There are those who are destined to be

"There are those who are destined to be good, but never to experience it. I believe I am one of them. " --- Evariste Galois (1811 -1832) 1

Mathematical Background: A Revision finite fields (FF) required for understanding n n AES Elliptic Curve Cryptography To study FF, we shall revise the concepts of n n n groups, rings, fields from abstract algebra Modular arithmetic and Euclidean Algorithm Finite fields of the form GF(p), where p is a prime number 2

Group Theory: History Groups: First used by Evariste Galois (b. 1811 - d. 1832) in his work, without defining a Group n Galois, a student of M. Vernier in 1827 and a contemporary of Cauchy, Poisson, Abel, Jacobi, Fourier, Gauss and Napolean (ruled during 1800 -1815) n He failed to join Ecole Polytechnique, though he appeared twice in the entrance tests. n An ardent Republican, he was sent to prison twice by the King. n 3

Quest for Academy Award n n n 1829: Galois (only 18 years old) submitted two papers to Académie des Sciences for publication in its ‘Memoirs’; Cauchy was the referee for the papers. Galois read a posthumous paper of Abel and found that there was an overlap between his and Abel’s work. So he consulted Cauchy (winner of Grand prix in 1816) advised him to rewrite it and submit it for Grand Prix. Feb 1830: Galois submitted the modified paper to Fourier for Grand Prix; Fourier died in April 1830 and the paper was lost; Abel and Jacobi got the Grand Prix prize. 4

Last Night n n n 1831: Galois again submitted to Académie des Sciences; Poisson was the Reviewer. He did not understand the paper and rejected it. night of 30 May 1832: injured at the duel with Perscheux d'Herbinville over the prison’s physician’s daughter named Stephanie-Felice du Motel: abandoned by both Perscheux as well as his seconds. A peasant took him to a hospital, where he died at the age of 21 in 1832. A story? : an injured Galois wrote notes on the rejected paper; a night of furious writings by Galois 5

First definitions Liouville, Galois’s elder brother, copied his papers and sent them to Gauss, Jacobi and others n 14 years later” 1846: Liouville got Galois' papers published n 1845: Cauchy defined a "conjugate system of substitutions“, another name of Groups. During 1845 -46, he wrote 25 papers on it. n 1854: The first person to try to give (not completely correct) an abstract definition of a group: Cayley. n 1863: Jordan’s commentary on Galois paper and his book used the term GROUP 6 n

Group Theory the first modern book n Walter Ledermann's book Introduction to theory of finite groups, published by publisher Oliver & Boyd in Edinburgh 1949 (when Ledermann was 38 years old, assistant lecturer at St Andrews ) n was based on Schur's lectures on group theory. 7

Group Theory and communism n n Ledermann wrote it in the British Museum Library (sitting in the same chair where Karl Marx wrote Das Capital) Ledermann came for a lecture on Group Theory at University of Notre Dame in the United States; the parcel of books was stopped by US Customs, who mistook it as a book of Communist groups, till the Head of Dept of Notre Dame personally spoke to Customs. 8

A note on types of numbers n n Positive integers and Integers Rational numbers: “A rational number is any number that can be written as a ratio of two integers. ” Reference: [1] http: //bing. search. sympatico. ca/? q=difference%20 between%20 a%20 real%20 number %20 and%20 a%20 rational%20 number&mkt=en-ca&set. Lang=en-CA n n Examples: Integers, fractions, mixed numbers, and decimals; together with their negative images. Examples of irrational numbers: √ 2, √ 3, √ 5, pi (π), e π = a mathematical constant whose value is the ratio of any circle 's circumference to its diameter =3. 14159265358979323846264338327950288419716939937510 e = base of the natural logarithm; known as Napier's constant; symbol honors Euler = 2. 71828459045235360287471352662497757…………. = is the unique number with the property that the area of the region bounded by the hyperbola y = 1/x, the x-axis, and the vertical lines x = 1 and x = e is 1. In other words e 1∫ (dx/x) = ln e = 1. 9

A note on types of numbers………………. . 2 n Real numbers: n n n Any number that can be found on the number line; a number required to label any point on the number line; a number whose absolute value names the distance of any point from 0. both rational and irrational numbers; Between any two rational numbers on the number line there is an irrational number. [1] Between any two irrational numbers there is a rational number [1] 10

A note on types of numbers………………. . 3 Complex numbers: Example: x + i y , where n n x and y: real numbers and i = √(-1) . The field of complex numbers includes the field of real numbers as a subfield. References: (i) http: //www. themathpage. com/a. Pre. Calc/rational-irrationalnumbers. htm (ii) http: //mathworld. wolfram. com/Complex. Number. html 11

Group DEFINITION: n a set of elements or “numbers” n with some operation whose result is also in the set (closure) (The operation is shown through the symbol “. ” in the examples below. ) n obeys: n n n associative law: (a. b). c = a. (b. c) has an identity element e so that for all a Є G, e. a = a. e = a For each a Є G, there exists an inverse element a-1 Є G, such that a. a-1 = e 12

Example of a group Example 1: N = a set of n distinct symbols = {1, 2, …. . , n} S = set of all permutations of the n symbols S is a Group, under the operation of permutation. Prove n Closure n Association n Existence of an identity element as a member of the group n Existence of an inverse for every member of the Group A Finite Group: if the number of members of the group is finite. An Infinite Group 13

