Mathematics of Cryptography Algebraic Structures ALGEBRAIC STRUCTURES
Cryptography requires sets of integers and specific operations that are defined for those sets. The combination of the set and the operations that are applied to the elements of the set is called an algebraic structure. We will define three common algebraic structures: groups, rings, and fields.
1 Groups 2 Rings 3 Fields
Groups
A group (G) is a set of elements with a binary operation (•) that satisfies four properties (or axioms). A commutative group satisfies an extra property, commutativity: ❏ Closure: If a and b are elements of G, then c=a•b is also an element of G. ❏ Associativity: a•(b•c)=(a•b)•c ❏ Commutativity: a•b=b•a ❏ Existence of identity: For all element a in G, there exists and element e, identity element, s. t. e•a=a•e=a ❏ Existence of inverse: for each a there exists an a’, inverse of a, s.t. a•a’=a’•a=e. Figure Group Application Although a group involves a single operation, the properties imposed on the operation allow the use of a pair of operations as long as they are inverses of each other.
Example The set of residue integers with the addition operator, G = < Zn , +>, is a commutative group. We can perform addition and subtraction on the elements of this set without moving out of the set.
Example
The set Zn* with the multiplication operator, G =
Figure Composition of permutation Example
Table Operation table for permutation group Example In the previous example, we showed that a set of permutations with the composition operation is a group. This implies that using two permutations one after another cannot strengthen the security of a cipher, because we can always find a permutation that can do the same job because of the closure property. Finite Group: A group is called a finite group if the set has a finite number of elements, else it is an infinite group.
Order of a Group: The number of elements in G
Subgroups: A subset H of a group G is a subgroup of G if H itself is a group w.r.t. operation on G. Or, If G=
Example
Is the group H =
Solution The answer is no. Although H is a subset of G, the operations defined for these two groups are different. The operation in H is addition modulo 10; the operation in G is addition modulo 12. Cyclic Subgroups If a subgroup of a group can be generated using the power of an element, the subgroup is called the cyclic subgroup. Example
Four cyclic subgroups can be made from the group G =
H2
H2 H4 Example Three cyclic subgroups can be made from the group G =
A cyclic group is a group that is its own cyclic subgroup.
Generator: The element that generates the cyclic subgroup can also generate the group itself. The elements of a finite cyclic group are: Example Three cyclic subgroups can be made from the group G =
Order of an Element The order of an element is the order of the cyclic subgroup it generates. Example a. In the group G =
A ring, R = <{…}, •, >, is an algebraic structure with two operations.
Ring Ring
A ring, R = <{…}, •, >, is an algebraic structure with two operations. The second operation must be distributed over the second.
Distributivity: for all a, b, and c elements of R, we have a (b•c) = (a b)•(a c) and (a•b) c = (a c)• (b c) Ring Example
The set Z with two operations, addition and multiplication, is a commutative ring. We show it by R =
A field, denoted by F = <{…}, •, > is a commutative ring in which the second operation satisfies all five properties defined for the first operation except that the identity of the first operation has no inverse.
Field Field
A Field supports two pairs of operations: addition/subtraction and multiplication/division, except that the division by zero is not allowed.
Field Finite Fields: are normally used in cryptography
Galois showed that for a field to be finite, the number of elements should be pn, where p is a prime and n is a positive integer.
Note A Galois field, GF(pn), is a finite field with pn elements. GF(p) Fields
When n = 1, we have GF(p) field. This field can be the set
Zp, {0, 1, …, p − 1}, with two arithmetic operations. Example
A very common field in this category is GF(2) with the set {0, 1} and two operations, addition and multiplication, as shown below.
GF(2) field Example A very common field in this category is GF(2) with the set {0, 1} and two operations, addition and multiplication, as shown below. GF(2) field
1. The set has only two elements – binary digits or bits 0 and 1.
2. Addition = XOR operation. Multiplication = AND
3. Addition and multiplication operations are same opn. XOR.
4. Multiplication and division are same AND operation. Example
We can define GF(5) on the set Z5 (5 is a prime) with addition and multiplication operators as shown below.
