Python 大数模矩阵求逆

我试图找到一种模矩阵求逆的方法。我找到了代码：

当我用小素数q和矩阵元素测试时，结果是正确的。但是，当我使用大素数q和由大元素（例如1024位）组成的矩阵进行测试时，它会输出错误：

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)

是因为np矩阵的数据类型吗？如果是这样，为什么“长”数据类型不能支持这种情况

---

你能告诉我怎么解决这个错误吗。提前感谢

我刚刚发现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。下面是我的代码

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

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手工做有什么理由吗？使用包中的numpy库。您尝试执行的许多操作已经在那里实现和优化。

dtype=long

已转换为固定宽度的整数dtype（例如，在我的机器上，

int64

）。您可能想要

dtype=object

。@jesuime我认为numpy不支持模块化矩阵inversion@MarkDickinson：谢谢。当我更改为dtype=object时，它会导致如下错误：“无法支持*：long和matrix:-A[I]=A[I]*因子%q，-Ainv[I]=Ainv[I]*因子%q，-A[j]=（A[j]-因子*A[I]）%q，-Ainv[j]=（Ainv[j]-因子*Ainv[I]）%q感兴趣。看起来这是NumPy中的一个bug。我可以用NumPy 1.6.2重现您的问题，但不能用NumPy 1.8.1重现。看见解决方法是使用

numpy.array

而不是

numpy.matrix

，这意味着您需要使用

numpy.dot

进行矩阵乘法。在NumPy中避免使用

矩阵

类型有很多原因；看起来你又找到了一个。：-）注意：

identitymatrix

在列表中缺少结束语

]

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