Math236 Discrete Maths with Applications

Transcription

1 Math236 Discrete Maths with Applications P. Ittmann UKZN, Pietermaritzburg Semester 1, 2012 Ittmann (UKZN PMB) Math / 1

2 Block Ciphers A block cipher is an encryption scheme in which the plaintext message is broken up into blocks of a fixed length d, each of which is then encrypted separately We examine one block cipher in particular: the permutation cipher Let d be a positive integer Divide the message M into blocks of length d Then take a permutation π of 1, 2, 3,..., d and apply π to each block Specifically, if the plaintext block is x 1 x 2 x d, then the corresponding ciphertext block is x π(1) x π(2) x π(d) Ittmann (UKZN PMB) Math / 1

3 Block Ciphers (cont.) Example Let d = 4 and π = (2413) Suppose that the message we want to encrypt is he is a great mathematician We remember which symbols are spaces We pad the message so that its length is a multiple of the block length, 4 So we get: he is a great mathematician Ittmann (UKZN PMB) Math / 1

4 Block Ciphers (cont.) Example Next we divide the message up into blocks of length 4 he i s a grea t ma them atic ian We now apply the permutation (2413) to each block This means that the first letter in the ciphertext block is the second letter in the plaintext block, the second letter in the ciphertext block is the fourth letter in the plaintext block, and so on Ittmann (UKZN PMB) Math / 1

5 Block Ciphers (cont.) Example Thus: plaintext he-i s-a- grea t-ma them atic ianciphertext EIH- SA RAGE -ATM HMTE TCAI A-IN To decrypt, we apply the inverse ( ) π = to each block of the ciphertext to recover the original message Ittmann (UKZN PMB) Math / 1

6 Block Ciphers (cont.) Permutation ciphers are more secure than simple substitution ciphers, but are still vulnerable to attack Ittmann (UKZN PMB) Math / 1

7 Modular arithmetic revisited Before we continue our discussion of polyalphabetic ciphers, we shall simplify matters by representing the letters of the alphabet by numbers For this to be truly useful, we recall modular arithmetic Example If a is a nonnegative integer and n a positive integer, then we define a mod n as the remainder when a is divided by n 11 mod 3 = 2, 15 mod 7 = 1, 6 mod 2 = 0 Ittmann (UKZN PMB) Math / 1

8 Modular arithmetic revisited (cont.) If a is negative, then we define a mod n in the following way Let k be the largest multiple of n that is less than or equal to a Then a mod n = a k Ittmann (UKZN PMB) Math / 1

9 Modular arithmetic revisited (cont.) Example The largest multiple of 5 which is less than or equal to 7 is 10 (which is 2 times 5) Therefore, 7 mod 5 = 7 ( 10) = 3 Similarly, 4 mod 2 = 0, 13 mod 22 = 9, and 20 mod 3 = 1 Notice that, whatever the value of a, the number a mod n is always in {0, 1,..., n 1} Ittmann (UKZN PMB) Math / 1

10 Modular arithmetic revisited (cont.) We now define a representation of all the letters of the alphabet: letter number letter number letter number letter number a 0 h 7 o 14 v 21 b 1 i 8 p 15 w 22 c 2 j 9 q 16 x 23 d 3 k 10 r 17 y 24 e 4 l 11 s 18 z 25 f 5 m 12 t 19 g 6 n 13 u 20 Ittmann (UKZN PMB) Math / 1

11 Modular arithmetic revisited (cont.) This representation allows us to add two letters together using arithmetic modulo 26 We abbreviate this as arithmetic ( mod 26) Recall that Z 26 = {0, 1,..., 25} For two numbers x, y Z 26 define x + y mod 26 to be the number (x + y) mod 26 That is, we add the two numbers x and y together and then find the remainder when the result is divided by 26 Ittmann (UKZN PMB) Math / 1

12 Modular arithmetic revisited (cont.) Example mod 26 = 32 mod 26 = 6 Similarly, mod 26 = 10 mod 26 = 10 Example Similarly, we define x y mod 26 to be the number (x y) mod mod 26 = 12 mod 26 = 14 Ittmann (UKZN PMB) Math / 1

