我正在尝试解决异或方程系统。例如:
A = [[1, 1, 1, 0, 0], [0, 1, 1, 1, 0], [0, 0, 1, 1, 1], [0, 1, 1, 0, 1], [0, 1, 0, 1, 1]]
s = [3, 14, 13, 5, 2]
m = 5 # len(s)
Ax = s => x = [12, 9, 6, 1, 10]
我试了两种方法:
- 第一种方法是高斯消去(~2.5秒),如图
- 第二种方法对模矩阵A(模2)进行逆变,然后与A逆变和s进行异或相乘(~7.5秒)
你能告诉我有一种方法或python库来加速吗?我甚至尝试使用gmpy2库,但它不能减少太多。下面我描述了python代码,以便您可以轻松理解。
使用高斯消去法:
def SolveLinearSystem (A, B, N):
for K in range (0, N):
if (A[K][K] == 0):
for i in range (K+1, N):
if (A[i][K]!=0):
for L in range (0, N):
s = A[K][L]
A[K][L] = A[i][L]
A[i][L] = s
s = B[i]
B[i] = B[K]
B[K] = s
break
for I in range (0, N):
if (I!=K):
if (A[I][K]):
#M = 0
for M in range (K, N):
A[I][M] = A[I][M] ^ A[K][M]
B[I] = B[I] ^ B[K]
SolveLinearSystem (A, s, 5)
使用反转def identitymatrix(n):
return [[long(x == y) for x in range(0, n)] for y in range(0, n)]
def multiply_vector_scalar (vector, scalar, q):
kq = []
for i in range (0, len(vector)):
kq.append (vector[i] * scalar %q)
return kq
def minus_vector_scalar(vector1, scalar, vector2, q):
kq = []
for i in range (0, len(vector1)):
kq.append ((vector1[i] - scalar * vector2[i]) %q)
return kq
def inversematrix(matrix, q):
n = len(matrix)
A =[]
for j in range (0, n):
temp = []
for i in range (0, n):
temp.append (matrix[j][i])
A.append(temp)
Ainv = identitymatrix(n)
for i in range(0, n):
factor = gmpy2.invert(A[i][i], q) #invert mod q
A[i] = multiply_vector_scalar(A[i],factor,q)
Ainv[i] = multiply_vector_scalar(Ainv[i],factor,q)
for j in range(0, n):
if (i != j):
factor = A[j][i]
A[j] = minus_vector_scalar(A[j], factor, A[i], q)
Ainv[j] = minus_vector_scalar(Ainv[j], factor, Ainv[i], q)
return Ainv
def solve_equation (A, y):
result = []
for i in range (0, m):
temp = 0
for j in range (0, m):
temp = (temp ^ A[i][j]* y[j])
result.append(temp)
return result
A_invert = inversematrix(A, 2)
print solve_equation (A_invert, s)
您提供的两种方法都使您执行三次位操作。有一些方法在渐近和实践中都更快。
第一步(这对您来说可能已经足够了)是使用一个32位整数(我相信它们在Python中被称为numpy.int32
)来存储一行的32个连续元素。在足够大的输入时,这将加快行减少的速度,接近32倍,并且在适度的输入时,可能会显著减少运行时间。
在您的特定代码中,有许多事情您可以简单地专门用于mod-2情况。搜索你的代码%
和inversemodp
和处理所有这些;这些额外的、无意义的操作对运行时毫无帮助。