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如何在表格中拟合长方程

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如何在表格中拟合长方程

我有一组数值优化的基准问题，我想将它们汇编成表格。该表由函数名称、函数定义、搜索范围和理论全局最优值组成。问题是方程式不适合表格。我使用的是 tabularx，如 MWE 中所示：

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{tabularx}
\usepackage{geometry}

\geometry{
    a4paper,
    left=30mm,
    right=25mm,
    top=30mm,
    bottom=30mm,
    asymmetric
}

\begin{document}

\begin{table}
    \begin{tabularx}{\textwidth}{|l|X|X|X|}
        \hline
        \textbf{Name} & \textbf{Definition} & \textbf{Search range} & \textbf{Global optimum}\\
        \hline
        Sphere & $f_1(\vec{x}) = \sum_{i=1}^{D} x_i^2$ & $[-5.12, 5.12]^D$ & $f_1(\vec{0}) = 0$\\
        \hline
        Rosenbrock & $f_2(\vec{x}) = \sum_{i=1}^{D-1} (100 (x_{i+1} - x_i^2)^2 + (x_i - 1)^2) $ & $[-2.048, 2.048]^D$ & $f_2(\vec{1}) = 0$\\
        \hline
        Rastrigin & $f_3(\vec{x}) = 10D + \sum_{i=1}^D \left(x_i^2 -10\cos(2\pi x_i)\right) $ & $[-5.12, 5.12]^D$ & $f_3(\vec{0}) = 0$\\
        \hline
        Griewank & $f_4(\vec{x}) = \sum_{i=1}^D \frac{x_i^2}{4000} - \prod_{i=1}^D \cos(\frac{x_i}{\sqrt{i}}) + 1 $ & $[-600, 600]^D$ & $f_4(\vec{0}) = 0$\\
        \hline
        Ackley & $f_5(\vec{x}) = -a\;\exp\left(-b \sqrt{\frac{1}{D} \sum_{i=1}^D x_i^2}\right) - \exp\left(\frac{1}{D} \sum_{i=1}^D \cos(c\;x_i)\right) + a + \exp(1) $& $[-32.768, 32.768]^D$ & $f_5(\vec{0}) = 0$\\
        \hline
        Schwefel & $f_6(\vec{x}) = 418.9829d - \sum_{i=1}^{D} x_i \sin(\sqrt{\lvert x_i \rvert}) $ & $[-500, 500]^D$ & $f_6(\vec{420.9687}) = 0$\\
        \hline
        Alpine & $f_7(\vec{x}) = \sum_{i=1}^{D} \lvert x_i \sin(x_i)+0.1x_i \rvert $ & $[-10, 10]^D$ & $f_7(\vec{0}) = 0$\\
        \hline
        Whitley & $f_8(\vec{x}) = \sum_{i=1}^D \sum_{j=1}^D \left(\frac{(100(x_i^2-x_j)^2 + (1-x_j)^2)^2}{4000} - \cos(100(x_i^2-x_j)^2 + (1-x_j)^2)+1\right) $ & $[-10, 10]^D$ & $f_8(\vec{1}) = 0$\\
        \hline
        Csendes & $f_9(\vec{x}) = \sum_{i=1}^D x_i^6\left( 2 + \sin \frac{1}{x_i}\right) $ & $[-1, 1]^D$ & $f_9(\vec{0}) = 0$\\
        \hline
        Dixon Price & $f_{10}(\vec{x}) = (x_1 - 1)^2 + \sum_{i = 2}^D i (2x_i^2 - x_{i - 1})^2 $ & $[-10, 10]^D$ & $f_{10}(x^*) = 0 \; x_i = 2^{-\frac{2^i - 2}{2^i}} $\\
        \hline
    \end{tabularx}
\end{table}

\end{document}

它产生了这样的混乱： 在此处输入图片描述

有没有办法让所有的数学运算都适合单元格，并使这个表格看起来更美观？过去几个小时我一直在为此烦恼。我还尝试使用 lscape 包将其置于横向，并将宽度设置为 24 厘米，但仍然不适合。任何帮助都将不胜感激。

