380 lines
12 KiB
TeX
380 lines
12 KiB
TeX
% Homework Template
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\documentclass[a4paper]{article}
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\usepackage{ctex}
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\usepackage{amsmath, amssymb, amsthm}
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\usepackage{moreenum}
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\usepackage{mathtools}
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\usepackage{url}
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\usepackage{bm}
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\usepackage{enumitem}
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\usepackage{graphicx}
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\usepackage{subcaption}
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\usepackage{booktabs} % toprule
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\usepackage[mathcal]{eucal}
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\usepackage[thehwcnt = 3]{iidef}
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\usepackage{listings}
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\usepackage{fontspec}
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\usepackage{xcolor}
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\usepackage{float}
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\usepackage{siunitx}
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\newfontfamily\codefont[Ligatures=ResetAll]{Fira Code}[Contextuals={Alternate}]
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\newfontfamily\cascadia{Cascadia Code}
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\lstset{
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basicstyle = \small\codefont,
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% ---
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tabsize = 4,
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showstringspaces = false,
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numbers = left,
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numberstyle = \codefont,
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breaklines = true,
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captionpos = t,
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% ---
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frame = l,
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flexiblecolumns,
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}
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\lstdefinestyle{Python}{
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language = Python, % 语言选Python
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keywordstyle = \color{blue},
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keywordstyle = [2] \color{teal},
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stringstyle = \color{orange!80!black},
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commentstyle = \color{red},
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identifierstyle = \color{blue!80!white},
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}
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\lstdefinestyle{Bash}{
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language = bash
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}
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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}}
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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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% 请在YOUR NAME处填写自己的姓名
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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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% 请在？处填写答案
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\subsection{\underline{D}}
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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{B}}
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\section{计算题（15 分）}
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\subsection{给定两个类别的样本分别为:
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\begin{align*}
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&\omega_1:\{(3,1),(2,2),(4,3),(3,2)\} \\
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&\omega_2:\{(1,3),(1,2),(-1,1),(-1,2)\}
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\end{align*}
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试利用LDA，将样本特征维数压缩为一维。
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}
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\begin{proof}[解]
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首先计算$\mu_1 = (3, 2), \mu_2 = (0, 2), \mu = (1.5, 2)$。因此
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\[S_1 = \frac{1}{4}
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\left(
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\begin{bmatrix}
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0 & 0\\
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0 & 1
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\end{bmatrix}
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+
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\begin{bmatrix}
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1 & 0\\
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0 & 0
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\end{bmatrix}
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+
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\begin{bmatrix}
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1 & 1\\
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1 & 1
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\end{bmatrix}
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+
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\begin{bmatrix}
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0 & 0\\
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0 & 0
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\end{bmatrix}
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\right)
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=
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\begin{bmatrix}
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0.5 & 0.25\\
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0.25 & 0.5
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\end{bmatrix}\]
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\[S_2 = \frac{1}{4}
