347 lines
12 KiB
TeX
347 lines
12 KiB
TeX
% Homework template for Inference and Information
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% UPDATE: September 26, 2017 by Xiangxiang
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language = bash
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\usepackage{booktabs} % toprule
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\usepackage[mathcal]{eucal}
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\usepackage[thehwcnt = 2]{iidef}
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\allowdisplaybreaks
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\thecourseinstitute{清华大学电子工程系}
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\thecoursename{\textbf{媒体与认知} \space 课堂2}
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\theterm{2023-2024学年春季学期}
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\hwname{作业}
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\begin{document}
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\courseheader
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\name{高艺轩}
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\vspace{3mm}
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\centerline{\textbf{\Large{理论部分}}}
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\section{单选题（15分）}
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\subsection{\underline{C}}
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\subsection{\underline{D}}
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\subsection{\underline{D}}
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\subsection{\underline{C}}
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\subsection{\underline{B}}
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\section{计算题（15 分）}
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\subsection{
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已知某卷积层的输入为$X$(该批量中样本数目为1，输入样本通道数为1)，采用一个卷积核$W$，即卷积输出通道数为1，卷积核尺寸为$2\times 2$，卷积的步长为1，无边界延拓，偏置量为$b$：
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$$X=\left[ \begin{array}{ccc}
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0.5 & -0.2 & 0.3 \\
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0.6 & 0.4 & -0.1 \\
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-0.4 & 0.5 & 0.2
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\end{array}\right],
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W=\left[ \begin{array}{cc}
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0.1 & -0.2 \\
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-0.3 & 0.4
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\end{array}\right], b=0.04$$
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}
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\subsubsection{请计算卷积层的输出$Y$。}
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\[\begin{cases}
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Y_{11} = 0.5 \times 0.1 + (-0.2) \times (-0.2) + 0.6 \times (-0.3) + 0.4 \times 0.4 + 0.04 = 0.11\\
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Y_{12} = (-0.2) \times 0.1 + 0.3 \times (-0.2) + 0.4 \times (-0.3) + (-0.1) \times 0.4 + 0.04 = -0.2\\
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Y_{21} = 0.6 \times 0.1 + 0.4 \times (-0.2) + (-0.4) \times (-0.3) + 0.5 \times 0.4 + 0.04 = 0.34\\
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Y_{22} = 0.4 \times 0.1 + (-0.1) \times (-0.2) + 0.5 \times (-0.3) + 0.2 \times 0.4 + 0.04 = 0.03
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\end{cases}\]
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\subsubsection{若训练过程中的目标函数为$L$，且已知$\frac{\partial L}{\partial Y}=\left[ \begin{array}{cc}
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0.3 & 0.1 \\
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-0.4 & 0.2
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\end{array} \right]$，请计算$\frac{\partial L}{\partial X}$。
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}
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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。
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\vspace{6mm}
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\begin{proof}[解]
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首先，
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\[\frac{\partial L}{\partial Y} = \begin{bmatrix}
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\frac{\partial L}{\partial Y_{11}} & \frac{\partial L}{\partial Y_{12}}\\
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\frac{\partial L}{\partial Y_{21}} & \frac{\partial L}{\partial Y_{22}}
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\end{bmatrix}\]
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\[\frac{\partial L}{\partial X} = \begin{bmatrix}
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\frac{\partial L}{\partial X_{11}} & \frac{\partial L}{\partial X_{12}} & \frac{\partial L}{\partial X_{12}}\\
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\frac{\partial L}{\partial X_{21}} & \frac{\partial L}{\partial X_{22}} & \frac{\partial L}{\partial X_{23}}\\
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\frac{\partial L}{\partial X_{31}} & \frac{\partial L}{\partial X_{32}} & \frac{\partial L}{\partial X_{33}}
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\end{bmatrix}\]
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同时，根据链式法则，
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\[\frac{\partial L}{\partial X_{11}} = \frac{\partial Y_{11}}{\partial X_{11}} \frac{\partial L}{\partial Y_{11}} + \frac{\partial Y_{12}}{\partial X_{11}} \frac{\partial L}{\partial Y_{12}} + \frac{\partial Y_{21}}{\partial X_{11}} \frac{\partial L}{\partial Y_{21}} + \frac{\partial Y_{22}}{\partial X_{11}} \frac{\partial L}{\partial Y_{22}}\]
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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，
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\[\frac{\partial L}{\partial X} = \sum_{i = 1}^2 \sum_{j = 1}^2
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\begin{bmatrix}
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\frac{\partial Y_{ij}}{\partial X_{11}} & \cdots & \frac{\partial Y_{ij}}{\partial X_{13}}\\
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\vdots & \ddots & \vdots\\
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\frac{\partial Y_{ij}}{\partial X_{31}} & \cdots & \frac{\partial Y_{ij}}{\partial X_{33}}