答案1

您可以考虑非表格布局

在此处输入图片描述

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\begin{description}
\item [sphere] Search range $[-5.12, 5.12]^D$,  Global optimum $f_1(\vec{0}) = 0$
\[f_1(\vec{x}) = \sum_{i=1}^{D} x_i^2\]
\item[Rosenbrock] Search Range $[-2.048, 2.048]^D$, Global optimum $f_2(\vec{1}) = 0$
\[f_2(\vec{x}) = \sum_{i=1}^{D-1} (100 (x_{i+1} - x_i^2)^2 + (x_i - 1)^2)\]
\item[Rastrigin] Search Range $[-5.12, 5.12]^D$, Global optimum  $f_3(\vec{0}) = 0$
\[f_3(\vec{x}) = 10D + \sum_{i=1}^D \left(x_i^2 -10\cos(2\pi x_i)\right) \]
\item[Griewank] Search Range $[-600, 600]^D$, Global optimum $f_4(\vec{0}) = 0$
\[ f_4(\vec{x}) = \sum_{i=1}^D \frac{x_i^2}{4000} - \prod_{i=1}^D \cos(\frac{x_i}{\sqrt{i}}) + 1 \]
\item[Ackley] Search Range $[-32.768, 32.768]^D$, Global optimum $f_5(\vec{0}) = 0$
\[ f_5(\vec{x}) = -a\;\exp\left(-b \sqrt{\frac{1}{D} \sum_{i=1}^D x_i^2}\right) - \exp\left(\frac{1}{D} \sum_{i=1}^D \cos(c\;x_i)\right) + a + \exp(1)  \]
\item[Schwefel] Search Range $[-500, 500]^D$, Global optimum $f_6(\vec{420.9687}) = 0$
\[ f_6(\vec{x}) = 418.9829d - \sum_{i=1}^{D} x_i \sin(\sqrt{\lvert x_i \rvert})  \]
\item[Alpine] Search Range $[-10, 10]^D$, Global optimum $f_7(\vec{0}) = 0$
\[ f_7(\vec{x}) = \sum_{i=1}^{D} \lvert x_i \sin(x_i)+0.1x_i \rvert  \]
\item[Whitley] Search Range $[-10, 10]^D$, Global optimum $f_8(\vec{1}) = 0$
\[ f_8(\vec{x}) = \sum_{i=1}^D \sum_{j=1}^D \left(\frac{(100(x_i^2-x_j)^2 + (1-x_j)^2)^2}{4000} - \cos(100(x_i^2-x_j)^2 + (1-x_j)^2)+1\right)  \]
\item[Csendes] Search Range $[-1, 1]^D$, Global optimum $f_9(\vec{0}) = 0$
\[ f_9(\vec{x}) = \sum_{i=1}^D x_i^6\left( 2 + \sin \frac{1}{x_i}\right) \]
\item[Dixon Price] Search Range $[-10, 10]^D$, Global optimum $f_{10}(x^*) = 0 \; x_i = 2^{-\frac{2^i - 2}{2^i}} $
\[ f_{10}(\vec{x}) = (x_1 - 1)^2 + \sum_{i = 2}^D i (2x_i^2 - x_{i - 1})^2  \]
\end{description}

\end{document}

答案2

以下是其他一些替代方案：

在此处输入图片描述

在此处输入图片描述

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{tabularx}
\usepackage{geometry}

\geometry{
    a4paper,
    left=30mm,
    right=25mm,
    top=30mm,
    bottom=30mm,
    asymmetric
}
\usepackage{pdflscape}
\usepackage{enumitem}
\usepackage{booktabs}

