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An Introduction to Binary Finite Fields GF(2 m ). By Francisco Rodríguez Henríquez.

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An Introduction to Binary Finite Fields GF(2m)

By Francisco Rodríguez Henríquez.

A field is a set of elements with two custom-defined arithmetic operations: most commonly, addition and multiplication. The elements of the field are an additive abelian group, and the non-zero elements of the field are a multiplicative abelian group. This means that all elements of the field have an additive inverse, and all non-zero elements have a multiplicative inverse.

A field is called finite if it has a finite number of elements. The most commonly used finite fields in cryptography are the field Fp (where p is a prime number) and the field F2m.

What is a Field?A finite field or Galois field denoted by GF(q=pn), is a field with characteristic p, and a number q of elements. As we have seen, such a finite field exists for every prime p and positive integer n, and contains a subfield having p elements. This subfield is called ground field of the original field.

For the rest of this class, we will consider only the two most used cases in cryptography: q=p, with p a prime and q=2m. The former case, GF(p), is denoted as the prime field, whereas the latter, GF(2m), is known as the finite field of characteristic two or simply binary field.

Finite FieldsA finite field is a field with a finite number of elements. The number of elements in a finite field is called the order of the field. Fields of the same order are isomorphic: they display exactly the same algebraic structure differing only in the representation of the elements.Finite Fields

‘Plegaria del Codificador teórico: Juro por Galois que seré leal a las nobles tradiciones de la teoría de códigos; que hablaré de ella en el secreto lenguaje sólo conocido por los contados iniciados; y que celosamente vigilaré la sagrada teoría de aquellos que quisieran profanarla para usarla en aplicaciones mundanas”.

J. L. Massey

The field F2mAlthough the description of the field F2m is complicated, this field is extremely

beautiful and also quite useful, because its computations can be done efficiently

when implemented in hardware. There are several ways to describe arithmetic in

F2m; the most common one is the so-called polynomial representation.

Here, we restrict our discussion to the numbers that belongs to the finite field F=GF(2m) over K=GF(2). K is also known as the characteristic field. The elements of F are polynomials of degree less than m, with coefficients in K; that is,

{am-1xm-1+am-2xm-2+...+a2x2+a1x+a0|ai= 0 or 1}.

These elements are frequently written in vector form as (am-1 ... a1 a0).

F has exactly 2m-1 nonzero elements plus the zero element.

Some definitionsA polynomial p in GF(2m) is irreducible if p is not a unit element and if p=fg then f or g must be a unit, that is, a constant polynomial.

Let us consider a finite field F=GF(2m) over K=GF(2).

Elements of F: Polynomials of degree less than m, with coefficients in K, such that,

{am-1xm-1+am-2xm-2+...+a2x2+a1x+a0|ai= 0 or 1}.

The Binary Field F2mFact: The field F has exactly q-1=2m-1 nonzero elements plus

the zero element.

Then, taking advantage of the fact that over GF(2) addition is equivalent to subtraction, we get the important relation Generating polynomial

The finite field F=GF(2m) is completely described by a monic irreducible

polynomial, often called generating polynomial, of the form

Where kiGF(2) for i=0,1,…,m-1. Let be a root of the monic irreducible

polynomial in (0), i.e., f() = 0, Then

Then, we define the polynomial or canonical basis of GF(2m) over GF(2) using the primitive element and its m first powers

{1, , 2,…, m-1},

which happen to be linearly independent over GF(2).

Generating polynomial and polynomial basisUsing the canonical basis we can uniquely represent any number

AF=GF(2m) as

Sometimes, it is more convenient to represent a field element using the

so-called coordinate representation,

Polynomial representationWhere all the coefficients aI's belong to the characteristic field GF(2). Elements of the field are m-bit strings. The rules for arithmetic in F can be defined by polynomial representation. Since F operates on bit strings, computers can perform arithmetic in this field very efficiently.Element’s Representation

By using the polynomial basis given in last equation, we can represent any

number AF=GF(2m) uniquely by

In fact, this is always the case for any finite field F=GF(2m) where we can always define the so-called polynomial basis of GF(2m) over GF(2) as as the linearly independent set of the first m powers of

{1, , 2,…, m-1}

Order definition- The order of an element in F, is defined as the smallest positive integer k
- such that k=1. Any finite field always contains at least one element, called
- a primitive element, which has order q-1. We say that f(x) is a primitive
- polynomial, if any one of its roots, say , is a primitive element in F. If f(x)
- is primitive, then all the q elements of F, can be expressed as the union of
- the zero element and the set of the first q-1 powers of ,

Example.Let K = GF(24), F = GF(2), with defining primitive polynomial f(x) given by

f(x) = x4 + x + 1

Then, if is a root of f(x), we have f()=0, which implies that

f() = 4 + + 1 = 0

This equation over GF(2), means that satisfies the following equation

4 = + 1.

