1 Cryptography and Network Security Third Edition by William Stallings Lecture slides by Lawrie Brown Chapter 4 – Finite Fields.

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Presentation transcript:

1 Cryptography and Network Security Third Edition by William Stallings Lecture slides by Lawrie Brown Chapter 4 – Finite Fields

2 Introduction will now introduce finite fields of increasing importance in cryptography –AES, Elliptic Curve, IDEA, Public Key concern operations on “numbers” –where what constitutes a “number” and the type of operations varies considerably start with concepts of groups, rings, fields from abstract algebra

3 Group a set of elements or “numbers” with some operation whose result is also in the set (closure) obeys: –associative law: (a.b).c = a.(b.c) –has identity e : e.a = a.e = a –has inverses a -1 : a.a -1 = e if commutative a.b = b.a –then forms an abelian group

4 Cyclic Group define exponentiation as repeated application of operator –example: a -3 = a.a.a and let identity be: e=a 0 a group is cyclic if every element is a power of some fixed element –ie b = a k for some a and every b in group a is said to be a generator of the group

5 Ring a set of “numbers” with two operations (addition and multiplication) which are: an abelian group with addition operation multiplication: –has closure –is associative –distributive over addition: a(b+c) = ab + ac if multiplication operation is commutative, it forms a commutative ring if multiplication operation has inverses and no zero divisors, it forms an integral domain

6 Field a set of numbers with two operations: –abelian group for addition –abelian group for multiplication (ignoring 0) –ring

7 Modular Arithmetic define modulo operator a mod n to be remainder when a is divided by n use the term congruence for: a ≡ b mod n –when divided by n, a & b have same remainder –eg. 100 ≡ 34 mod 11 b is called the residue of a mod n –since with integers can always write: a = qn + b usually have 0 <= b <= n mod 7 ≡ -5 mod 7 ≡ 2 mod 7 ≡ 9 mod 7

8 Modulo 7 Example

9 Divisors say a non-zero number b divides a if for some m have a=mb ( a,b,m all integers) that is b divides into a with no remainder denote this b|a and say that b is a divisor of a And a mod b=0 OR a ≡0 mod b eg. all of 1,2,3,4,6,8,12,24 divide 24

10 Modular Arithmetic Operations is 'clock arithmetic' uses a finite number of values, and loops back from either end modular arithmetic is when do addition & multiplication and modulo reduce answer can do reduction at any point, ie –a+b mod n = [a mod n + b mod n] mod n

11 Modular Arithmetic can do modular arithmetic with any group of integers: Z n = {0, 1, …, n-1} form a commutative ring for addition with a multiplicative identity note some peculiarities –if (a+b)≡(a+c) mod n then b≡c mod n –but (ab)≡(ac) mod n then b≡c mod n only if a is relatively prime to n

12 Modulo 8 Example

13 Greatest Common Divisor (GCD) a common problem in number theory GCD (a,b) of a and b is the largest number that divides evenly into both a and b –eg GCD(60,24) = 12 often want no common factors (except 1) and hence numbers are relatively prime –eg GCD(8,15) = 1 –hence 8 & 15 are relatively prime

14 Euclid's GCD Algorithm an efficient way to find the GCD(a,b) uses theorem that: –GCD(a,b) = GCD(b, a mod b) Euclid's Algorithm to compute GCD(a,b): –A=a, B=b –while B>0 R = A mod B A = B, B = R –return A

15 Example GCD(1970,1066) 1970 = 1 x gcd(1066, 904) 1066 = 1 x gcd(904, 162) 904 = 5 x gcd(162, 94) 162 = 1 x gcd(94, 68) 94 = 1 x gcd(68, 26) 68 = 2 x gcd(26, 16) 26 = 1 x gcd(16, 10) 16 = 1 x gcd(10, 6) 10 = 1 x gcd(6, 4) 6 = 1 x gcd(4, 2) 4 = 2 x gcd(2, 0)

16 Galois Fields finite fields play a key role in cryptography can show number of elements in a finite field must be a power of a prime p n known as Galois fields denoted GF(p n ) in particular often use the fields: –GF(p) –GF(2 n )

17 Galois Fields GF(p) GF(p) is the set of integers {0,1, …, p-1} with arithmetic operations modulo prime p these form a finite field –since have multiplicative inverses hence arithmetic is “well-behaved” and can do addition, subtraction, multiplication, and division without leaving the field GF(p)

18 Example GF(7)

19 Prime Numbers prime numbers only have divisors of 1 and self –they cannot be written as a product of other numbers –note: 1 is prime, but is generally not of interest eg. 2,3,5,7 are prime, 4,6,8,9,10 are not prime numbers are central to number theory list of prime number less than 200 is:

20 Prime Factorisation to factor a number n is to write it as a product of other numbers: n=a × b × c note that factoring a number is relatively hard compared to multiplying the factors together to generate the number the prime factorisation of a number n is when its written as a product of primes –eg. 91=7×13 ; 3600=2 4 ×3 2 ×5 2

21 Relatively Prime Numbers & GCD two numbers a, b are relatively prime if have no common divisors apart from 1 –eg. 8 & 15 are relatively prime since factors of 8 are 1,2,4,8 and of 15 are 1,3,5,15 and 1 is the only common factor conversely can determine the greatest common divisor by comparing their prime factorizations and using least powers –eg. 300=2 1 ×3 1 ×5 2 18=2 1 ×3 2 hence GCD(18,300)=2 1 ×3 1 ×5 0 =6

22 Fermat's Theorem a p-1 mod p = 1 –where p is prime and gcd(a,p)=1 also known as Fermat’s Little Theorem useful in public key and primality testing

23 Euler Totient Function ø(n) when doing arithmetic modulo n complete set of residues is: 0..n-1 reduced set of residues is those numbers (residues) which are relatively prime to n –eg for n=10, –complete set of residues is {0,1,2,3,4,5,6,7,8,9} –reduced set of residues is {1,3,7,9} number of elements in reduced set of residues is called the Euler Totient Function ø(n)

24 Euler Totient Function ø(n) to compute ø(n) need to count number of elements to be excluded in general need prime factorization, but –for p (p prime) ø(p) = p-1 –for p.q (p,q prime) ø(p.q) = (p-1)(q-1) eg. –ø(37) = 36 –ø(21) = (3–1)×(7–1) = 2×6 = 12

25 Euler's Theorem a generalisation of Fermat's Theorem a ø(n) mod N = 1 –where gcd(a,N)=1 eg. –a=3;n=10; ø(10)=4; –hence 3 4 = 81 = 1 mod 10 –a=2;n=11; ø(11)=10; –hence 2 10 = 1024 = 1 mod 11