Abelian Group If in addition to the three properties stated in slide 2, the property of commutation is satisfied, G is said to be an abelian group. Commutative: if for all a, b Є G, a. b = b. a Examples: 2. Prove that S, as defined in Example 1, is not an Abelian group. 3. Prove that the set of integers (positive, negative and zero) is an Abelian group under addition. Hint: Identity element = 0, Inverse element of X is –X. 14

Some Definitions and the definition of a Cyclic Group n n n Exponentiation: defined as repeated application of an operator. n example: a 3 = a. a. a Identity Element : e=a 0 If a’ be the inverse of a, a-n = (a’)n A Group is cyclic if every member of the Group is generated by a single element “a”, (called the Generator) through exponentiation. “a” is a member of the Group. A cyclic group is Abelian. n 15

Cyclic Group (continued) Cyclic group: b = ak For some integer value of k, b should stand for every member of the Group n n n A cyclic Group may be finite or infinite. Subgroups of a cyclic group are also cyclic. A cyclic group may have more than one generator element. Example 4 a: A group of integers, under the operation of addition, is a cyclic group. Both 1 and – 1 are the generators. 16

Cyclic Groups of Finite Group Order n A cyclic group of finite group order n is denoted as Cn with a generator element a and an identity element e such that e = an. The operations of such a group may be defined mod n. Example 4 b: Zn is a finite cyclic group of integers 0, 1, 2……(n-1), under the operation of “addition mod n”, with a generator element of 1 and an identity element of 0 17

Generator of a Field n GENERATOR: an element whose successive powers take on every element of the field except the zero n For Prime number fields: a = gj modp n n n Not every element of a field is a generator. For every 0<j<=(p-1), a different element is obtained. ORDER of a generator element: the smallest exponent j (< p), that gets the identity element. gj mod p = 1 18

Example of a generator and order Examples 1: Modulo 13: 4 and 5 are NOT generator elements. a = 2 is a generator element. Its order is 12. exponent, b 1 2 3 4 5 6 7 8 9 10 11 12 ab mod 13 2 4 8 3 6 12 11 9 5 10 7 1 19

Another Example: a generator and order Examples 2: Modulo 11: 2, 6, 7 and 8 are examples of generator elements. Order of 2, 6, 7 and 8: 10. 20

Ring Consider a set of “numbers” with two binary operations, called addition and multiplication. n n If the set constitutes an Abelian group with addition operation, and, if with multiplication operation, the set: has closure: For a, b Є G, a. b Є G n is associative: For a, b, c Є G, (a. b). c = a. (b. c) n distributive over addition: a. (b+c) = a. b + a. c the set constitutes a Ring. In a Ring, we can do multiplication, addition and subtraction without leaving the Ring. 21 n

Commutative Ring Ex 5: The set of all square matrices is a Ring over addition and multiplication. For a Ring, if multiplication operation is commutative, the set forms a commutative ring. Examples : n Ex 6: The set of matrices of Ex 5 is NOT a commutative Ring. Ex 7: The set S 2 of even integers ( positive, negative and 0), under the operations of addition and multiplication, is a Commutative Ring. 22

Integral Domain n A commutative ring R is said to constitute an Integral Domain if, n multiplication operation has an identity: a. 1 = 1. a for all a Є R, and if, n for a, b Є R, if a. b = 0, then either a = 0 or b = 0. Ex 8: S 3, the set of integers (positive, negative and 0) under the operations of addition and multiplication is an Integral domain. 23

Field a Field: a set of elements F, with two binary operations, called addition and multiplication, such that n F is an Integral Domain, and, n For each a Є F, except 0, there is an element a-1 in F such that a. a-1 = a-1. a = 1 (Existence of multiplicative inverse) 24

Field (continued) Thus in a Field, we can do addition, subtraction, multiplication and division without leaving the set. Ex 9. The set of all integers S 3 is not a Field. 10. The following are Fields: n n n The set of Rational Numbers The set of real numbers The set of complex numbers. All of the above examples of Fields have infinite number of elements. We shall see that Fields can be finite also. 25

Group, Ring and Field [A 1] closure under addition: [A 2] Associativity of addition: [A 3] Additive identity: Group Abelian Group Ring Commutative Ring Integral domain Field [A 4] Additive inverse: [A 5] Commutativity of addition: [M 1] closure under multiplication: [M 2] Associativity of multiplication: [M 3] Distributive laws: [M 4] Commutativity of multiplication [M 5] Multiplicative identity: [M 6] No zero divisors: [M 7] Multiplicative inverse: 26

Mathematical properties 1 A 1: If a and b belong to S, then a + b is also in S A 2: a + (b+c) = (a+b) + c for all a, b, c in S A 3: There is an element 0 in R such that a + 0 = 0 + a = a for all a in S A 4: For each a in S there is an element –a in S such that a + (-a) = (-a) + a = 0 A 5: a + b = b + a for all a, b in A M 1: If a and b belong to S, then ab is also in S M 2: a (bc) = (ab) c for all a, b, c in S 27

Mathematical properties 2 M 3: a(b+c) = ab + ac for all a, b, c in S (a+b)c = ac + bc for all a, b, c in S M 4: ab = ba for all a, b in S M 5: There is an element 1 is S such that a 1 = 1 a = a for all a in S M 6: If a , b in S and ab = 0, then either a = 0 or b = 0 M 7: If a belongs to S and a 0, there is an element a-1 in S such that a. a-1 = a-1. a = 1 28

