Diagonal lines in table cell

Question

I need this table:

(each row and column should have the same height and length, last cell should be divided by diagonal line). I tried the following code:

\documentclass[11pt]{article}
\usepackage[T1]{fontenc}
\usepackage{array}
\usepackage{makecell}
\newcolumntype{x}[1]{>{\centering\let\newline\\\arraybackslash\hspace{0pt}}p{#1}}
\begin{document}
\setlength{\extrarowheight}{0.1cm}
\begin{tabular}{|x{0.5cm}|x{0.5cm}|x{0.5cm}|x{0.5cm}|x{0.5cm}|}\hline
&&&&20\\ \hline
&&&&30\\ \hline
&&&&45\\ \hline
15&12&18&50&\diaghead(-3,2){\hskip \hsize}{$a_i$}{$b_j$}\\ \hline
\end{tabular}
\end{document}

but text in the last cell is displayed incorrectly and cells are not the same.

How can I change that?

Answer 1

Editor's note: The solution below has an inaccuracy because of the line width, refer to https://tex.stackexchange.com/a/89840/250119 for a fix.

---

Exact solution with TikZ:

\documentclass[11pt]{article}
\usepackage[T1]{fontenc}
\usepackage{array}
\usepackage{makecell}
\newcolumntype{x}[1]{>{\centering\arraybackslash}p{#1}}
\usepackage{tikz}
\newcommand\diag[4]{%
  \multicolumn{1}{p{#2}|}{\hskip-\tabcolsep
  $\vcenter{\begin{tikzpicture}[baseline=0,anchor=south west,inner sep=#1]
  \path[use as bounding box] (0,0) rectangle (#2+2\tabcolsep,\baselineskip);
  \node[minimum width={#2+2\tabcolsep},minimum height=\baselineskip+\extrarowheight] (box) {};
  \draw (box.north west) -- (box.south east);
  \node[anchor=south west] at (box.south west) {#3};
  \node[anchor=north east] at (box.north east) {#4};
 \end{tikzpicture}}$\hskip-\tabcolsep}}
\begin{document}
\setlength{\extrarowheight}{0.1cm}
\begin{tabular}{|x{0.5cm}|x{0.5cm}|x{0.5cm}|x{0.5cm}|x{0.5cm}|}\hline
&&&&20\ \hline
&&&&30\ \hline
&&&&45\ \hline
15&12&18&50&\diag{.1em}{.5cm}{$a_i$}{$b_j$}\ \hline
\end{tabular}
\end{document}

Also it is a reimplementation of \diaghead.

Answer 2

Okay as this seems to be the question that all duplicates link to eventually, I want to include @Leo Liu's solution from here as well as this seems to solve the problem in the easiest way.

In short: Use the @Leo Liu's package diagbox.

MWE (coming from a different context and does not exactly answer to this question) shamelessly copied from Leo's answer (link above):

\documentclass{article}

\usepackage{diagbox}

\begin{document}

\begin{tabular}{|l|c|c|}\hline
\diagbox[width=10em]{Diag\\Column Head I}{Diag Column\\Head II}&
  Second & Third \\ \hline
& foo & bar \\ \hline
\end{tabular}

\end{document}

Resulting in

If you like this solution make sure you go ahead and show Leo's original answer some love.

---

Admittedly this solution doesn't work so well (out of the box) with the provided MWE but if one uses normal column types instead of the custom ones, one gets a really good result:

\documentclass{standalone}
\usepackage{diagbox}

\begin{document}
    \begin{tabular}{|c|c|c|c|c|}
        \hline
           &    &    &    &           20           \\ \hline
           &    &    &    &           30           \\ \hline
           &    &    &    &           45           \\ \hline
        15 & 12 & 18 & 50 & \diagbox{$a_i$}{$b_j$} \\ \hline
    \end{tabular}
\end{document}

Results in

Answer 3

It might be possible to draw a diagonal line which fits exactly in a table cell, but it might be easier to draw the whole table as a picture.

Here my attempt using tikz. For large tables the need for nodes for each cell might be quite an effort, but it should be OK for smaller ones.