\begin{document}
\begin{landscape}
\begin{table}
    \begin{tabular}{|l|l|l|l|}
        \hline
        \textbf{Name} & \textbf{Definition} & \textbf{Search range} & \textbf{Global optimum}\\
        \hline
        Sphere & $f_1(\vec{x}) = \sum_{i=1}^{D} x_i^2$ & $[-5.12, 5.12]^D$ & $f_1(\vec{0}) = 0$\\
        \hline
        Rosenbrock & $f_2(\vec{x}) = \sum_{i=1}^{D-1} (100 (x_{i+1} - x_i^2)^2 + (x_i - 1)^2) $ & $[-2.048, 2.048]^D$ & $f_2(\vec{1}) = 0$\\
        \hline
        Rastrigin & $f_3(\vec{x}) = 10D + \sum_{i=1}^D \left(x_i^2 -10\cos(2\pi x_i)\right) $ & $[-5.12, 5.12]^D$ & $f_3(\vec{0}) = 0$\\
        \hline
        Griewank & $f_4(\vec{x}) = \sum_{i=1}^D \frac{x_i^2}{4000} - \prod_{i=1}^D \cos(\frac{x_i}{\sqrt{i}}) + 1 $ & $[-600, 600]^D$ & $f_4(\vec{0}) = 0$\\
        \hline
        Ackley & $f_5(\vec{x}) = -a\;\exp\left(-b \sqrt{\frac{1}{D} \sum_{i=1}^D x_i^2}\right) - \exp\left(\frac{1}{D} \sum_{i=1}^D \cos(c\;x_i)\right) + a + \exp(1) $& $[-32.768, 32.768]^D$ & $f_5(\vec{0}) = 0$\\
        \hline
        Schwefel & $f_6(\vec{x}) = 418.9829d - \sum_{i=1}^{D} x_i \sin(\sqrt{\lvert x_i \rvert}) $ & $[-500, 500]^D$ & $f_6(\vec{420.9687}) = 0$\\
        \hline
        Alpine & $f_7(\vec{x}) = \sum_{i=1}^{D} \lvert x_i \sin(x_i)+0.1x_i \rvert $ & $[-10, 10]^D$ & $f_7(\vec{0}) = 0$\\
        \hline
        Whitley & $f_8(\vec{x}) = \sum_{i=1}^D \sum_{j=1}^D \left(\frac{(100(x_i^2-x_j)^2 + (1-x_j)^2)^2}{4000} - \cos(100(x_i^2-x_j)^2 + (1-x_j)^2)+1\right) $ & $[-10, 10]^D$ & $f_8(\vec{1}) = 0$\\
        \hline
        Csendes & $f_9(\vec{x}) = \sum_{i=1}^D x_i^6\left( 2 + \sin \frac{1}{x_i}\right) $ & $[-1, 1]^D$ & $f_9(\vec{0}) = 0$\\
        \hline
        Dixon Price & $f_{10}(\vec{x}) = (x_1 - 1)^2 + \sum_{i = 2}^D i (2x_i^2 - x_{i - 1})^2 $ & $[-10, 10]^D$ & $f_{10}(x^*) = 0 \; x_i = 2^{-\frac{2^i - 2}{2^i}} $\\
        \hline
    \end{tabular}
\end{table}
\end{landscape}


\begin{table}
\renewcommand{\arraystretch}{1.5}
    \begin{tabular}{ll}
        \toprule
        Name & Definition, Search Range, Global optimum \\
        \midrule
        Sphere & $f_1(\vec{x}) = \sum_{i=1}^{D} x_i^2$ \\
               & $[-5.12, 5.12]^D$                     \\
               & $f_1(\vec{0}) = 0$\\
        \addlinespace
        Rosenbrock & $f_2(\vec{x}) = \sum_{i=1}^{D-1} (100 (x_{i+1} - x_i^2)^2 + (x_i - 1)^2) $ \\
                   & $[-2.048, 2.048]^D$ \\
                   & $f_2(\vec{1}) = 0$\\
        \addlinespace
        Rastrigin & $f_3(\vec{x}) = 10D + \sum_{i=1}^D \left(x_i^2 -10\cos(2\pi x_i)\right) $ \\
                  & $[-5.12, 5.12]^D$ \\
                  & $f_3(\vec{0}) = 0$\\
        \addlinespace
        Griewank & $f_4(\vec{x}) = \sum_{i=1}^D \frac{x_i^2}{4000} - \prod_{i=1}^D \cos(\frac{x_i}{\sqrt{i}}) + 1 $ \\
                 & $[-600, 600]^D$ \\
                 & $f_4(\vec{0}) = 0$\\
        \addlinespace
        Ackley & $f_5(\vec{x}) = -a\;\exp\left(-b \sqrt{\frac{1}{D} \sum_{i=1}^D x_i^2}\right) - \exp\left(\frac{1}{D} \sum_{i=1}^D \cos(c\;x_i)\right) + a + \exp(1) $ \\
               & $[-32.768, 32.768]^D$ \\
               & $f_5(\vec{0}) = 0$\\
        \addlinespace
        Schwefel & $f_6(\vec{x}) = 418.9829d - \sum_{i=1}^{D} x_i \sin(\sqrt{\lvert x_i \rvert}) $ \\
                 & $[-500, 500]^D$ \\
                 & $f_6(\vec{420.9687}) = 0$\\
        \addlinespace
        Alpine & $f_7(\vec{x}) = \sum_{i=1}^{D} \lvert x_i \sin(x_i)+0.1x_i \rvert $ \\
               & $[-10, 10]^D$ \\
               & $f_7(\vec{0}) = 0$\\
        \addlinespace
        Whitley & $f_8(\vec{x}) = \sum_{i=1}^D \sum_{j=1}^D \left(\frac{(100(x_i^2-x_j)^2 + (1-x_j)^2)^2}{4000} - \cos(100(x_i^2-x_j)^2 + (1-x_j)^2)+1\right) $ \\
                & $[-10, 10]^D$ \\
                & $f_8(\vec{1}) = 0$\\
        \addlinespace
        Csendes & $f_9(\vec{x}) = \sum_{i=1}^D x_i^6\left( 2 + \sin \frac{1}{x_i}\right) $ \\
                & $[-1, 1]^D$ \\
                & $f_9(\vec{0}) = 0$\\
        \addlinespace
        Dixon Price & $f_{10}(\vec{x}) = (x_1 - 1)^2 + \sum_{i = 2}^D i (2x_i^2 - x_{i - 1})^2 $ \\
                    & $[-10, 10]^D$ \\
                    & $f_{10}(x^*) = 0 \; x_i = 2^{-\frac{2^i - 2}{2^i}} $\\
        \bottomrule
    \end{tabular}
\end{table}


