Documentation |
[quot,remd] = gfdeconv(b,a)
[quot,remd] = gfdeconv(b,a,p)
[quot,remd] = gfdeconv(b,a,field)
Note: This function performs computations in GF(p^{m}), where p is prime. To work in GF(2^{m}), use the deconv function with Galois arrays. For details, see Multiplication and Division of Polynomials. |
The gfdeconv function divides polynomials over a Galois field. (To divide elements of a Galois field, use gfdiv instead.) Algebraically, dividing polynomials over a Galois field is equivalent to deconvolving vectors containing the polynomials' coefficients, where the deconvolution operation uses arithmetic over the same Galois field.
[quot,remd] = gfdeconv(b,a) computes the quotient quot and remainder remd of the division of b by a in GF(2).
[quot,remd] = gfdeconv(b,a,p) divides the polynomial b by the polynomial a over GF(p) and returns the quotient in quot and the remainder in remd. p is a prime number. b, a, quot, and remd are row vectors that give the coefficients of the corresponding polynomials in order of ascending powers. Each coefficient is between 0 and p-1.
[quot,remd] = gfdeconv(b,a,field) divides the polynomial b by the polynomial a over GF(p^{m}) and returns the quotient in quot and the remainder in remd. Here p is a prime number and m is a positive integer. b, a, quot, and remd are row vectors that list the exponential formats of the coefficients of the corresponding polynomials, in order of ascending powers. The exponential format is relative to some primitive element of GF(p^{m}). field is the matrix listing all elements of GF(p^{m}), arranged relative to the same primitive element. See Representing Elements of Galois Fields for an explanation of these formats.
The code below shows that
$$(x+{x}^{3}+{x}^{4})\xf7(1+x)=1+{x}^{3}\text{Remainder}2$$
in GF(3). It also checks the results of the division.
p = 3; b = [0 1 0 1 1]; a = [1 1]; [quot, remd] = gfdeconv(b,a,p) % Check the result. bnew = gfadd(gfconv(quot,a,p),remd,p); if isequal(bnew,b) disp('Correct.') end;
The output is below.
quot = 1 0 0 1 remd = 2 Correct.
Working over GF(3), the code below outputs those polynomials of the form x^{k} - 1 (k = 2, 3, 4,..., 8) that 1 + x^{2} divides evenly.
p = 3; m = 2; a = [1 0 1]; % 1+x^2 for ii = 2:p^m-1 b = gfrepcov(ii); % x^ii b(1) = p-1; % -1+x^ii [quot, remd] = gfdeconv(b,a,p); % Display -1+x^ii if a divides it evenly. if remd==0 multiple{ii}=b; gfpretty(b) end end
The output is below.
4 2 + X 8 2 + X
In light of the discussion in Algorithms on the gfprimck reference page, along with the irreducibility of 1 + x^{2} over GF(3), this output indicates that 1 + x^{2} is not primitive for GF(9).