Layout of complex table

The MWE

\documentclass{article}

\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}

\usepackage{amsmath}
\usepackage{tabularx}
\usepackage{colortbl}
\usepackage{multirow}
\usepackage{ragged2e}

\newcolumntype{C}{>{\Centering\arraybackslash}X}

\renewcommand{\arraystretch}{1.7}

\definecolor{ColDiag}{gray}{0.7}
\definecolor{ColBelow}{gray}{0.9}

\newcommand{\NA}{\cellcolor{ColDiag}}
\newcommand{\B}{\cellcolor{ColBelow}}

\begin{document}

\rowcolor{ColHead} & & \multicolumn{6}{c|}{$\vec{x}$} \\ \hline
\rowcolor{ColHead} & & $\vec{a}$ & $\vec{b}$ & $\vec{c}$ & $\vec{d}$ & $\vec{e}$ & $\vec{f}$ \\ \hline
\multirow{6}{*}{$\vec{y}$}
& $\vec{a}$ & \NA & $-\tfrac{1}{2}$ & & & & \\ \cline{2-8}
& $\vec{b}$ & \B $-2$ & \NA & & & & \\ \cline{2-8}
& $\vec{c}$ & \B & \B & \NA & & & \\ \cline{2-8}
& $\vec{d}$ & \B & \B & \B & \NA & & $-\tfrac{1}{3}$ \\ \cline{2-8}
& $\vec{e}$ & \B & \B & \B & \B & \NA & \\ \cline{2-8}
& $\vec{f}$ & \B & \B & \B & \B $-3$ & \B & \NA \\ \hline
\end{tabularx}

\end{document}

produces the following result. There are four problems:

1. The $vec{y}$ does not show in the leftmost multi-row cell.
2. It seems that some of the lines (such as marked in the image) are too short.
3. I have not managed to merge the top-left four cells using a combination of \multirow and \multicol.
4. I would like to achieve that the cells are exact squares. So far I have fiddled around with the tabularx width and the \arraystretch factor.

Might it be better to use a tikzpicture instead of a tabularx environment in such a scenario?

Edit

With the TikZ code

\documentclass{article}

\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}

\usepackage{amsmath}
\usepackage{tikz}

\definecolor{ColDiag}{gray}{0.7}
\definecolor{ColBelow}{gray}{0.9}

\begin{document}

\begin{tikzpicture}

\fill [fill=ColHead] (0, 0) rectangle (8, -2);
\fill [fill=ColHead] (0, -2) rectangle (2, -8);

\foreach \i in {1, ..., 6}
\fill [fill=ColDiag] (\i + 1, -1 - \i) rectangle (\i + 2, -2 - \i);

\foreach \i/\j in {2/1, 3/1, 3/2, 4/1, 4/2, 4/3, 5/1, 5/2, 5/3, 5/4, 6/1, 6/2, 6/3, 6/4, 6/5}
\fill [fill=ColBelow] (\j + 1, -1 - \i) rectangle (\j + 2, -2 - \i);

\draw (0, 0) rectangle (8, -8);
\draw (0, -2) -- (8, -2);
\draw (2, 0) -- (2, -8);
\draw (2, -1) -- (8, -1);
\draw (1, -2) -- (1, -8);

\node (x) at (5, -0.5) {$\vec{x}$};
\node (y) at (0.5, -5) {$\vec{y}$};

\foreach \i in {2, ..., 6}
{
\draw (1, -1 - \i) -- (8, -1 - \i);
\draw (1 + \i, -1) -- (1 + \i, -8);
}

\foreach \i/\v in {1/a, 2/b, 3/c, 4/d, 5/e, 6/f}
{
\node (ch\i) at (\i + 1.5, -1.5) {$\vec{\v}$};
\node (rw\i) at (1.5, -1.5 - \i) {$\vec{\v}$};
}

\node (12) at (3.5, -2.5) {$-\tfrac{1}{2}$};
\node (21) at (2.5, -3.5) {$-2$};
\node (46) at (7.5, -5.5) {$-\tfrac{1}{3}$};
\node (64) at (5.5, -7.5) {$-3$};

\end{tikzpicture}

\end{document}

I achieve the desired result. The TikZ coding does not seem to be quite elegant though. Comapared to the tabular version, it is hard to see what the source code is meant to result in.

This is a slight variation of leandriis' nice answer in which the cells get filled programmatically. This is done with a style

colorize cells/.style={/utils/exec={%
\pgfmathsetmacro{\mycolor}{ifthenelse(\the\pgfmatrixcurrentrow>\the\pgfmatrixcurrentcolumn,
"gray!20",ifthenelse(\the\pgfmatrixcurrentrow==\the\pgfmatrixcurrentcolumn,
"gray!50","white"))]}},fill=\mycolor}

which looks at the current column and row indices, which are stored in the counts \pgfmatrixcurrentcolumn and \pgfmatrixcurrentrow, respectively, and selects the fill color depending on whether the column index is larger or smaller than, or equal to the row index.