\begin{itemize}[leftmargin=*]
\item Sphere
  \begin{itemize}
  \item Definition: $f_1(\vec{x}) = \sum_{i=1}^{D} x_i^2$ 
  \item Search range: $[-5.12, 5.12]^D$ 
  \item Global optimum: $f_1(\vec{0}) = 0$
  \end{itemize}
\item  Rosenbrock
  \begin{itemize}
  \item Definition: $f_2(\vec{x}) = \sum_{i=1}^{D-1} (100 (x_{i+1} - x_i^2)^2 + (x_i - 1)^2) $ 
  \item Search range: $[-2.048, 2.048]^D$ 
  \item Global optimum: $f_2(\vec{1}) = 0$
  \end{itemize}
\item  Rastrigin
  \begin{itemize}
  \item Definition: $f_3(\vec{x}) = 10D + \sum_{i=1}^D \left(x_i^2 -10\cos(2\pi x_i)\right) $
  \item Search range: $[-5.12, 5.12]^D$
  \item Global optimum:  $f_3(\vec{0}) = 0$
  \end{itemize}
\item  Griewank
  \begin{itemize}
  \item Definition: $f_4(\vec{x}) = \sum_{i=1}^D \frac{x_i^2}{4000} - \prod_{i=1}^D \cos(\frac{x_i}{\sqrt{i}}) + 1 $
  \item Search range: $[-600, 600]^D$
  \item Global optimum: $f_4(\vec{0}) = 0$
  \end{itemize}
\item  Ackley
  \begin{itemize}
  \item Definition: $f_5(\vec{x}) = -a\;\exp\left(-b \sqrt{\frac{1}{D} \sum_{i=1}^D x_i^2}\right) - \exp\left(\frac{1}{D} \sum_{i=1}^D \cos(c\;x_i)\right) + a + \exp(1) $
  \item Search range: $[-32.768, 32.768]^D$
  \item Global optimum: $f_5(\vec{0}) = 0$ 
  \end{itemize}
\item Schwefel
  \begin{itemize}
  \item Definition: $f_6(\vec{x}) = 418.9829d - \sum_{i=1}^{D} x_i \sin(\sqrt{\lvert x_i \rvert}) $
  \item Search range: $[-500, 500]^D$ 
  \item Global optimum: $f_6(\vec{420.9687}) = 0$
  \end{itemize}
\item  Alpine
  \begin{itemize}
  \item Definition: $f_7(\vec{x}) = \sum_{i=1}^{D} \lvert x_i \sin(x_i)+0.1x_i \rvert $  
  \item Search range: $[-10, 10]^D$
  \item Global optimum: $f_7(\vec{0}) = 0$
  \end{itemize}
\item Whitley
  \begin{itemize}
  \item Definition: $f_8(\vec{x}) = \sum_{i=1}^D \sum_{j=1}^D \left(\frac{(100(x_i^2-x_j)^2 + (1-x_j)^2)^2}{4000} - \cos(100(x_i^2-x_j)^2 + (1-x_j)^2)+1\right) $
  \item Search range: $[-10, 10]^D$
  \item Global optimum: $f_8(\vec{1}) = 0$
  \end{itemize}
\item Csendes
  \begin{itemize}
  \item Definition: $f_9(\vec{x}) = \sum_{i=1}^D x_i^6\left( 2 + \sin \frac{1}{x_i}\right) $
  \item Search range: $[-1, 1]^D$
  \item Global optimum: $f_9(\vec{0}) = 0$
  \end{itemize}
\item  Dixon Price
  \begin{itemize}
  \item Definition: $f_{10}(\vec{x}) = (x_1 - 1)^2 + \sum_{i = 2}^D i (2x_i^2 - x_{i - 1})^2 $
  \item Search range: $[-10, 10]^D$
  \item Global optimum: $f_{10}(x^*) = 0 \; x_i = 2^{-\frac{2^i - 2}{2^i}} $
  \end{itemize}
\end{itemize}

