Commit abfd66ae by Lenovo

架构已经搭好,需要李昕闻补充

parent 76a6eacc
......@@ -75,6 +75,7 @@ wangwenhao11@nudt.edu.cn).
\input{main/introduction}
\input{main/background}
\input{main/nonergodic}
\input{main/2048isNonergodic}
%\input{main/nonergodicity}
%\input{main/paradox}
......
\section{Non-ergodicity of 2048}
\begin{theorem}
2048 game is non-ergodic between non-absorbing states.
\end{theorem}
\begin{IEEEproof}
\end{IEEEproof}
......@@ -54,9 +54,9 @@ reaching the leftmost or rightmost node where it terminates.
The terminal states are usually called absorbing states.
The transition probobility matrix
of random walk with absorbing states
$P_{\text{absorbing}}$ is defined as follows:
$P_{\text{ab}}$ is defined as follows:
\[
P_{\text{absorbing}}\dot{=}\begin{array}{c|ccccccc}
P_{\text{ab}}\dot{=}\begin{array}{c|ccccccc}
&\text{T} & \text{A} & \text{B} & \text{C} & \text{D} & \text{E} \\\hline
\text{T} & 1 & 0 & 0 & 0 & 0 & 0 \\
\text{A} & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 & 0 \\
......@@ -122,6 +122,7 @@ where $Q$ is the matrix of transition probabilities between
is
\begin{equation}
N\dot{=} \sum_{i=0}^{\infty}Q^i=(I_{n-1}-Q)^{-1},
\label{definitionN}
\end{equation}
where $I_{n-1}$ is the $(n-1)\times(n-1)$ identity matrix.
It is now easy to define whether the non-absorbing states
......@@ -139,15 +140,15 @@ Assume that $N$ exists for any policy $\pi$
For random walk with absorbing states,
\[
P_{\text{absorbing}} =
P_{\text{ab}} =
\begin{bmatrix}
Q_{\text{absorbing}} & R_{\text{absorbing}} \\
0 & I_{\text{absorbing}}
Q_{\text{ab}} & R_{\text{ab}} \\
0 & I_{\text{ab}}
\end{bmatrix},
\]
where
\[
Q_{\text{absorbing}}\dot{=}\begin{array}{c|ccccc}
Q_{\text{ab}}\dot{=}\begin{array}{c|ccccc}
& \text{A} & \text{B} & \text{C} & \text{D} & \text{E} \\\hline
\text{A} & 0 & \frac{1}{2} & 0 & 0 & 0 \\
\text{B} & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\
......@@ -175,7 +176,7 @@ Q_{\text{absorbing}}\dot{=}\begin{array}{c|ccccc}
Then,
\[
N_{\text{absorbing}}=(I_5-Q_{\text{absorbing}})^{-1}=\begin{array}{c|ccccc}
N_{\text{ab}}=(I_5-Q_{\text{ab}})^{-1}=\begin{array}{c|ccccc}
& \text{A} & \text{B} & \text{C} & \text{D} & \text{E} \\\hline
\text{A} & \frac{5}{3} & \frac{4}{3} & 1 & \frac{2}{3} & \frac{1}{3} \\
\text{B} & \frac{4}{3} & \frac{8}{3} & 2 & \frac{4}{3} & \frac{2}{3} \\
......
......@@ -62,7 +62,7 @@ Figure \ref{TruncatedPetersburg} is a truncated version
of the St. Petersburg paradox. The transition probabilities between
non-absorbing states are as follows:
\[
Q_{\text{truncated}}\dot{=}\begin{array}{c|ccccc}
Q_{\text{st}}\dot{=}\begin{array}{c|ccccc}
& \text{S}_1 & \text{S}_2 & \text{S}_3 & \text{S}_4 & \text{S}_5 \\\hline
\text{S}_1 & 0 & \frac{1}{2} & 0 & 0 & 0 \\
\text{S}_2 & 0 & 0 & \frac{1}{2} & 0 & 0 \\
......@@ -73,7 +73,7 @@ Q_{\text{truncated}}\dot{=}\begin{array}{c|ccccc}
\]
Then,
\[
N_{\text{truncated}}=(I_5-Q_{\text{truncated}})^{-1}=\begin{array}{c|ccccc}
N_{\text{st}}=(I_5-Q_{\text{st}})^{-1}=\begin{array}{c|ccccc}
& \text{S}_1 & \text{S}_2 & \text{S}_3 & \text{S}_4 & \text{S}_5 \\\hline
\text{S}_1 & 1 & \frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \frac{1}{16} \\
\text{S}_2 & 0 & 1 & \frac{1}{2} & \frac{1}{4} & \frac{1}{8} \\
......@@ -86,6 +86,71 @@ Bases on Definition \ref{definition3},
the truncated St. Petersburg paradox
is non-ergodic between non-absorbing states.
\subsection{A sufficient condition for non-ergodicity between non-absorbing states}
Based on the truncated St. Petersburg paradox,
it is easy to provide a sufficient condition for non-ergodicity between non-absorbing states.
\begin{theorem}[A sufficient condition for non-ergodicity between non-absorbing states]
Given a Markov chain with absorbing states,
suppose the size of the non-absorbing states $|S\setminus\{\text{T}\}|\geq 2$.
If the transition probabilities $Q$ between non-absorbing states satifies,
$\forall i,j \in S\setminus\{\text{T}\}$,
\begin{equation}
Q_{i,j}=\begin{cases}
0, & \text{if } i\leq j; \\
\geq 0, & \text{otherwise.}
\end{cases}
\label{condition}
\end{equation}
Then, the Markov chain is non-ergodic between non-absorbing states.
\end{theorem}
\begin{IEEEproof}
Based on the assumption, the $Q$ matrix is an upper triangular matrix.
The product of two upper triangular matrices is still an upper triangular matrix.
Furthermore, the sum of two upper triangular matrices
is still an upper triangular matrix.
Based on Definition \ref{definitionN},
the $N$ matrix is product and sum of upper triangular matrices.
Then, the $N$ matrix is an upper triangular matrix.
The claim now follows based on Definition \ref{definition3}.
\end{IEEEproof}
......
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