Finite field arithmetic
From Wikipedia, the free encyclopedia
Arithmetic in a finite field is different from standard integer arithmetic. There are a limited number of elements in the finite field; all operations performed in the finite field result in an element within that field.
Finite fields are used in a variety of applications, including linear block codes such as BCH and RS codes in classical coding theory and in cryptography algorithms such as the Rijndael encryption algorithm. In computer science applications, the operations are simplified for finite fields of characteristic 2, also called GF(2n) Galois fields, making these fields especially popular choices for applications.
There are infinitely many different finite fields; however, their number of elements (or cardinal) is necessarily of the form pn where p is a prime number, the characteristic of the field.
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[edit] Effective polynomial representation
GF(p), where p is a prime number, is simply the ring of integers modulo p. That is, one can perform operations (addition, subtraction, multiplication) using the usual operation on integers, followed by reduction modulo p. For instance, in GF(5), 4+3=7 is reduced to 2 modulo 5. Division is multiplication by the inverse modulo p, which may be computed using the extended Euclidean algorithm.
A particular case is GF(2), where addition is exclusive OR (XOR) and multiplication is AND. Since the only inversible element is 1, division is the identity function.
Elements of GF(pn) may be represented as polynomials of degree strictly less than n over GF(p). Operations are then performed modulo R where R is an irreducible polynomial of degree n over GF(p), for instance using polynomial long division. The addition of two polynomials P and Q is done as usual; multiplication may be done as follows: compute W =P.Q as usual, then compute the remainder modulo R (there exist better ways to do this).
When the prime is 2, it is conventional to express them as binary numbers, with each term in a polynomial represented by one bit in the corresponding element's binary expression. Braces ( "{" and "}" ) or similar delimiters are commonly added to binary numbers, or to their hexadecimal equivalents, to indicate that the value is an element of a field. For example, the following are equivalent representations of the same value in a characteristic 2 finite field:
- Polynomial: x6 + x4 + x + 1
- Binary: {01010011}
- Hexadecimal: {53}
[edit] Addition and subtraction
Addition and subtraction are performed by adding or subtracting two of these polynomials together, and reducing the result modulo the characteristic.
In a finite field with characteristic 2, addition and subtraction are identical, and are accomplished using the XOR operator. Thus,
- Polynomial: (x6 + x4 + x + 1) + (x7 + x6 + x3 + x) = x7 + x4 + x3 + 1
- Binary: {01010011} + {11001010} = {10011001}
- Hexadecimal: {53} + {CA} = {99}
Notice that under regular addition of polynomials, the sum would contain a term 2x6, but that this term becomes 0x6 and is dropped when the answer is reduced modulo 2.
Here is a table with both the normal algebraic sum and the characteristic 2 finite field sum of a few polynomials:
| p1 | p2 | p1 + p2 (normal algebra) | p1 + p2 in GF(2n) |
| x3 + x + 1 | x3 + x2 | 2x3 + x2 + x + 1 | x2 + x + 1 |
| x4 + x2 | x6 + x2 | x6 + x4 + 2x2 | x6 + x4 |
| x + 1 | x2 + 1 | x2 + x + 2 | x2 + x |
| x3 + x | x2 + 1 | x3 + x2 + x + 1 | x3 + x2 + x + 1 |
| x2 + x | x2 + x | 2x2 + 2x | 0 |
[edit] Multiplication
Multiplication in a finite field is multiplication modulo an irreducible reducing polynomial used to define the finite field. (I.e., it is multiplication followed by division using the reducing polynomial as the divisor—the remainder is the product.) The symbol "•" may be used to denote multiplication in a finite field.
[edit] Rijndael's finite field
Rijndael uses a characteristic 2 finite field with 8 terms, which can also be called the Galois field GF(28). It employs the following reducing polynomial for multiplication:
- x8 + x4 + x3 + x + 1.
