( A, B )---5*30*2---( 1, 0 )( 0, 1 )
让网络的输入只有5个节点，AB训练集各由5张二值化的图片组成，让A 中有5个点，B全是0，排列组合，统计迭代次数并排序。
其中有4个结构
| 5-x | 差值结构 | 迭代次数 | 4+1 | ||||||||||||
| 6 | 1 | 1 | 1 | 0 | 9139.653 | 2 | 4 | 9 | 10 | ||||||
| 6 | 0 | 0 | 0 | 1 | 9139.653 | 2 | 4 | 9 | 10 | ||||||
| 6 | 0 | 0 | 1 | 0 | 9139.653 | 2 | 4 | 9 | 10 | ||||||
| 6 | 0 | 0 | 0 | 0 | 9139.653 | 2 | 4 | 9 | 10 | ||||||
| 6 | 0 | 0 | 0 | 0 | 9139.653 | 2 | 4 | 9 | 10 | ||||||
| 6 | 2 | 4 | 9 | 10 | |||||||||||
| 7 | 1 | 1 | 0 | 0 | 9453.382 | 2 | 7 | 9 | 13 | ||||||
| 7 | 0 | 0 | 0 | 1 | 9453.382 | 2 | 7 | 9 | 13 | ||||||
| 7 | 0 | 0 | 1 | 0 | 9453.382 | 2 | 7 | 9 | 13 | ||||||
| 7 | 0 | 1 | 0 | 0 | 9453.382 | 2 | 7 | 9 | 13 | ||||||
| 7 | 0 | 0 | 0 | 0 | 9453.382 | 2 | 7 | 9 | 13 | ||||||
| 7 | 2 | 7 | 9 | 13 | |||||||||||
| 14 | 1 | 1 | 0 | 0 | 16046.96 | 2 | 9 | 11 | 15 | ||||||
| 14 | 0 | 1 | 0 | 0 | 16046.96 | 2 | 9 | 11 | 15 | ||||||
| 14 | 0 | 0 | 1 | 1 | 16046.96 | 2 | 9 | 11 | 15 | ||||||
| 14 | 0 | 0 | 0 | 0 | 16046.96 | 2 | 9 | 11 | 15 | ||||||
| 14 | 0 | 0 | 0 | 0 | 16046.96 | 2 | 9 | 11 | 15 | ||||||
| 14 | 2 | 9 | 11 | 15 | |||||||||||
| 22 | 0 | 0 | 0 | 0 | 25055.74 | 2 | 9 | 14 | |||||||
| 22 | 0 | 0 | 0 | 0 | 25055.74 | 2 | 9 | 14 | |||||||
| 22 | 0 | 1 | 1 | 0 | 25055.74 | 2 | 9 | 14 | |||||||
| 22 | 0 | 0 | 0 | 1 | 25055.74 | 2 | 9 | 14 | |||||||
| 22 | 1 | 1 | 0 | 0 | 25055.74 | 2 | 9 | 14 | |||||||
| 22 | 2 | 9 | 14 | ||||||||||||
5a6,5a7,5a14,5a22
5个点的结构总可以用4个点+1来表示
已知
| 4-x | 差值结构 | 迭代次数 | 4-x | 差值结构 | 迭代次数 | |||||||
| 2 | - | - | - | 13474.83 | 10 | - | - | - | - | 32007.82 | ||
| 2 | 1 | 1 | - | 13474.83 | 10 | - | - | - | - | 32007.82 | ||
| 2 | - | - | 1 | 13474.83 | 10 | - | 1 | 1 | 1 | 32007.82 | ||
| 2 | - | 1 | - | 13474.83 | 10 | 1 | - | - | - | 32007.82 | ||
| 2 | - | - | - | 13474.83 | 10 | - | - | - | - | 32007.82 | ||
| 2 | 13474.83 | 10 | 32007.82 | |||||||||
| 4 | - | - | - | 16553.19 | 11 | - | - | - | 35038.88 | |||
| 4 | - | - | - | 16553.19 | 11 | - | - | 1 | 35038.88 | |||
| 4 | 1 | 1 | 1 | 16553.19 | 11 | 1 | 1 | - | 35038.88 | |||
| 4 | - | 1 | - | 16553.19 | 11 | - | - | 1 | 35038.88 | |||
| 4 | - | - | - | 16553.19 | 11 | - | - | - | 35038.88 | |||
| 4 | 16553.19 | 11 | 35038.88 | |||||||||
| 7 | - | - | 1 | 23100.62 | 13 | - | - | - | 1 | 39571.72 | ||
| 7 | - | - | - | 23100.62 | 13 | - | - | 1 | - | 39571.72 | ||
| 7 | - | - | 1 | 23100.62 | 13 | - | 1 | - | - | 39571.72 | ||
| 7 | 1 | - | - | 23100.62 | 13 | 1 | - | - | - | 39571.72 | ||
| 7 | - | 1 | - | 23100.62 | 13 | - | - | - | - | 39571.72 | ||
| 7 | 23100.62 | 13 | 39571.72 | |||||||||
| 9 | - | - | 1 | - | 29623.51 | 14 | - | - | - | 44485.74 | ||
| 9 | - | - | - | - | 29623.51 | 14 | - | - | - | 44485.74 | ||
| 9 | - | - | - | - | 29623.51 | 14 | - | 1 | 1 | 44485.74 | ||
| 9 | 1 | 1 | - | - | 29623.51 | 14 | 1 | - | 1 | 44485.74 | ||
| 9 | - | - | - | 1 | 29623.51 | 14 | - | - | - | 44485.74 | ||
| 9 | 29623.51 | 14 | 44485.74 | |||||||||
| 15 | - | - | - | - | 52026.59 | |||||||
| 15 | - | - | - | - | 52026.59 | |||||||
| 15 | - | 1 | - | 1 | 52026.59 | |||||||
| 15 | 1 | - | 1 | - | 52026.59 | |||||||
| 15 | - | - | - | - | 52026.59 | |||||||
| 15 | 52026.59 | |||||||||||
4a2,4a4,4a7,4a9,4a10,4a11,4a13,4a14,4a15的迭代次数，则得到加法关系
5a6=4a2+1=4a4+1=4a9+1=4a10+1
5a7=4a2+1=4a7+1=4a9+1=4a13+1
5a14=4a2+1=4a9+1=4a11+1=4a15+1
5a22=4a2+1=4a9+1=4a14+1
这4个结构中都有公共特征4a2和4a9，把这个两个特征去掉
5a6=4a4+1=4a10+1
5a7=4a7+1=4a13+1
则5a6的剩余特征就是4a4和4a10，5a7的剩余特征就是4a7和4a13.因为迭代次数4a4<4a7,4a10<4a13,所以5a6的迭代次数小于5a7
5a7=4a7+1=4a13+1
5a14=4a11+1=4a15+1
同样5a7和5a14的的剩余特征4a7<4a11,4a13<4a15，因此迭代次数5a7<5a14
比较5a14和5a22如果有4a11+4a15<4a14+4a14,则可以得到5a14<5a22.
因此迭代次数表达的相似性顺序5a6<5a7<5a14<5a22.