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\end{bmatrix} \frac{\partial L}{\partial Y_{ij}} = \sum_{i = 1}^2 \sum_{j = 1}^2 \frac{\partial Y_{ij}}{\partial X} \frac{L}{\partial Y_{ij}}\]
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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
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\begin{align*}
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\frac{\partial L}{\partial X} & =
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\begin{bmatrix}
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0.1 & -0.2 & 0\\
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-0.3 & 0.4 & 0\\
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0 & 0 & 0
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\end{bmatrix} \frac{\partial L}{\partial Y_{11}}
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+
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\begin{bmatrix}
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0 & 0.1 & -0.2\\
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0 & -0.3 & 0.4\\
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0 & 0 & 0
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\end{bmatrix} \frac{\partial L}{\partial Y_{12}}\\
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& \quad +
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\begin{bmatrix}
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0 & 0 & 0\\
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0.1 & -0.2 & 0\\
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-0.3 & 0.4 & 0
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\end{bmatrix} \frac{\partial L}{\partial Y_{21}}
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+
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\begin{bmatrix}
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0 & 0 & 0\\
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0 & 0.1 & -0.2\\
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0 & -0.3 & 0.4
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\end{bmatrix} \frac{\partial L}{\partial Y_{22}}\\
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& = \mathrm{zeropad}(W) \ast \frac{\partial L}{\partial Y}\\
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& =
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\begin{bmatrix}
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0.03 & -0.06 & 0\\
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-0.09 & 0.12 & 0\\
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0 & 0 & 0
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\end{bmatrix}
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+
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\begin{bmatrix}
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0 & 0.01 & -0.02\\
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0 & -0.03 & 0.04\\
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0 & 0 & 0
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\end{bmatrix}\\
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& \quad +
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\begin{bmatrix}
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0 & 0 & 0\\
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-0.04 & 0.08 & 0\\
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0.12 & -0.16 & 0
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\end{bmatrix}
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+
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\begin{bmatrix}
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0 & 0 & 0\\
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0 & 0.02 & -0.04\\
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0 & -0.06 & 0.08
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\end{bmatrix}\\
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& =
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\begin{bmatrix}
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0.03 & -0.05 & -0.02\\
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-0.13 & 0.19 & 0\\
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0.12 & -0.22 & 0.08
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\end{bmatrix} \qedhere
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\end{align*}
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\end{proof}
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\centerline{\textbf{\Large{编程部分}}}
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\vspace{3mm}
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% 请根据是否选择自选课题的情况选择“编程作业报告”或“自选课题开题报告”中的一项完成
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\section{编程作业报告}
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\subsection{探究batch normalization和dropout的作用}
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\begin{enumerate}
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\item 使用默认配置训练模型：
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\begin{lstlisting}[style=Bash]
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python train.py --ckpt_path checkpoints/default
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\end{lstlisting}
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{img/models/default/loss_and_acc.jpg}
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\end{figure}
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之后测试得到的正确率为90.8\%。
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\item 启用batch normalization：
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\begin{lstlisting}[style=Bash]
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python train.py --ckpt_path checkpoints/bn --bn
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\end{lstlisting}
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{img/models/bn/loss_and_acc.jpg}
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\end{figure}
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测试得到的正确率为95.9\%。
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\item 启用dropout并设置概率为0.3：
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\begin{lstlisting}[style=Bash]
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python train.py --ckpt_path checkpoints/dropout --dropout 0.3
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\end{lstlisting}
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{img/models/dropout/loss_and_acc.jpg}
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\end{figure}