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\left(
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\begin{bmatrix}
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0 & 0\\
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0 & 1
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\end{bmatrix}
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+
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\begin{bmatrix}
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1 & 0\\
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0 & 0
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\end{bmatrix}
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+
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\begin{bmatrix}
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1 & 1\\
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1 & 1
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\end{bmatrix}
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+
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\begin{bmatrix}
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1 & 0\\
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0 & 0
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\end{bmatrix}
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\right)
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=
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\begin{bmatrix}
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0.75 & 0.25\\
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0.25 & 0.5
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\end{bmatrix}\]
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进一步地，
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\[S_w = \frac{1}{2} (S_1 + S_2) =
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\begin{bmatrix}
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0.625 & 0.25\\
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0.25 & 0.5
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\end{bmatrix}\]
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\[S_b = \frac{1}{2} \left(
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\begin{bmatrix}
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2.25 & 0\\
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0 & 0
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\end{bmatrix}
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+
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\begin{bmatrix}
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2.25 & 0\\
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0 & 0
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\end{bmatrix}
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\right)
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=
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\begin{bmatrix}
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2.25 & 0\\
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0 & 0
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\end{bmatrix}\]
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广义特征值分解得到$\lambda = 4.5$，$v = (0.8944, -0.4472)$。投影后的样本为
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\[\omega_1: \left\{2.2360, 0.8944, 2.2360, 1.7888\right\}\]
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\[\omega_2: \left\{-0.4472, 0, -1.3416, -1.7888\right\}\]
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\end{proof}
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\vspace{3mm}
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\subsection{模型训练通常需要大量的数据，假设某采集的数据集包含80\%的有效数据和20\%的无效数据。采用一种算法判断数据是否有效，其中无效数据被成功判别为无效数据的概率为90\%，而有效数据被误判为无效数据的概率为5\%。如果某条数据经过该算法被判别为无效数据，则根据贝叶斯定理，这条数据是无效数据的概率是多少？(提示：全概率公式$P(Y)=\sum^{N}_{i=1}P(Y|X_i)P(X_i)$)\\}
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\begin{proof}[解]
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\begin{align*}
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& P(\text{无效数据} \mid \text{判定无效})\\
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= & \frac{p(\text{判定无效} \mid \text{无效数据})p(\text{无效数据})}{p(\text{判定无效} \mid \text{无效数据})p(\text{无效数据}) + p(\text{判定无效} \mid \text{有效数据})p(\text{有效数据})}\\
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= & \frac{0.9 \times 0.2}{0.9 \times 0.2 + 0.05 \times 0.8}\\
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= & \frac{0.18}{0.18 + 0.04}\\
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= & \frac{9}{11}
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\end{align*}
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\end{proof}
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\vspace{3mm}
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\subsection{设有两类正态分布的样本集，第一类均值为$\mu_1=[2,-1]^T$，第二类均值为$\mu_2=[1,1]^T$。两类样本集的协方差矩阵和出现的先验概率都相等：$\Sigma_1=\Sigma_2=\Sigma=\left[ \begin{array}{cc}
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4 & 2 \\
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2 & \frac{4}{3}
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\end{array} \right]$，$p(\omega_1)=p(\omega_2)$。试计算分类界面，并对特征向量$x=[6,2]^T$分类。}
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\begin{proof}[解]
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\[\Sigma^{-1} = \begin{bmatrix}
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1 & -1.5\\
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-1.5 & 3
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\end{bmatrix}\]
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决策方程
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\[g_{LDF1} = \Sigma^{-1} \mu_1 \boldsymbol{x} + -\frac{1}{2} \mu_1^T \Sigma^{-1} \mu_1 = (3.5, -1) \boldsymbol{x} - 6.5\]
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类似地可以得到
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\[g_{LDF2} = (-0.5, 1.5) \boldsymbol{x} - 0.5\]
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因此分类界面为
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\begin{align*}
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(3.5, -1) \boldsymbol{x} - 6.5 & = (-0.5, 1.5) \boldsymbol{x} - 0.5\\
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(4, -2.5) \boldsymbol{x} & = 6
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\end{align*}
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对于$(6, 2)$，计算$g_{LDF1}((6, 2)) = 12.5$，$g_{LDF2}((6, 2)) = -0.5$，因此属于第一类。
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\end{proof}
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\vspace{3mm}