For example, {53} • {CA} = {01} in Rijndael's field because
(x6 + x4 + x + 1)(x7 + x6 + x3 + x) =
x13 + x12 + x9 + x7 + x11 + x10 + x7 + x5 + x8 + x7 + x4 + x2 + x7 + x6 + x3 + x =
x13 + x12 + x9 + x11 + x10 + x5 + x8 + x4 + x2 + x6 + x3 + x =
x13 + x12 + x11 + x10 + x9 + x8 + x6 + x5 + x4 + x3 + x2 + x
and
x13 + x12 + x11 + x10 + x9 + x8 + x6 + x5 + x4 + x3 + x2 + x modulo x8 + x4 + x3 + x + 1 (= 11111101111110 mod 100011011) = 1, which can be demonstrated through long division (shown using binary notation, since it lends itself well to the task. Notice that exclusive OR is applied in the example and not arithmetic subtraction, as one might use in grade-school long division.):
111101
100011011)11111101111110
^100011011
1110000011110
^100011011
110110101110
^100011011
10101110110
^100011011
0100011010
^000000000
100011010
^100011011
00000001
(The elements {53} and {CA} happen to be multiplicative inverses of one another since their product is 1.)
Multiplication in this particular finite field can also be done as follows:
- Take two eight-bit numbers, a and b, and an eight-bit product p. The reason for eight bits is because this finite field can only have eight elements in the polynomial.
- Set the product to zero.
- Make a copy of a and b, which we will simply call a and b in the rest of this algorithm
- Run the following loop eight times:
- If the low bit of b is set, exclusive or the product p by the value of a
- Keep track of whether the high (leftmost) bit of a is set to one
- Rotate a one bit to the left, discarding the high bit, and making the low bit have a value of zero
- If a's high bit had a value of one prior to this rotation, exclusive or a with the hexadecimal number 0x1b (27 in decimal). 0x1b corresponds to the irreducible polynomial x8 + x4 + x3 + x + 1.
- Rotate b one bit to the right, discarding the low bit, and making the high (leftmost) bit have a value of zero.
- The product p now has the product of a and b
This algorithm generalizes easily to multiplication over other fields of characteristic 2, changing the lengths of a, b, and p and the value 0x1b appropriately.
[edit] Multiplicative inverse
The multiplicative inverse for an element n of a finite field can be calculated a number of different ways:
- By multiplying n by every number in the field until the product is one. This is a Brute-force search
- By using the Extended Euclidean algorithm
- By making a logarithm table of the finite field, and performing subtraction in the table. Subtraction of logarithms is the same as division.
[edit] Primitive finite fields
A finite field is considered a primitive finite field if the element x is a generator for the finite field. In other words, if the powers of x assume every nonzero value in the field, it is a primitive finite field. As it turns out, the GF(28) finite field with the reducing polynomial x8 + x4 + x3 + x + 1 is not primitive, although x + 1 is a generator in this field. The GF(28) finite field with the reducing primitive polynomial x8 + x4 + x3 + x2 + 1, however, is a primitive field.
Primitive finite fields are used, for example, by Linear feedback shift registers.
[edit] Program examples
Here is some C code which will add, subtract, and multiply numbers in Rijndael's finite field:
/* Add two numbers in a GF(2^8) finite field */
uint8_t gadd(uint8_t a, uint8_t b) {
return a ^ b;
}
/* Subtract two numbers in a GF(2^8) finite field */
uint8_t gsub(uint8_t a, uint8_t b) {
return a ^ b;
}
/* Multiply two numbers in the GF(2^8) finite field defined
* by the polynomial x^8 + x^4 + x^3 + x + 1 */
uint8_t gmul(uint8_t a, uint8_t b) {
uint8_t p = 0;
uint8_t counter;
uint8_t hi_bit_set;
for(counter = 0; counter < 8; counter++) {
if((b & 1) == 1)
p ^= a;
hi_bit_set = (a & 0x80);
a <<= 1;
if(hi_bit_set == 0x80)
a ^= 0x1b; /* x^8 + x^4 + x^3 + x + 1 */
b >>= 1;
}
return p;
}
Note that this code is vulnerable to timing attacks when used for cryptography.
