Chapter 10 AsymmetricKey Cryptography


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1 Chapter 10 AsymmetricKey Cryptography Copyright The McGrawHill Companies, Inc. Permission required for reproduction or display. 10.1
2 Chapter 10 Objectives To distinguish between two cryptosystems: symmetrickey and asymmetrickey To introduce trapdoor oneway functions and their use in asymmetrickey cryptosystems To introduce the knapsack cryptosystem as one of the first ideas in asymmetrickey cryptography To discuss the RSA cryptosystem 10.2
3 INTRODUCTION Symmetric and asymmetrickey cryptography will exist in parallel and continue to serve the community. We actually believe that they are complements of each other; the advantages of one can compensate for the disadvantages of the other. Topics discussed in this section: Keys General Idea Need for Both Trapdoor OneWay Function Knapsack Cryptosystem
4 101 INTRODUCTION Symmetric and asymmetrickey cryptography will exist in parallel and continue to serve the community. We actually believe that they are complements of each other; the advantages of one can compensate for the disadvantages of the other. Note Symmetrickey cryptography is based on sharing secrecy; asymmetrickey cryptography is based on personal secrecy. 10.4
5 Keys Asymmetric key cryptography uses two separate keys: one private and one public. Figure 10.1 Locking and unlocking in asymmetrickey cryptosystem 10.5
6 General Idea Figure 10.2 General idea of asymmetrickey cryptosystem 10.6
7 Continued Plaintext/Ciphertext Unlike in symmetrickey cryptography, plaintext and ciphertext are treated as integers in asymmetrickey cryptography. Encryption/Decryption C = f (K public, P) P = g(k private, C) 10.7
8 Need for Both There is a very important fact that is sometimes misunderstood: The advent of asymmetrickey cryptography does not eliminate the need for symmetrickey cryptography. 10.8
9 Trapdoor OneWay Function The main idea behind asymmetrickey cryptography is the concept of the trapdoor oneway function. Functions Figure 10.3 A function as rule mapping a domain to a range 10.9
10 Continued OneWay Function (OWF) 1. f is easy to compute. 2. f 1 is difficult to compute. Trapdoor OneWay Function (TOWF) 3. Given y and a trapdoor, x can be computed easily
11 Continued Example When n is large, n = p q is a oneway function. Given p and q, it is always easy to calculate n ; given n, it is very difficult to compute p and q. This is the factorization problem. Example When n is large, the function y = x k mod n is a trapdoor oneway function. Given x, k, and n, it is easy to calculate y. Given y, k, and n, it is very difficult to calculate x. This is the discrete logarithm problem. However, if we know the trapdoor, k such that k k = 1 mod f(n), we can use x = y k mod n to find x
12 Knapsack Cryptosystem Definition a = [a 1, a 2,, a k ] and x = [x 1, x 2,, x k ]. Given a and x, it is easy to calculate s. However, given s and a it is difficult to find x. Superincreasing Tuple a i a 1 + a a i
13 Continued
14 Continued Example As a very trivial example, assume that a = [17, 25, 46, 94, 201,400] and s = 272 are given. Table 10.1 shows how the tuple x is found using inv_knapsacksum routine in Algorithm In this case x = [0, 1, 1, 0, 1, 0], which means that 25, 46, and 201 are in the knapsack
15 Continued Secret Communication with Knapsacks. Figure 10.4 Secret communication with knapsack cryptosystem 10.15
16 Continued Example This is a trivial (very insecure) example just to show the procedure
17 Continued Example In the previous example, the following rules hold: 1. b = [b 1, b 2,, b k ] is a superincreasing set. 2. n > b 1 + b b k. 3. r is relatively prime to n (1 < r < n1). 4. t = [t 1, t 2,, t k ], t i = r b i (mod n)
18 102 RSA CRYPTOSYSTEM The most common publickey algorithm is the RSA cryptosystem, named for its inventors (Rivest, Shamir, and Adleman). Topics discussed in this section: Introduction Procedure Some Trivial Examples Attacks on RSA Recommendations Optimal Asymmetric Encryption Padding (OAEP) Applications
19 Introduction Figure 10.5 Complexity of operations in RSA 10.19
20 Procedure Figure 10.6 Encryption, decryption, and key generation in RSA 10.20
21 Continued
22 Continued Encryption 10.22
23 Continued Decryption 10.23
24 Continued Proof of RSA 10.24
25 Some Trivial Examples Example Bob chooses 7 and 11 as p and q and calculates n = 77. The value of f(n) = (7 1)(11 1) or 60. Now he chooses two exponents, e and d, from Z 60. If he chooses e to be 13, then d is 37. Note that e d mod 60 = 1 (they are inverses of each Now imagine that Alice wants to send the plaintext 5 to Bob. She uses the public exponent 13 to encrypt 5. Bob receives the ciphertext 26 and uses the private key 37 to decipher the ciphertext: 10.25
26 Some Trivial Examples Example Now assume that another person, John, wants to send a message to Bob. John can use the same public key announced by Bob (probably on his website), 13; John s plaintext is 63. John calculates the following: Bob receives the ciphertext 28 and uses his private key 37 to decipher the ciphertext: 10.26
27 Some Trivial Examples Example Jennifer creates a pair of keys for herself. She chooses p = 397 and q = 401. She calculates n = She then calculates f(n) = She then chooses e = 343 and d = Show how Ted can send a message to Jennifer if he knows e and n. Suppose Ted wants to send the message NO to Jennifer. He changes each character to a number (from 00 to 25), with each character coded as two digits. He then concatenates the two coded characters and gets a fourdigit number. The plaintext is Figure 10.7 shows the process
28 Continued Figure 10.7 Encryption and decryption in Example
29 Continued Example Here is a more realistic example. We choose a 512bit p and q, calculate n and f(n), then choose e and test for relative primeness with f(n). We then calculate d. Finally, we show the results of encryption and decryption. The integer p is a 159digit number
30 Continued Example Continued The modulus n = p q. It has 309 digits. f(n) = (p 1)(q 1) has 309 digits
31 Continued Example Continued Bob chooses e = (the ideal is 65537) and tests it to make sure it is relatively prime with f(n). He then finds the inverse of e modulo f(n) and calls it d
32 Continued Example Continued Alice wants to send the message THIS IS A TEST, which can be changed to a numeric value using the encoding scheme (26 is the space character). The ciphertext calculated by Alice is C = P e, which is 10.32
33 Continued Example Continued Bob can recover the plaintext from the ciphertext using P = C d, which is The recovered plaintext is THIS IS A TEST after decoding
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