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theoretical part

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2024-04-05 17:43:17 +08:00
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hw2/report/main.tex
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@@ -49,20 +49,20 @@
\hwname{作业}
\begin{document}
\courseheader
\name{YOUR NAME}
\name{高艺轩}
\vspace{3mm}
\centerline{\textbf{\Large{理论部分}}}
\section{单选题（15分）}
\subsection{\underline{?}}
\subsection{\underline{A}}
\subsection{\underline{?}}
\subsection{\underline{D}}
\subsection{\underline{?}}
\subsection{\underline{D}}
\subsection{\underline{?}}
\subsection{\underline{D}}
\subsection{\underline{?}}
\subsection{\underline{B}}
\section{计算题（15 分）}
\subsection{
@@ -78,6 +78,12 @@ W=\left[ \begin{array}{cc}
\end{array}\right], b=0.04$$
}
\subsubsection{请计算卷积层的输出$Y$。}
\[\begin{cases}
Y_{11} = 0.5 \times 0.1 + (-0.2) \times (-0.2) + 0.6 \times (-0.3) + 0.4 \times 0.4 + 0.04 = 0.11\\
Y_{12} = (-0.2) \times 0.1 + 0.3 \times (-0.2) + 0.4 \times (-0.3) + (-0.1) \times 0.4 + 0.04 = -0.2\\
Y_{21} = 0.6 \times 0.1 + 0.4 \times (-0.2) + (-0.4) \times (-0.3) + 0.5 \times 0.4 + 0.04 = 0.34\\
Y_{22} = 0.4 \times 0.1 + (-0.1) \times (-0.2) + 0.5 \times (-0.3) + 0.2 \times 0.4 + 0.04 = 0.03
\end{cases}\]
\subsubsection{若训练过程中的目标函数为$L$，且已知$\frac{\partial L}{\partial Y}=\left[ \begin{array}{cc}
0.3 & 0.1 \\
@@ -88,6 +94,85 @@ W=\left[ \begin{array}{cc}
注：本题的计算方式不限，但需要提供计算过程以及各步骤的结果。
\vspace{6mm}
\begin{proof}[解]
首先，
\[\frac{\partial L}{\partial Y} = \begin{bmatrix}
\frac{\partial L}{\partial Y_{11}} & \frac{\partial L}{\partial Y_{12}}\\
\frac{\partial L}{\partial Y_{21}} & \frac{\partial L}{\partial Y_{22}}
\end{bmatrix}\]
\[\frac{\partial L}{\partial X} = \begin{bmatrix}
\frac{\partial L}{\partial X_{11}} & \frac{\partial L}{\partial X_{12}} & \frac{\partial L}{\partial X_{12}}\\
\frac{\partial L}{\partial X_{21}} & \frac{\partial L}{\partial X_{22}} & \frac{\partial L}{\partial X_{23}}\\
\frac{\partial L}{\partial X_{31}} & \frac{\partial L}{\partial X_{32}} & \frac{\partial L}{\partial X_{33}}
\end{bmatrix}\]
同时，根据链式法则，
\[\frac{\partial L}{\partial X_{11}} = \frac{\partial Y_{11}}{\partial X_{11}} \frac{\partial L}{\partial Y_{11}} + \frac{\partial Y_{12}}{\partial X_{11}} \frac{\partial L}{\partial Y_{12}} + \frac{\partial Y_{21}}{\partial X_{11}} \frac{\partial L}{\partial Y_{21}} + \frac{\partial Y_{22}}{\partial X_{11}} \frac{\partial L}{\partial Y_{22}}\]
其它的$\frac{\partial L}{X_{12}}, \dots, \frac{\partial L}{\partial X_{33}}$的计算方式也是类似的。因此，
\[\frac{\partial L}{\partial X} = \sum_{i = 1}^2 \sum_{j = 1}^2
\begin{bmatrix}
\frac{\partial Y_{ij}}{\partial X_{11}} & \cdots & \frac{\partial Y_{ij}}{\partial X_{13}}\\
\vdots & \ddots & \vdots\\
\frac{\partial Y_{ij}}{\partial X_{31}} & \cdots & \frac{\partial Y_{ij}}{\partial X_{33}}
\end{bmatrix} \frac{\partial L}{\partial Y_{ij}} = \sum_{i = 1}^2 \sum_{j = 1}^2 \frac{\partial Y_{ij}}{\partial X} \frac{L}{\partial Y_{ij}}\]
式中的$\frac{\partial Y_{ij}}{\partial X}$与对应元是由哪几个$X$中的元素卷积得到有关，它们是$W$在$3 \times 3$矩阵中的平移。综合起来，有
\begin{align*}
\frac{\partial L}{\partial X} & =
\begin{bmatrix}
0.3 & 0.1 & 0\\
-0.4 & 0.2 & 0\\
0 & 0 & 0
\end{bmatrix} \frac{\partial L}{\partial Y_{11}}
+
\begin{bmatrix}
0 & 0.3 & 0.1\\
0 & -0.4 & 0.2\\
0 & 0 & 0
\end{bmatrix} \frac{\partial L}{\partial Y_{12}}\\
& \quad +
\begin{bmatrix}
0 & 0 & 0\\
0.3 & 0.1 & 0\\
-0.4 & 0.2 & 0
\end{bmatrix} \frac{\partial L}{\partial Y_{21}}
+
\begin{bmatrix}
0 & 0 & 0\\
0 & 0.3 & 0.1\\
0 & -0.4 & 0.2
\end{bmatrix} \frac{\partial L}{\partial Y_{22}}\\
& =
\begin{bmatrix}
0.09 & 0.03 & 0\\
-0.12 & 0.06 & 0\\
0 & 0 & 0
\end{bmatrix}
+
\begin{bmatrix}
0 & 0.03 & 0.01\\
0 & -0.04 & 0.02\\
0 & 0 & 0
\end{bmatrix}\\
& \quad +
\begin{bmatrix}
0 & 0 & 0\\
-0.12 & -0.04 & 0\\
0.16 & -0.08 & 0
\end{bmatrix}
+
\begin{bmatrix}
0 & 0 & 0\\
0 & 0.06 & 0.02\\
0 & -0.08 & 0.04
\end{bmatrix}\\
& =
\begin{bmatrix}
0.09 & 0.06 & 0.01\\
-0.24 & 0.04 & 0.04\\
0.16 & -0.16 & 0.04
\end{bmatrix} \qedhere
\end{align*}
\end{proof}
\centerline{\textbf{\Large{编程部分}}}
\vspace{3mm}