\documentclass{article}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{matrix}
\begin{document}

\begin{tikzpicture}[nodes in empty cells,
colorize cells/.style={/utils/exec={%
\pgfmathsetmacro{\mycolor}{ifthenelse(\the\pgfmatrixcurrentrow>\the\pgfmatrixcurrentcolumn,
"gray!20",ifthenelse(\the\pgfmatrixcurrentrow==\the\pgfmatrixcurrentcolumn,
"gray!50","white"))]}},fill=\mycolor}]
\matrix(mat)[matrix of nodes,column 1/.style={nodes={fill=gray!50}},
row 1/.style={nodes={fill=gray!50}},
row sep =-\pgflinewidth,column sep = -\pgflinewidth,
nodes={anchor=center,draw,text height=2ex,text depth=0.25ex,
minimum width=1cm,colorize cells}]
{
|[fill=white,draw=none]| & $\vec{a}$ & $\vec{b}$ & $\vec{c}$ & $\vec{d}$ & $\vec{e}$ & $\vec{f}$\\
$\vec{a}$ &  & $-\tfrac{1}{2}$ & & & &  \\
$\vec{b}$ & $-2$ &  & & & &  \\
$\vec{c}$ & & &  & & &  \\
$\vec{d}$ & & & &  & & $-\tfrac{1}{3}$ \\
$\vec{e}$ & & & & &  & \\
$\vec{f}$ & & & & $-3$ & &  \\
};
\draw[fill=gray!50] ([yshift=-\pgflinewidth]mat-1-2.north west) rectangle ([yshift=4ex]mat-1-7.north
east) coordinate(TR) node[midway]{$\vec x$};
\draw[fill=gray!50] ([xshift=\pgflinewidth]mat-2-1.north west) rectangle ([xshift=-1cm]mat-7-1.south
west) coordinate(BL) node[midway]{$\vec y$};
\draw[fill=gray!50] ([yshift=-\pgflinewidth,xshift=\pgflinewidth]mat-2-2.north west) rectangle
(BL|-TR);
\end{tikzpicture}
\end{document} A version thereof which is easier to generalize may be something in which one stores the colors in a list and then defines a function that computes the list index from the column and row indices, in this case one may choose

colorize cells/.style={
/utils/exec=\pgfmathsetmacro{\mycolor}{{\LstColors}[1-sign(\the\pgfmatrixcurrentrow-\the\pgfmatrixcurrentcolumn)]},fill=\mycolor}

together with \edef\LstColors{"gray!50","gray!20","white"}. This one also has minimum height=1cm, as you suggest.

\documentclass{article}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{matrix}
\begin{document}

\begin{tikzpicture}[nodes in empty cells,
colorize cells/.style={
/utils/exec=\pgfmathsetmacro{\mycolor}{{\LstColors}[1-sign(\the\pgfmatrixcurrentrow-\the\pgfmatrixcurrentcolumn)]},
fill=\mycolor}]
\edef\LstColors{"gray!50","gray!20","white"}
\matrix(mat)[matrix of nodes,column 1/.style={nodes={fill=gray!50}},
row 1/.style={nodes={fill=gray!50}},
row sep =-\pgflinewidth,column sep = -\pgflinewidth,
nodes={anchor=center,draw,minimum height=1cm,%text height=2ex,text depth=0.25ex,
minimum width=1cm,colorize cells}]
{
|[fill=white,draw=none]| & $\vec{a}$ & $\vec{b}$ & $\vec{c}$ & $\vec{d}$ & $\vec{e}$ & $\vec{f}$\\
$\vec{a}$ &  & $-\tfrac{1}{2}$ & & & &  \\
$\vec{b}$ & $-2$ &  & & & &  \\
$\vec{c}$ & & &  & & &  \\
$\vec{d}$ & & & &  & & $-\tfrac{1}{3}$ \\
$\vec{e}$ & & & & &  & \\
$\vec{f}$ & & & & $-3$ & &  \\
};
\draw[fill=gray!50] ([yshift=-\pgflinewidth]mat-1-2.north west) rectangle
([yshift=1cm]mat-1-7.north east) coordinate(TR) node[midway]{$\vec x$};
\draw[fill=gray!50] ([xshift=\pgflinewidth]mat-2-1.north west) rectangle ([xshift=-1cm]mat-7-1.south
west) coordinate(BL) node[midway]{$\vec y$};
\draw[fill=gray!50] ([yshift=-\pgflinewidth,xshift=\pgflinewidth]mat-2-2.north west) rectangle
(BL|-TR);
\end{tikzpicture}
\end{document} but one does not have to stop there:

