在一维的世界里寻找迭代次数的影子
( A, B )---1*30*2---( 1, 0 )( 0, 1 )
讓網絡的輸入只有1個節點，AB各由3張二值化的圖片組成，排列組合A和B的所有可能性，固定收斂誤差，統計收斂迭代次數
| A | B | 迭代次數 | ||||||||||
| 0 | 0 | 0 | 1 | 1 | 1 | 1b | 1b | 1b | 0*0*0-1*1*1 | 27152.97 |
如網絡0*0*0-1*1*1，A是0，0，0，B是1，1，1.差值結構是1b，1b，1b，當收斂誤差為7e-4的時候平均迭代次數為27152次。
其余各組數據如下
| A | B | 7.00E-04 | ||||||||||
| 0 | 0 | 0 | 1 | 1 | 1 | 1b | 1b | 1b | 0*0*0-1*1*1 | 27152.97487 | ||
| 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1*1*1-0*0*0 | 27300.39698 | ||
| 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 1*1*0-0*0*0 | 28585.72362 | ||
| 0 | 0 | 0 | 1 | 0 | 1 | 1b | 0 | 1b | 0*0*0-1*0*1 | 28613.12563 | ||
| 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 1*0*1-0*0*0 | 28617.34171 | ||
| 0 | 0 | 0 | 1 | 1 | 0 | 1b | 1b | 0 | 0*0*0-1*1*0 | 28675.66834 | ||
| 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 0*1*1-0*0*0 | 28685.56281 | ||
| 0 | 0 | 0 | 0 | 1 | 1 | 0 | 1b | 1b | 0*0*0-0*1*1 | 28721.68844 | ||
| 1 | 1 | 1 | 0 | 1 | 0 | 1 | k | 1 | 1*1*1-0*1*0 | 46507.47739 | ||
| 1 | 1 | 1 | 1 | 0 | 0 | k | 1 | 1 | 1*1*1-1*0*0 | 46618.54271 | ||
| 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | k | 1*1*1-0*0*1 | 46639.72864 | ||
| 0 | 0 | 1 | 1 | 1 | 1 | 1b | 1b | k | 0*0*1-1*1*1 | 46759.24121 | ||
| 1 | 0 | 0 | 1 | 1 | 1 | k | 1b | 1b | 1*0*0-1*1*1 | 46799.26131 | ||
| 0 | 1 | 0 | 1 | 1 | 1 | 1b | k | 1b | 0*1*0-1*1*1 | 46822.42714 | ||
| 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1b | 0*0*0-0*0*1 | 63014.05025 | ||
| 0 | 0 | 0 | 1 | 0 | 0 | 1b | 0 | 0 | 0*0*0-1*0*0 | 63260 | ||
| 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1b | 0 | 0*0*0-0*1*0 | 63339.60804 | ||
| 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1*0*0-0*0*0 | 63340.37688 | ||
| 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0*0*1-0*0*0 | 63340.75879 | ||
| 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0*1*0-0*0*0 | 63485.92462 | ||
| 1 | 1 | 1 | 1 | 0 | 1 | k | 1 | k | 1*1*1-1*0*1 | 121408.5427 | ||
| 1 | 1 | 1 | 1 | 1 | 0 | k | k | 1 | 1*1*1-1*1*0 | 121973.3668 | ||
| 0 | 1 | 1 | 1 | 1 | 1 | 1b | k | k | 0*1*1-1*1*1 | 121988.8392 | ||
| 1 | 0 | 1 | 1 | 1 | 1 | k | 1b | k | 1*0*1-1*1*1 | 122891.6533 | ||
