diff --git a/document.tex b/document.tex index f2f4237..d23d748 100644 --- a/document.tex +++ b/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} diff --git a/main/2048pic/0.png b/main/2048pic/0.png new file mode 100644 index 0000000..f1e7f79 Binary files /dev/null and b/main/2048pic/0.png differ diff --git a/main/2048pic/2_0.png b/main/2048pic/2_0.png new file mode 100644 index 0000000..fa40579 Binary files /dev/null and b/main/2048pic/2_0.png differ diff --git a/main/2048pic/2_1.png b/main/2048pic/2_1.png new file mode 100644 index 0000000..cc9a6a5 Binary files /dev/null and b/main/2048pic/2_1.png differ diff --git a/main/2048pic/2_2.png b/main/2048pic/2_2.png new file mode 100644 index 0000000..5cbdb15 Binary files /dev/null and b/main/2048pic/2_2.png differ diff --git a/main/2048pic/3_0.png b/main/2048pic/3_0.png new file mode 100644 index 0000000..69bb6f0 Binary files /dev/null and b/main/2048pic/3_0.png differ diff --git a/main/2048pic/3_1.png b/main/2048pic/3_1.png new file mode 100644 index 0000000..8355006 Binary files /dev/null and b/main/2048pic/3_1.png differ diff --git a/main/2048pic/3_2.png b/main/2048pic/3_2.png new file mode 100644 index 0000000..b049b93 Binary files /dev/null and b/main/2048pic/3_2.png differ diff --git a/main/nonergodicity.tex b/main/nonergodicity.tex index a9abafd..892bb1b 100644 --- a/main/nonergodicity.tex +++ b/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} + + + + diff --git a/main/paradox.tex b/main/paradox.tex index 05e507d..3339ab4 100644 --- a/main/paradox.tex +++ b/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} + diff --git a/main/theorem.tex b/main/theorem.tex index c0ead7d..57822ed 100644 --- a/main/theorem.tex +++ b/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} ] + diff --git a/pic/2048epsilon-greedy.pdf b/pic/2048epsilon-greedy.pdf new file mode 100644 index 0000000..2e2f49c Binary files /dev/null and b/pic/2048epsilon-greedy.pdf differ diff --git a/pic/2048example.tex b/pic/2048example.tex new file mode 100644 index 0000000..fd2a7ab --- /dev/null +++ b/pic/2048example.tex @@ -0,0 +1,53 @@ +\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 diff --git a/pic/migongeps-greedy.pdf b/pic/migongeps-greedy.pdf new file mode 100644 index 0000000..7535168 Binary files /dev/null and b/pic/migongeps-greedy.pdf differ