\documentclass{article}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{matrix}
\begin{document}

\begin{tikzpicture}[nodes in empty cells,
colorize cells/.style={
/utils/exec=\pgfmathsetmacro{\mygraylevel}{40-10*\the\pgfmatrixcurrentrow
+10*\the\pgfmatrixcurrentcolumn},
fill=gray!\mygraylevel}]
\matrix(mat)[matrix of nodes,column 1/.style={nodes={fill=gray!50}},
row 1/.style={nodes={fill=gray!50}},
row sep =-\pgflinewidth,column sep = -\pgflinewidth,
nodes={anchor=center,draw,text height=2ex,text depth=0.25ex,
minimum width=1cm,colorize cells}]
{
|[fill=white,draw=none]| & $\vec{a}$ & $\vec{b}$ & $\vec{c}$ & $\vec{d}$ & $\vec{e}$ & $\vec{f}$\\
$\vec{a}$ &  & $-\tfrac{1}{2}$ & & & &  \\
$\vec{b}$ & $-2$ &  & & & &  \\
$\vec{c}$ & & &  & & &  \\
$\vec{d}$ & & & &  & & $-\tfrac{1}{3}$ \\
$\vec{e}$ & & & & &  & \\
$\vec{f}$ & & & & $-3$ & &  \\
};
\draw[fill=gray!50] ([yshift=-\pgflinewidth]mat-1-2.north west) rectangle ([yshift=4ex]mat-1-7.north
east) coordinate(TR) node[midway]{$\vec x$};
\draw[fill=gray!50] ([xshift=\pgflinewidth]mat-2-1.north west) rectangle ([xshift=-1cm]mat-7-1.south
west) coordinate(BL) node[midway]{$\vec y$};
\draw[fill=gray!50] ([yshift=-\pgflinewidth,xshift=\pgflinewidth]mat-2-2.north west) rectangle
(BL|-TR);
\end{tikzpicture}
\end{document} • Very nice solution! I would just add minimum height = 1cm in order to obtain square cells. – Matthias Jun 22 at 14:56
• @Matthias Thanks! I added a version with this. (One could generalize this method in many ways, e.g. let not only the fill depend on the column and row but also a path picture or the cell content.) – user121799 Jun 22 at 15:03
• The flexible solution using the list clashes with the babel package. Do you have any idea how to fix this problem? – Matthias Jun 22 at 15:14
• @Matthias Does adding \usetikzlibrary{babel} solve the problem? If not, could you please tell me how you precisely load babel, i.e. all options? (For instance, \usepackage{amsmath} \usepackage[spanish]{babel} \usepackage{tikz} \usetikzlibrary{matrix,babel} only works after loading the babel library.) – user121799 Jun 22 at 15:36
• Yes, \usetikzlibrary{babel} does solve the problem without any further adaptions. – Matthias Jun 22 at 17:00
\documentclass{article}
\usepackage[T1]{fontenc}

\usepackage{amsmath}
\usepackage[table]{xcolor}
\definecolor{ColDiag}{gray}{0.7}
\definecolor{ColBelow}{gray}{0.9}
\newcommand{\NA}{\cellcolor{ColDiag}}
\newcommand{\B}{\cellcolor{ColBelow}}
\usepackage{ragged2e}
\usepackage{multirow, tabularx}
\newcolumntype{C}{>{\Centering}X}

\begin{document}
*{6}{C<{\rule[-0.75em]{0pt}{2.5em}}|}}
\hline
\multicolumn{2}{c|}{}
& \multicolumn{6}{c|}{$\vec{x}$}    \\ \cline{3-8}
\multicolumn{2}{c|}{}
& $\vec{a}$ & $\vec{b}$ & $\vec{c}$ & $\vec{d}$ & $\vec{e}$ & $\vec{f}$ \\ \cline{2-8}
& $\vec{a}$ & \NA & $-\tfrac{1}{2}$ & & & & \\ \cline{2-8}
& $\vec{b}$ & \B $-2$ & \NA & & & & \\ \cline{2-8}
& $\vec{c}$ & \B & \B & \NA & & & \\ \cline{2-8}
& $\vec{d}$ & \B & \B & \B & \NA & & $-\tfrac{1}{3}$ \\ \cline{2-8}
& $\vec{e}$ & \B & \B & \B & \B & \NA & \\ \cline{2-8}
\multirow{-12}{*}{$\vec{y}$}
& $\vec{f}$ & \B & \B & \B & \B $-3$ & \B & \NA \\ \hline
\end{tabularx}