Agenda After defining Rings and Fields: n Modular arithmetic n Divisors, GCD, Euclid’s theorem n prime numbers n Fields of type Zp n Finite Fields, Extended Euclid’s Theorem for finding multiplicative inverse Polynomial arithmetic 29

Modular Arithmetic: Definitions modulo operator: a mod n = b where b is the remainder when a is divided by n; b is called the residue of a mod n. n a = q. n + b 0 <= b < n; q = a/n where x is the largest integer less than or equal to x n Example 13: a = (b+c)mod 8 In the next slide, b is the element given in the first column (outside the box). c is the element given in the top row (outside the box). The values of a are given in the box. 30

Modulo 8 Example 31

Congruency mod n If a mod n = b mod n, a and b are said to be congruent mod n. The above statement may be written as, a=b mod n n n reducing k modulo n: The process of finding the smallest Non-negative integer, to which k is congruent 32

Modular Arithmetic: A Revision (continued) n Modular Arithmetic: n a = qn + r. r 0 1. n 2. n q. n a (q+1). n r 0 -q. n a -(q-1). n -3. n -2. n -n Thus 11 = 1. 7 + 4 r = 4 = 11 mod 7 -11 = -2. 7 + 3 r = 3 =-11 mod 7 33

k mod m 11 mod 7 = 4 n (-11) mod 7 = 3 n In general, If r = k mod m, ( - k) mod m = m - r if r ≠ 0; But ( - k) mod m = 0 if r = 0. i. e. k mod m may or may not be equal to (-k) mod m. r = k mod m = k mod (-m) = k mod(lml) n 34

Reducing k modulo 7: Example 12. . . -21 -20 -19 -18 -17 -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 7 14 21 28. . . 8 15 22 29 9 16 23 30 3 10 17 24 31 4 5 11 18 25 32 12 19 26 33 6 Reduced values 13 20 27 34 All the elements in a column are congruent mod 7 n n [O] = {…. , -21, -14, -7, 0, 7, 14…. } is called a Residue Class. (Every column constitutes a Residue Class. ) The Smallest Non-negative integer of the class is used to represent the class. 35

Modular Arithmetic: [a mod n + b mod n] mod n = (a + b)mod n n [a mod n - b mod n] mod n = (a - b)mod n n [a mod n x b mod n] mod n = (a x b)mod n Ex 14 of Exponentiation: To evaluate 1211 mod 7: 122 mod 7 = 4; 128 mod 7 = 44 mod 7 = 4; 12 x 128 mod 7= 5 x 4 mod 7 = 3 n 36

“Note that the positions of primes constitute just about the most fundamental, inarguable, nontrivial information available to our consciousness. This transcends history, culture, and opinion. It would appear to exist 'outside' space and time and yet to be accessible to any consciousness with some sense of repetition, rhythm, or counting. ” -- Matthew R. Watkins, School of Mathematical Sciences at Exeter University, UK http: //www. maths. ex. ac. uk/%7 Emwatkins/zeta/ss-b. htm, as of November 3, 2007 37

Modular Arithmetic Additive and multiplicative inverses additive inverse: Let c be the inverse of a. Then a + c = 0 mod n. Example 15: Additive inverse of 5 mod 8: 5 + c = 0 mod 8. Therefore c = 3 multiplicative inverse: Let c be the inverse of a. Then a x c = 1 mod n. Example 16: Multiplicative inverse of 5 mod 8: 5 x c = 1 mod 8. Therefore c = 5, 13, …. 38

Relatively Prime Numbers Two integers are said to be relatively prime if their only common positive integer factor is 1. n In Example 16, 5 and 8 are relatively prime. n n Consider the case where ‘a’ and ‘n’ have a common factor other than 1 (i. e. the case where ‘a’ and ‘n’ are not relatively prime) 39

Multiplicative Inverse (continued…) n Example 17: a=6 & n=8 6. c = 1 mod 8 No value of c, that satisfies the above, can be found. In general an integer has a multiplicative inverse in Zn if that integer is relatively prime to n. n 40

Inverses for modulo 8 a Additive Inverse of a Multiplicative Inverse of a 0 0 - 1 7 1 2 6 - 3 5 3 4 4 - 5 3 5 6 2 - 7 1 7 41

Multiplicative Inverse: Table 2 a 6. a mod 8 5. a mod 8 0 0 0 1 6 5 2 4 2 3 2 7 4 0 4 5 6 1 6 4 6 7 2 3 a =5 is the multiplicative inverse of 5 mod 8. 42

Multiplicative Inverse: Table 2 Continued a 6. a mod 8 5. a mod 8 8 0 0 9 6 5 10 4 2 11 2 7 12 0 4 13 6 1 14 4 6 15 2 3 a =13 is the multiplicative inverse of 5 mod 8. 43

Multiplicative Inverse Let c be the Multiplicative Inverse of b mod n. b. c = 1 mod n = k. n + 1 Therefore b. (c + n) = (k + b). n + 1 = k 1. n + 1 Thus c, c + n, c + 2 n……. are all multiplicative inverses of c. However for a field Zp, with members as 0, 1, 2, 3……. (p-1), the smallest positive number would be said to be the Multiplicative Inverse. n 44

