Question

我有N个10维向量，其中每个元素的值可以是0,1或2。例如，vector v=(0,1,1,2,0,1,2,0,1,1)是向量之一。是否存在将这些向量压缩为最小数量的笛卡尔积的算法（最好是在python中）。如果不是完美的解决方案，是否存在至少提供良好压缩的算法。

示例：两个＆＃34;笛卡尔向量＆＃34; ([1,2], 0, 1, 0, 0, 0, 1, 1, [0,1], 0])（给出4个向量）和(0, 1, 0, 2, 0, 0, [0,2], 2, 0, 1)（给出2个向量）给出N = 6向量的最优解：

1,0,1,0,0,0,1,1,0,0
2,0,1,0,0,0,1,1,0,0
1,0,1,0,0,0,1,1,1,0
2,0,1,0,0,0,1,1,1,0
0,1,0,2,0,0,0,2,0,1
0,1,0,2,0,0,2,2,0,1

Answer 1

如果您的向量是（0,1） - 值，那么您将面临最小的DNF问题，which is（如果我错了，请纠正我）NP-hard。具有（0,1,2）值的向量不会使问题变得更简单。

Answer 2

贪心算法可能很有用：

Let S = <V1, V2, ... Vn>

has_merge_ops = True

while has_merge_ops:
    has_merge_ops = False

    S' = []
    for v in S:
        if there exits v' which differs from v by just only ONE element:
            V = merge v and v' by merging the different values at index k
            S'.append(V)
            S.remove(v)
            S.remove(v')
            has_merge_ops = True

    S = S'

return S

所以它的工作原理如下：

after sorting, S = 
0,1,0,2,0,0,0,2,0,1
0,1,0,2,0,0,2,2,0,1
1,0,1,0,0,0,1,1,0,0
1,0,1,0,0,0,1,1,1,0
2,0,1,0,0,0,1,1,0,0
2,0,1,0,0,0,1,1,1,0

After first pass:
0,1,0,2,0,0,[0,2],2,0,1
1,0,1,0,0,0,1,1,[0,1],0
2,0,1,0,0,0,1,1,[0,1],0

After 2nd pass:
0,1,0,2,0,0,[0,2],2,0,1
[1,2],0,1,0,0,0,1,1,[0,1],0

Algorithm stops since no more merging is possible.

将载体压缩成笛卡尔积

2 个答案: