# gfsub

Subtract polynomials over Galois field

## Syntax

c = gfsub(a,b,p)
c = gfsub(a,b,p,len)
c = gfsub(a,b,field)

## Description

Note

This function performs computations in GF(pm), where p is prime. To work in GF(2m), apply the - operator to Galois arrays of equal size. For details, see Example: Addition and Subtraction.

c = gfsub(a,b,p) calculates a minus b, where a and b represent polynomials over GF(p) and p is a prime number. a, b, and c are row vectors that give the coefficients of the corresponding polynomials in order of ascending powers. Each coefficient is between 0 and p-1. If a and b are matrices of the same size, the function treats each row independently. Alternatively, a and b can be represented as polynomial character vectors.

c = gfsub(a,b,p,len) subtracts row vectors as in the syntax above, except that it returns a row vector of length len. The output c is a truncated or extended representation of the answer. If the row vector corresponding to the answer has fewer than len entries (including zeros), extra zeros are added at the end; if it has more than len entries, entries from the end are removed.

c = gfsub(a,b,field) calculates a minus b, where a and b are the exponential format of two elements of GF(pm), relative to some primitive element of GF(pm). p is a prime number and m is a positive integer. field is the matrix listing all elements of GF(pm), arranged relative to the same primitive element. c is the exponential format of the answer, relative to the same primitive element. See Representing Elements of Galois Fields for an explanation of these formats. If a and b are matrices of the same size, the function treats each element independently.

## Examples

collapse all

Calculate $\left(2+3x+{x}^{2}\right)-\left(4+2x+3{x}^{2}\right)$ over GF(5).

x = gfsub([2 3 1],[4 2 3],5)
x = 1×3

3     1     3

Subtract the two polynomials and display the first two elements.

y = gfsub([2 3 1],[4 2 3],5,2)
y = 1×2

3     1

For prime number p and exponent m, create a matrix listing all elements of GF(p^m) given primitive polynomial $2+2x+{x}^{2}$.

p = 3;
m = 2;
primpoly = [2 2 1];
field = gftuple((-1:p^m-2)',primpoly,p);

Subtract ${A}^{4}$ from ${A}^{2}$. The result is ${A}^{7}$.

g = gfsub(2,4,field)
g = 7