Python求解行列式

COUNT=0
res = []
def perm(n,begin,end):
    global COUNT
    global res
    if begin>=end:
        # print(n,end='\t')
        # print(inverse(n))
        res.append((n.copy(),inverse(n)))
        # print("res:{}".format(res))
        # print()
        COUNT +=1
    else:
        i=end-1
        for num in range(begin,end):
            n[num],n[i]=n[i],n[num]
            perm(n,begin,end-1)
            n[num],n[i]=n[i],n[num]
 
def inverse(arr):
    count = 0
    for i,n in enumerate(arr):
        for j in arr[:i]:
            if j>n:
                count+=1
    return count


def det(matrix):
    length = len(matrix)
    perm(list(range(length)),0,length)
    Sum = 0
    for series,num in res:
        product = 1
        for i,mark in enumerate(series):
            product *= matrix[i][mark]
        #     print("{}*".format(matrix[i][mark]),end='')
        # print("={}".format(product))
        Sum += ((-1)**num)*product
        # print('Sum:{}'.format(Sum))
    return Sum

def main():
    n=[[1,1,1],[1,1,1],[6,4,8]]
    print(det(n))
    # print(res)

if __name__ == '__main__':
    main()