Some properties of modulo operator some peculiarities n n n if (a+b)≡(a+c) mod n then b≡c mod n but if (a. b)≡(a. c) mod n then b≡c mod n only if a is relatively prime to n Proof: n Given (a+b) = (a+c) mod n n Add -a (the additive inverse of a) to both sides. [-a +a+b] = [-a +a+c] mod n b = c mod n 45

properties of modulo operator: Proof n Proof: n Given (a x b) = (a x c) mod n n Multiply with a-1 (Multiplicative inverse of a) on both sides: a-1 (a x b) = [a-1 (a x c)] mod n b = c mod n n n REVISION: However the multiplicative inverse of ‘a’ exists only if ‘a’ and ‘n‘ are relatively prime. a ≡ b mod n if n|(a-b) 46

Agenda After studying examples of modular arithmetic: n Modular arithmetic n Divisors, GCD, Euclid’s theorem n prime numbers n Fields of type Zp n Finite Fields, Extended Euclid’s Theorem for finding multiplicative inverse Polynomial arithmetic 47

Divisors If for some m, a=mb (a, b, m all integers), that is b divides into a with no remainder , n denote this as b|a n and say that b is a divisor of a n eg. all of 1, 2, 3, 4, 6, 8, 12, 24 are the divisors of 24. n 48

Properties of Divisors If a|1, then a = 1. n If a|b and b|a, then a = b. n Any b 0, divides 0. n If b|g and b|h, then b|(mg + nh) for arbitrary integers m and n n 49

Greatest Common Divisor n gcd(a, b) = max [k, such that k|a and k|b] n Properties: 1. gcd is required to be positive. gcd(a, b) = gcd(a, -b) = gcd(-a, -b) = gcd(|a|, |b|) 2. gcd(a, 0) = |a| 3. If gcd(a, b) = 1, a and b are relatively prime. 50

Properties of gcd function contd… Assume that a › b. 4. gcd(a, b) = gcd (b, a mod b) called a Theorem on the next slide Proof: n let d = gcd(a, b) n Then d|a and d|b ( i. e. a = k 1 d and b = k 2 d ) If (a mod b) = r, n a = kb + r or r = a – kb = k 1. d – k. k 2 d This proves d|r. n n n Thus (4) can be repetitively used to find d. 51

Greatest Common Divisor: 2 definitions n c = gcd(a, b) is the largest number that divides evenly into both a and b n n eg gcd(60, 24) = 12 Positive integer c is gcd of two positive integers a and b if n n c is a divisor of a and b; Any divisor of a and b is a divisor of c. Theorem: gcd(a, b) = gcd (b, a mod b) RHS may be a simpler function if a>b. n 52

Euclid’s algorithm n n Stated in his book “Elements”, written in 300 BC. Historians believe that the algorithm was devised ~200 years earlier an efficient way to find gcd(a, b) derived from the observation: If a & b have a common factor d (ie a=m. d & b=n. d), then d is also a factor in any difference between them, a-p. b = (m. d)-p. (n. d) = d. (m-p. n). uses successive instances of theorem: n gcd(a, b) = gcd(b, a mod b) Note: This MUST always terminate by giving gcd since eventually we get a mod b = 0 (no remainder). 53

Euclid's GCD Algorithm Euclid's Algorithm to compute gcd(a, b): n n A a, B b while B>0 n n n R = A mod B A B, B R return A = gcd(a, b) The example on the next slide uses Euclid’s algorithm. Even more useful: Extended Euclid’s Algorithm: Used for finding out the Multiplicative Inverse 54

Example GCD(1970, 1066) 1970 = 1 x 1066 + 904 1066 = 1 x 904 + 162 904 = 5 x 162 + 94 162 = 1 x 94 + 68 94 = 1 x 68 + 26 68 = 2 x 26 + 16 26 = 1 x 16 + 10 16 = 1 x 10 + 6 10 = 1 x 6 + 4 6 = 1 x 4 + 2 4 = 2 x 2 + 0 Hence gcd(1970, 1066) = 2 gcd(1066, 904) gcd(904, 162) gcd(162, 94) gcd(94, 68) gcd(68, 26) gcd(26, 16) gcd(16, 10) gcd(10, 6) gcd(6, 4) gcd(4, 2) gcd(2, 0) 55

Agenda After the Euclid’s theorem: n Modular arithmetic n Divisors, GCD, Euclid’s theorem n prime numbers n Fields of type Zp n Finite Fields, Extended Euclid’s Theorem for finding multiplicative inverse Polynomial arithmetic 56

Prime Numbers n A prime number p: an integer, whose only integer factors are itself and 1. Aug 6, 2002: Manindra Agrawal, Neeraj Kayal, Nitin Saxena of IIT Kanpur: Theorem: There is a deterministic polynomialtime algorithm for determining whether a number is a prime or a composite. n Odd Primes: all prime numbers except 2 n The magical prime: 2, used in cryptography n 57

Prime Numbers sequence Reference: http: //www. maths. ex. ac. uk/%7 Emwatkins/zeta/ss-b. htm Here the sequence of primes is presented graphically in terms of a step function or counting function which is traditionally denoted as (x). (Note: this has nothing to do with the value =3. 14159. . . ) The height of the graph at horizontal position x indicates the number of primes less than or equal to x. Hence at each prime value of x, we see a vertical jump of one unit. 58

