Question

我试图找到模块化矩阵求逆的方法。我找到了代码here：

def generalizedEuclidianAlgorithm(a, b):
    if b > a:
        #print a, b
        return generalizedEuclidianAlgorithm(b,a);
    elif b == 0:
        return (1, 0);
    else:
        #print a,b
        (x, y) = generalizedEuclidianAlgorithm(b, a % b);
        return (y, x - (a / b) * y)

def inversemodp(a, p):
    a = a % p
    if (a == 0):
        print "a is 0 mod p"
        return 0
    (x,y) = generalizedEuclidianAlgorithm(p, a % p);
    return y % p

def identitymatrix(n):
    return [[long(x == y) for x in range(0, n)] for y in range(0, n)]

def inversematrix(matrix, q):
    n = len(matrix)
    A = np.matrix([[ matrix[j, i] for i in range(0,n)] for j in range(0, n)], dtype = long)
    Ainv = np.matrix(identitymatrix(n), dtype = long)
    for i in range(0, n):
        factor = inversemodp(A[i,i], q)
        A[i] = A[i] * factor % q
        Ainv[i] = Ainv[i] * factor % q
        for j in range(0, n):
            if (i != j):
                factor = A[j, i]
                A[j] = (A[j] - factor * A[i]) % q
                Ainv[j] = (Ainv[j] - factor * Ainv[i]) % q
                # print A, Ainv
                # print i, j, factor
    return Ainv

当我使用小素数q和矩阵元素进行测试时，结果是正确的。但是，当我用大质量q和由大元素组成的矩阵（例如1024位）进行测试时，它打印出错误：

A = np.matrix([[ matrix[j, i] for i in range(0,n)] for j in range(0, n)], dtype = long)

File "C:\Python27\lib\site-packages\numpy\matrixlib\defmatrix.py", line 257, in __new__

arr = N.array(data, dtype=dtype, copy=copy)

是不是因为np.matrix的dtype？如果是这样，为什么“long”数据类型不支持这种情况？

你能否告诉我如何解决这个错误。提前致谢

Answer 1

我刚才发现np.array会导致很多问题。一种解决方案是将矩阵视为二维列表。以下是我的代码

def generalizedEuclidianAlgorithm(a, b):
    if b > a:
        return generalizedEuclidianAlgorithm(b,a);
    elif b == 0:
        return (1, 0);
    else:
        (x, y) = generalizedEuclidianAlgorithm(b, a % b);
        return (y, x - (a / b) * y)

def inversemodp(a, p):
    a = a % p
    if (a == 0):
        print "a is 0 mod p"
        return 0
    (x,y) = generalizedEuclidianAlgorithm(p, a % p);
    return y % p

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_scalar1(vector1, scalar, vector2, q):
    kq = []
    for i in range (0, len(vector1)):
        kq.append ((vector1[i] - scalar * vector2[i]) %q)
    return kq

def inversematrix1(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 = inversemodp(A[i][i], 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_scalar1(A[j],factor,A[i],q)
                Ainv[j] = minus_vector_scalar1(Ainv[j],factor,Ainv[i],q)
    return Ainv

模数矩阵求逆数大

1 个答案: