12

How can I ensure that consecutive diagonal arrows in a tikz-cd commutative diagram spanning many rows and columns lie on a straight line (as is automatically the case for consecutive horizontal arrows)?

I suspect it would be enough to force all the center nodes to lie on a regular grid, but I don't know how to achieve that.

Most minimal example I can think of where this occurs:

\begin{tikzcd}
  A  \arrow{dr}{\alpha} & B \arrow{dr}{\beta} & C & D \\
  A & B \times B \times B \arrow{dr}{\gamma} & C \arrow{dr}{\delta} & D \\
  A & B & C & D
\end{tikzcd}
3
  • 1
    Do you have an MWE that you can provide? Jan 3, 2013 at 10:57
  • I cannot install tikz-cd to test this, but, perhaps if you add the option [on grid] to your tikzcd environment that would do it?
    – JLDiaz
    Jan 3, 2013 at 11:06
  • Sadly [on grid] has no effect.
    – Dan
    Jan 3, 2013 at 11:16

2 Answers 2

9

The option on grid won’t work because internally the diagrams are built with a \matrix.

But the original /tikz/column sep (and /tikz/row sep) styles understand the option between origins.

The problem is that tikz-cd does not provide a (good) hook to add between origins to the already defined separators or to an arbitrary distance (e.g. column sep={between origins}, column sep={normal,between origins}, or column sep={10ex,between origins}).

My solution provides

  • a re-definition of the original \tikzcd@sep#1#2 macro that is used by tikz cd to set the separators and to sort out the likes of column sep=<named distance>;
  • a re-definition of the original /tikz/commutative diagrams/column sep and /tikz/commutative diagrams/row sep keys that accomodate the changes to \tikzcd@sep; and
  • /tikz/commutative diagrams/bo row sep and
  • /tikz/commutative diagrams/bo col sep keys that works (kind of) like the original.

“Kind of” because I opted for a factor of 1.7 to the named distances (tiny, small, scriptsize, normal, large, and huge) because the distances are now smaller due to the fact that the node widths do not affect the separator (just like on grid would).

Code

\documentclass[tikz]{standalone}
\usepackage{tikz-cd}
\makeatletter
\pgfqkeys{/tikz/commutative diagrams}{
  row sep/.code={\tikzcd@sep{row}{#1}{}},
  column sep/.code={\tikzcd@sep{column}{#1}{}},
  bo row sep/.code={\tikzcd@sep{row}{#1}{between origins}},
  bo column sep/.code={\tikzcd@sep{column}{#1}{between origins}},
  bo column sep/.default=normal,
  bo row sep/.default=normal,
}
\def\tikzcd@sep#1#2#3{% re-defintion of original package macro!
  \pgfkeysifdefined{/tikz/commutative diagrams/#1 sep/#2}%
    {\pgfkeysalso{/tikz/#1 sep={\ifx\\#3\\1*\else1.7*\fi\pgfkeysvalueof{/tikz/commutative diagrams/#1 sep/#2},#3}}}%
    {\pgfkeysalso{/tikz/#1 sep={#2,#3}}}}
\makeatother
\begin{document}
\begin{tikzcd}[bo column sep]
  A \arrow{dr}{\alpha} & B \arrow{dr}{\beta}                    & C                    & D \\
  A                    & B \times B \times B \arrow{dr}{\gamma} & C \arrow{dr}{\delta} & D \\
  A                    & B                                      & C                    & D
\end{tikzcd}
\end{document}

Output

enter image description here

A manual way perhaps?

A few other options in this example are

  • left- and right-lapping the biggest node (mathtools required):

    \mathllap{B \times} B \mathrlap{\times B}
    
  • settings a fixed node width that is the maximum of all:

    text width=\widthof{$B \times B \times B$}
    

Code

\documentclass[tikz]{standalone}
\usepackage{tikz-cd} \usepackage{mathtools}
\begin{document}

\begin{tikzcd}
  A  \arrow[shorten >=8pt]{dr} & B \arrow{dr} & C & D \\
  A & \mathllap{B \times} B \mathrlap{\times B} \arrow{dr} & C \arrow{dr} & D \\
  A & B & C & D
\end{tikzcd}

\begin{tikzcd}[column sep=-1ex,cells={nodes={align=center,text width=\widthof{$B \times B \times B$}}}]
  A  \arrow{dr} & B \arrow{dr} & C & D \\
  A & \mathllap{B \times} B \mathrlap{\times B} \arrow{dr} & C \arrow{dr} & D \\
  A & B & C & D
\end{tikzcd}
\end{document}

Output

enter image description here

enter image description here

6
  • What happens if you put every matrix/.append style={...} in the optional argument to the \begin{tikzcd}? Does that provide the right hook? Jan 3, 2013 at 12:00
  • @AndrewStacey No, it does not. The problem, I guess, is that between origins is not even recognized in /tikz/commutative diagrams/column sep={10ex,between origins}. Jan 3, 2013 at 12:07
  • At least you can make it local with \makeatletter \tikzset{enforce orig/.code={\let\pgf@matrix@fixedfalse\pgf@matrix@fixedtrue}} \makeatother and use that option.
    – percusse
    Jan 3, 2013 at 13:10
  • This works well with labels on the arrows too. Just out of interest, why does the explicit &[between origins] version require different column/row spacing settings to get the same result as \let\pgf@matrix@fixedfalse\pgf@matrix@fixedtrue?
    – Dan
    Jan 3, 2013 at 13:31
  • @Dan I can’t follow. Both solutions had [column sep=10ex]] added to the tikzcd environment? Anyway, I have updated my answer with a hopefully more appropriate way to set between origins. (The manual way &[between origins] still works.) Jan 3, 2013 at 13:40
4

Try specifying the text width, etc., and alignment for the nodes. To stop this affecting labels on the arrows, you should put this inside the specification for the cells. For example, as follows:

enter image description here

\documentclass{article}

\usepackage{tikz,tikz-cd}

\begin{document}

\begin{tikzcd}[/tikz/cells={/tikz/nodes={shape=asymmetrical
  rectangle,text width=1.5cm,text height=2ex,text depth=0.3ex,align=center}}]
  A  \arrow{dr}{t} & B_X \arrow{dr}{p} & C & D \\
  A & B \arrow{dr} \times B \times B & C^Y \arrow{dr} & D \\
  A & B & C & D
\end{tikzcd}

\end{document}
3
  • I guess that to do this automatically one would have to measure all the columns, store that information in the aux file, and read it back in at the start of the table. Jan 3, 2013 at 11:50
  • Thanks! This is a great improvement, but it causes problems when the arrows are labelled, as the position of the label text is affected. (I tried to add an example here but it seems that's not possible in a comment.)
    – Dan
    Jan 3, 2013 at 11:55
  • OK I have updated my answer so only the cells get affected. Jan 3, 2013 at 12:28

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