Prime Numbers sequence Reference: http: //www. maths. ex. ac. uk/%7 Emwatkins/zeta/ss-e. htm Now zooming out by a factor of 2500, we get the above graph. Senior Max Planck Institute mathematician Don Zagier, in his article "The first 50 million primes" [Mathematical Intelligencer, 0 (1977) 1 -19] states: "For me, the smoothness with which this curve climbs is one of the most astonishing facts in mathematics. " 59

Prime Number Factors of a number n Unique factors of any integer a > 1: a n a = p p where P is the set of prime numbers p P and where ap is the degree of p n c = a. b cp = (ap+bp) for all p. Ex: 33033 = 3 x 7 x 112 X 13; 85833 = 3 x 3 x 3 x 11 x 172 c 3 = 3+1 =4, c 7 = 1, c 11 = 2 +1 = 3, c 13 = 1, c 17 = 2 gcd(33033, 85833) = 3 x 11 =33 n d|b dp bp for all p; Thus if d = 143, 143|33033 Calculating the prime factors of a large number is a difficult task. So prime number factorization NOT used for evaluation of a. b or of the greatest common divisor (gcd) of a and b. 60

Agenda After discussing prime numbers: n Modular arithmetic n Divisors, GCD, Euclid’s theorem n prime numbers n Fields of type Zp n Finite Fields, Extended Euclid’s Theorem for finding multiplicative inverse Polynomial arithmetic n with coefficient obeying modulo n arithmetic n with modulo m(x) and with coefficient obeying modulo n arithmetic 61

Modular Arithmetic n Consider the set of non – negative integers: n n n Zp = { 0, 1, 2, 3………(p-1) } Each element of Zp represents a residue class modulo ‘p’ where ‘p’ is a prime number. Properties of Modular Arithmetic for Integers in Zp are given in table 4. 2 (Stallings) 4 th Ed. 62

Table 4. 2 Reference: Page 105 Stallings, 4 th Edition Properties Expressions Commutative Laws (w+x) mod p = (x+w) mod p (w. x) mod p = (x. w) mod p Associative laws [(w+x) + y] mod p = [w+(x+y)] mod p [(w. x). y] mod p = [w. (x. y)] mod p Distributive Laws [w. (x + y)] mod p = [w. x + w. y] mod p Identities (0 +w)mod p = w mod p (1. w) mod p = w mod p Additive inverse (-w) Multiplicative Inverse (w-1) For each w Zp , there exists a z such that w+z 0 mod n For each w Zp , there exists a z such that w. z = 1 mod p 63

Agenda After discussing Fields of type Zp: n Modular arithmetic n Divisors, GCD, Euclid’s theorem n prime numbers n Fields of type Zp n Finite Fields, Extended Euclid’s Theorem for finding multiplicative inverse Polynomial arithmetic 64

Order of a Finite Field n n Order of a Finite Field: the number of elements in the field For n Zp = { 0, 1, 2, 3………(p-1) } Order = p 65

Galois Fields Galois Field GF(pn): A finite field of order pn For p: any prime integer and n: any integer, greater than or equal to 1, there is a unique field with pn elements, denoted by GF(pn). Unique: Any two fields with the same number of elements must be essentially the same, except perhaps for giving the elements of the field different names. An interesting fact 66

Galois fields of interest in cryptography: GF(p) GF(2 n). Let us first consider GF(p) = {0, 1, 2, …. (p-1)}, with arithmetic operations modulo p. 67

Galois Fields GF(p): Some Properties Every element in GF(p): relatively prime to p every element has a multiplicative inverse. Hence GF(p) is a Field. CHARACTERISTIC of a Field: The number of times a multiplicative identity can be added to itself before you get to zero. For GF(p), Characteristic = the number of elements in the field = p. A Field of characteristic p: Fp 68

Mutiplicative Inverse Algorithm finding the multiplicative inverse of b, such that b. b-1 = 1: Given that b <m Extended Euclid (m, b) Algorithm: To find c such that c. b = 1 mod m 69

Finding Inverses for m>>b EXTENDED EUCLID(m, b) ALGORITHM 1. (A 1, A 2, A 3) (1, 0, m); (B 1, B 2, B 3) (0, 1, b) 2. if B 3 = 0, return A 3 = gcd(m, b); no inverse 3. if B 3 = 1 return B 3 = gcd(m, b); B 2 = b– 1 mod m i. e. B 2: multiplicative inverse of b 4. Q = A 3/B 3 5. (T 1, T 2, T 3) (A 1 – Q B 1, A 2 – Q B 2, A 3 – Q B 3) 6. (A 1, A 2, A 3) (B 1, B 2, B 3) 7. (B 1, B 2, B 3) (T 1, T 2, T 3) 8. goto 2 70

Example: Inverse of 550 in GF(1759) Ti = Ai – Bi x Q Hence 355 is multiplicative inverse of 550 mod 1759. 71 If B 2 be –ve, subtract it from m to get the answer.

