GF(2 m ) arithmetic: summary
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1 GF(2 m ) arithmetic: summary EE 387, Notes 18, Handout #32 Addition/subtraction: bitwise XOR (m gates/ops) Multiplication: bit serial (shift and add) bit parallel (combinational) subfield representation log and anti-log tables software Division (reciprocal): Euclidean algorithm lookup tables sequential search time-memory tradeoff exponentiation subfield representation log and anti-log tables software Division is performed as multiplication by reciprocal: a/b = a b 1. Single-cycle division is not needed because decoders use few divides. Most multiplication and division methods take O(m 2 ) bit operations. EE 387, November 6, 2015 Notes 18, Page 1
2 Multiplication by constants GF(2 m ) is a vector space over GF(2) of dimension m. We choose a basis {1,α,α 2,...,α m 1 } for some α (usually primitive). Multiplication by constant b = b 0 +b 1 α+ +b m 1 α m 1 is a linear transformation. It can be described by an m m matrix B over GF(2). To derive the components of B, use the distributive law: a b = (a 0 +a 1 α+ +a m 1 α m 1 ) b = a 0 b+a 1 (αb)+ +a m 1 (α m 1 b). The products b i = α i b for 0 i < m can be precomputed. The components of these vectors can be stored in a binary matrix b 0 b 0,0 b 0,1 b 0,m 1 b B = 1. = b 1,0 b 1,1 b 1,m b m 1 b m 1,0 b m 1,1 b m 1,m 1 EE 387, November 6, 2015 Notes 18, Page 2
3 Multiplication by constants (cont.) The product a b is the vector-matrix product b 0,0 b 0,1 b 0,m 1 b y = a b = ab = [a 0 a 1 a m 1 ] 1,0 b 1,1 b 1,m b m 1,0 b m 1,1 b m 1,m 1 Each bit y j of product y is the inner product of row a with column j of B: y j = m 1 i=0 a i b i,j = a i i:b i,j =1 In the above formula sum is XOR, product is logical AND. Any GF(2 m ) scaler can be built using m(m 1) 2-input XOR gates. The typical scaler uses 1 2 m2 XOR gates. These gate count estimates do not include the use of common subexpressions. Finding the minimum circuit is an NP-complete problem. EE 387, November 6, 2015 Notes 18, Page 3
4 Example: scaler in GF(2 6 ) GF(2 6 ) = binary polynomials modulo x 6 +x+1 (a primitive polynomial). Multiplication by b = [110001] is defined by the matrix B shown below. The rows of B are x i b (i = 0,...,5), msb on right. Successive rows of B are obtained by shifting previous row right using the feedback pattern [110000] corresponding to 1+x+x B = = y 0 = a 0 a 1 a 5 y 1 = a 0 a 2 a 5 y 2 = a 1 a 3 y 3 = a 2 a 4 y 4 = a 3 a 5 y 5 = a 0 a 4 The coefficients y 0,y 1,...,y 5 of the product y = a b = ab can be read from columns of B. EE 387, November 6, 2015 Notes 18, Page 4
5 Matrix of multiplier matrices The matrix M b corresponding to multiplier b = x j mod p(x) consists of m consecutive rows x j mod p(x),...,x j+m 1 mod p(x) from the matrix of powers of primitive element x. Since p(x) is primitive, the matrix of powers of x mod p(x) is same as matrix of powers of primitive element α. For GF(2 4 ), the (15+3) 4 matrix to the right (lsb first) contains all multiplier matrices. For example, M α 7 consists of rows 7 to 10 from this table M α 7 = α i i EE 387, November 6, 2015 Notes 18, Page 5
6 General multiplication Let GF(2 m ) be polynomials over GF(2) modulo prime p(x) of degree m. Let t i = i j=0 a jb i j be the coefficient of x i in a(x)b(x). a(x)b(x) = (t 0 +t 1 x+ +t m 1 x m 1 +t m x m + +t 2m 2 x 2m 2 ) mod p(x) = t 0 +t 1 x+ +t m 1 x m 1 + t m (x m mod p(x))+ +t 2m 2 (x 2m 2 mod p(x)) We can precompute x m mod p(x),...,x 2m 2 mod p(x) and store them as rows of (m 1) m binary matrix: x m mod p(x) α m x m+1 mod p(x) T =. = α m+1. x 2m 2 mod p(x) α 2m 2 Rows of T are obtained by shifts with feedback corresponding to p(x). EE 387, November 6, 2015 Notes 18, Page 6