Using the above equation, one can now express each one of the 15 nonzero elements of K over F as is shown in the next table.

An exampleFinite fields: definitions and operations

F2m finite field operations : Addition, Squaring, multiplication and inversion

The irreducible generatingpolynomial used for these sample calculations is again f(x) =x4+x+1.

Notice that all thecoefficients are reduced modulo 2!!

Addition

(0110)+(0101)=(0011).

Multiplication

(1101)(1001)

= (x3+x2+1)(x3+1) mod f(x)

= x6+x5+2x3+x2+1 mod f(x)

= x6+x5+x2+1 mod f(x)

= (x4+x+1)(x2+x)+(x3+x2+x+1) mod f(x)

= x3+x2+x+1

= (1111).

Arithmetic in the field F2mExponentiation

To compute (0010)4, first find

(0010)2 = (0010)(0010)

= x x mod f(x)

= x2

= (0100).

Then

(0010)4 = (0010)2(0010)2

= (0100)(0100)

= x2x2 mod f(x)

= (x4+x+1)(1)+(x+1) mod f(x)

= x + 1

= (0011).

Arithmetic in the field F2mMultiplicative Inversion

The multiplicative identity for the field is 0 = (0001). The multiplicative inverse of 7 = (1011) is

-7 mod 15=8 mod 15=(0101).

To verify this, see that,

(1011)(0101) = (x3+x+1) (x2+1) mod f(x)

= x5+x2+x+1 mod f(x)

= (x4+x+1)(x)+(1) mod f(x)

= 1

= (0001)

Which is the multiplicative identity.

Arithmetic in the field F2mIn most algorithms the modular product is computed in two steps: polynomial multiplication followed by modular reduction. Let A(x), B(x) and (x)GF(2m) and P(x) be the irreducible field generator polynomial.

In order to compute the modular product we first obtain the product polynomial C(x), of degree at most 2m-2, as

Polynomial product

2m-1 coordinates

Reduction step

m coordinates

Two-steps Multipliers- Then, in the second step, a reduction operation is performed in order
- to obtain the m-1 degree polynomial C’(x) is defined as

In most algorithms the modular product is computed in two steps: polynomial multiplication followed by modular reduction. Let A(x)GF(2m) be an arbitrary element in the field and P(x) be the irreducible field generator polynomial.

In order to compute the modular square of the element A(x) we first obtain the polynomial product C(x), of degree at most 2m-2, as

Polynomial product

2m-1 coordinates

Reduction step

m coordinates

GF(2m) Squarer- Then, in a second step, a reduction operation is performed in order
- to obtain the m-1 degree polynomial C’(x) defined as

Let A be an element of the finite field F=GF(25). Then, the square of A is given as,

a40a30 a20a10a0

In general, for an arbitrary element A in the field F=GF(25), we have,

a4a3a2a1a0 * a4a3a2a1a0

Squaring: ExampleSquaring: Software Solution

rct_word sqr_table_low[256] = {

0, 1, 4, 5, 16, 17, 20, 21, 64 65, 68, 69, 80, 81, 84, 85,

256, 257, 260, 261, 272, 273, 276, 277, 320, 321, 324, 325, 336, 337, 340, 341,

1024, 1025, 1028, 1029, 1040, 1041, 1044, 1045, 1088, 1089, 1092, 1093, 1104, 1105, 1108, 1109,

1280, 1281, 1284, 1285, 1296, 1297, 1300, 1301, 1344, 1345, 1348, 1349, 1360, 1361, 1364, 1365,

4096, 4097, 4100, 4101, 4112, 4113, 4116, 4117, 4160, 4161, 4164, 4165, 4176, 4177, 4180, 4181,

4352, 4353, 4356, 4357, 4368, 4369, 4372, 4373, 4416, 4417, 4420, 4421, 4432, 4433, 4436, 4437,

5120, 5121, 5124, 5125, 5136, 5137, 5140, 5141, 5184, 5185, 5188, 5189, 5200, 5201, 5204, 5205,

5376, 5377, 5380, 5381, 5392, 5393, 5396, 5397, 5440, 5441, 5444, 5445, 5456, 5457, 5460, 5461,