Finite Field GF(2) A B A+B A-B 0 0 0 1 1 1 1 0 0 Thus in GF(2), a+b = a-b is an XOR operation. a. b is an AND operation. A. B 0 0 0 1 72

Agenda Polynomial arithmetic (Ordinary polynomial algebra is of no interest in cryptography. ) n with coefficients obeying modulo n arithmetic n Prime polynomials and polynomial gcd n with modulo m(x) and with coefficient obeying modulo n arithmetic 73

Polynomial Arithmetic n Consider a polynomial: n n n A zero-th degree polynomial is a ‘constant polynomial’. A nth degree polynomial is called a MONIC polynomial, if an = 1. several alternatives available n n n ordinary polynomial arithmetic: Not used in cryptography poly arithmetic: with coeff arithmetic as mod p: called polynomial basis over a finite field poly arithmetic with coeff mod p and polynomials mod M(x) 74

A Revision: Group, Ring and Field [A 1] closure under addition: [A 2] Associativity of addition: [A 3] Additive identity: Group Abelian Group Ring Commutative Ring Integral domain Field [A 4] Additive inverse: [A 5] Commutativity of addition: [M 1] closure under multiplication: [M 2] Associativity of multiplication: [M 3] Distributive laws: [M 4] Commutativity of multiplication [M 5] Multiplicative identity: [M 6] No zero divisors: [M 7] Multiplicative inverse: 75

Polynomial Arithmetic with Modulo Coefficients Poly arithmetic is based on the fact that powers of x are linearly independent Let coefficients be elements of a Field GF(p). n The set of such polynomials forms a polynomial ring. Difference between a Field and a Ring: Consider two elements a and b. n n n Field: a/b = a. b-1 is also an element of the field. Ring: (that is not a Field): b-1 may not exist as an element of the Ring. ( a/b may not result in an exact division. ) Even if the coeff are the elements of a Field, the division of polynomials may leave a remainder. 76

Polynomials over GF(2) n In cryptography, we are interested in mod 2 n n all coefficients are 0 or 1 The coeff use modulo 2 arithmetic EXAMPLE: f(x) = x 3 + x 2 and g(x) = x 2 + x + 1 ADDITION: f(x) + g(x) = x 3 + x + 1 Addition of polynomials: requires XOR of coeffs MULTIPLICATION: multiplication of g(x) with x 3: x 5 + x 4 + x 3 multiplication of g(x) with x 2: x 4 + x 3 + x 2 f(x). g(x) = x 5 + x 2 77

Polynomials over GF(2) Multiplication and Addition n f(x): 1100 g(x): 0111 Addition: XOR process yields: 1011 Multiplication: Uses shifting and XOR: multiplication of g(x) with x 3: 111000 Lshift by 3 multiplication of g(x) with x 2: 011100 Lshift by 2 f(x). g(x) = 100100 78

Agenda Polynomial arithmetic (Ordinary polynomial algebra is of no interest in cryptography. ) n with coefficients obeying modulo n arithmetic n Prime polynomials and polynomial gcd n with modulo m(x) and with coefficient obeying modulo n arithmetic 79

Modulo m(x): A preliminary view Multiplication: increases the degree of the resultant polynomial. To ensure that the degree remains ‘the same’, we may consider: ( f(x). g(x) ) mod m(x). If a(x) = f(x). g(x), a(x) = q(x). m(x) + r(x), ( f(x). g(x) ) mod m(x) may be said to be equal to r(x) The degree of r(x) <= that of m(x). 80

A Prime Polynomial n n can write any polynomial in the form: n a(x) = q(x) m(x) + r(x) if the remainder is zero, m(x) divides a(x) If f(x), over a Field F, has no divisors other than itself & 1, it is called an irreducible (or prime) polynomial. n Another definition: f(x), over a Field F, is irreducible, iff f(x) cannot be expressed as a product of two polynomials, both of degree lower than that of f(x). n 81

Polynomial GCD Definition: c(x) is the greatest common divisor of a(x) and b(x) if n n c(x) divides both a(x) and b(x). Any divisor of a(x) and b(x) is a divisor of c(x). Euclid’s Algorithm to find polynomial gcd: Based on gcd[a(x), b(x)] = gcd[b(x), a(x) mod b(x)] with the assumption that the degree of a(x) > the degree of b(x). 82

Euclid’s Algorithm to find gcd[a(x), b(x)] -- similar to Extended Euclid(m, b) Algorithm gcd[a(x), b(x)]; Assume: the degree of a(x) > the degree of b(x). 1. A(x) a(x); B(x) b(x) 2. if B(x) = 0 return A(x) = gcd[a(x), b(x)] 3. R(x) = A(x) mod B(x) 4. A(x) B(x) 5. B(x) R(x) 6. goto 2 83

Euclid’s Algorithm to find gcd[a(x), b(x)] An Example Given: a(x) = x 6+x 5+x 4+x 3+x 2+x+1 b(X) = x 4 +x 2 +x+1 Euclid’s Algorithm A x 6+x 5+x 4+x 3+x 2+x 1+x+1 x 4 +x 2 +x+1 B x 4 +x 2 +x+1 x 3 +x 2+1 R x 3 +x 2+1 0 Q x 2 +x x + 1 gcd[a(x), b(x)] = A(x) = x 3 +x 2+1 0 84

Agenda Polynomial arithmetic (Ordinary polynomial algebra is of no interest in cryptography. ) n with coefficients obeying modulo n arithmetic n Prime polynomials and polynomial gcd n with modulo m(x) and with coefficient obeying modulo n arithmetic 85

Polynomials over GF(2) Polynomial arithmetic modulo an irreducible polynomial forms a Field. By analogy with modulo operations studied earlier, if a and b are relatively prime, the multiplicative inverse exists. We shall look at an extended Euclid algorithm to evaluate the multiplicative inverse of a(x) modulo b(x), where b(x) is an irreducible polynomial. n On the coefficients, the arithmetic is modulo 2. n 86

