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如何使游戏树中的分支平行？

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如何使游戏树中的分支平行？

我创建了以下游戏树： 在此处输入图片描述 这是我的代码：

\documentclass[leqno, oneside, 12pt]{book}
\usepackage{amsmath}
\usepackage{istgame}
\usepackage{caption}
\usepackage{makecell}
\begin{document}
\begin{figure}[ht]
    \centering
    \begin{istgame}[font=\footnotesize]
        \xtdistance{30mm}{90mm}
        \setxtinfosetstyle{dashed}
        \xtShowEndPoints
        \istroot[right](0)<above>{0}
            \istbA{0.95}[above,sloped]
            \istbA{0.05}[above,sloped]
        \endist
        \xtdistance{30mm}{30mm}
        \istroot[right](1)(0-2)<above>{1.1}
            \istbA{g_1\quad(\beta)}[above,sloped]
            \istbA{f_1\quad(1-\beta)}[above,sloped]{0,0}
        \endist
        \xtOwner(1){\makecell{$\langle\alpha\rangle$}}[below]
        \istroot[right](2)(0-1)<above>{1.1}
            \istbA(2){g_1\quad(\beta)}[above,sloped]
            \istbA{f_1\quad(1-\beta)}[above,sloped]{0,0}
        \endist
        \xtOwner(2){\makecell{$\langle1-\alpha\rangle$}}[below]
        \istroot[right](3)(1-1)<above>{2.2}
            \istbA{g_2\quad(\gamma)}[above,sloped]
            \istbA{f_2\quad(1-\gamma)}[above,sloped]{-1,5}
        \endist
        \xtOwner(3){\makecell{$\langle1\rangle$}}[below]
        \istroot[right](4)(3-1)<above>{1.3}
            \istbA{g_3\quad(\epsilon)}[above,sloped]
            \istbA{f_3\quad(1-\epsilon)}[above,sloped]{4,4}
        \endist
        \xtOwner(4){\makecell{$\langle\delta\rangle$}}[below]
        \istroot[right](5)(2-1)<above>{1.3}
            \istbA(2){g_3\quad(\epsilon)}[above,sloped]{8,8}
            \istbA{f_3\quad(1-\epsilon)}[above,sloped]{4,4}
        \endist
        \xtOwner(5){\makecell{$\langle1-\delta\rangle$}}[below]
        \istroot[right](6)(4-1)<above>{2.4}
            \istbA{g_4\quad(\zeta)}[above,sloped]{8,8}
            \istbA{f_4\quad(1-\zeta)}[above,sloped]{3,9}
        \endist
        \xtOwner(6){\makecell{$\langle1\rangle$}}[below]
        \xtInfosetO(1)(2)(1.2cm)
        \xtInfosetO(3)(3)(1.2cm)
        \xtInfosetO(4)(5)(1.2cm)
        \xtInfosetO(6)(6)(1.2cm)
    \end{istgame}
    \caption{}
\end{figure}
\end{document}

我想让每个向上的分支彼此平行，每个向下的分支也彼此平行。（如您所见，现在向下的线并不平行。）同时，两个 1.1 节点需要对齐，两个 1.3 节点也需要对齐。我该怎么做？非常感谢任何帮助！

答案1

我对这个包不是很熟悉，但是我查看了文档，似乎没有办法指定替代（或弧）的角度\istbA；只有像你所做的那样，一种方法来指定子项的扩展。

为了诊断目的，我制作了下图： 原树，剥去

\documentclass[tikz,margin=10pt]{standalone}
\usepackage{istgame}
\usepackage{makecell}

\begin{document}
    \begin{istgame}
        \xtdistance{30mm}{90mm}
        \setxtinfosetstyle{dashed}
        \xtShowEndPoints

        \setistgrowdirection'{east}
        
        \istroot(0)<above>{0}
            \istbA{to:0-1}[above,sloped]
            \istbA{to:0-2}[above,sloped]
        \endist
        
        \xtdistance{30mm}{30mm}
        
        \istroot(1)(0-1)<above>{1}
            \istbA{to:1-1}[above,sloped]{end}
            \istbA{to:1-2}[above,sloped]
        \endist
        \xtOwner(1){\makecell{1label}}[below]
        
        \istroot(2)(0-2)<above>{2}
            \istbA{to:2-1}[above,sloped]{end}
            \istbA(2){to:2-2}[above,sloped]
        \endist
        \xtOwner(2){\makecell{2label}}[below]
        
        \istroot(3)(1-2)<above>{3}
            \istbA{to:3-1}[above,sloped]{end}
            \istbA{to:3-2}[above,sloped]
        \endist
        \xtOwner(3){\makecell{3label}}[below]
        
        \istroot(4)(3-2)<above>{4}
            \istbA{to:4-1}[above,sloped]{end}
            \istbA{to:4-2}[above,sloped]
        \endist
        \xtOwner(4){\makecell{4label}}[below]
        
        \istroot(5)(2-2)<above>{5}
            \istbA{to:5-1}[above,sloped]{end}
            \istbA(2){to:5-2}[above,sloped]{end}
        \endist
        \xtOwner(5){\makecell{5label}}[below]
        
