Divide polynomials over Galois field
[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 |
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). a
and b
can
be either polynomial character vectors or
numeric vectors.
[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).