winogard一维运算参数详细推导

来源：榕意旅游网

对于winogard算法而言，我初始不太明白几个参数矩阵的含义及具体过程，于是进行了详细的证明和参数分析如下。

一维卷积的定义

经典的例子：输入信号为: $d=[d{_{0}},d{_{1}},d{_{2}},d{_{3}}]^{T} \,$
卷积核为： $g=[g{_{0}},g{_{1}},g{_{2}}]^{T} \,$
卷积核指的是图像处理时，给定输入图像，输入图像中一个小区域中像素加权后成为输出图像中的每个对应的元素，权值由一个函数定义，这个函数就是卷积核。

根据一维wingoard卷积运算的定义可知：
$\ F(2,3)=\left[\begin{matrix}d_{_0}&d_{_1}&d_{_2}\\ d_{_1}&d_{_2}&d_{_3}\end{matrix} \right]\left[\begin{matrix} g_{_0}\\ g_{_1}\\g_{_2} \end{matrix} \right]=\left[ \begin{matrix} d_{_0}g_{_0}+d_{_1}g_{_1}+d_{_2}g_{_2}\\d_{_1}g_{_0}+d_{_2}g_{_1}+d_{_3}g_{_2} \end{matrix} \right]=\left[\begin{matrix}r_{_0}\\r_{_1} \end{matrix}\right] \,$

第一步推导

首先证明
$\ F(2,3)=\left[\begin{matrix}d_{_0}g_{_0}+d_{_1}g_{_1}+d_{_2}g_{_2}\\d_{_1}g_{_0}+d_{_2}g_{_1}+d_{_3}g_{_2}\end{matrix}\right]=\left[\begin{matrix} m_{_1}+m_{_2}+m_{_3}\\m_{_2}-m_{_3}-m_{_4}\end{matrix}\right] \,$
其中：
$\ m_{1}=(d_{0}-d_{2})g_{0} \quad\quad\quad m_{2}=(d_{1}+d_{2})\frac{g_{0}+g_{1}+g_{2}}{2}\\ m_{3}=(d_{2}-d_{1})\frac{g_{0}-g_{1}+g_{2}}{2} \quad m_{4}=(d_{1}-d_{3})g_{2} \,$

则 $\ m_{1}+m_{2}+m_{3}=\\(d_{0}-d_{2})g_{0}+(d_{1}+d_{2})\frac{g_{0}+g_{1}+g_{2}}{2}+(d_{2}-d_{1})\frac{g_{0}-g_{1}+g_{2}}{2}=\\g_{0}(d_{0}-d_{2}+\frac{d_{1}+d_{2}}{2}+\frac{d_{2}-d_{1}}{2})+g_{1}(\frac{d_{1}+d_{2}}{2}-\frac{d_{2}-d_{1}}{2})+g_{2}(\frac{d_{1}+d_{2}}{2}+\frac{d_{2}-d_{1}}{2})\\=g_{0}d_{0}+g_{1}d_{1}+g_{2}d_{2} \,$
$\ m_{2}-m_{3}-m_{4}=\\(d_{1}+d_{2})\frac{g_{0}+g_{1}+g_{2}}{2} -(d_{2}-d_{1})\frac{g_{0}-g_{1}+g_{2}}{2}-(d_{1}-d_{3})g_{2} =\\g_{0}(\frac{d_{1}+d_{2}}{2}-\frac{d_{2}-d_{1}}{2})+ g_{1}(\frac{d_{1}+d_{2}}{2}+\frac{d_{2}-d_{1}}{2})+ g_{2}(\frac{d_{1}+d_{2}}{2}-\frac{d_{2}-d_{1}}{2}-(d_{1}-d_{3}))\\=g_{0}d_{1}+g_{1}d_{2}+g_{2}d_{3} \,$

由此可以证明： $\\\ F(2,3)=\left[\begin{matrix}d_{_0}g_{_0}+d_{_1}g_{_1}+d_{_2}g_{_2}\\d_{_1}g_{_0}+d_{_2}g_{_1}+d_{_3}g_{_2}\end{matrix}\right]=\left[\begin{matrix} m_{_1}+m_{_2}+m_{_3}\\m_{_2}-m_{_3}-m_{_4}\end{matrix}\right] \,$

参数矩阵的推导

首先推导 $A^{T}\,$

$F(2,3)=A^{T}.M, \,$ 其中 $\ M=\left[\begin{matrix} m_{1}\\m_{2}\\m_{3}\\m_{4}\end{matrix}\right]\,$ ;
由此可以推出：
$\ F(2,3)=\left[\begin{matrix} m_{1}+m_{2}+m_{3}\\ m_{2}-m_{3}-m_{4}\end{matrix}\right] =\left[\begin{matrix}1 \quad1\quad1\quad0\\0\quad1\quad-1\quad-1\end{matrix}\right]\left[\begin{matrix} m_{1}\\m_{2}\\m_{3}\\m_{4}\end{matrix}\right]=A^{T}.M \,$
则
$\ A^{T}=\left[\begin{matrix} 1 \quad1\quad1\quad0\\0\quad1\quad-1\quad-1\end{matrix}\right]\,$ ;

推导 $G和B^{T}\,$

又 $\ M=[(G\cdot g)\odot(B^{T}\cdot d)] \,$
且 $\ M=\left[\begin{matrix} m_{1}=(d_{0}-d_{2})g_{0} \\ m_{2}=(d_{1}+d_{2})\frac{g_{0}+g_{1}+g_{2}}{2}\\ m_{3}=(d_{2}-d_{1})\frac{g_{0}-g_{1}+g_{2}}{2} \\ m_{4}=(d_{1}-d_{3})g_{2}\\ \end{matrix}\right] \,$
则 $\ G\cdot g=\left[\begin{matrix} g_{0} \\ \frac{g_{0}+g_{1}+g_{2}}{2}\\ \frac{g_{0}-g_{1}+g_{2}}{2} \\ g_{2}\\ \end{matrix}\right]=\left[\begin{matrix} 1 \quad0\quad0\\\frac{1}{2} \quad\frac{1}{2}\quad\frac{1}{2}\\\frac{1}{2} \quad\frac{-1}{2}\quad\frac{1}{2} \\ 0 \quad0\quad1 \end{matrix}\right]\left[\begin{matrix} g_{0}\\g_{1}\\g_{2}\end{matrix}\right] \,$
由此可得：
$\ G=\left[\begin{matrix} 1 \quad0\quad0\\\frac{1}{2} \quad\frac{1}{2}\quad\frac{1}{2}\\\frac{1}{2} \quad\frac{-1}{2}\quad\frac{1}{2} \\ 0 \quad0\quad1 \end{matrix}\right]\,$

同理可以推导 $\ B\,$ 参数如下：
$\ B^{T}\cdot d=\left[\begin{matrix}d_{0}-d_{2} \\ d_{1}+d_{2}\\d_{2}-d_{1} \\ d_{1}-d_{3} \end{matrix}\right]=\left[\begin{matrix} 1 \quad0\quad-1\quad0\\0\quad1\quad1\quad0 \\0\quad-1\quad1\quad0 \\0\quad1\quad0\quad-1 \end{matrix}\right] \left[\begin{matrix} d_{0}\\d_{1}\\d_{2}\\d_{3}\end{matrix}\right] \,$
由此可得：
$\ B^{T}=\left[\begin{matrix} 1 \quad0\quad-1\quad0\\0\quad1\quad1\quad0 \\0\quad-1\quad1\quad0 \\0\quad1\quad0\quad-1 \end{matrix}\right]\,$

综上完成了一维winogard算法的证明以及参数的含义和推导。

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