( 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3个点，B全为0，排列组合A，统计迭代次数并排序。
其中的7组数据
| 差值结构 | 迭代次数 | 构造平均列 | L | E | |||||||
| 0 | 1 | 0 | 30302 | 0 | 2 | 0 | 1.8 | 8.1 | 8.1 | ||
| 0 | 1 | 0 | 30302 | 0 | 1.5 | 0 | 1.8 | 8.1 | |||
| 0 | 1 | 0 | 30302 | 0 | 2 | 0 | 1.8 | 8.1 | |||
| 0 | 0 | 0 | 30302 | 0 | 0 | 0 | 0 | 8.1 | |||
| 0 | 0 | 0 | 30302 | 0 | 0 | 0 | 0 | 8.1 | |||
| 0 | 8.1 | ||||||||||
| 0 | 0 | 1 | 30392 | 0 | 0 | 2 | 2 | 6 | 6 | ||
| 0 | 1 | 0 | 30392 | 0 | 1 | 0 | 1 | 6 | |||
| 0 | 0 | 1 | 30392 | 0 | 0 | 2 | 2 | 6 | |||
| 0 | 0 | 0 | 30392 | 0 | 0 | 0 | 0 | 6 | |||
| 0 | 0 | 0 | 30392 | 0 | 0 | 0 | 0 | 6 | |||
| 0 | 6 | ||||||||||
| 0 | 0 | 1 | 31656 | 0 | 0 | 2 | 2 | 7 | 7 | ||
| 0 | 0 | 1 | 31656 | 0 | 0 | 2 | 2 | 7 | |||
| 0 | 1 | 0 | 31656 | 0 | 1 | 0 | 1 | 7 | |||
| 0 | 0 | 0 | 31656 | 0 | 0 | 0 | 0 | 7 | |||
| 0 | 0 | 0 | 31656 | 0 | 0 | 0 | 0 | 7 | |||
| 0 | 7 | ||||||||||
| 0 | 0 | 1 | 31954 | 0 | 0 | 2 | 1.8 | 5.94 | 5.94 | ||
| 0 | 0 | 1 | 31954 | 0 | 0 | 2 | 1.8 | 5.94 | |||
| 0 | 0 | 0 | 31954 | 0 | 0 | 0 | 0 | 5.94 | |||
| 0 | 0 | 1 | 31954 | 0 | 0 | 1.18 | 1.8 | 5.94 | |||
| 0 | 0 | 0 | 31954 | 0 | 0 | 0 | 0 | 5.94 | |||
| 0 | 5.94 | ||||||||||
| 0 | 0 | 1 | 32131 | 0 | 0 | 2 | 2 | 5.67 | 5.67 | ||
| 0 | 0 | 1 | 32131 | 0 | 0 | 2 | 2 | 5.67 | |||
| 0 | 0 | 0 | 32131 | 0 | 0 | 0 | 0 | 5.67 | |||
| 0 | 1 | 0 | 32131 | 0 | 1 | 0 | 1 | 5.67 | |||
| 0 | 0 | 0 | 32131 | 0 | 0 | 0 | 0 | 5.67 | |||
| 0 | 5.67 | ||||||||||
| 1 | 0 | 0 | 32399 | 2 | 0 | 0 | 2 | 5.17 | 5.17 | ||
| 1 | 0 | 0 | 32399 | 2 | 0 | 0 | 2 | 5.17 | |||
| 0 | 0 | 0 | 32399 | 0 | 0 | 0 | 0 | 5.17 | |||
| 0 | 0 | 0 | 32399 | 0 | 0 | 0 | 0 | 5.17 | |||
| 0 | 1 | 0 | 32399 | 0 | 1 | 0 | 1 | 5.17 | |||
| 0 | 5.17 | ||||||||||
| 0 | 1 | 0 | 32770 | 0 | 2 | 0 | 2 | 4.33 | 4.33 | ||
| 1 | 0 | 0 | 32770 | 1 | 0 | 0 | 1 | 4.33 | |||
| 0 | 0 | 0 | 32770 | 0 | 0 | 0 | 0 | 4.33 | |||
| 0 | 1 | 0 | 32770 | 0 | 2 | 0 | 2 | 4.33 | |||
| 0 | 0 | 0 | 32770 | 0 | 0 | 0 | 0 | 4.33 | |||
| 0 | 4.33 | ||||||||||
收敛误差为7e-4，收敛199次，取迭代次数平均值
其中特别值得注意的是
| 0 | 0 | 1 | 31656 | 0 | 0 | 2 | ||
| 0 | 0 | 1 | 31656 | 0 | 0 | 2 | ||
| 0 | 1 | 0 | 31656 | 0 | 1 | 0 | ||
| 0 | 0 | 0 | 31656 | 0 | 0 | 0 | ||
| 0 | 0 | 0 | 31656 | 0 | 0 | 0 | ||
| 0 | 0 | 1 | 31954 | 0 | 0 | 2 | ||
| 0 | 0 | 1 | 31954 | 0 | 0 | 2 | ||
| 0 | 0 | 0 | 31954 | 0 | 0 | 0 | ||
| 0 | 0 | 1 | 31954 | 0 | 0 | 1.18 | ||
| 0 | 0 | 0 | 31954 | 0 | 0 | 0 | ||
| 0 | 0 | 1 | 32131 | 0 | 0 | 2 | ||
| 0 | 0 | 1 | 32131 | 0 | 0 | 2 | ||
| 0 | 0 | 0 | 32131 | 0 | 0 | 0 | ||
| 0 | 1 | 0 | 32131 | 0 | 1 | 0 | ||
| 0 | 0 | 0 | 32131 | 0 | 0 | 0 |
1*1*0*1*0这个结构，因为这个结构夹在1*1*2*0*0和1*1*0*2*0两个结构之间，因此1*1*0*1*0的平均列应该更接近2，2，0，1.18，0 E=5.96.与之对应的等价的平均列是1.8，1.8，0，1.8，0.
也就是当3个1在一列的情况下把1的面值修正为1.8.
| 0 | 1 | 0 |
| 0 | 1 | 0 |
| 0 | 1 | 0 |
| 0 | 0 | 0 |
| 0 | 0 | 0 |
这样结构2*2*2*0*0的平均列就是1.8，1.8，1.8，0，0. E=8.1
这个方法就是假定上下连着的两个1的排斥能为4是单位，用这个单位去修正其他结构的面值。
得到图
排斥能和迭代次数成反比。
