昕闻画出了两个2048游戏图

main/2048isAcyclic.tex

@@ -77,12 +77,28 @@ For all possible board,
 For any transition $\langle B_1, a, B_1', B_2\rangle$ in the 2048 game,
 we have 
 $sum(B_1)=sum(B_1')$  regardless of whether at least two tiles merge.
-\highlight{需要昕闻画2048的状态迁移图，同一个状态包括两种后续情况，
-一种是发生合并，一种是没发生合并的}
-
-
 Due to a new generated 2-tile or 4-tile in board $B_2$,
 $sum(B_2)>sum(B_1')$, that is $sum(B_2)>sum(B_1)$.
+Figure \ref{2048merge1} and Figure \ref{2048merge2} show transition
+examples with and without a merge. 
+
+\begin{figure}[ht]
+    \begin{center}
+	\include{pic/2048merge1}
+	\caption{2048 transition  example with a merge, where $sum(B_1)=sum(B_1')=46$,
+	 $sum(B_2)=48>sum(B_1')$.}
+    \label{2048merge1}
+    \end{center}
+\end{figure}
+\begin{figure}[ht]
+    \begin{center}
+	\include{pic/2048merge2}
+	\caption{2048 transition  example without a merge, where $sum(B_1)=sum(B_1')=62$,
+	 $sum(B_2)=64>sum(B_1')$.}
+    \label{2048merge2}
+    \end{center}
+\end{figure}
+
 
 Based on (\ref{size}) and (\ref{utility}),
 we have  $u(B_2)>u(B_1)$.

pic/2048merge1.tikz → pic/2048merge1.tex

-\begin{tikzpicture}[scale=0.75,transform shape]
+\begin{tikzpicture}[scale=0.5,transform shape]
     % Draw the grid
     \foreach \i in {0,...,3} {
         \foreach \j in {0,...,3} {
@@ -43,7 +43,7 @@
     \draw[thick, black,->] (2,4) -- (2,5) -- (8,5) -- (8,4) ;
     \draw[thick, black,->] (8,0) -- (8,-1) -- (14,-1) -- (14,0) ;
 
-    \node at (5,4.5) {$r=32$};
+    \node at (5,4.5) {$a=$\textit{right}, $r=32$};
     \node at (11.5,-0.5) {random tile};
 
     \node at (-0.5,2) {$B_1=$};

pic/2048merge2.tikz → pic/2048merge2.tex

-\begin{tikzpicture}[scale=0.75,transform shape]
+\begin{tikzpicture}[scale=0.5,transform shape]
     % Draw the grid
     \foreach \i in {0,...,3} {
         \foreach \j in {0,...,3} {
@@ -43,7 +43,7 @@
     \draw[thick, black,->] (2,4) -- (2,5) -- (8,5) -- (8,4) ;
     \draw[thick, black,->] (8,0) -- (8,-1) -- (14,-1) -- (14,0) ;
 
-    \node at (5,4.5) {$r=0$};
+    \node at (5,4.5) {$a=$\textit{right}, $r=0$};
     \node at (11.5,-0.5) {random tile};
 
     \node at (-0.5,2) {$B_1=$};