Discrete Mathematics with Applications MATH236 Dr. Hung P. - - PowerPoint PPT Presentation

Discrete Mathematics with Applications MATH236

• Dr. Hung P. Tong-Viet

School of Mathematics, Statistics and Computer Science University of KwaZulu-Natal Pietermaritzburg Campus

Semester 1, 2013

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Table of contents

1

The birth of public key cryptography

2

The RSA cryptosystem Key generation Encryption and Decryption

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The birth of public key cryptography

The birth of cryptography

Diffie and Hellman proposed the idea of a public-key cryptosystem in a paper in 1976. Such a system would work in the following fashion: A public-key cryptosystem is an asymmetric-key cryptosystem, i.e., it is not computationally feasible to compute a decryption key KD from the corresponding encryption key KE Suppose Alice wants to use a (hypothetical) public-key

• cryptosystem. She begins by generating a pair of keys, which we

denote pri(Alice) and pub(Alice) Anyone who wants to send her a message can encrypt it using her public key. But only Alice, using her private key, can decrypt such a message Suppose that Bob, another user of the system, wants to send Alice an encrypted message.

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The birth of public key cryptography

Example

1 Bob looks up Alice’s public key, pub(Alice) 2 Bob encrypts the message using pub(Alice) and send the encrypted

message to Alice using an open channel (e.g., by email)

3 Alice decrypts the encrypted message using her private key pri(Alice) Tong-Viet (UKZN) MATH236 Semester 1, 2013 4 / 14

The birth of public key cryptography

public key cryptograph

We can think of the encryption function in a public key cryptosystem as a trapdoor one-way function with trapdoor information being the private key Someone who knows pri(Alice) can decrypt a message encrypted using pub(Alice), but without this knowledge the task is computationally infeasible The security of a public-key crypto system depends on the function E and D used for encryption and decryption, respectively. They should have the following properties:

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The birth of public key cryptography

Properties of encryption and decryption functions

1 For a given plaintext P, and public key pub(A), it should be easy to

compute the corresponding ciphertext C = Epub(A)(P)

2 If only the ciphertext C is known, it should be computationally

infeasible to find the plaintext P

3 If the ciphertext C and the private key pri(A) are known, it should be

easy to compute the plaintext P = Dpri(A)(C)

4 It should be easy to generate pairs (pub(A),pri(A)) of public and

private keys so that too many such pairs exist for an enemy to construct a look-up table

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The RSA cryptosystem

Introduction

The RSA cryptosystem was invented by Rives, Shamir and Adleman in 1976 They are researchers at MIT Notice that Clifford Cocks, a British mathematician, came up with the same idea in 1973 The RSA is not widely used for general purposes of encryption and decryption of long message as the processes of encryption and decryption in public-key cryptosystems are slower than the corresponding operations in a symmetric-key cryptosystem The public-key cryptosystems are widely used to encrypt keys for symmetric-key cryptosystems-like DES

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The RSA cryptosystem Key generation

The mechanics of RSA: key generation

Suppose that Alice wants to use the RSA system. She must first generate a key-pair-pri(Alice) and pub(Alice).

1 Alice picks two large primes p and q of roughly the same size. She

computes their product n = pq (public modulus) and also the Euler function φ(n) = (p − 1)(q − 1)

2 Alice then selects a random integer e, 1 < e < φ(n) such that

gcd(e, φ(n)) = 1

3 She computes the inverse d of e in Zφ(n), i.e., ed ≡ 1 (mod φ(n)) 4 Finally, she sets pri(Alice)=(n,d) and pub(Alice)=(n,e) 5 Since the modulus n in Alice’s private key is the same n in her public

key, we shall think of the private key as just the number d and write pri(Alice)=d.

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The RSA cryptosystem Key generation

The mechanics of RSA: key generation

Example

1 Suppose that Alice chooses p = 47 and q = 59 2 Then n = pq = 2773 and φ(n) = (p − 1)(q − 1) = 2668 3 Alice now chooses a number e such that 1 < e < 2668 and

gcd(e, 2668) = 1

4 She chooses e = 17, a small prime number, with 17 ∤ 2668 5 Find e−1 = 17−1 in Z2668 6 d = e−1 = 157 (using Extended Division Algorithm) 7 Thus pri(Alice)= d = 157 and pub(Alice)= (2773, 17). Tong-Viet (UKZN) MATH236 Semester 1, 2013 9 / 14

The RSA cryptosystem Encryption and Decryption

The mechanics of RSA: encryption and decryption

Suppose that Bob wants to send an encrypted message to Alice

1 Bob looks up pub(Alice)=(n,e) 2 Bob represents his message, M, as an integer in the interval

[0, n − 1] If M is too large, he divides it into blocks and then encrypts each block separately

3 Bob encrypts M into ciphertext C by the rule C = Me mod n 4 To decrypt the ciphertext C Alice receives from Bob, she uses her

private key pri(Alice)=d to find M = C d mod n.

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The RSA cryptosystem Encryption and Decryption

Representing the alphabet with numbers

Table: An alphabet with spaces

letter number letter number letter number letter number space 00 g 07 n 14 u 21 a 01 h 08

• 15

v 22 b 02 i 09 p 16 w 23 c 03 j 10 q 17 x 24 d 04 k 11 r 18 y 25 e 05 l 12 s 19 z 26 f 06 m 13 t 20

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The RSA cryptosystem Encryption and Decryption

The mechanics of RSA: encryption and decryption

Suppose that Bob wants to send the message ‘pay me’to Alice He first encodes the message as numbers to keep the spaces, he use the coding scheme as in the previous thable The message is encoded as: 160125001305 Alice’s public key is (n, e) = (2773, 17) In order to use the RSA system, Bob breaks the message into blocks so that each block is an element of Z2773, so Bob breaks the message up into blocks of length 4 1601 2500 1305

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The RSA cryptosystem Encryption and Decryption

The mechanics of RSA: encryption and decryption

Now Bob encrypts each block M separately, each time producing an encrypted block C according to the rule C = Me mod n The first block is 1601, so we compute 160117 mod 2773 = 160124+1 mod 2773 = (16012)23 · 1601 mod 2773 = (9492)22 · 1601 mod 2773 = (21492)2 · 1601 mod 2773 = (11562) · 1601 mod 2773 = 2523 · 1601 mod 2773 = 1835 mod 2773

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The RSA cryptosystem Encryption and Decryption

The mechanics of RSA: encryption and decryption

The process is repeated for the other blocks in the message, yielding the ciphertext: 1835 0521 0813 Bob sends Alice this ciphertext After receiving the message from Bob, Alice uses her private key pri(Alice)=d=157 to decrypt For each block C, she computes M = C d mod n For example, the first block 1835157 mod 2773=1601 mod 2773.

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