| 1 | 1 | 1 | 0 | 1 | 1 | 1 | k | k | 1*1*1-0*1*1 | 123237.3869 | ||
| 1 | 1 | 0 | 1 | 1 | 1 | k | k | 1b | 1*1*0-1*1*1 | 123529.2915 | ||
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0*0*0-0*0*0 | 400000 | ||
| 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | k | 0*0*1-0*0*1 | 400000 | ||
| 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1b | 1 | 0*0*1-0*1*0 | 400000 | ||
| 0 | 0 | 1 | 1 | 0 | 0 | 1b | 0 | 1 | 0*0*1-1*0*0 | 400000 | ||
| 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 1b | 0*1*0-0*0*1 | 400000 | ||
| 0 | 1 | 0 | 0 | 1 | 0 | 0 | k | 0 | 0*1*0-0*1*0 | 400000 | ||
| 0 | 1 | 0 | 1 | 0 | 0 | 1b | 1 | 0 | 0*1*0-1*0*0 | 400000 | ||
| 0 | 1 | 1 | 0 | 1 | 1 | 0 | k | k | 0*1*1-0*1*1 | 400000 | ||
| 0 | 1 | 1 | 1 | 0 | 1 | 1b | 1 | k | 0*1*1-1*0*1 | 400000 | ||
| 0 | 1 | 1 | 1 | 1 | 0 | 1b | k | 1 | 0*1*1-1*1*0 | 400000 | ||
| 1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 1b | 1*0*0-0*0*1 | 400000 | ||
| 1 | 0 | 0 | 0 | 1 | 0 | 1 | 1b | 0 | 1*0*0-0*1*0 | 400000 | ||
| 1 | 0 | 0 | 1 | 0 | 0 | k | 0 | 0 | 1*0*0-1*0*0 | 400000 | ||
| 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1b | k | 1*0*1-0*1*1 | 400000 | ||
| 1 | 0 | 1 | 1 | 0 | 1 | k | 0 | k | 1*0*1-1*0*1 | 400000 | ||
| 1 | 1 | 0 | 0 | 1 | 1 | 1 | k | 1b | 1*1*0-0*1*1 | 400000 | ||
| 1 | 1 | 0 | 1 | 0 | 1 | k | 1 | 1b | 1*1*0-1*0*1 | 400000 | ||
| 1 | 1 | 0 | 1 | 1 | 0 | k | k | 0 | 1*1*0-1*1*0 | 400000 | ||
| 1 | 1 | 1 | 1 | 1 | 1 | k | k | k | 1*1*1-1*1*1 | 400000 | ||
| 1 | 0 | 1 | 1 | 1 | 0 | k | 1b | 1 | 1*0*1-1*1*0 | 400023.7889 | ||
| 0 | 0 | 1 | 0 | 1 | 1 | 0 | 1b | k | 0*0*1-0*1*1 | 400502.5126 | ||
| 0 | 0 | 1 | 1 | 0 | 1 | 1b | 0 | k | 0*0*1-1*0*1 | 400502.5126 | ||
| 0 | 0 | 1 | 1 | 1 | 0 | 1b | 1b | 1 | 0*0*1-1*1*0 | 400502.5126 | ||
| 0 | 1 | 0 | 0 | 1 | 1 | 0 | k | 1b | 0*1*0-0*1*1 | 400502.5126 | ||
| 0 | 1 | 0 | 1 | 0 | 1 | 1b | 1 | 1b | 0*1*0-1*0*1 | 400502.5126 | ||
| 0 | 1 | 0 | 1 | 1 | 0 | 1b | k | 0 | 0*1*0-1*1*0 | 400502.5126 | ||
| 0 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | k | 0*1*1-0*0*1 | 400502.5126 | ||
| 0 | 1 | 1 | 0 | 1 | 0 | 0 | k | 1 | 0*1*1-0*1*0 | 400502.5126 | ||