Extracts from earlier slides n n n If a mod 7 = b mod 7, a and b are said to be congruent mod 7. [O] = {…. , -21, -14, -7, 0, 7, 14…. } is called a Residue Class Mod 7. The Smallest Non-negative integer of the class is used to represent the class. To find the smallest Non-negative integer, to which k is congruent, is called reducing k modulo n Zp = { 0, 1, 2, 3………(p-1) } n Each element of Zp represents a residue class modulo ‘p’ where ‘p’ is a prime number. 87

Set of Residues modulo m(x) n n m(x): nth degree polynomial Example: residue class (x+1), modulo m(x) consists of all such polynomials a(x) such that a(x) = (x+1)mod m(x) Or all the polynomials, which satisfy a(x) mod m(x) = x+1. For m(X) = x 3 +x+1, one possible value of a(x) is x 4 +x 2 +1. 88

GF (pn) with an irreducible polynomial b(x) Set of residues: n n consisting of p elements. n Each of these elements represented by one of the pn polynomials of degree m<n Example: GF (23) with an irreducible polynomial b(x) = x 3 +x+1 The set of residues are {0, 1, x, (x+1), x 2, (x 2 +1), (x 2 + x), (x 2+x+1)} n Finding Multiplicative inverse of b(x) modulo m(x): Assume: degree of b(x) < degree of m(x) gcd[m(x), b(x)] = 1 89

23 elements of finite polynomial field GF(23) Decimal number 0 1 2 3 4 5 6 7 Binary number 000 001 010 011 100 101 110 111 Polynomial 0 1 x x+1 x 2+1 x 2+x+1 Choose m(x)=(x 3+x+1) as the irreducible polynomial. 90

Example GF(23) 91

Multiplicative Inverse: a(x). b(x) mod (x 3 +x+1) = 1 a(x) x x + 1 x 2 + x + 1 1 b(x) = a-1(x) x 2 +1 x 2 + x + 1 x 2 1 92

Additive and Multiplicative Inverses in GF (23) w 0 1 2 3 4 5 6 7 Additive Inverse -w 0 1 2 3 4 5 6 7 Multiplicative Inverse w-1 1 5 6 7 2 3 4 If mult results in a polynomial a(x) of degree greater than 2 (ie n-1 for pn or a degree greater than or equal to n), reduce it to a polynomial, r(x), of degree less than or equal to 2 by using r(x) = a(x) mod(x 3+x+1). 93

Multiplicative inverse Extended Euclid[m(x), b(x)] Algorithm 1. 2. 3. (A 1, A 2, A 3) (1, 0, m); (B 1, B 2, B 3) (0, 1, b) If B 3 = 0, return A 3 = gcd(m, b); no inverse If B 3 = 1 return B 2 as the multiplicative inverse of B (i. e. b(x). B 2 = 1 mod m(x) ) 7. Q = A 3/B 3 (T 1, T 2, T 3) (A 1 – Q B 1, A 2 – Q B 2, A 3 – QB 3) (A 1, A 2, A 3) (B 1, B 2, B 3) (T 1, T 2, T 3) 8. Go to 2 4. 5. 6. 94

Modular Polynomial Arithmetic n can compute in field GF(2 n) n n n polynomials with coefficients modulo 2 The elements of GF are polynomials, whose degree is less than n hence must reduce modulo an irreducible poly of degree n (for multiplication only) The polynomials form a finite field. The number of elements in the field is 2 n. For every element of the field, a multiplicative inverse can always be found by using Euclid’s Inverse algorithm. 95

ARITHMETIC OPERATIONS: GF(28) with m(x) = (x 8+x 4+x 3+x+1) AES uses GF(28) and an irreducible polynomial (x 8+x 4+x 3+x+1). In binary, it is 100011011 In HEX, the polynomial: 0 x 11 B Justification: The first out of the 30 irreducible polynomials of degree 8, given in Lidl, R. , Niederreiter, H. ‘Introduction to Finite Fields and Their Applications’, Cambridge University Press, 1994 96

MULTIPLICATIVE INVERSE: To find c(x) such that: (x 7+x+1). c(x) = 1 mod(x 8+x 4+x 3+x+1) A 1 1 0 1 x 3+ x 2+1 A 2 0 1 x 4+x 3+ x+1 A 3 x 8+x 4+x 3+x+1 x 7+x+1 x 4+x 3+ x 2+1 x B 1 0 1 x 3+ x 2+1 x 6+x 2+ x+1 B 2 1 x 4+x 3+ x+1 x 7 B 3 x 7+x+1 x 4+x 3+ x 2+1 x 1 Q - x 3+ x 2+1 x 3+ x 2+x Answer: The Multiplicative Inverse of (x 7+x+1) mod(x 8+x 4+x 3+x+1) = c(x) = x 7 97

"Genius is condemned by a malicious social organization to an eternal denial of justice in favor of fawning mediocrity" -- Evariste Galois 98

Representation A polynomial with coeff, obeying modulo 2 arithmetic, can be represented by a binary or a HEX number. Example : 0 x 11 B = 100011011 represents x 8+x 4+x 3+x+1. This is an irreducible polynomial. A polynomial in GF (28), a(x) = a 7 x 7+a 6 x 6+…+a 1 x+a 0 can be represented as ( a 7 a 6 a 5………. … a 1 a 0 ) Addition of two polynomials a(x) and b(x): Use XOR operation on two bit arrays: ( a 7 a 6 a 5…. . … a 1 a 0 ) ( b 7 b 6 b 5… …. . b 1 b 0 ) 99

