Tikz：将多个边连接到矩形节点的一侧

Question

以下是我正在尝试做的事情的快速近似/破解：

但我希望顶部看起来更像这样

问题是我不知道矩形的顶部在哪里，也不知道如何将节点附加到顶部。也许还有其他方法可以做到这一点，或者有一种在 tikz 中制作字符串图的通用方法？我希望解决方案是相当模块化的，因为我将制作一个很多并将它们粘合在一起。矩形中的证明是使用总线证明（使用选项）制作的\EnableBpAbbreviations。

图1：

  \begin{tikzpicture}
    \node (g1) at (0, 5) [label = above: $\Gamma_1$] {};
    \node (g2) at (4, 5) [label = above: $\Gamma_2$] {};
    \node (atb) at (2,5) [label = above: $A \otimes B$] {};
    \node (t) at (2,4) [circle,draw,inner sep = 0pt] {$\otimes$};
    \node (box) at (2,2) [rectangle,draw] {
      \AXC{$D_i$}
      \UIC{$f:\Gamma_1 , A , B , \Gamma_2 \vdash \Delta $}
      \DisplayProof
    };
    \node (d) at (2,0) [label = below: $\Delta$] {};

    \draw  (g1) to [out = 270, in = 150] (box) ;
    \draw  (g2) to [out = 270, in = 30] (box) ;
    \draw  (atb) to (t) ;
    \draw (t) to [out = 225, in = 120] (box);
    \draw (t) to [out = 315, in = 60] (box);
    \draw  (box) to (d) ;
  \end{tikzpicture}

图2：

  \begin{tikzpicture}
    \node (g1) at (0, 2) [label = above: $\Gamma_1$] {};
    \node (g2) at (4, 2) [label = above: $\Gamma_2$] {};
    \node (atb) at (2,2) [label = above: $A \otimes B$] {};
    \node (t) at (2,1) [circle,draw,inner sep = 0pt] {$\otimes$};
    \node (g1x) at (0,0) {};
    \node (g2x) at (4,0) {};
    \node (a) at (1,0) {};
    \node (b) at (3,0) {};

    \draw  (g1) to (g1x) ;
    \draw  (g2) to (g2x) ;
    \draw  (atb) to (t) ;
    \draw (t) to [out = 225, in = 90] (a);
    \draw (t) to [out = 315, in = 90] (b);
  \end{tikzpicture}

Answer 1

我将先从盒子开始，然后使用库绘制相对于它的其余部分positioning。

\documentclass[border=3.14mm,tikz]{standalone}
\usetikzlibrary{positioning}
\usepackage{bussproofs}
\EnableBpAbbreviations 
\begin{document}
\begin{tikzpicture}
% start with the box
    \node (box) at (2,2) [rectangle,draw] {
      \AXC{$D_i$}
      \UIC{$f:\Gamma_1 , A , B , \Gamma_2 \vdash \Delta $}
      \DisplayProof
    };
% add the rest relative to the box  
    \node[above=1.5cm of box] (t) [circle,draw,inner sep = 0pt] {$\otimes$};
% as you are working with labels, you could replace the nodes by coordinates    
    \path (box.north west) -- (box.north east) coordinate[pos=0.25] (t1)
    coordinate[pos=0.75] (t2) coordinate[pos=0.02] (t0) coordinate[pos=0.98] (t3);
    \draw (t0) -- ++(0,3cm) coordinate[label = above: $\Gamma_1$] (g1){};
    \draw (t3) -- ++(0,3cm) coordinate[label = above: $\Gamma_2$] (g2){};
    \node[above=2.5cm of box] (atb) [label = above: $A \otimes B$] {};
    \node[below=1.5cm of box] (d)  [label=below:$\Delta$] {};
    \draw  (atb) to (t) ;
    \draw (t) to [out = 225, in = 90,looseness=1.2] (t1);
    \draw (t) to [out = 315, in = 90,looseness=1.2] (t2);
    \draw  (box) to (d) ;
\end{tikzpicture}
\end{document}  

Answer 2

另一种方法是使用lualatex其内置的元帖子图书馆，有unicode-math。

\documentclass[border=5mm]{standalone}
\usepackage{luatex85}
\usepackage{luamplib}
\usepackage{unicode-math}
\usepackage{bussproofs}
\EnableBpAbbreviations
\begin{document}
\mplibtextextlabel{enable}
\begin{mplibcode}
beginfig(1);
    z0 = origin;
    x0 = x12 = x4;
    y12 = y31 = y32 = -y4 = 55;
    -x31 = x32 = 42;

    z1 = 5/8 [z0, z12];

    picture P[];

    P0  = thelabel("\AXC{$D_i$}\UIC{$f : Γ_1, A, B, Γ_2 ⊢ Δ$}\DisplayProof", z0);
    P12 = thelabel("$A ⊗ B$", z12);
    P31 = thelabel("$Γ_1$", z31);
    P32 = thelabel("$Γ_2$", z32);
    P4  = thelabel("$Δ$", z4);

    interim bboxmargin := 4pt;
    drawoptions(withcolor 1/4[1/2 blue, white]);

    draw point 1/2 of bbox P32 {down} .. point 15/7 of bbox P0 {down};
    draw point 1/2 of bbox P31 {down} .. point 20/7 of bbox P0 {down};

    draw point 1/2 of bbox P12 .. z1 {down} .. point 16/7 of bbox P0 {down};
    draw point 1/2 of bbox P12 .. z1 {down} .. point 19/7 of bbox P0 {down};

    draw point 1/2 of bbox P0 .. point 5/2 of bbox P4;

    draw bbox P0;

    unfill fullcircle scaled 10 shifted z1;
    draw fullcircle scaled 10 shifted z1;
    drawoptions();

    draw P0; 
    draw P12;
    draw P31;
    draw P32;
    draw P4;
    label("$⊗$", z1);
endfig;
\end{mplibcode}
\end{document}