7 General multiplication (cont.) The product y = a b can be expressed in matrix notation: y = (a 0,...,a m 1 ) (b 0,...,b m 1 ) = [t 0,t 1,...,t m 1 ] + [t m,...,t 2m 2 ]T [ ] I = [t 0,t 1,...,t 2m 2 ] = [t 0,t 1,...,t 2m 2 ] T where I is the m m identity matrix and T is (m 1) m. 1 α. α 2m 2 Low-level computational formula for product bits y j for j = 0,...,m 1: m 2 y j = t j + t m+i T ij = i=0 j l=0 m 2 m 1 a l b j l + T ij i=0 l=i+1 a l b m+i l. Each bit of the product vector is a sum of a subset of the produt terms a i b j. The product vector consists of m bilinear functions of a and b. EE 387, November 6, 2015 Notes 18, Page 7
8 General multiplication example GF(2 4 ) can be defined by any of three prime polynomials over GF(2): p 1 (x) = x 4 +x+1, p 2 (x) = x 4 +x 3 +1, p 3 (x) = x 4 +x 3 +x 2 +x+1 The respective T matrices are T 1 = , T 2 = , T 3 = The equations for the product defined using T 3 : [y 0,y 1,y 2,y 3 ] = [t 0,t 1,t 2,t 3 ]+[t 4,t 5,t 6 ] = [t 0 +t 4 +t 5, t 1 +t 4 +t 6, t 2 +t 4, t 3 +t 4 ]. Even though p 3 (x) is not primitive, it is prime and can be used to define GF(2 4 ). EE 387, November 6, 2015 Notes 18, Page 8
9 General multiplication example (cont.) Expanding the matrix form yields these Boolean equations: t 0 = a 0 b 0 t 1 = a 0 b 1 +a 1 b 0 t 2 = a 0 b 2 +a 1 b 1 +a 2 b 0 t 3 = a 0 b 3 +a 1 b 2 +a 2 b 1 +a 3 b 0 t 4 = a 1 b 3 +a 2 b 2 +a 3 b 1 t 5 = a 2 b 3 +a 3 b 2 t 6 = a 3 b 3 y 0 = t 0 +t 4 +t 5 y 1 = t 1 +t 4 +t 6 y 2 = t 2 +t 4 y 3 = t 3 +t 4 Every product a i b j appears in at least one equation, hence m 2 AND gates. Also (m 1) 2 XOR gates are needed to compute {t 0,t 1,...,t 2m 2 }. The number of 1s in T counts the XOR gates needed to compute {y 0,y 1,...,y m 1 } from {t 0,t 1,...,t 2m 2 }. Polynomials like x 6 +x+1 result in T matrices with few 1s. We ignore common subexpression simplification. EE 387, November 6, 2015 Notes 18, Page 9
10 Multiplication using subfield representation Galois fields can be represented using subfields larger than the field integers. A common case: GF(2 2m ) = pairs from GF(2 m ) mod degree 2 prime polynomial over GF(2 m ). Good choice: prime quadratic of form x 2 +αx+1, where α is in GF(2 m ). a b = (a 0 +a 1 x)(b 0 +b 1 x) = a 0 b 0 + (a 0 b 1 +a 1 b 0 )x + a 1 b 1 x 2 = a 0 b 0 + (a 0 b 1 +a 1 b 0 )x + a 1 b 1 (αx+1) = (a 0 b 0 +a 1 b 1 ) + (a 0 b 1 +a 1 b 0 +αa 1 b 1 )x Final answer: components of (y 0,y 1 ) = (a 0,a 1 ) (b 0,b 1 ) are y 0 = a 0 b 0 +a 1 b 1 y 1 = a 0 b 1 +a 1 b 0 +αa 1 b 1 The only common subexpression is a 1 b 1, which affects both y 0 and y 1. EE 387, November 6, 2015 Notes 18, Page 10
11 Multiplication using subfield representation (cont.) Multiplication in GF(2 2m ) uses operations in the subfield GF(2 m ). In circuit below, there are 4 multipliers, one scaler, and 3 adders. a 0 a 1 y 0 α b 0 y 1 b 1 Multiplications can be performed in parallel, sequentially, or two at a time. Thus we can trade off between time and gates. EE 387, November 6, 2015 Notes 18, Page 11
12 Circuits for reciprocal in GF(2 m ) Since a/b = a b 1, division requires computation of multiplicative inverses. Methods for calculating reciprocals: Euclidean algorithm table lookup sequential search time-memory tradeoff exponentiation subfield representation recursive combinational circuit Single-cycle division is not usually needed. EE 387, November 6, 2015 Notes 18, Page 12
13 Reciprocals: Euclidean algorithm Let GF(Q) is represented by polynomials modulo prime p(x). Extended Euclidean algorithm for gcd(r(x), p(x)) finds a(x), b(x) such that a(x)r(x)+b(x)p(x) = gcd(r(x),p(x)) = 1 If deg p(x) = m, algorithm takes O(m) operations on m-digit registers. Implementation is straightforward. Drawback: O(m) clocks. Remainders r i (x) and coefficients a i (x) can share memory. p(x) r(x) a(x) The final remainder r is a constant, and the reciprocal is r 1 a(x). EE 387, November 6, 2015 Notes 18, Page 13