\documentclass[11pt]{article}
\usepackage[T1]{fontenc}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}[x=.75cm,y=.5cm]
    \draw (0,0) grid [step=1] (5,4);
    \node at (0.5,0.5) {15};
    \node at (1.5,0.5) {12};
    \node at (2.5,0.5) {18};
    \node at (3.5,0.5) {50};
    \node at (4.5,3.5) {20};
    \node at (4.5,2.5) {30};
    \node at (4.5,1.5) {45};
    \draw (4,1) -- (5,0);
    \node at (5.0,1.0) [below left,inner sep=1pt] {\small$a_i$};
    \node at (4.0,0.0) [above right,inner sep=1pt] {\small$b_j$};
\end{tikzpicture}
\end{document}

Answer 4

Here's a solution using TikZ:

\documentclass[11pt]{article}
\usepackage[T1]{fontenc}
\usepackage{tikz}
\usetikzlibrary{matrix}

\begin{document}

\begin{tikzpicture}%[thick]
  \matrix (mat) [%
    matrix of nodes,
    nodes in empty cells,
    text width=0.8cm,
    text height=10pt,
    text depth=2pt,
    text badly centered
  ] 
  {%
     & & & & 20 \\
     & & & & 30  \\
     & & & & 45  \\
    15 & 12 & 18 & 50 & \raisebox{5pt}{$a_i$}\hspace{-15pt}\llap{\raisebox{-1pt}{$b_j$}}\\
  };
  % horizontal lines
  \foreach \i in {1,2,3,4}
        \draw (mat-\i-1.north west) -- (mat-\i-5.north east);
  \draw (mat-4-1.south west) -- (mat-4-5.south east);
  % vertical lines
  \foreach \j in {1,2,3,4,5}
    \draw (mat-1-\j.north west) -- (mat-4-\j.south west);
  \draw (mat-1-5.north east) -- (mat-4-5.south east);
  % diagonal line    
  \draw (mat-4-5.north west) -- (mat-4-5.south east);
\end{tikzpicture}

\end{document}

EDIT: in the solution I initially gave, the last cell contents was not what the OP wanted. I corrected it.

Answer 5

A solution with {NiceTabular} of nicematrix which has a built-in command \diagbox. Moreover, you can draw all the rules with only one key hvlines.

\documentclass{article}
\usepackage{nicematrix}
\begin{document}
\setlength{\extrarowheight}{1mm}
\begin{NiceTabular}{ccccc}[hvlines]
   &    &    &    &           20           \ 
   &    &    &    &           30           \ 
   &    &    &    &           45           \ 
15 & 12 & 18 & 50 & \diagbox{$a_i$}{$b_j$} \ 
\end{NiceTabular}
\end{document}

Answer 6

Here's another solution using Tikz and based on a given solution by Heiko Oberdiek.

\documentclass{article}
\pagestyle{empty}% for cropping
\usepackage{array}
\usepackage{makecell}
\newcolumntype{x}[1]{>{\centering\arraybackslash}p{#1}}

\usepackage{tikz}
\usetikzlibrary{calc}
\usepackage{zref-savepos}

\newcounter{DiagonalizedEntry}
\renewcommand*{\theDiagonalizedEntry}{NTE-\the\value{DiagonalizedEntry}}

\newcommand*{\diagonalize}[2]{%
  \multicolumn{1}{@{}c@{}|}{%
    \stepcounter{DiagonalizedEntry}%
    \vadjust pre{\zsavepos{\theDiagonalizedEntry t}}% top
    \vadjust{\zsavepos{\theDiagonalizedEntry b}}% bottom
    \zsavepos{\theDiagonalizedEntry l}% left
    \hspace{0pt plus 1filll}%
    \zsavepos{\theDiagonalizedEntry r}% right
    \tikz[overlay]{%
      \draw[red]
        let
          \n{llx}={\zposx{\theDiagonalizedEntry l}sp-\zposx{\theDiagonalizedEntry r}sp}, % x left
          \n{urx}={0}, % x right
          \n{lly}={\zposy{\theDiagonalizedEntry b}sp-\zposy{\theDiagonalizedEntry r}sp}, % y bottom
          \n{ury}={\zposy{\theDiagonalizedEntry t}sp-\zposy{\theDiagonalizedEntry r}sp} %  y top
    in
        (\n{llx}, \n{ury}) -- (\n{urx}, \n{lly})
    node[anchor=south west] at (\n{llx}, \n{lly}) {#1}
    node[anchor=north east] at (\n{urx}, \n{ury}) {#2}
    ;
    }% 
  }%
}

\begin{document}
\renewcommand{\arraystretch}{2}
\renewcommand\tabcolsep{10pt}
\begin{tabular}{|x{0.5cm}|x{0.5cm}|x{0.5cm}|x{0.5cm}|x{0.5cm}|}\hline
&&&&20\\ \hline
&&&&30\\ \hline
&&&&45\\ \hline
15&12&18&50&\diagonalize{$a_i$}{$b_j$} \\ \hline
\end{tabular}
\end{document}