| 0 | 1 | 1 | 1 | 0 | 0 | 1b | 1 | 1 | 0*1*1-1*0*0 | 400502.5126 | ||
| 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1b | 1b | 1*0*0-0*1*1 | 400502.5126 | ||
| 1 | 0 | 0 | 1 | 0 | 1 | k | 0 | 1b | 1*0*0-1*0*1 | 400502.5126 | ||
| 1 | 0 | 0 | 1 | 1 | 0 | k | 1b | 0 | 1*0*0-1*1*0 | 400502.5126 | ||
| 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | k | 1*0*1-0*0*1 | 400502.5126 | ||
| 1 | 0 | 1 | 0 | 1 | 0 | 1 | 1b | 1 | 1*0*1-0*1*0 | 400502.5126 | ||
| 1 | 0 | 1 | 1 | 0 | 0 | k | 0 | 1 | 1*0*1-1*0*0 | 400502.5126 | ||
| 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1b | 1*1*0-0*0*1 | 400502.5126 | ||
| 1 | 1 | 0 | 0 | 1 | 0 | 1 | k | 0 | 1*1*0-0*1*0 | 400502.5126 | ||
| 1 | 1 | 0 | 1 | 0 | 0 | k | 1 | 0 | 1*1*0-1*0*0 | 400502.5126 |
這些數據可分成6組，其中5組如下
| A | B | 迭代次數 | 等位點數值差 | |||||||||||
| 0 | 0 | 0 | 1 | 1 | 1 | 1b | 1b | 1b | 0*0*0-1*1*1 | 27152.97 | 3 | |||
| 0 | 0 | 0 | 1 | 1 | 0 | 1b | 1b | 0 | 0*0*0-1*1*0 | 28675.67 | 2 | |||
| 0 | 0 | 1 | 1 | 1 | 1 | 1b | 1b | k | 0*0*1-1*1*1 | 46759.24 | 2 | |||
| 0 | 0 | 0 | 1 | 0 | 0 | 1b | 0 | 0 | 0*0*0-1*0*0 | 63260 | 1 | |||
| 0 | 1 | 1 | 1 | 1 | 1 | 1b | k | k | 0*1*1-1*1*1 | 121988.8 | 1 |
這5組數據大體上是符合迭代次數與等位點數值差的反比假設的，
| 0 | 0 | 0 | 1 | 1 | 0 | 1b | 1b | 0 | 0*0*0-1*1*0 | 28675.67 | 2 | |||
| 0 | 0 | 1 | 1 | 1 | 1 | 1b | 1b | k | 0*0*1-1*1*1 | 46759.24 | 2 |
這兩組的等位點數值差都是2但是1b，1b，k的迭代次數要大些，同樣
| 0 | 0 | 0 | 1 | 0 | 0 | 1b | 0 | 0 | 0*0*0-1*0*0 | 63260 | 1 | |||
| 0 | 1 | 1 | 1 | 1 | 1 | 1b | k | k | 0*1*1-1*1*1 | 121988.8 | 1 |
這兩組的等位點數值差也是相同的，但是1b，k，k的迭代次數要大些。
所以只要假設k是一個大于-1且小于0的數，就可以解釋所有這5組數據迭代次數的大小關系。
| A | B | 迭代次數 | 等位點數值差 | |||||||||||
| 0 | 0 | 0 | 1 | 1 | 1 | 1b | 1b | 1b | 0*0*0-1*1*1 | 27152.97 | 3 | |||
| 0 | 0 | 0 | 1 | 1 | 0 | 1b | 1b | 0 | 0*0*0-1*1*0 | 28675.67 | 2 | |||
| 0 | 0 | 1 | 1 | 1 | 1 | 1b | 1b | k | 0*0*1-1*1*1 | 46759.24 | 2+k | |||
| 0 | 0 | 0 | 1 | 0 | 0 | 1b | 0 | 0 | 0*0*0-1*0*0 | 63260 | 1 | |||
| 0 | 1 | 1 | 1 | 1 | 1 | 1b | k | k | 0*1*1-1*1*1 | 121988.8 | 1+k+k | |||
| -1<k<0 |
還有第6組數據，如果迭代400000次還未收斂則終止收斂，表明A或B中如果不含0，0，0，或者1，1，1則網絡似乎無法收斂。
總結
以上是生活随笔為你收集整理的在一维的世界里寻找迭代次数的影子的全部內容，希望文章能夠幫你解決所遇到的問題。
- 上一篇： Axiom3D学习日记 4.地形,天空,
- 下一篇： 简单井字棋设计