GF(5) field Summary
Summary GF(2n) FIELDS
In cryptography, we often need to use four operations (addition, subtraction, multiplication, and division). In other words, we need to use fields. We can work in GF(2n) and uses a set of 2n elements. The elements in this set are n-bit words.
1 Polynomials 2 Using A Generator 3 Summary Example Let us define a GF(22) field in which the set has four 2-bit words: {00, 01, 10, 11}. We can redefine addition and multiplication for this field in such a way that all properties of these operations are satisfied, as shown below.
An example of GF(22) field Example Let us define a GF(22) field in which the set has four 2-bit words: {00, 01, 10, 11}. We can redefine addition and multiplication for this field in such a way that all properties of these operations are satisfied, as shown below.
An example of GF(22) field Each word is a additive inverse of itself.
Every word except 00 has a multiplicative inverse.
Addition and multiplication are defined in terms of polynomials. Polynomials
A polynomial of degree n − 1 is an expression of the form
i where x is called the ith term and ai is called coefficient of the ith term. Example
Figure below shows how we can represent the 8-bit word (10011001) using a polynomials.
Representation of an 8-bit word by a polynomial Example
To find the 8-bit word related to the polynomial x5 + x2 + x, we first supply the omitted terms. Since n = 8, it means the polynomial is of degree 7. The expanded polynomial is
This is related to the 8-bit word 00100110. GF(2n) Fields
Note
Any operation on polynomials actually involves two operations: on coefficients, on two polynomials. Coefficients 0 and 1 belong to GF(2). The polynomials belong to GF(2n). GF(2n) Fields
Note
Polynomials representing n-bit words use two fields: GF(2) and GF(2n). Modulus
Addition of two polynomials never creates a polynomial out of the set.
Multiplication of two polynomials may create a polynomial with degree more than n-1. Need to divide the result by a modulus and keep the remainder. For the sets of polynomials in GF(2n), a group of polynomials of degree n is defined as the modulus.
Modulus
The modulus in this case acts as the prime polynomial, that is no polynomial in this set can divide this polynomial.
A prime polynomial cannot be factored into a polynomial with degree less than n.
Such polynomials are referred to as irreducible polynomials. Modulus For the sets of polynomials in GF(2n), a group of polynomials of degree n is defined as the modulus. Such polynomials are referred to as irreducible polynomials.
List of irreducible polynomials Addition Add the coefficients in GF(2).
Adding two polynomials of degree n-1 always creates a polynomial of degree n-1.
Additive identity: zero polynomial – all coeffn set to zero. Additive inverse: of a polynomial in GF(2) is the polynomial itself. Addition and subtraction operations are the same operation. Example
Let us do (x5 + x2 + x) (x3 + x2 + 1) in GF(28). We use the symbol to show that we mean polynomial addition. The following shows the procedure: Example
There is also another short cut. Because the addition in GF(2) means the exclusive-or (XOR) operation. So we can exclusive-or the two words, bits by bits, to get the result. In the previous example, x5 + x2 + x is 00100110 and x3 + x2 + 1 is 00001101. The result is 00101011 or in polynomial notation x5 + x3 + x + 1. Multiplication Sum of multiplication of each term of the first polynomial with each term of the second polynomial.
1. The coefficient multiplication is done in GF(2).
2. The multiplying xi by xj results in xi+j.
3. The multiplication may create terms with degree more than n − 1, which means the result needs to be reduced using a modulus polynomial. Multiplication
Multiplicative Identity: The multiplicative identity is always 1, in GF(2^8) it is 00000001.
