整数 NxN 矩阵的精确行列式

Question

行列式定义只有加法、减法和乘法。所以具有整数元素的矩阵的行列式必须是整数。

但是numpy.linalg.det() returns一个“稍微偏离”的浮点数：

>>> import numpy
>>> M = [[-1 if i==j else 1 for j in range(7)] for i in range(7)]
>>> numpy.linalg.det(M)
319.99999999999994

对于更大的矩阵，情况会变得更糟：

>>> M = [[-1024 if i==j else 1024 for j in range(7)] for i in range(7)]
>>> numpy.linalg.det(M)
3.777893186295698e+23
>>> "%.0f" % numpy.linalg.det(M)
'377789318629569805156352'

这是错误的！我确定正确答案是：

>>> 320 * 1024**7
377789318629571617095680

当然，对于一个大矩阵，它可能是一个相当长的整数。但是 python 内置了长整数。

如何获得行列式的精确整数值而不是近似浮点值？

Answer 1

计算整数矩阵行列式的一种简单实用的方法是Bareiss algorithm。

def det(M):
    M = [row[:] for row in M] # make a copy to keep original M unmodified
    N, sign, prev = len(M), 1, 1
    for i in range(N-1):
        if M[i][i] == 0: # swap with another row having nonzero i's elem
            swapto = next( (j for j in range(i+1,N) if M[j][i] != 0), None )
            if swapto is None:
                return 0 # all M[*][i] are zero => zero determinant
            M[i], M[swapto], sign = M[swapto], M[i], -sign
        for j in range(i+1,N):
            for k in range(i+1,N):
                assert ( M[j][k] * M[i][i] - M[j][i] * M[i][k] ) % prev == 0
                M[j][k] = ( M[j][k] * M[i][i] - M[j][i] * M[i][k] ) // prev
        prev = M[i][i]
    return sign * M[-1][-1]

这个算法相当快（O(N³) 复杂度）。

这是一个整数保留算法。它确实有一个部门。但只要M的所有元素都是整数，中间的计算也都是整数（除法余数为零）。

作为奖励，如果您删除 assert 行并将整数除法 // 替换为常规除法 /.

PS: Another alternative is to use sympy instead of numpy:

>>> import sympy
>>> sympy.Matrix([ [-1024 if i==j else 1024 for j in range(7)] for i in range(7) ]).det()
377789318629571617095680

But somewhy that is MUCH slower than the above det() function.

# Performance test: `numpy.linalg.det(M)` vs `det(M)` vs `sympy.Matrix(M).det()`
import timeit
def det(M):
    ...
M = [[-1024 if i==j else 1024 for j in range(7)] for i in range(7)]
print(timeit.repeat("numpy.linalg.det(M)", setup="import numpy; from __main__ import M", number=100, repeat=5))
#: [0.0035009384155273, 0.0033931732177734, 0.0033941268920898, 0.0033800601959229, 0.0033988952636719]
print(timeit.repeat("det(M)", setup="from __main__ import det, M", number=100, repeat=5))
#: [0.0171120166778564, 0.0171020030975342, 0.0171608924865723, 0.0170948505401611, 0.0171010494232178]
print(timeit.repeat("sympy.Matrix(M).det()", setup="import sympy; from __main__ import M", number=100, repeat=5))
#: [0.9561479091644287, 0.9564781188964844, 0.9539868831634521, 0.9536828994750977, 0.9546608924865723]

Summary:

det(M) is 5+ times slower than numpy.linalg.det(M),
det(M) is ~50 times faster than sympy.Matrix(M).det()

It becomes even faster without the assert line.

Answer 2

@pycoder 的回答是首选方案；为了比较，我使用 Fraction class 编写了一个高斯消去函数，它允许对有理数进行精确运算。在相同的基准测试中，它比 Bareiss 算法慢了大约 11 倍。

from fractions import Fraction

def det(matrix):
    matrix = [[Fraction(x, 1) for x in row] for row in matrix]
    n = len(matrix)
    d, sign = 1, 1
    for i in range(n):
        if matrix[i][i] == 0:
            j = next((j for j in range(i + 1, n) if matrix[j][i] != 0), None)
            if j is None:
                return 0
            matrix[i], matrix[j] = matrix[j], matrix[i]
            sign = -sign
        d *= matrix[i][i]
        for j in range(i + 1, n):
            factor = matrix[j][i] / matrix[i][i]
            for k in range(i + 1, n):
                matrix[j][k] -= factor * matrix[i][k]
    return int(d) * sign

整数 NxN 矩阵的精确行列式

Precise determinant of integer NxN matrix

python

algorithm

math

linear-algebra