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\subsection{给定异或的样本集$D=\left\{\left((0,0)^T,-1\right),\left((0,1)^T,1\right),\left((1,0)^T,1\right),\left((1,1)^T,-1\right)\right\}$该样本集是线性不可分的，可采用如下所示的多项式函数$\phi(\mathbf{x})$将样本$D=\left\{(\mathbf{x}_n,y_n)\right\}$映射为$D_\phi=\left\{(\phi(\mathbf{x}_n),y_n)\right\}$，其中$\phi(\mathbf{x})$满足
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\begin{equation*}
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\begin{aligned}
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\phi_1(\mathbf{x})&=2(x_1-0.5) \\
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\phi_2(\mathbf{x})&=4(x_1-0.5)(x_2-0.5)
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\end{aligned}
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\end{equation*}
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\\
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\qquad(1) 给出映射后的样本集；\\
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\qquad(2) 在映射后的样本集中，设计一个线性SVM分类器，给出支持向量及分类界面。
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}
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\begin{proof}[解]
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映射后的样本集
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\[D_{\phi} = \left\lbrace\left((-1, 1)^T, -1\right), \left((-1, -1)^T, 1\right), \left((1, -1)^T, 1\right), \left((1, 1)^T, -1\right)\right\rbrace\]
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待优化的问题为
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\[L(\boldsymbol{\alpha}) = \sum_{i = 1}^4 \alpha_i - \frac{1}{2} \sum_{i = 1}^4 \sum_{j = 1}^4 \alpha_i \alpha_j y_i y_j \boldsymbol{x}_i^T \boldsymbol{x}_j\]
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因此
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\begin{align*}
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\frac{\partial L}{\partial \alpha_1} & = 1 - \frac{1}{2}\sum_{i \neq 1}^4 \alpha_i y_1 y_i \boldsymbol{x}_1^T \boldsymbol{x}_i - 2 \alpha_1 y_1 y_1 \boldsymbol{x}_1^T \boldsymbol{x}_1\\
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& = 1 - 2 \alpha_3 - 4 \alpha_1\\
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\frac{\partial L}{\partial \alpha_2} & = 1 - 2\alpha_4 - 4 \alpha_2\\
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\frac{\partial L}{\partial \alpha_3} & = 1 - 2 \alpha_1 - 4 \alpha_3\\
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\frac{\partial L}{\partial \alpha_4} & = 1 - 2 \alpha_3 - 4 \alpha_4
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\end{align*}
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令四个偏导数均为0，得到$\alpha_1 = \alpha_2 = \alpha_3 = \alpha_4 = \frac{1}{6}$。全部的点均为支持向量。因此
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\[\boldsymbol{w} = \sum_{i = 1}^4 \alpha_i y_i \boldsymbol{x}_i = \left(0, -\frac{2}{3}\right)\]
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为求偏置量，带入$\boldsymbol{x}_1$：
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\[(-1) (\boldsymbol{w}^T \boldsymbol{x}_1 + b) = 1\]
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得到$b = -\frac{1}{3}$。
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分类界面$\boldsymbol{w}^T \boldsymbol{x} + b = 0$，即
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\[\begin{bmatrix}
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0\\-\frac{2}{3}
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\end{bmatrix} \boldsymbol{x} - \frac{1}{3} = 0\]
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得到$x_2 = \frac{1}{2}$，因此在原空间中，
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\[4(x_1 - 0.5)(x_2 - 0.5) = 0.5\]
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\end{proof}
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\vspace{3mm}
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\subsection{使用KMeans算法对2维空间中的6个点$(0,2)$,$(2,0)$,$(2,3)$,$(3,2)$,$(4,0)$,$(5,4)$进行聚类，距离函数选择欧氏距离$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$。\\
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\qquad (1)起始聚类中心选择(0,0)和(4,3)，计算聚类中心；\\
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\qquad (2)起始聚类中心选择(1,4)和(3,1)，计算聚类中心。\\
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}
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\begin{proof}[解]
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中心选择$(0, 0), (4, 3)$，第一次分为$(0, 2), (2,0)$与$(2, 3), (3, 2), (4, 0), (5, 4)$，更新后的中心为$(1, 1)$与$\left(\frac{7}{2}, \frac{9}{4}\right)$。收敛。
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中心选择$(1, 4)$与$(3, 1)$，第一次分为$(0, 2), (2, 3)$与$(2, 0), (4, 0), (3, 2), (5, 4)$，更新后中心为$(1, \frac{5}{2})$与$(\frac{7}{2}, \frac{3}{2})$，收敛。
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\end{proof}
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\vspace{3mm}
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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“自选课题进度汇报”中的一项完成
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\section{编程作业报告}
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\subsection{程序验证}
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与助教给出的图片相比，我写出的程序PCA得到的结果的xy坐标都在$[-1, 1]$之间，不利于之后的分类。我将所有的PCA之后的坐标都扩大了20倍。
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运行\lstinline{check.py}进行检查：
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{img/check/check.png}
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\end{figure}
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\subsection{数据预处理}
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运行
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\begin{lstlisting}[style=Bash]
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python data_preprocess.py
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\end{lstlisting}
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得到的输出为
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\begin{figure}[H]
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\centering
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\begin{subfigure}[t]{.45\linewidth}
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\includegraphics[width=\textwidth]{img/preprocess/preprocess_train.png}
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\caption{训练集preprocess结果}
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\end{subfigure}
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\hspace{0.5cm}
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\begin{subfigure}[t]{.45\linewidth}
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\includegraphics[width=\textwidth]{img/preprocess/preprocess_val.png}