        \istroot(6)(4-2)<above>{6}
            \istbA{to:6-1}[above,sloped]{end}
            \istbA{to:6-2}[above,sloped]{end}
        \endist
        \xtOwner(6){\makecell{6label}}[below]
        
        \xtInfosetO(1)(2)(1.2cm)
        \xtInfosetO(3)(3)(1.2cm)
        \xtInfosetO(4)(5)(1.2cm)
        \xtInfosetO(6)(6)(1.2cm)
    \end{istgame}
\end{document}

有一种方法可以添加“缺失”子元素，简称为\istb<missing>或\istbm。不幸的是，很难说出 的角度是如何\istbA(<level>)计算的，因此即使使用这种简化的模型，尝试通过添加这些缺失的子元素来对齐底行也显得太过繁琐。

相反，如果你愿意牺牲对称性，我提供这个解决方案： 替代树，剥离

\documentclass[tikz,margin=10pt]{standalone}
\usepackage{istgame}
\usepackage{makecell}

\begin{document}
    \begin{istgame}
        \xtdistance{30mm}{60mm}
        \setxtinfosetstyle{dashed}
        \xtShowEndPoints

        \setistgrowdirection'{east}
        
        \istroot(0)<above>{0}
            \istbA{to:0-1}[above,sloped]
            \istbA{to:0-2}[above,sloped]
        \endist
        
        \xtdistance{30mm}{30mm}
        
        \istroot(1)(0-1)<above>{1}
            \istbA{to:1-1}[above,sloped]{end}
            \istbA{to:1-2}[above,sloped]
            \istbm
        \endist
        \xtOwner(1){\makecell{1label}}[below]
        
        \istroot(2)(0-2)<above>{2}
            \istbA{to:2-1}[above,sloped]{end}
            \istbA(2){to:2-2}[above,sloped]
            \istbm
        \endist
        \xtOwner(2){\makecell{2label}}[below]
        
        \istroot(3)(1-2)<above>{3}
            \istbA{to:3-1}[above,sloped]{end}
            \istbA{to:3-2}[above,sloped]
            \istbm
        \endist
        \xtOwner(3){\makecell{3label}}[below]
        
        \istroot(4)(3-2)<above>{4}
            \istbA{to:4-1}[above,sloped]{end}
            \istbA{to:4-2}[above,sloped]
            \istbm
        \endist
        \xtOwner(4){\makecell{4label}}[below]
        
        \istroot(5)(2-2)<above>{5}
            \istbA{to:5-1}[above,sloped]{end}
            \istbA(2){to:5-2}[above,sloped]{end}
            \istbm
        \endist
        \xtOwner(5){\makecell{5label}}[below]
        
        \istroot(6)(4-2)<above>{6}
            \istbA{to:6-1}[above,sloped]{end}
            \istbA{to:6-2}[above,sloped]{end}
            \istbm
        \endist
        \xtOwner(6){\makecell{6label}}[below]
        
        \xtInfosetO(1)(2)(1.2cm)
        \xtInfosetO(3)(3)(1.2cm)
        \xtInfosetO(4)(5)(1.2cm)
        \xtInfosetO(6)(6)(1.2cm)
    \end{istgame}
\end{document}

填写后看起来像 在此处输入图片描述

同样，这不是理想的选择，但如果您的目标是并行，那么这样做就可以了。我建议添加[scale=1.2]或类似于的\begin{istgame}扩展，但这是个人喜好。在下面的 MWE 中，您会注意到我删除了标签[right]并添加了标签\setistgrowdirection'{east}，这会使节点在代码中以合理的方式从上到下排列，至少在我看来是这样。

希望这能有点帮助！

梅威瑟：

\documentclass[tikz,margin=10pt]{standalone}
\usepackage{istgame}
\usepackage{makecell}

\begin{document}
    \begin{istgame}
        \xtdistance{30mm}{60mm}
        \setxtinfosetstyle{dashed}
        \xtShowEndPoints

        \setistgrowdirection'{east}
        
        \istroot(0)<above>{0}
            \istbA{0.05}[above,sloped]
            \istbA{0.95}[above,sloped]
        \endist
        
        \xtdistance{30mm}{30mm}
        
        \istroot(1)(0-1)<above>{1.1}
            \istbA{f_1\quad(1-\beta)}[above,sloped]{0,0}
            \istbA{g_1\quad(\beta)}[above,sloped]
            \istbm
        \endist
        \xtOwner(1){\makecell{$\langle\alpha\rangle$}}[below]
        
        \istroot(2)(0-2)<above>{1.1}
            \istbA{f_1\quad(1-\beta)}[above,sloped]{0,0}
            \istbA(2){g_1\quad(\beta)}[above,sloped]
            \istbm
        \endist
        \xtOwner(2){\makecell{$\langle1-\alpha\rangle$}}[below]
        
        \istroot(3)(1-2)<above>{2.2}
            \istbA{f_2\quad(1-\gamma)}[above,sloped]{-1,5}
            \istbA{g_2\quad(\gamma)}[above,sloped]
            \istbm
        \endist
        \xtOwner(3){\makecell{$\langle1\rangle$}}[below]
        