Multiplicative Inverse: It can be found using the extended Euclidean algorithm. Example
Find the result of (x5 + x2 + x) ⊗ (x7 + x4 + x3 + x2 + x) in GF(28) with irreducible polynomial (x8 + x4 + x3 + x + 1). Note that we use the symbol ⊗ to show the multiplication of two polynomials.
Solution
To find the final result, divide the polynomial of degree 12 by the polynomial of degree 8 (the modulus) and keep only the remainder. Polynomial division with coefficients in GF(2) Example
In GF (24), find the inverse of (x2 + 1) modulo (x4 + x + 1).
Using the Extended Euclidean Algorithm Example
Find the multiplicative inverse of 11 in Z26.
Solution
The gcd (26, 11) is 1; the inverse of 11 is -7 or 19. Example In GF (24), find the inverse of (x2 + 1) modulo (x4 + x + 1).
Solution The answer is (x3 + x + 1) as shown below.
Using Extended Euclidean Algorithm Example In GF(28), find the inverse of (x5) modulo (x8 + x4 + x3 + x + 1).
Solution The answer is (x5 + x4 + x3 + x) as shown below.
Euclidean algorithm for above example Multliplication Using Computer The computer implementation uses a better algorithm, repeatedly multiplying a reduced polynomial by x. Example
5 2 7 4 3 Find the result of multiplying P1 = (x + x + x) by P2 = (x + x + x + x2 + x) in GF(28) with irreducible polynomial (x8 + x4 + x3 + x + 1) Solution We first find the partial result of multiplying x0, x1, x2, x3, x4, and 5 x by P2. Note that although only three terms are needed, the m product of x ⊗ P2 for m from 0 to 5 because each calculation depends on the previous result. Example Continued
5 2 7 4 3 Find the result of multiplying P1 = (x + x + x) by P2 = (x + x + x + x2 + x) in GF(28) with irreducible polynomial (x8 + x4 + x3 + x + 1)
An efficient algorithm Example Repeat above Example using bit patterns of size 8. Solution We have P1 = 000100110, P2 = 10011110, modulus = 100011010 (nine bits). We show the exclusive or operation by .
An efficient algorithm for multiplication using n-bit words Example
The GF(23) field has 8 elements.
We use the irreducible polynomial (x3 + x2 + 1) and show the addition and multiplication tables for this field.
We show both 3-bit words and the polynomials.
Note that there are two irreducible polynomials for degree 3.
The other one, (x3 + x + 1), yields a totally different table for multiplication. Example Continued Addition table for GF(2^3) Example Continued Multiplication table for GF(23) Using a Generator
Sometimes it is easier to define the elements of the GF(2n) field using a generator.
In this Field with the irreducible polynomial f(x), an element in the field, a, must satisfy the relation f(a) = 0.
In particular, if g is a generator of the Field, then f(g)=0.
It can be proved that the elements can be generated as: Example
Generate the elements of the field GF(24) using the irreducible polynomial ƒ(x) = x4 + x + 1.
Solution The elements 0, g0, g1, g2, and g3 can be easily generated, because they are the 4-bit representations of 0, 1, x2, and x3. Elements g4 through g14, which represent x4 though x14 need to be divided by the irreducible polynomial. To avoid the polynomial division, the relation ƒ(g) = g4 + g + 1 = 0 can be used. Example Continued Inverses Additive inverses: Additive inverse of each element is the element itself because addition and subtraction in this field are same. –g^3 = g^3
Multiplicative inverses: To find the multiplicative inverses g3 (g3)-1 = g-3 = g12 = g3+g2+g+1
Note the exponents are calculated mudulo 2^n-1=15. See g3 x g12 = g15 = g0 = 1 So g3 and g12 are multiplicative inverses of each other. Example
The following show the results of addition and subtraction operations: Example
The following show the result of multiplication and division operations: Summary
The finite field GF(2n) can be used to define four operations of addition, subtraction, multiplication and division over n-bit words. The only restriction is that division by zero is not defined.