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20240414IEEETG
Commit 6ebe2bb2 by LiXinwen
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parent 691fa138
Showing with 235 additions and 4 deletions
document.tex
......@@ -70,10 +70,10 @@ wangwenhao11@nudt.edu.cn).
\end{IEEEkeywords}
\input{main/background}
%\input{main/introduction}
%\input{main/nonergodicity}
%\input{main/paradox}
%\input{main/theorem}
\input{main/introduction}
\input{main/nonergodicity}
\input{main/paradox}
\input{main/theorem}
......
main/nonergodicity.tex
......@@ -7,6 +7,114 @@ We assume that the state-process is ergodic — i.e. all states
are reachable under any policy from the current state after
sufficiently many steps. \cite{majeed2018q}
% ABCDE的随机游走的状态矩阵
\[
P = \begin{pmatrix}
1 & 0 & 0 & 0 & 0 & \\
\frac{1}{2} & 0 & \frac{1}{2} & 0 & 0\\
0 & \frac{1}{2} & 0 & \frac{1}{2} & 0\\
0 & 0 & \frac{1}{2} & 0 & \frac{1}{2}\\
0 & 0 & 0 & 0 & 1
\end{pmatrix}
\]
%可重启的随机游走
\[
P = \begin{pmatrix}
0 & 0 & 1 & 0 & 0 & \\
\frac{1}{2} & 0 & \frac{1}{2} & 0 & 0\\
0 & \frac{1}{2} & 0 & \frac{1}{2} & 0\\
0 & 0 & \frac{1}{2} & 0 & \frac{1}{2}\\
0 & 0 & 1 & 0 & 0
\end{pmatrix}
\]
% 计算平稳分布
\[
\begin{cases}
\pi_1 = \pi_3 \\
\frac{1}{2}\pi_1 + \frac{1}{2}\pi_3 = \pi_2 \\
\frac{1}{2}\pi_2 + \frac{1}{2}\pi_4 = \pi_3 \\
\frac{1}{2}\pi_3 + \frac{1}{2}\pi_5 = \pi_4 \\
\pi_3 = \pi_5 \\
\pi_1 + \pi_2 + \pi_3 + \pi_4 + \pi_5 = 1
\end{cases}
\]
%随机游走pic
\begin{tikzpicture}
\node[draw, rectangle, fill=gray!50] (DEAD) at (-2,0) ;
\node[draw, rectangle, fill=gray!50] (DEAD2) at (10,0) ;
\node[draw, circle] (A) at (0,0) {A};
\node[draw, circle] (B) at (2,0) {B};
\node[draw, circle] (C) at (4,0) {C};
\node[draw, circle] (D) at (6,0) {D};
\node[draw, circle] (E) at (8,0) {E};
\draw[->] (A) -- (DEAD);
\draw[->] (B) -- (A);
\draw[->] (B) to [bend left=30] (C);
\draw[->] (C) to [bend left=30] (B);
\draw[->] (C) to [bend left=30] (D);
\draw[->] (D) to [bend left=30] (C);
\draw[->] (D) -- (E);
\draw[->] (E) -- (DEAD2);
\draw[->] ([yshift=4ex]C.north) -- ([yshift=4.5ex]C.south);
\end{tikzpicture}
设两个上三角矩阵为( A ) 和 ( B )，它们的形式分别为：
% 两个上三角矩阵乘积求和为上三角矩阵
A = \begin{pmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
0 & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & a_{nn}
\end{pmatrix}, \quad
B = \begin{pmatrix}
b_{11} & b_{12} & \cdots & b_{1n} \\
0 & b_{22} & \cdots & b_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & b_{nn}
\end{pmatrix}
[
c_{ij}=\sum_{k=1}^{n} a_{ik}b_{kj}
]
当 ( i>j ) 时，有 ( c_{ij}=0 )，因为在此情况下，( a_{ik}=0 ) 或 ( b_{kj}=0 )，乘积中至少有一项为 0。
所以 ( C ) 也是一个上三角矩阵。
因此，证明了两个上三角矩阵的乘积还是一个上三角矩阵。
% N矩阵
$N=1+Q^1+Q^2……$
% “重启”随机游走 pic
\begin{tikzpicture}
\node[draw, circle] (A) at (0,0) {A};
\node[draw, circle] (B) at (2,0) {B};
\node[draw, circle] (C) at (4,0) {C};
\node[draw, circle] (D) at (6,0) {D};
\node[draw, circle] (E) at (8,0) {E};
\draw[->] (A.north) to [bend left=30] (C.north)
\draw[->] (B) -- (A);
\draw[->] (B) to [bend left=30] (C);
\draw[->] (C) to [bend left=30] (B);
\draw[->] (C) to [bend left=30] (D);
\draw[->] (D) to [bend left=30] (C);
\draw[->] (D) -- (E);
\draw[->] (E.south) to [bend left=30] (C.south)
\end{tikzpicture}
......
main/paradox.tex
......@@ -41,6 +41,61 @@ of a potentially substantial prize, the actual expected
Consequently, in the long run, participants could face
an infinite monetary loss.