13 Modular arithmetic revisited (cont.) We can use modular arithmetic to implement the shift cipher Convert the letters in the plaintext to numbers, as per the table above Then add the shift or key (modulo 26) to each number To decrypt, we convert the ciphertext to numbers Then subtract the shift or key (modulo 26) from each number Ittmann (UKZN PMB) Math / 1

14 Modular arithmetic revisited (cont.) Example For a shift of 7, let s encrypt the message penguinofdeath Plaintext (letters) p e n g u i n o f d e a t h Plaintext (numbers) Ciphertext (numbers) Ciphertext (letters) W L U N B P U V M K O H A O We decipher by subtracting 7 from each number in the ciphertext (modulo 26) Ittmann (UKZN PMB) Math / 1

15 One-way functions Let S, T be sets A one-way function f : S T is a function for which For each x S, the value f (x) is easy to compute For almost every y T, it is computationally infeasible to find x S such that y = f (x) Ittmann (UKZN PMB) Math / 1

16 One-way functions (cont.) Example Let p be a large prime number and f (x) a polynomial of high degree, where f : Z p Z p It is easy to calculate f (x) for all x Z p, but usually hard to solve f (x) = y for x The function f is a one-way function For example We choose f (x) = 2x x 1471 x and p = It is very difficult to find x Z for which f (x) (mod 17957) Ittmann (UKZN PMB) Math / 1

17 One-way functions (cont.) Example Choose N 1 and N 2 to be large prime numbers Let S consist of all ordered pairs (p, q) of prime numbers with N 1 p q N 2 Define f : S Z by the rule f (p, q) = pq It is easy to calculate f (p, q) for all (p, q) S Suppose we are given the number pq (without being told the factors p and q) It is (in general) computationally infeasible to find p and q Ittmann (UKZN PMB) Math / 1

18 One-way functions (cont.) Example For example, the number is the product of two (relatively small) 4-digit primes p and q If you wish to gauge the difficulty of finding p and q, try to factor Easy to do on a computer But extremely time-consuming by hand Ittmann (UKZN PMB) Math / 1

19 One-way functions (cont.) Example Instead of using two 4-digit prime numbers, consider two 300-digit prime numbers Their product will be a number that even a computer will have tremendous difficulty factoring This difficulty is the basis of the RSA cryptosystem Ittmann (UKZN PMB) Math / 1

20 One-way functions (cont.) Suppose we are given a one-way function f We could create a table of all the pairs (f (x), x) That is, work our way through the set S, calculating f (x) for each x S Re-order the results by increasing f (x) Using this table, we can quickly look up an x for any given f (x) However, when S is large enough, this approach is not feasible It requires too much memory to store the table Ittmann (UKZN PMB) Math / 1

21 The password problem One of the first applications of one-way functions was to solving the problem of the security of computer passwords Suppose a group of users has access to a computer Each user logs in to the computer by supplying a user name u and password p(u) The computer then checks the entered password against the one it has on file for the user u to determine whether u should be allowed to login or not Ittmann (UKZN PMB) Math / 1

22 The password problem (cont.) It is dangerous to store a list {(u i, p(u i ))} of user names with their passwords in unencrypted form in a file on the computer If a hacker gains access to the system, they can make a copy of the file and gain access to all of the user accounts Let f be a one-way bijection whose domain is the set of all possible passwords Suppose that instead of storing a list of pairs (u, p(u)), the computer stores the list of pairs (u, f (p(u))) Ittmann (UKZN PMB) Math / 1

23 The password problem (cont.) Each time a user logs in The user enters a name u and a password p The computer calculates f (p ) The computer checks the entered name-password pair, (u, f (p )), against the stored name-password pair (u, f (p(u))) If f (p ) = f (p(u)), then, since f is a bijection, p = p(u) The computer allows the user to login Otherwise, the computer denies the user access Ittmann (UKZN PMB) Math / 1

24 The password problem (cont.) An intruder who gains access to the list of pairs (u, f (p(u))) obtains no useful information To login as a user u, they must know the password p(u) However, all they know is the encrypted form of the password, f (p(u)) Since f is a one-way function, it is (for all practical purposes) impossible to determine p(u) from f (p(u)) Ittmann (UKZN PMB) Math / 1