( A, B )---3*30*2---( 1, 0 )( 0, 1 )
让网络的输入只有3个节点,AB训练集各由5张二值化的图片组成,让A中有4个1,B全是0,排列组合,统计迭代次数并排序。
其中有6组数据4a6,4a8,4a12,4a13,4a15,4a16
3x | 差值结构 |
迭代次数 |
4x | 差值结构 |
迭代次数 |
3+1 |
2+2 |
|||||||||||
1 |
0 |
1 |
1 |
23302 |
6 |
- |
- |
- |
22075 |
2 |
11 |
|||||||
0 |
1 |
0 |
23302 |
- |
1 |
- |
22075 |
2 |
11 |
|||||||||
0 |
0 |
0 |
23302 |
- |
1 |
- |
22075 |
2 |
11 |
|||||||||
0 |
0 |
0 |
23302 |
- |
1 |
- |
22075 |
2 |
11 |
|||||||||
0 |
0 |
0 |
23302 |
- |
1 |
- |
22075 |
2 |
11 |
|||||||||
22075 |
2 |
11 |
||||||||||||||||
2 |
0 |
1 |
0 |
30302 |
8 |
- |
- |
1 |
28209 |
3 |
11 |
22 |
||||||
0 |
1 |
0 |
30302 |
- |
- |
1 |
28209 |
3 |
11 |
22 |
||||||||
0 |
1 |
0 |
30302 |
- |
1 |
- |
28209 |
3 |
11 |
22 |
||||||||
0 |
0 |
0 |
30302 |
- |
- |
- |
28209 |
3 |
11 |
22 |
||||||||
0 |
0 |
0 |
30302 |
- |
1 |
- |
28209 |
3 |
11 |
22 |
||||||||
28209 |
3 |
11 |
22 |
|||||||||||||||
3 |
0 |
0 |
1 |
30392 |
12 |
- |
- |
- |
39014 |
1 |
11 |
22 |
33 |
|||||
0 |
1 |
0 |
30392 |
- |
- |
- |
39014 |
1 |
11 |
22 |
33 |
|||||||
0 |
0 |
1 |
30392 |
- |
- |
- |
39014 |
1 |
11 |
22 |
33 |
|||||||
0 |
0 |
0 |
30392 |
- |
1 |
1 |
39014 |
1 |
11 |
22 |
33 |
|||||||
0 |
0 |
0 |
30392 |
- |
1 |
1 |
39014 |
1 |
11 |
22 |
33 |
|||||||
39014 |
1 |
11 |
22 |
33 |
||||||||||||||
4 |
1 |
0 |
1 |
43725 |
13 |
- |
- |
- |
1 |
39572 |
5 |
22 |
||||||
0 |
0 |
0 |
43725 |
- |
- |
1 |
- |
39572 |
5 |
22 |
||||||||
0 |
0 |
0 |
43725 |
- |
1 |
- |
- |
39572 |
5 |
22 |
||||||||
0 |
0 |
0 |
43725 |
1 |
- |
- |
- |
39572 |
5 |
22 |
||||||||
0 |
1 |
0 |
43725 |
- |
- |
- |
- |
39572 |
5 |
22 |
||||||||
39572 |
5 |
22 |
||||||||||||||||
5 |
1 |
0 |
0 |
49778 |
15 |
- |
- |
- |
- |
52027 |
4 |
22 |
33 |
|||||
0 |
1 |
0 |
49778 |
- |
- |
- |
- |
52027 |
4 |
22 |
33 |
|||||||
0 |
0 |
1 |
49778 |
- |
1 |
- |
1 |
52027 |
4 |
22 |
33 |
|||||||
0 |
0 |
0 |
49778 |
1 |
- |
