结构加法比较4个结构的顺序"/>
用结构加法比较4个结构的顺序
( 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。统计迭代次数并排序。
其中5-x有4组数据
| 5-x | 差值结构 | 迭代次数 | 4+1 | 4-x | 差值结构 | 迭代次数 | ||||||||
| 19 | 0 | 0 | 1 | 23659.18 | 5 | 6 | 5 | - | - | 1 | 20004.91 | |||
| 0 | 1 | 0 | 23659.18 | 5 | 6 | 1 | - | - | 20004.91 | |||||
| 0 | 1 | 0 | 23659.18 | 5 | 6 | 1 | - | - | 20004.91 | |||||
| 0 | 1 | 0 | 23659.18 | 5 | 6 | - | - | - | 20004.91 | |||||
| 0 | 1 | 0 | 23659.18 | 5 | 6 | 1 | - | - | 20004.91 | |||||
| 5 | 6 | 20004.91 | ||||||||||||
| 20 | 0 | 0 | 1 | 24182.31 | 5 | 7 | 6 | - | - | - | 22074.99 | |||
| 0 | 0 | 1 | 24182.31 | 5 | 7 | - | 1 | - | 22074.99 | |||||
| 0 | 0 | 1 | 24182.31 | 5 | 7 | - | 1 | - | 22074.99 | |||||
| 0 | 1 | 0 | 24182.31 | 5 | 7 | - | 1 | - | 22074.99 | |||||
| 1 | 0 | 0 | 24182.31 | 5 | 7 | - | 1 | - | 22074.99 | |||||
| 5 | 7 | 22074.99 | ||||||||||||
| 27 | 1 | 0 | 0 | 29172.19 | 5 | 8 | 7 | - | - | 1 | 23100.62 | |||
| 1 | 0 | 0 | 29172.19 | 5 | 8 | - | - | - | 23100.62 | |||||
| 1 | 0 | 0 | 29172.19 | 5 | 8 | - | - | 1 | 23100.62 | |||||
| 0 | 1 | 0 | 29172.19 | 5 | 8 | 1 | - | - | 23100.62 | |||||
| 0 | 1 | 0 | 29172.19 | 5 | 8 | - | 1 | - | 23100.62 | |||||
| 5 | 8 | 23100.62 | ||||||||||||
| 31 | 1 | 0 | 0 | 33687.66 | 7 | 8 | 8 | - | - | 1 | 28209.13 | |||
| 0 | 1 | 0 | 33687.66 | 7 | 8 | - | - | 1 | 28209.13 | |||||
| 0 | 0 | 1 | 33687.66 | 7 | 8 | - | 1 | - | 28209.13 | |||||
| 0 | 1 | 0 | 33687.66 | 7 | 8 | - | - | - | 28209.13 | |||||
| 0 | 0 | 1 | 33687.66 | 7 | 8 | - | 1 | - | 28209.13 | |||||
| 7 | 8 | 28209.13 | ||||||||||||
收敛误差为7e-4，收敛199次，取迭代次数平均值
得到5-x的加法
5a19=4a5+1=4a6+1
5a20=4a5+1=4a7+1
5a27=4a5+1=4a8+1
5a31=4a7+1=4a8+1
在结构5a19，5a20，5a27中都有共同的特征4a5.把这个特征约掉。5a19，5a20，5a27的剩余特征就是4a6，4a7，4a8.而迭代次数有4a6<4a7<4a8，因此有迭代次数5a19<5a20<5a27
比较5a27和5a31
5a27=4a5+1=4a8+1
5a31=4a7+1=4a8+1
这两个结构中都有共同特征4a8，把4a8作为共同特征去掉后，两个结构的剩余特征为4a5，4a7.而迭代次数4a5<4a7,因此迭代次数5a27<5a31.因此有总的顺序
5a19<5a20<5a27<5a31.
更多推荐
用结构加法比较4个结构的顺序
发布评论