ARITHMETIC OPERATIONS: MULTIPLICATION for GF(28) with m(x) = (x 8+x 4+x 3+x+1) Reduction: Example 1: x 8 mod m (x) = m (x) – x 8 = x 4 + x 3 + x + 1 Note: x 4 + x 3 + x + 1 can be represented as 0 x 1 B. In general : xn mod m (x) = m (x) – xn Multiplication: Let b(x) = b 7 x 7+ b 6 x 6+…+ b 1 x+ b 0 Example 2: Consider multiplication of b (x) with x : x. b (x) mod m (x) if b 7 = 0, x b (x) is in the reduced form. If b 7 = 1 using results of Example 1, (b 6 x 7+…+b 1 x 2+b 0 x) (x 4 + x 3 + x + 1) 100

ARITHMETIC OPERATIONS: MULTIPLICATION: Generalized Result This multiplication x. b (x) mod m (x) is done as follows x. b (x) mod m (x) = b 6 b 5 b 4 b 3 b 2 b 1 b 00 if b 7 = 0 = (b 6 b 5 b 4 b 3 b 2 b 1 b 00) (00011011) if b 7 = 1 Multiplication by a higher power can be achieved by a repeated application of Step 2. Example 3: r (x) = b (x). a (x) mod m (x) =(x 6 + x 4 + x 2 + x + 1). (x 7 + x + 1) mod (x 8+x 4 + x 3 + x + 1) 101

ARITHMETIC OPERATIONS: MULTIPLICATION: Example 3 To get r (x), Step 1 (x 6+x 4 + x 2 + x + 1). x mod m (x) (0101 0111). (0000 0010) Shift left 1010 1110 step 2 (x 6+x 4 + x 2 + x + 1). x 2 mod m (x) (0101 0111). (0000 0100) = (1010 1110). (0000 0010) ( 0001 1011) = (0101 1100) (0001 1011) = (0100 0111) 102

ARITHMETIC OPERATIONS: MULTIPLICATION Example (continued) Step 3 (x 6 + x 4 + x 2 + x + 1). x 3 mod m (x) (0101 0111). (0000 1000) = (0100 0111). (0000 0010) = 1000 1110 Step 4 Multiplication of b (x) by x 4 mod m (x) (0101 0111). (0001 0000) = (1000 1110). (0000 0010) (0001 1011) = (0001 1100) (0001 1011) = (0000 0111) 103

ARITHMETIC OPERATIONS: MULTIPLICATION Example (continued) Step 5 Multiplication of b (x) by x 5 mod m (x) (0101 0111). (0010 0000) = (0000 0111). (0000 0010) = 0000 1110 Step 6 Multiplication of b (x) by x 6 mod m (x) Result = 0001 1100 Step 7 Multiplication of b (x) by x 7 mod m (x) Result = 0011 1000 104

ARITHMETIC OPERATIONS: MULTIPLICATION Example (continued) Step 8 b (x). a (x) mod m (x) where a (x) = x 7 + x + 1 (0011 1000) (1010 1110) ( 0101 0111) = 1100 0001 Hence b (x). a (x) mod m (x) = (x 6+x 4 + x 2 + x + 1). (x 7 + x + 1) mod (x 8+x 4 + x 3 + x + 1) = x 7+x 6+ 1 105

Computational Considerations n n n Since coefficients are 0 or 1, any such polynomial can be represented as a bit string. Addition becomes XOR of the bit strings. Multiplication is shift or “shift & XOR”. n n n cf long-hand multiplication See, again, the line in red, five slides back. Modulo reduction done by repeatedly applying the rule of that slide. 106

Use of the bit notation for polynomials: Ex: for GF(28) with m(x) = x 8+x 4+x 3+x+1. Example: rc 1(x) = 1 rcj(x) = x. rcj-1(x) mod m(x) for j = 2 to 10 n Denoted by RC(1) = 1 RC(j) = 2. RC(j-1) for j = 2 to 10 For GF(28), the number of members of the finite group are 256, starting from 0 to 255. Thus RC(2) = 2, ………………RC(8) = 128 n rc 9(x) = x 8 mod m(x) = x 4+x 3+x+1 RC(9) = 1 B n RC(10) = 0011 0110 = 3616 = x 5+x 4+x 2+x – obtained by shifting RC(9) to the left 107

Win thousands of dollars! Solve problems in Number theory, Graph theory and Combinatorics-- and WIN! Paul Erdos, the great Hungarian problem solver, is the purser of all of the problems. (The purser is the final judge and arbiter of prize-winning solutions. The award only goes to the person who solves a problem first, and the purser is the arbiter of that too. ) Volunteer Advisor for solvers: greg@math. berkeley. edu References: 1. “A Tribute to Paul Erdos”, Cambridge University Press, 1990, pp. 467 -477. 2. “Paths, Flows, and VLSI Layout”, Springer-Verlag, 1980, pp. 35 -45. 3. “Erdos on graphs, his legacy of unsolved problems”, Fan Chung & Ron. Graham, AK Peters 1998 4. http: //www. math. upenn. edu/~chung/ 108