( 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个点,B全是0。统计迭代次数并排序。
其中有5组数据
|-----|---|---|---|---------|---|----|---|---|---|---|---------|----|----|----|----|----|
| | | | | | | | | 差值结构 ||| 迭代次数 | 11 | 12 | 13 | 22 | 23 |
| | | | | | | 1 | | - | - | - | 5579.44 | 11 | | | | 23 |
| | | | | | | | | 1 | 1 | - | 5579.44 | 11 | | | | 23 |
| | | | | | | | | 1 | - | - | 5579.44 | 11 | | | | 23 |
| | | | | | | | | - | 1 | - | 5579.44 | 11 | | | | 23 |
| | | | | | | | | - | - | - | 5579.44 | 11 | | | | 23 |
| | | | | | | | | | | | 5539.78 | 11 | | | | 23 |
| | | | | | | 2 | | - | - | - | 13474.8 | | 12 | | | 23 |
| | | | | | | | | 1 | 1 | - | 13474.8 | | 12 | | | 23 |
| | | | | | | | | - | - | 1 | 13474.8 | | 12 | | | 23 |
| | | | | | | | | - | 1 | - | 13474.8 | | 12 | | | 23 |
| | | | | | | | | - | - | - | 13474.8 | | 12 | | | 23 |
| | | | | | | | | | | | 13474.8 | | 12 | | | 23 |
| | 差值结构 || | 迭代次数 | | 4 | | - | - | - | 16553.2 | | | 13 | | 23 |
| 2a1 | 0 | 0 | 0 | 48757 | | | | - | - | - | 16553.2 | | | 13 | | 23 |
| | 0 | 0 | 0 | 48757 | | | | 1 | 1 | 1 | 16553.2 | | | 13 | | 23 |
| | 0 | 0 | 0 | 48757 | | | | - | 1 | - | 16553.2 | | | 13 | | 23 |
| | 0 | 0 | 1 | 48757 | | | | - | - | - | 16553.2 | | | 13 | | 23 |
| | 0 | 0 | 1 | 48757 | | | | | | | 16553.2 | | | 13 | | 23 |
| | | | | | | 9 | - | - | 1 | - | 29623.5 | | | | 22 | 23 |
| 2a2 | 0 | 0 | 0 | 66503.8 | | | - | - | - | - | 29623.5 | | | | 22 | 23 |
| | 0 | 0 | 0 | 66503.8 | | | - | - | - | - | 29623.5 | | | | 22 | 23 |
| | 0 | 0 | 0 | 66503.8 | | | 1 | 1 | - | - | 29623.5 | | | | 22 | 23 |
| | 0 | 1 | 0 | 66503.8 | | | - | - | - | 1 | 29623.5 | | | | 22 | 23 |
| | 0 | 0 | 1 | 66503.8 | | | | | | | 29623.5 | | | | 22 | 23 |
| | | | | | | 10 | - | - | - | - | 32007.8 | | | | | 23 |
| 2a3 | 0 | 0 | 0 | 85401.9 | | | - | - | - | - | 32007.8 | | | | | 23 |
| | 0 | 0 | 0 | 85401.9 | | | - | 1 | 1 | 1 | 32007.8 | | | | | 23 |
| | 0 | 0 | 0 | 85401.9 | | | 1 | - | - | - | 32007.8 | | | | | 23 |
| | 0 | 0 | 0 | 85401.9 | | | - | - | - | - | 32007.8 | | | | | 23 |
| | 0 | 1 | 1 | 85401.9 | | | | | | | 32007.8 | | | | | 23 |
比如第1组数据4a1,可以理解为由两个2a1合成,或者由一个2a2和一个2a3合成。因此有加法关系
4a1=2a1+2a1
4a1=2a2+2a3
同样由结构4a2可以得到加法关系
4a2=2a1+2a2
4a2=2a2+2a3
|---|---|---|
| - | 1 | - |
| 1 | 1 | - |
- | - | 1 |
---|---|---|
- | - | - |
尽管结构4a2也可以看作
4a2=2a2+2a2,但这里的两个2a2是90度相交的,假设2a2和2a2只能是平行的关系,4a2不考虑这个加法关系。
4a4有加法关系
4a4=2a1+2a3
4a4=2a2+2a3
综合考虑4a1和4a2,4a4的加法关系
4a1=2a1+2a1
4a1=2a2+2a3
4a2=2a1+2a2
4a2=2a2+2a3
4a4=2a1+2a3
4a4=2a2+2a3
他们都有2a2和2a3的特征,把这个公共特征去掉
4a1=2a1+2a1
4a2=2a1+2a2
4a4=2a1+2a3
这3个结构剩余的特征中仍然有2a1作为公共特征,再把2a1的公共特征去掉则4a1和4a2,4a4的独立特征就是2a1,2a2,2a3,而2a1,2a2,2a3的迭代次数依次变大,这和4a1,4a2,4a4的变化关系相同。
4a9的加法是
4a9=2a2+2a2
4a9=2a2+2a3
4a10的加法可以看作是
4a10=2a2+2a3
4a10=2a2+2a3
所以4a9和4a10的独立特征就是2a2和2a3,所以4a10的迭代次数更大。
如果有关系2a2+2a2>2a1+2a3 ,则所有这5组迭代次数的顺序都可以用差值结构的加法去解释。