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测试后得到的正确率为94.1\%。
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\end{enumerate}
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\subsection{探究数据增广的作用}
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考虑到在不同的视角下，交通标志可能有旋转或者变形，因此使用
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\begin{lstlisting}[style=Python]
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transforms.RandomAffine(degrees=30,shear=10)
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\end{lstlisting}
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来对数据进行随机的形变与旋转；另外，考虑到可能在不同的光线条件下导致对比度变化，因此使用
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\begin{lstlisting}[style=Python]
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transforms.RandomAutocontrast()
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\end{lstlisting}
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来对数据进行随机的对比度调整。
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执行
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\begin{lstlisting}[style=Bash]
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python unit_test.py data_loader
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\end{lstlisting}
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得到
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{img/augmentation.jpg}
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\caption{数据增广后的结果}
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\end{figure}
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训练最优模型使用的命令为
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\begin{lstlisting}[style=Bash]
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python train.py --ckpt_path checkpoints/bn_aug --bn --augment --epoch 20
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\end{lstlisting}
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{img/models/bn_aug/loss_and_acc.jpg}
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\end{figure}
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测试得到的正确率为96.0\%，略微高于不使用数据增强时的结果。
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\subsection{探究空间变换网络（STN）的作用}
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运行
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\begin{lstlisting}[style=Bash]
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python train.py --ckpt_path checkpoints/stn --bn --stn
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\end{lstlisting}
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{img/models/stn/loss_and_acc.jpg}
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\end{figure}
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测试得到的正确率为94.6\%。正确率比不使用stn反而有所降低，可能是设计的网络结构不够理想导致的。
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\subsection{可视化}
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\begin{enumerate}
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\item 可视化各层卷积核：
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{img/filter/filter_layer_0.jpg}
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\caption{第0层的卷积核}
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\end{figure}
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{img/filter/filter_layer_1.jpg}
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\caption{第1层的卷积核}
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\end{figure}
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{img/filter/filter_layer_2.jpg}
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\caption{第2层的卷积核}
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\end{figure}
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{img/filter/filter_layer_3.jpg}
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\caption{第3层的卷积核}
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\end{figure}
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{img/filter/filter_layer_4.jpg}
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\caption{第4层的卷积核}
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\end{figure}
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\item 可视化各层卷积层的输出特征图
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{img/feature/layer_0/feature_map.jpg}
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\caption{第0层的卷积核特征图}
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\end{figure}
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{img/feature/layer_1/feature_map.jpg}
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\caption{第1层的卷积核特征图}
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\end{figure}
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{img/feature/layer_2/feature_map.jpg}
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\caption{第2层的卷积核特征图}
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\end{figure}
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{img/feature/layer_3/feature_map.jpg}
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\caption{第3层的卷积核特征图}
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\end{figure}
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{img/feature/layer_4/feature_map.jpg}
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\caption{第4层的卷积核特征图}
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\end{figure}
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\item t-SNE可视化最后一层隐藏层的输出特征
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{img/tsne/tsne.jpg}
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\end{figure}
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t-SNE最后一层的隐藏层的输出证明，不同类别的输入已经被通过非线性变换分类到了不同的聚类。
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\item STN学习到的变换
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{img/stn/stn.jpg}
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\end{figure}
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网络尽可能将所有的路牌都变换到了同样的倾斜角度。
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\end{enumerate}
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\section{遇到的问题与解决办法}
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在自定义STN网络的时候，我最开始使用了比较小的卷积核，使得STN的效果很差，使用之后会使得正确率只有80\%；之后，我分析认为STN主要要感知整个图片的倾斜以及旋转情况，需要较大的视野，因此选择了较大的卷积核，之后得到了比较理想的效果。
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完成作业没有使用大模型。
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% \section{自选课题工作进度汇报}
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\end{document}
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