内部有方程式的 TikZ 矩形

Question

我不太精通 Ti钾Z. 我想做一些我认为非常基本的事情：6 个矩形，里面有公式，跨越 2 列文档中的 2 列。我想要这样的东西

经过几次尝试，我接受了在框外添加文字，我走到了这一步（不是太远）

使用此代码

 \documentclass[conference]{IEEEtran}
    \IEEEoverridecommandlockouts
    % The preceding line is only needed to identify funding in the first footnote. If that is unneeded, please comment it out.
    \usepackage{amsmath,amssymb,amsfonts}
    \usepackage{amsthm}
    \usepackage{algorithmic}
    \usepackage{graphicx}
    \usepackage{textcomp}
    \usepackage{xcolor}
    \usepackage{tikz}
    \usepackage{adjustbox}
    \usetikzlibrary{calc,positioning,mindmap,trees,decorations.pathreplacing}
    \usepackage{dblfloatfix}
    \usepackage{placeins}  
\begin{document}
\begin {figure*}%[!hbtp]
    \centering
    \begin{adjustbox}{width=\textwidth}
        \begin {tikzpicture}[-latex ,auto,node distance =1.8cm and 1.5cm ,on grid,semithick,
        state/.style ={ circle ,top color =white , bottom color = processblue!20,
            draw,processblue , text=blue , minimum width =1 cm}]
    \node[rectangle,draw] (A) 
    {
        $   I_{x}=  \int_{x_1}^{x_2} \int_{x_3}^{x_4} f(|x-x^{\prime}|)~dx^{\prime}dx \,$
    };
    \node[rectangle,draw] (B) [below =of A]  {$\begin{aligned}\label{eq:int_hatx}
    I_{x} &   =\int_{x_1}^{x_2} \int_{x - x_4} ^{x - x_3} f(\widehat{x})~ d\widehat{x}~dx \, .
            \end{aligned}$};
    \node[rectangle,draw] (C) [right =of A] {$\begin{aligned}\label{eq:int_p_single_case1a}
            &\int_{x_1}^{x_2} \int_{x - x_4} ^{x - x_3}f(\widehat{x})~d\widehat{x}~dx = \\
            & \int_{x_1 - x_4} ^{x_2 - x_4} f(\widehat{x}) (\widehat{x} + x_4 - x_1)~ d\widehat{x}  \nonumber \\
            &+(x_2 - x_1) \int_{x_2 - x_4} ^{x_1 - x_3} f(\widehat{x}) ~d\widehat{x} \nonumber \\
            & + \int_{x_1 - x_3} ^{x_2 - x_3} f(\widehat{x})(x_2 - \widehat{x} - x_3)~d\widehat{x} \, \nonumber .  
        \end{aligned}$};
    \node[rectangle,draw] (D) [below =of C] {$blablabla$};
    \node[rectangle,draw] (E) [right =of C] {$blablabla$};
    \node[rectangle,draw] (F) [below =of E] {$blablabla$};
    \path (A) edge [] node {Change of variable $x^{\prime}\to\widehat{x} = x - x^{\prime}$} (B);
    \path (B) edge [] node {Integrate directing with respect to $x$} (C);
        {\tiny }
    \end{tikzpicture}
    \end{adjustbox}
    \end{figure*}
    \end{document}

如能提供一点帮助我将非常感激。

Answer 1

通过使用nccmath（for \medsizemath）和mathtools（for multlinedand \mathrclap）包sk，用于节点定位positioningTi钾Z 库：

\documentclass[conference]{IEEEtran}
\IEEEoverridecommandlockouts
\usepackage{nccmath, mathtools}

\usepackage{tikz}
\usetikzlibrary{arrows.meta,
                positioning,
                quotes}

%---------------- Show page layout. Don't use in a real document!
\usepackage{showframe}
\renewcommand\ShowFrameLinethickness{0.15pt}
\renewcommand*\ShowFrameColor{\color{red}}
%---------------------------------------------------------------%
\usepackage{lipsum}% For dummy text. Don't use in a real document
\begin{document}
    \begin{figure*}
    \centering
\begin{tikzpicture}[
node distance = 12mm and 12mm,
   arr/.style = {draw, -Straight Barb, thick},
   lbl/.style = {font=\footnotesize, align=left, pos=0.75, right},
     N/.style = {draw, semithick, minimum size=22mm},
every edge/.style = {arr},
every edge quotes/.style = {auto, font=\footnotesize, align=left}
                        ]
\node[N] (A)
{
$\medmath{I_{x} = \int_{x_1}^{x_2} 
                    \mathrlap{\int_{x_3}^{x_4}}
                    \quad f(|x-x'|)~dx'~dx}
$
};
\node[N] (B) [below=of A]  
{
$\medmath{I_{x} = \int_{x_1}^{x_2} 
                    \mathrlap{\int_{x-x_4}^{x-x_3}}
                        \quad f(\hat{x})~ d\hat{x}~dx.}
$
};

\node[N] (C) [right=of A] 
{
$\medmath{\begin{multlined}%\label{eq:int_p_single_case1a}
\int_{x_1}^{x_2}
\mathrlap{\int_{x-x_4}^{x-x_3}}
    \quad f(\hat{x}) ~d\hat{x}~dx 
 = \mathrlap{\int_{x_1-x_4}^{x_2-x_4} }
    \quad f(\hat{x}) (\hat{x} + x_4 - x_1)~ d\hat{x}    \\ 
    {} + (x_2 - x_1) 
        \mathrlap{\int_{x_2-x_4}^{x_1-x_3}}
        \quad f(\hat{x}) ~d\hat{x} +
        \mathrlap{\int_{x_1-x_3}^{x_2-x_3}}
        \quad f(\hat{x})(x_2 - \hat{x} - x_3)~d\hat{x} .
    \end{multlined}}
$
};

\node[N] (D) [below=of C] {$\medmath{blablabla}$};
\node[N] (E) [right=of C] {$\medmath{blablabla}$};
\node[N] (F) [below=of E] {$\medmath{blablabla}$};
%%%% arrows
\path   (A) edge["Change\\ $x'\to\hat{x} = x - x'$"] (B)
        (C) edge["???"] (D)
        (E) edge["???"] (F);
\coordinate[left=8mm of C.west] (aux1);
\coordinate[left=8mm of E.west] (aux2);
\draw[arr]  (B) -| (aux1) node[lbl] {Integrate directing\\ 
                                    with respect to $x$}
                -- (C);
\draw[arr]  (D) -| (aux2)
                |- (E);

\end{tikzpicture}
\end{figure*}
\end{document}