14 Reciprocals: table lookup Store precomputed reciprocals in 2 m m ROM. For gate arrays, one bit of ROM costs about 1/8 of 2-input NAND gate. Reciprocal table for GF(2 8 ) costs 256 gates. Combinational multiplier for GF(2 8 ) uses 64 ANDs + 66 XORs 260 gates. However, lookup tables are not as feasible for larger fields. For example, the gate equivalent of a reciprocal table for GF(2 10 ) is about = 1280 This is much larger than about 400 gates for a combinational multiplier. The lookup table size can be reduced by using precomputation to transform the input to a value whose reciprocal is known, then postcomputation to adjust the inverse. For example, if the inverse of βα i is δ (obtained from a table) then the inverse of β is δα i. EE 387, November 6, 2015 Notes 18, Page 14
15 Reciprocals: sequential search Reciprocal of a can be found by testing a b = 1 for each b in GF(2 m ). All nonzero values b can be generated using a maximum-length linear feedback shift register. E.g., when field is defined by p(x) = x 5 +x 2 +1: initial value: a final value: 1 initial value: 1 final value: a 1 Shifting a register multiplies the contents by the primitive element α. Left shift register is loaded with a while right shift register is loaded with 1. The registers are shifted simultaneously until the left shift register reaches 1. After every shift, the ratio of the left register to the right register is a. If i is the number of shifts needed, then a α i = 1, so the value α i in the right shift register is the reciprocal of a. EE 387, November 6, 2015 Notes 18, Page 15
16 Time-memory tradeoff An associative memory (hash table) can be used to reduce the number of clocks needed to find the reciprocal without using a complete lookup table. Suppose we store the reciprocals of α 16i for i = 0,1,...,(2 m )/16, The following program fragment finds reciprocal of a in at most 16 steps: for (i = 0; i < 16; i++) { if (a α i is in reciprocal table) { return α i reciprocal(a α i ); } } The search time can be decreased by using a larger associative memory. The same approach can be used to reduce the storage needed for computing the discrete logarithm from 2 m entries for a direct table lookup to (2 m )/c entries if c lookups are used. EE 387, November 6, 2015 Notes 18, Page 16
17 Reciprocals: exponentiation If β is a nonzero element of GF(q) then β q 1 = 1. Therefore β 1 = β q 2. Powers of β can be computed efficiently by squaring and multiplying by β. Binary representation of 2 m 2 contains m bits: square initial value: 1 β The successive values of the storage element are 1, β 2, β 6, β 14,..., β 2m 2. We obtain β 2m 2 = β 1 in m 1 clocks, one multiply/squaring per clock. In GF(2 m ) squaring is linear and represented by m m binary matrix. EE 387, November 6, 2015 Notes 18, Page 17
18 Reciprocals: subfield representation Suppose GF(2 2m ) = pairs from GF(2 m ) modulo prime x 2 +αx+1 over GF(2 m ). Special case: x+b. Use Euclidean algorithm to compute (x+b) 1 : x 2 +αx+1 = (x+b)(x+(α+b)) + (b 2 +αb+1) (x+b) 1 = x+(α+b) b 2 +αb+1 = α+b b 2 +αb b 2 +αb+1 x In general, the reciprocal of an arbitrary element ax+b with a 0: (ax+b) 1 = a 1 (x+b/a) 1 x+(α+b/a) = a((b/a) 2 +α(b/a)+1) = ax+(αa+b) b 2 +αab+a 2 = αa+b b 2 +αab+a 2 + a b 2 +αab+a 2 x Denominator 0 because a 0 and x 2 +αx+1 is prime. Computation of (ax+b) 1 uses operations from the subfield GF(2 m ): one inverse, one scaler, one squaring, and three multiplications. EE 387, November 6, 2015 Notes 18, Page 18
19 Reciprocals: recursive combinational circuit If β is in GF(2 2m ), then (β 2m +1 ) 2m 1 = β (2m +1)(2 m 1) = β 22m 1 = 1. Since order of β 2m +1 divides 2 m 1, it belongs to small subfield GF(2 m ). Its reciprocal can be computed using a small circuit. Example: Reciprocal circuit for GF(2 8 ) = GF((2 4 ) 2 ): β 16-th power β β17 inverse β in GF(16) β 1 Circuit for β 2m is linear, 1 2 m2 XOR gates. Multipliers use O(m 2 ) gates. Subfield reciprocal unit is small. Overall cost of circuit 3 multipliers. This method was discovered by Itoh and Tsujii in Hardware implementations of Galois field arithmetic are presented in Christof Paar s 1994 University of Essen doctoral thesis, Efficient VLSI Architectures for Bit-Parallel Computation in Galois Fields. EE 387, November 6, 2015 Notes 18, Page 19
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