16384, 16385, 16388, 16389, 16400, 16401, 16404, 16405, 16448, 16449, 16452, 16453, 16464, 16465, 16468, 16469, 16640, 16641, 16644, 16645, 16656, 16657, 16660, 16661, 16704, 16705, 16708, 16709, 16720, 16721, 16724, 16725, 17408, 17409, 17412, 17413, 17424, 17425, 17428, 17429, 17472, 17473, 17476, 17477, 17488, 17489, 17492, 17493, 17664, 17665, 17668, 17669, 17680, 17681, 17684, 17685, 17728, 17729, 17732, 17733, 17744, 17745, 17748, 17749, 20480, 20481, 20484, 20485, 20496, 20497, 20500, 20501, 20544, 20545, 20548, 20549, 20560, 20561, 20564, 20565, 20736, 20737, 20740, 20741, 20752, 20753, 20756, 20757, 20800, 20801, 20804, 20805, 20816, 20817, 20820, 20821, 21504, 21505, 21508, 21509, 21520, 21521, 21524, 21525, 21568, 21569, 21572, 21573, 21584, 21585, 21588, 21589, 21760, 21761, 21764, 21765, 21776, 21777, 21780, 21781,

21824, 21825, 21828, 21829, 21840, 21841, 21844, 21845

};

Squaring: Software Implementation

void rce_FieldSqr2k_Random(rct_word *ax, rct_word *tx, rce_context *cntxt, rct_octet *offsetptr)

{

rct_index i;

rct_word C, S;

rct_index wlen, blen_p;

rct_word *tmp;

wlen = cntxt->ecp->wlen;

blen_p = cntxt->ecp->blen_p;

tmp = (rct_word *) offsetptr;

tmp[0]=0; tmp[1]=0;

for (i=0; i<wlen; i++) {

S = sqr_table_low[(ax[i]&0xff)];

S ^= (sqr_table_low[(ax[i]>>8)&0xff]<<16);

C = sqr_table_low[(ax[i]>>16)&0xff];

C ^= (sqr_table_low[(ax[i]>>24)&0xff]<<16);

tmp[i*2] = S; tmp[i*2+1] = C;

}

RCE_FIELD_REDUC2K(cntxt) (tmp, blen_p, cntxt->ecp->poly);

//rce_residue2k(tmp, blen_p, cntxt->ecp->poly);

for (i=0; i<wlen; i++) tx[i] = tmp[i];

}

Problem: Given the polynomial product C(x) with at most, 2m-1, obtain the modular product C' with m coordinates, using the generating irreducible polynomial P(x). Second step: reduction

- Notice that since we are interested in the polynomial reminder of the
- above equation, we can safely add any multiple of P(x) to C(x) without
- altering the desired result. This simple observation suggest the following
- algorithm that can reduce k bits of the polynomial product C at once.

Let us assume that the m+1 and 2m-1 coordinates of P(x) and C(x), respectively, are distributed as follows:

Then, there always exists a k-bit constant scalar S, such that

where 0 < k <m. Notice that all the k MSB of SP become identical to the corresponding ones of the number C. By left shifting the number SP

exactly Shift = 2m-2-k-1 positions, we effectively reduce the number in

C by k bit.

Second step: reductionSoftware reduction implementation

Addition operations < 4wlen;

SHIFT operations < 4wlen;

Comparisons = 2wlen.

Software

Modular multiplication for software applications- 1. Polynomial multiplication:
- Look-up tables
- Karatsuba
- Karatsuba/Look-up tables

- 2. Reduction step:
- Standard reduction
- trinomials & pentanomials
- General irreducible polynomials
- Montgomery reduction
- trinomials & pentanomials
- General irreducible polynomials

Karatsuba's algorithm is based on the idea that the polynomial product C=AB can be written as,Polynomial multiplication: Karatsuba Multipliers

- It can be computed with 3 poly mults and 4 poly additions.

- By using this idea recursively, one can obtain O(mlog23) space complexities.

- Best results obtained by using a combination of classic and Karatsuba
- strategies.

2kn-bit Karatsuba Multipliers

- There are some asymptotically faster methods for polynomial multiplications, such as the Karatsuba-Ofman algorithm.
- Discovered in 1962, it was the first algorithm able to accomplish polynomial multiplication under O(m2) operations.
- Karatsuba's algorithm is based on the idea that the polynomial product C=AB can be written as,

2kn-bit Karatsuba Multipliers

- last equation can be carried out at the cost of only 3 polynomial multiplications and four polynomial additions.
- Of course, Karatsuba strategy can be applied recursively to the three polynomial multiplications of last equation.
- By applying this strategy recursively, it is possible to achieve a polynomial complexity of
- Best results can be obtained by combining classical method with Karatsuba strategy.