1 |
- |
52027 |
4 |
22 |
33 |
|||||||
0 |
0 |
0 |
49778 |
- |
- |
- |
- |
52027 |
4 |
22 |
33 |
|||||||
52027 |
4 |
22 |
33 |
|||||||||||||||
6 |
1 |
1 |
1 |
76204 |
16 |
- |
- |
- |
- |
69983 |
6 |
33 |
||||||
0 |
0 |
0 |
76204 |
- |
- |
- |
- |
69983 |
6 |
33 |
||||||||
0 |
0 |
0 |
76204 |
- |
- |
- |
- |
69983 |
6 |
33 |
||||||||
0 |
0 |
0 |
76204 |
1 |
1 |
1 |
1 |
69983 |
6 |
33 |
||||||||
0 |
0 |
0 |
76204 |
- |
- |
- |
- |
69983 |
6 |
33 |
||||||||
69983 |
6 |
33 |
因为差值结构中只有4个点,所以4-x可以认为是由3-x+1构成的,因此得到加法关系
4a6=3a2+1
4a8=3a3+1
4a12=3a1+1
4a13=3a5+1
4a15=3a4+1
4a16=3a6+1
尽管3a6+1这个操作可以得到3个不同的结果,4a16只是3种可能之一,但是4a16-1却只能唯一的得到3a6.其余5个结构4a6-4a15执行-1的操作也都只能得到唯一的一个结构。比如
4a15无论减去哪个1最终得到的都是一个行分布为00012,列分布为111的结构3a4。
4a6-4a16这6个结构还可以分解成2+2
1 |
0 |
0 |
0 |
48757 |
0 |
0 |
0 |
48757 |
|
0 |
0 |
0 |
48757 |
|
0 |
0 |
1 |
48757 |
|
0 |
0 |
1 |
48757 |
|
2 |
0 |
0 |
0 |
66504 |
0 |
0 |
0 |
66504 |
|
0 |
0 |
0 |
66504 |
|
0 |
1 |
0 |
66504 |
|
0 |
0 |
1 |
66504 |
|
3 |
0 |
0 |
0 |
85402 |
0 |
0 |
0 |
85402 |
|
0 |
0 |
0 |
85402 |
|
0 |
0 |
0 |
85402 |
|
0 |
1 |
1 |
85402 |
因此有加法关系
4a6=2a1+2a1
4a8=2a1+2a1
4a8=2a2+2a2
4a12=2a1+2a1
4a12=2a2+2a2
4a12=2a3+2a3
4a13=2a2+2a2
4a15=2a2+2a2
4a15=2a3+2a3
4a16=2a3+2a3
首先比较前面2组
4a6=2a1+2a1
4a6=2a1+2a1
4a8=2a1+2a1
4a8=2a2+2a2
这两组都有2a1+2a1作为共同特征,而独立特征2a1的迭代次数小于2a2,因此4a6的迭代次数小于4a8.
4a8=2a1+2a1
4a8=2a2+2a2
4a8=2a2+2a2
4a12=2a1+2a1
4a12=2a2+2a2
4a12=2a3+2a3
同样4a8和4a12有公共特征2a1+2a1和2a2+2a2,而独立特征2a2的迭代次数更小因而4a8的迭代次数小于4a12.
同样4a13,4a15,4a16也可以用这种办法比较
4a13=2a2+2a2
4a13=2a2+2a2
4a15=2a2+2a2
4a15=2a3+2a3
4a16=2a3+2a3
4a16=2a3+2a3
因为2a2的迭代次数小于2a3,因此迭代次数4a13<4a15<4a16.
比较4a12和4a13
4a12=2a1+2a1
4a12=2a2+2a2
4a12=2a3+2a3
4a13=2a2+2a2
4a13=2a2+2a2
4a13=2a2+2a2
因为这两项的迭代次数其实相差不大,所以如果
2a1+2a1+2a3+2a3≈2a2+2a2+2a2+2a2
或者2a2+2a2+2a2+2a2-(2a1+2a1+2a3+2a3)=δ,
2a2+2a2+2a2+2a2仅仅只是比2a1+2a1+2a3+2a3稍大一些,就可以从差值结构加法的角度解释这6组的分布顺序。