问题描述
我正在尝试为 GF(2^8) 中的 2 个多项式创建欧几里得算法(以解决 Bezout 关系)。
我目前有这个代码用于我的不同操作
class ReedSolomon:
gfSize = 256
genPoly = 285
log = [0]*gfSize
antilog = [0]*gfSize
def _genLogAntilogArrays(self):
self.antilog[0] = 1
self.log[0] = 0
self.antilog[255] = 1
for i in range(1,255):
self.antilog[i] = self.antilog[i-1] << 1
if self.antilog[i] >= self.gfSize:
self.antilog[i] = self.antilog[i] ^ self.genPoly
self.log[self.antilog[i]] = i
def __init__(self):
self._genLogAntilogArrays()
def _galPolynomialDivision(self,dividend,divisor):
result = dividend.copy()
for i in range(len(dividend) - (len(divisor)-1)):
coef = result[i]
if coef != 0:
for j in range(1,len(divisor)):
if divisor[j] != 0:
result[i + j] ^= self._galMult(divisor[j],coef) # équivalent result[i + j] += -divisor[j] * coef car dans un champ GF(2) addition <=> substraction <=> XOR
remainderIndex = -(len(divisor)-1)
return result[:remainderIndex],result[remainderIndex:]
def _galMultiplicationPolynomiale(self,x,y):
result = [0]*(len(x)+len(y)-1)
for i in range(len(x)):
for j in range(len(y)):
result[i+j] ^= self._galMult(x[i],y[j])
return result
def _galMult(self,y):
if ((x == 0) or (y == 0)):
val = 0
else:
val = self.antilog[(self.log[x] + self.log[y])%255]
return val
def _galPolynomialAddition(self,a,b):
polSum = [0] * max(len(a),len(b))
for index in range(0,len(a)):
polSum[index + len(polSum) - len(a)] = a[index]
for index in range(0,len(b)):
polSum[index + len(polSum) - len(b)] ^= b[index]
return (polSum)
这是我的欧几里得算法:
def _galEuclideanAlgorithm(self,b):
r0 = a.copy()
r1 = b.copy()
u0 = [1]
u1 = [0]
v0 = [0]
v1 = [1]
while max(r1) != 0:
print(r1)
q,r = self._galPolynomialDivision(r0,r1)
r0 = self._galPolynomialAddition(self._galMultiplicationPolynomiale(q,r1),r)
r1,r0 = self._galPolynomialAddition(r0,self._galMultiplicationPolynomiale(q,r1)),r1.copy()
u1,u0 = self._galPolynomialAddition(u0,u1)),u1.copy()
v1,v0 = self._galPolynomialAddition(v0,v1)),v1.copy()
return r1,u1,v1
我不明白我的算法循环的问题,这是我测试的剩余输出:
rs = ReedSolomon()
a = [1,15,7,8,11]
b = [1,0]
print(rs._galEuclideanAlgorithm(b,a))
#Console output
'''
[1,11]
[0,82,37,120,11,105]
[1,11]
'''
我知道我似乎在抛出一些代码只是在期待答案,但我真的是在寻找错误。
提前致谢!
解决方法
我创建了一个名为 galois 的 Python 包来执行此操作。 galois
扩展了 NumPy 数组以对 Galois 字段进行操作。代码是用 Python 编写的,但为了速度使用 Numba 进行了 JIT 编译。除了数组算术,它还支持伽罗瓦域上的多项式。 ...而且 Reed-Solomon codes 也实现了 :)
用于求解 GF(2^8)
中两个多项式的 Bezout 恒等式的扩展欧几里得算法将以这种方式求解。下面是一段简短的源代码。您可以查看我的完整源代码 here。
def poly_egcd(a,b):
field = a.field
zero = Poly.Zero(field)
one = Poly.One(field)
r2,r1 = a,b
s2,s1 = one,zero
t2,t1 = zero,one
while r1 != zero:
q = r2 / r1
r2,r1 = r1,r2 - q*r1
s2,s1 = s1,s2 - q*s1
t2,t1 = t1,t2 - q*t1
# Make the GCD polynomial monic
c = r2.coeffs[0] # The leading coefficient
if c > 1:
r2 /= c
s2 /= c
t2 /= c
return r2,s2,t2
这是一个使用 galois
库和示例中的多项式的完整示例。 (我假设最高阶系数是第一?)
In [1]: import galois
In [2]: GF = galois.GF(2**8)
In [3]: print(GF.properties)
GF(2^8):
characteristic: 2
degree: 8
order: 256
irreducible_poly: x^8 + x^4 + x^3 + x^2 + 1
is_primitive_poly: True
primitive_element: x
In [4]: a = galois.Poly([1,15,7,8,11],field=GF); a
Out[4]: Poly(x^5 + 15x^4 + 7x^3 + 8x^2 + 11,GF(2^8))
In [5]: b = galois.Poly([1,0],field=GF); b
Out[5]: Poly(x^6,GF(2^8))
In [6]: d,s,t = galois.poly_egcd(a,b); d,t
Out[6]:
(Poly(1,GF(2^8)),Poly(78x^5 + 7x^4 + 247x^3 + 74x^2 + 152,Poly(78x^4 + 186x^3 + 45x^2 + x + 70,GF(2^8)))
In [7]: a*s + b*t == d
Out[7]: True