\bigskip
slightly better:

\begin{tabularx}{20em}{ C|
*{6}{C<{\rule[-0.75em]{0pt}{2.5em}}|}}
\multicolumn{2}{c}{}
& \multicolumn{6}{c}{$\vec{x}$}    \\ \cline{3-8}
\multicolumn{2}{>{\cellcolor{white}}c|}{}
& $\vec{a}$
& $\vec{b}$
& $\vec{c}$
& $\vec{d}$
& $\vec{e}$
& $\vec{f}$ \\ \cline{2-8}
& $\vec{a}$ & \NA       & $-\frac{1}{2}$
&   &   &   &           \\ \cline{2-8}
& $\vec{b}$ & \B $-2$   &\NA&   &   &   &           \\ \cline{2-8}
& $\vec{c}$ & \B        &\B &\NA&   &   &           \\ \cline{2-8}
& $\vec{d}$ & \B        &\B &\B &\NA&   & $-\tfrac{1}{3}$ \\ \cline{2-8}
& $\vec{e}$ & \B        &\B &\B &\B &\NA&           \\ \cline{2-8}
\multirow{-12}{*}{$\vec{y}$}
& $\vec{f}$ & \B        &\B &\B &\B $-3$
&\B & \NA       \\ \cline{2-8}
\end{tabularx}
\end{document}

gives Please observe that multirow cells is moved to the last row (with negative number of spanned lines in other rows). For exact squares I add rule with zero thick and height 2.5 em (what is 20em/8)

However, far better result you would obtain, if you will draw your table as picture, for example with tikz package. In this case you will not have problems with visibility of lines in colored cells (as you have now).

Here is a tikz solution (that is most likely still in need of some improvement): \documentclass{article}
\usepackage{tikz}
\usetikzlibrary{fit,matrix}
\usepackage{amsmath}
\usepackage{multirow}
\usepackage{tabulary}
\colorlet{mylightgray}{gray!20}
\colorlet{mygray}{gray!50}

\begin{document}

\begin{tikzpicture}[cell/.style={rectangle,draw=black}, nodes in empty cells]
\matrix(table)[
matrix of nodes,
row sep =-\pgflinewidth,
column sep = -\pgflinewidth,
nodes={anchor=center,text height=2ex,text depth=0.25ex},
column 1/.style = {nodes={cell, minimum width=1cm, fill=white, draw=none}},
column 2/.style = {nodes={cell, minimum width=1cm, fill=mygray}},
column 3/.style = {nodes={cell, minimum width=1cm}},
column 4/.style = {nodes={cell, minimum width=1cm}},
column 5/.style = {nodes={cell, minimum width=1cm}},
column 6/.style = {nodes={cell, minimum width=1cm}},
column 7/.style = {nodes={cell, minimum width=1cm}},
column 8/.style = {nodes={cell, minimum width=1cm}},
row 1/.style={nodes={cell, minimum height=1cm, fill=white, draw=none}},
row 2/.style={nodes={cell, minimum height=1cm, fill=mygray}},
row 3/.style={nodes={cell, minimum height=1cm}},
row 4/.style={nodes={cell, minimum height=1cm}},
row 5/.style={nodes={cell, minimum height=1cm}},
row 6/.style={nodes={cell, minimum height=1cm}},
row 7/.style={nodes={cell, minimum height=1cm}},
row 8/.style={nodes={cell, minimum height=1cm}},
]
{ & & & & & & & & \\
|[fill=white,draw=none]| & |[fill=white,draw=none]| & $\vec{a}$ & $\vec{b}$ & $\vec{c}$ & $\vec{d}$ & $\vec{e}$ & $\vec{f}$\\
|[draw=none]| & $\vec{a}$ & |[fill=mygray]| & $-\tfrac{1}{2}$ & & & &  \\
|[draw=none]| & $\vec{b}$ & |[fill=mylightgray]| $-2$ & |[fill=mygray]| & & & &  \\
|[draw=none]| & $\vec{c}$ & |[fill=mylightgray]| & |[fill=mylightgray]| & |[fill=mygray]| & & &  \\
|[draw=none]| & $\vec{d}$ & |[fill=mylightgray]| & |[fill=mylightgray]| & |[fill=mylightgray]| & |[fill=mygray]| & & $-\tfrac{1}{3}$ \\
|[draw=none]| & $\vec{e}$ & |[fill=mylightgray]| & |[fill=mylightgray]| & |[fill=mylightgray]| & |[fill=mylightgray]| & |[fill=mygray]| & \\
|[draw=none]| & $\vec{f}$ & |[fill=mylightgray]| & |[fill=mylightgray]| & |[fill=mylightgray]| & |[fill=mylightgray]| $-3$ & |[fill=mylightgray]| & |[fill=mygray]| \\
};
\node[fit=(table-1-5)(table-1-6)]{$\vec{x}$};
\node[fit=(table-5-1)(table-6-1)]{$\vec{y}$};
\end{tikzpicture}

\end{document}