%圣彼得堡悖论期望
[
E(X)=\sum_{n}x(n)p(n) = \frac{1}{2}\times 2 + \frac{1}{4}\times 4 + \frac{1}{8}\times 8 + \cdots = \infty
]
% 圣彼得堡悖论状态转移矩阵
\[
P = \begin{pmatrix}
0 & \frac{1}{2} & 0 & 0 & 0 & ... & ... & \frac{1}{2} \\
0 & 0 & \frac{1}{2} & 0 & 0 & ... & ... & \frac{1}{2} \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\
1 & 0 & 0 & 0 & 0 & ... & ... & 0
\end{pmatrix}
\]
% 圣彼得堡悖论Q矩阵
\[
Q = \begin{pmatrix}
0 & \frac{1}{2} & 0 & 0 & 0 & ... & ... \\
0 & 0 & \frac{1}{2} & 0 & 0 & ... & ... \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \ddots \\
\end{pmatrix}
\]
% N矩阵
$N=1+Q^1+Q^2……$
% 圣彼得堡悖论的N矩阵
[
N = \begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \frac{1}{16} & \dots \
0 & 1 & \frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \dots \
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots
\end{pmatrix}
]
% 带截断的圣彼得堡悖论
\begin{table}[ht]
\centering
\begin{tabular}{|c|c|c|}
\hline
截断长度（期望） & 100000次试验结果平均 & 偏差 \\
\hline
5 & 4.99 & -0.01 \\
10 & 9.89 & -0.11 \\
15 & 14.66 & -0.34 \\
20 & 16.83 & -3.17 \\
25 & 15.74 & -9.26 \\
30 & 186.15 & +156.15 \\
\hline
\end{tabular}
\end{table}
......
main/theorem.tex
......@@ -81,6 +81,20 @@ P可以分解为Q R I 0，那么$N=(I-Q)^{-1}$，即描述了非吸收态之间
这样的话，相当于我们提出了一种满足非遍历性的充分条件吧？
似乎论文可以从这方面下手！
% 2048游戏局面编码
$p=2^{64} \cdot \sum_{m=0}^{15} I(B_m \neq 0) \cdot 2^{B_m} + \sum_{m=0}^{15} (1 \ll 4m) \cdot B_m$
% 马尔可夫链标准形式
[
P = \begin{bmatrix}
Q & R \
0 & I
\end{bmatrix}
]
% 带策略的马尔可夫链标准形式
[ P_\pi = \begin{pmatrix} Q_\pi & R_\pi \ 0 & I \end{pmatrix} ]
......
pic/2048example.tex 0 → 100644
\begin{figure*}
\centering
\tikzstyle{place}=[circle,draw=blue!50,fill=blue!20,thick,minimum size=10mm]
\tikzstyle{transition}=[rectangle,draw=black!50,fill=black!20,thick,minimum size=10mm]
\begin{tikzpicture}[scale=0.6]
\node[place] (st) at ( 0,29) {$S_i$};
\node[transition] (as2) at ( 4,25) {$S_{i^'}$};
\node[transition] (as1) at ( -4,25) {$S_{i^'}$};
\foreach \y in {1,2}
\foreach \x in {1,2}
\node[place] (s\y\x) at (8*\y+4*\x-18,21) {$S_i$};
\node[circle,draw=black!50,fill=red!60,thick,minimum size=10mm] (sd) at (-6,21) {$S_j$};
\foreach \y in {1,2}
\draw [->] (st.south) -- (as\y.north);
\foreach \y in {1,2}
\foreach \x in {1,2}
\draw [->] (as\y.south) -- (s\y\x.north);
\node (p1) at(-10,16) {\includegraphics[height=3.0cm,width=3cm]{main/2048pic/2_1.png}};
\node (p2) at(-3,16) {\includegraphics[height=3.0cm,width=3cm]{main/2048pic/2_2.png}};
\node (p3) at(3,16) {\includegraphics[height=3.0cm,width=3cm]{main/2048pic/3_1.png}};
\node (p4) at(10,16) {\includegraphics[height=3.0cm,width=3cm]{main/2048pic/3_2.png}};
\draw [-] (p1.north) -- (s11.south);
\draw [-] (p2.north) -- (s12.south);
\draw [-] (p3.north) -- (s21.south);
\draw [-] (p4.north) -- (s22.south);
\node (p5) at(-7,30) {\includegraphics[height=3.0cm,width=3cm]{main/2048pic/0.png}};
\node (p6) at(10,25) {\includegraphics[height=3.0cm,width=3cm]{main/2048pic/2_0.png}};
\node (p7) at(-10,25) {\includegraphics[height=3.0cm,width=3cm]{main/2048pic/2_0.png}};
\draw [-] (st.west) -- (p5.east);
\draw [-] (as1.west) -- (p7.east);
\draw [-] (as2.east) -- (p6.west);
\end{tikzpicture}
\caption{ 2048的一个例子}
\label{sttree}
\end{figure*}
\ No newline at end of file
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