Input: Two elements A ,BЄGF(2m) with m=rn=2kn, and where A, B can be expressed as,

Output: A polynomial C=AB with up to 2m-1 coordinates, where C=xmCH+CL..

2kn-bit Karatsuba Multipliers

It can be shown that the space and time complexities of a m=2kn-bit Karatsuba multiplier combined with a classical method are given as,

Space complexity of hybrid Karatsuba multipliers for arbitrary m using n=1, 2, 3

Binary Karatsuba Multipliers

- Problem: Find an efficient Karatsuba strategy for the multiplication of two polynomials A, B GF(2m), such that m = 2k + d, d 0.
- Basic Idea: Pretend that both operands are polynomials with degree m’ = 2(k+1), and use normal Karatsuba approach for two of the three required polynomial multiplications, i.e., given

Binary Karatsuba Multipliers

- Compute the two 2k-bit polynomial multiplications:
- While the remaining d-bit polynomial multiplication AHBH can be computed using a -bit Karatsuba multiplier in a recursive manner (since the leftover d bits can be expressed as,

d = 2k1+d1).

Binary Karatsuba Multipliers

- The above outlined strategy yields a Binary Karatsuba scheme where the hamming weight of the original m will determine the number of recursive iterations to be used by the algorithm.

An Example

- As a design example, let us consider the polynomial multiplication of the elements A and B GF(2193). Since (193)2 = 11000001, the Hamming weight of m is h = 3.
- This will imply that we need a total of three iterations in order to compute the multiplication using the generalized m-bit binary Karatsuba multiplier. Additionally we notice that for this case, m = 193 =27+65.

An Example

- Where we have assumed that the above circuit has been implemented using a 1.2 CMOS technology, where we have that the time delays associated to the AND, XOR logic gates are given as: TA Tx=0.5 nS.
- Next slide shows a comparison between the proposed binary Karatsuba approach and the more traditional hybrid approach discussed previously.

Field Multiplication

Preliminary results yield a time delay of 50-70 Sec and 9K Slices of hardware resources utilization.

Problem: Given the polynomial product C(x) with at most, 2m-1, obtain the modular product C' with m coordinates, using the generating irreducible polynomial P(x). Second step: reduction

- The computational complexity of the reduction operation is linearly
- proportional to the Hamming weight (the number of nonzero terms) of
- the generating irreducible polynomial.

Field multipliers using special irreducible polynomials

Field multipliers

•

Equally-spaced polynomials

trinomials

pentanomials

There exist for only 81

degrees m, less than

1024 ( 8%)

There exist for only 468

degrees m, less than

1024 (45%)

There exists at least

one for any degree

m>3

the total time required for execution (speed) and;Performance criteria and element representation

- The most important measures of the performance for software
- implementations of the arithmetic operations in the Galois
- field GF(2m) are,

- the amount of memory required for the algorithm
- (memory requirements)

Problem: Given the polynomial product C(x) with at most, 2m-1, obtain the modular product C' with m coordinates, using the generating irreducible polynomial P(x). Second step: reduction

- Notice that since we are interested in the polynomial reminder of the
- above equation, we can safely add any multiple of P(x) to C(x) without
- altering the desired result. This simple observation suggest the following
- algorithm that can reduce k bits of the polynomial product C at once.

Let us assume that the m+1 and 2m-1 coordinates of P(x) and C(x), respectively, are distributed as follows:

Then, there always exists a k-bit constant scalar S, such that

where 0 < k <m. Notice that all the k MSB of SP become identical to the corresponding ones of the number C. By left shifting the number SP

exactly Shift = 2m-2-k-1 positions, we effectively reduce the number in

C by k bit.

Second step: reductionStandard reduction for trinomials and pentanomials

Addition operations < 4wlen;

SHIFT operations < 4wlen;

Comparisons = 2wlen.

0) Consider the polynomial

Find if F=GF(55) constructed using f as a generating polynomial, is a field or not.

1) Consider the polynomial

a) Show that P(x) forms a field in GF(2m).

b) Find whether P() is a primitive root or not.

c) Find a primitive element in the field.

Exercises2) Consider the polynomial

a) Show that P(x) forms a field in GF(2m).

b) Is P(x) a primitive polynomial?

c) Find 47 as a polynomial of degree less or equal to 5.

d) Find the positive number k that satisfies:

Exercises
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