\end{document}

答案3

以表格形式呈现（用于练习，不推荐）：

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{mathtools}
\usepackage{makecell, tabularx}
\renewcommand\theadfont{\small\bfseries}
\usepackage{geometry}

\geometry{
    a4paper,
    left=30mm,
    right=25mm,
    top=30mm,
    bottom=30mm,
    asymmetric
}

\begin{document}

\begin{table}
\setlength\tabcolsep{4pt}
\setcellgapes{3pt}
\makegapedcells
    \begin{tabularx}{\textwidth}{|l|X|c|l|}
        \hline
\thead{Name} & \thead{Definition} & \thead{Search\\ range} & \thead{Global\\ optimum}\\
        \hline
Sphere & $f_1(\vec{x}) = \sum\limits_{i=1}^{D} x_i^2$ & $[-5.12, 5.12]^D$ & $f_1(\vec{0}) = 0$\\
        \hline
Rosenbrock & $f_2(\vec{x}) = \sum\limits_{i=1}^{D-1} (100 (x_{i+1} - x_i^2)^2 + (x_i - 1)^2) $ & $[-2.048, 2.048]^D$ & $f_2(\vec{1}) = 0$\\
        \hline
Rastrigin & $f_3(\vec{x}) = 10D + \sum\limits_{i=1}^D \left(x_i^2 -10\cos(2\pi x_i)\right) $ & $[-5.12, 5.12]^D$ & $f_3(\vec{0}) = 0$\\
        \hline
Griewank & $f_4(\vec{x}) = \sum\limits_{i=1}^D \frac{x_i^2}{4000} - \prod\limits_{i=1}^D \cos(\frac{x_i}{\sqrt{i}}) + 1 $ & $[-600, 600]^D$ & $f_4(\vec{0}) = 0$\\
        \hline
Ackley & $ \begin{multlined}
            f_5(\vec{x}) = -a\;\exp\left(-b \sqrt{\frac{1}{D} \sum_{i=1}^D x_i^2}\right)\\
            - \exp\left(\frac{1}{D} \sum_{i=1}^D \cos(c\;x_i)\right) + a + \exp(1) 
            \end{multlined}$
                & $[-32.768, 32.768]^D$ 
                    & $f_5(\vec{0}) = 0$\\
        \hline
Schwefel & $f_6(\vec{x}) = 418.9829d - \sum\limits_{i=1}^{D} x_i \sin(\sqrt{\lvert x_i \rvert}) $ & $[-500, 500]^D$ & $f_6(\vec{420.9687}) = 0$\\
        \hline
Alpine & $f_7(\vec{x}) = \sum\limits_{i=1}^{D} \lvert x_i \sin(x_i)+0.1x_i \rvert $ & $[-10, 10]^D$ & $f_7(\vec{0}) = 0$\\
        \hline
Whitley & $\begin{multlined}[t]
    f_8(\vec{x}) = \\
    \sum_{i=1}^D \sum_{j=1}^D \Bigl(\frac{(100(x_i^2-x_j)^2 + (1-x_j)^2)^2}{4000}\\ 
                    - \cos(100(x_i^2-x_j)^2 + (1-x_j)^2)+1\Bigr)
            \end{multlined}$ 
                & $[-10, 10]^D$              
                    & $f_8(\vec{1}) = 0$\\        
        \hline
Csendes & $f_9(\vec{x}) = \sum\limits_{i=1}^D x_i^6\left( 2 + \sin \frac{1}{x_i}\right) $ & $[-1, 1]^D$ & $f_9(\vec{0}) = 0$\\
        \hline
Dixon Price & $f_{10}(\vec{x}) = (x_1 - 1)^2 + \sum\limits_{i = 2}^D i (2x_i^2 - x_{i - 1})^2 $ 
                & $[-10, 10]^D$ 
                    & \makecell{$f_{10}(x^*) = 0$,\\ 
                                $x_i = 2^{-\frac{2^i - 2}{2^i}} $}\\
        \hline
    \end{tabularx}
\end{table}

\end{document}

在此处输入图片描述

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