        \istroot(4)(3-2)<above>{1.3}
            \istbA{f_3\quad(1-\epsilon)}[above,sloped]{4,4}
            \istbA{g_3\quad(\epsilon)}[above,sloped]
            \istbm
        \endist
        \xtOwner(4){\makecell{$\langle\delta\rangle$}}[below]
        
        \istroot(5)(2-2)<above>{1.3}
            \istbA{f_3\quad(1-\epsilon)}[above,sloped]{4,4}
            \istbA(2){g_3\quad(\epsilon)}[above,sloped]{8,8}
            \istbm
        \endist
        \xtOwner(5){\makecell{$\langle1-\delta\rangle$}}[below]
        
        \istroot(6)(4-2)<above>{2.4}
            \istbA{f_4\quad(1-\zeta)}[above,sloped]{3,9}
            \istbA{g_4\quad(\zeta)}[above,sloped]{8,8}
            \istbm
        \endist
        \xtOwner(6){\makecell{$\langle1\rangle$}}[below]
        
        \xtInfosetO(1)(2)(1.2cm)
        \xtInfosetO(3)(3)(1.2cm)
        \xtInfosetO(4)(5)(1.2cm)
        \xtInfosetO(6)(6)(1.2cm)
    \end{istgame}
\end{document}

答案2

我想使用\istb而不是使用\istbA。

你可能想替换\istbA(2)为

\istbA(2)<grow=-27,level distance=67mm>

或者简单地使用

\istb<grow=-27,level distance=67mm>。

这种改变将产生您想要的（近似）结果。

在此处输入图片描述

\documentclass[leqno, oneside, 12pt]{book}
\usepackage{amsmath}
\usepackage{istgame}
\usepackage{caption}
\usepackage{makecell}

\begin{document}
\begin{figure}[ht]
\centering
\begin{istgame}[font=\footnotesize]
    %%% some calculation:
    \pgfmathparse{90-atan(30/15)}  
        \node(0,0)[draw,yshift=10mm]{\pgfmathresult};
    \pgfmathparse{sqrt(30^2+15^2)} 
        \node(0,0)[draw,yshift=15mm]{\pgfmathresult};
    %%% tree direction
    \setistgrowdirection{east}     %%% instead of using [right] in \istroot
    %\setistgrowdirection'{east}   %%% (swap version) looks more convenient

    %%% \istb (instead of \istbA)
    \xtdistance{30mm}{90mm}
    \setxtinfosetstyle{dashed}
    \xtShowEndPoints
    \istroot(0)<above>{0}
        \istb{0.95}[above,sloped]
        \istb{0.05}[above,sloped]
    \endist
    \xtdistance{30mm}{30mm}
    \istroot(1)(0-2)<above>{1.1}
        \istb{g_1\quad(\beta)}[above,sloped]
        \istb{f_1\quad(1-\beta)}[above,sloped]{0,0}
    \endist
    \xtOwner(1){\makecell{$\langle\alpha\rangle$}}[below]
    \istroot(2)(0-1)<above>{1.1}
        \istb<grow=-27,level distance=67mm>[blue,thick]  %%% CHANGED
             {g_1\quad(\beta)}[above,sloped]
        \istb{f_1\quad(1-\beta)}[above,sloped]{0,0}
    \endist
    \xtOwner(2){\makecell{$\langle1-\alpha\rangle$}}[below]
    \istroot(3)(1-1)<above>{2.2}
        \istb{g_2\quad(\gamma)}[above,sloped]
        \istb{f_2\quad(1-\gamma)}[above,sloped]{-1,5}
    \endist
    \xtOwner(3){\makecell{$\langle1\rangle$}}[below]
    \istroot(4)(3-1)<above>{1.3}
        \istb{g_3\quad(\epsilon)}[above,sloped]
        \istb{f_3\quad(1-\epsilon)}[above,sloped]{4,4}
    \endist
    \xtOwner(4){\makecell{$\langle\delta\rangle$}}[below]
    \istroot(5)(2-1)<above>{1.3}
        \istb<grow=-27,level distance=67mm>[blue,thick]  %%% CHANGED
             {g_3\quad(\epsilon)}[above,sloped]{8,8}
        \istb{f_3\quad(1-\epsilon)}[above,sloped]{4,4}
    \endist
    \xtOwner(5){\makecell{$\langle1-\delta\rangle$}}[below]
    \istroot(6)(4-1)<above>{2.4}
        \istb{g_4\quad(\zeta)}[above,sloped]{8,8}
        \istb{f_4\quad(1-\zeta)}[above,sloped]{3,9}
    \endist
    \xtOwner(6){\makecell{$\langle1\rangle$}}[below]
    \xtInfosetO(1)(2)(1.2cm)
    \xtInfosetO(3)(3)(1.2cm)
    \xtInfosetO(4)(5)(1.2cm)
    \xtInfosetO(6)(6)(1.2cm)
\end{istgame}
\caption{}
\end{figure}
\end{document}

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