8  311 mod 17 Ξ 7 Equally likely……! Discrete Logarithm problem (one way function) Is to find the exponent given the resultant value Eg : given 7 to find the exponent of 3 You may find it easy with smaller prime numbers…. It would take years to find if the prime number was 100 digits long….. Strength of this one way function is the time needed to compute
6. 6. Friday, 28 March 20146 Sender Intruder/eavesdropper Receiver Public distribution of generator ‘g’ and prime number ‘p’ gn mod p Ξ c 315 mod 17 Ξ 6 313 mod 17 Ξ 12 3, 17 3, 17 3,17 g = 3 p = 17 Spr= 15 Rpu= 12 Rpr= 13 Spu= 6 Sender selects his private ‘n’ to generate his public key and distributes it to all Receiver keeps sender’s public key and selects his private key ‘n’ and sends back his public key to all THE CRUX NOW!!Sender = Rpu Spr mod 17 = Actual Secret Key 10 Receiver = Spu Rpr mod 17 = Actual Secret Key 10
7. 7. Friday, 28 March 20147 The Story behind the logic 1215 mod 17 Ξ 10 12 Ξ 313 mod 17 313^15 mod 17 Ξ 10 Spr= 15 Rpu= 12 Rpr= 13 Spu= 6 Sender Receiver 613 mod 17 Ξ 10 6 Ξ 315 mod 17 315^13 mod 17 Ξ 10 Without any of the private keys, intruder or eavesdropper cannot access the secret key. It requires huge computation power to find it
8. 8.  Thanks to Diffie – Hellman  Who devised the algorithm to share keys in public Public Key Cryptography Thus the sharing of the keys between any unknown person is made Friday, 28 March 20148  Complexities  What if there are multiple receivers  Key management problem  Computation overhead
9. 9. Friday, 28 March 20149  Ronald Rivest  Adi Shamir  Leonard Adleman You know us all by name
10. 10. Friday, 28 March 201410 Sender Receiver Intruder R wants to send ‘89’ p1 = 53 p2 = 59 n = 53 * 59 n = 3127 (n) = 3016 by totient e = 3 d = 2011 Exponent e - Odd - Not a factor of (n) Calculated by modular Inverse using Euclidean Algorithm d = (k * (n) + 1)/e Hide eth except n = 3127 e = 3 n = 3127 e = 3 n = 3127 e = 3 n = 3127 e = 3 c = 1394 1394d Ξ 89 mod 3127 Message = 89 893 mod 3127 Ξ 1394 Crypt c = 1394 c = 1394
11. 11. Friday, 28 March 201411  In 1970 James Alice, a British Engineer, devise d a plan.  ‘A’ Sends an open lock to all, let those who wish to send message may lock it and send back.  ‘A’ opens all locks with his one and only key RSA Foundation of RSA To resolve computational complexityy Key Management
12. 12. Friday, 28 March 201412  Multiplication is easy to perform  (computer takes less than seconds to do it)  Instead finding factors of given numbers is hard.  (for larger numbers days and years) factorizor RSA Foundation of RSA To resolve computational complexityy Key Management
13. 13. Friday, 28 March 201413   PHI function is to measure the breakability of the number;  Where it is less than ‘n’ and not a factor of ‘n’  (Prime) = P-1  (7) =6 RSA Foundation of RSA To resolve computational complexityy Key Management
14. 14. Totient of a number  When n is a product of two primes, in arithmetic operations modulo n, the exponents behave modulo the totient φ(n) of n  15 = 3 x 5  φ(15) = 8  Relation (43)5 mod 15 Ξ 4(3x5)mod 8 mod 15 Ξ 47 mod 15 Friday, 28 March 201414
15. 15. Connection bw PHI φ function and Modular exponentiation  Connection mφ(n) Ξ 1 mod n let us assume any two numbers such that they do not share any common factors m = 5 and n = 8 5φ(8) Ξ 1 mod 8 Friday, 28 March 201415
16. 16. Connection bw PHI φ function and Modular exponentiation  Breakthrough a. 1k = 1 b. m* 1k = m based on the above propositions mφ(n) Ξ 1 mod n can be written as mk*φ(n) Ξ 1 mod n using (a) m*mk*φ(n) Ξ m mod n using (b) Finally we get mk*φ(n)+1 Ξ m mod n Friday, 28 March 201416
17. 17. Friday, 28 March 201417 Sender Receiver Intruder R wants to send ‘89’ p1 = 53 p2 = 59 n = 53 * 59 n = 3127 (n) = 3016 by totient e = 3 d = 2011 Exponent e - Odd - Not a factor of (n) Calculated by modular Inverse using Euclidean Algorithm d = (k * (n) + 1)/e Hide eth except n = 3127 e = 3 n = 3127 e = 3 n = 3127 e = 3 n = 3127 e = 3 c = 1394 1394d Ξ 89 mod 3127 Message = 89 893 mod 3127 Ξ 1394 Crypt c = 1394 c = 1394
18. 18. QUERIES ? Just a sec … Khan Videos on Cryptography https://engineering.purdue.ed u/kak/compsec/NewLectures/ Lecture12.pdf Friday, 28 March 201418
19. 19. Friday, 28 March 201419  Confidential communication  An individual can use (e,n) and (d, n) as public and private keys respectively.  If the message is long, it could be used as block cipher to reduce the size Usage of RSA Choice of values, keys, primes
20. 20. THANKS EVERYONE prassanna john paul prasaugus@gmail.com Friday, 28 March 2014 20