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\caption{验证集preprocess结果}
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\end{subfigure}\\[2ex]
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\begin{subfigure}[t]{.45\linewidth}
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\includegraphics[width=\textwidth]{img/preprocess/preprocess_test.png}
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\caption{测试集preprocess结果}
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\end{subfigure}
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\end{figure}
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\subsection{训练、验证及测试}
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\begin{figure}[H]
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\centering
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\begin{subfigure}[t]{.45\linewidth}
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\includegraphics[width=\textwidth]{img/train/default/loss.png}
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\end{subfigure}
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\hspace{0.5cm}
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\begin{subfigure}[t]{.45\linewidth}
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\includegraphics[width=\textwidth]{img/train/default/train_accu.png}
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\end{subfigure}\\[2ex]
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\begin{subfigure}[t]{.45\linewidth}
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\includegraphics[width=\textwidth]{img/train/default/sv.png}
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\end{subfigure}
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\hspace{0.5cm}
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\begin{subfigure}[t]{.45\linewidth}
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\includegraphics[width=\textwidth]{img/train/default/val.png}
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\end{subfigure}\\[2ex]
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\begin{subfigure}[t]{.8\linewidth}
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\includegraphics[width=\textwidth]{img/train/default/test.png}
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\end{subfigure}
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\end{figure}
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\subsection{调整正则化系数}
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\subsubsection{C = \num{1e-6}}
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\begin{figure}[H]
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\centering
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\begin{subfigure}[t]{.45\linewidth}
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\includegraphics[width=\textwidth]{img/train/1e-6/loss.png}
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\end{subfigure}
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\hspace{0.5cm}
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\begin{subfigure}[t]{.45\linewidth}
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\includegraphics[width=\textwidth]{img/train/1e-6/accu.png}
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\end{subfigure}\\[2ex]
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\begin{subfigure}[t]{.45\linewidth}
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\includegraphics[width=\textwidth]{img/train/1e-6/sv.png}
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\end{subfigure}
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\hspace{0.5cm}
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\begin{subfigure}[t]{.45\linewidth}
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\includegraphics[width=\textwidth]{img/train/1e-6/val.png}
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\end{subfigure}\\[2ex]
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\begin{subfigure}[t]{.8\linewidth}
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\includegraphics[width=\textwidth]{img/train/1e-6/test.png}
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\end{subfigure}
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\end{figure}
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可以看到出现了严重的欠拟合，分类界面超出了绘图的范围。这是因为C过小，导致不能正确地分辨合适的分类界面。
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\subsubsection{C = 1}
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\begin{figure}[H]
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\centering
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\begin{subfigure}[t]{.45\linewidth}
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\includegraphics[width=\textwidth]{img/train/1/loss.png}
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\end{subfigure}
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\hspace{0.5cm}
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\begin{subfigure}[t]{.45\linewidth}
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\includegraphics[width=\textwidth]{img/train/1/accu.png}
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\end{subfigure}\\[2ex]
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\begin{subfigure}[t]{.45\linewidth}
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\includegraphics[width=\textwidth]{img/train/1/sv.png}
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\end{subfigure}
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\hspace{0.5cm}
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\begin{subfigure}[t]{.45\linewidth}
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\includegraphics[width=\textwidth]{img/train/1/val.png}
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\end{subfigure}\\[2ex]
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\begin{subfigure}[t]{.8\linewidth}
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\includegraphics[width=\textwidth]{img/train/1/test.png}
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\end{subfigure}
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\end{figure}
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发生了过拟合，直线被交界面的点限制，斜率不是最优。
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\end{document}
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%%% Local Variables:
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%%% mode: late\rvx
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%%% TeX-master: t
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%%% End:
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