8

I would like adaptively sized delimiters, but only up to \bigl/\bigr-size. That is, there is variation only between ordinary/normal size and one size higher.

\documentclass{article}
\usepackage{amsmath}

\newcommand*{\abs}[1]{\left\lvert#1\right\rvert}


\begin{document}

\(\abs{M} + \abs{M^2} + \abs{\abs{M^2}^2}\)
  % the outer delimiters on the right are of the size of \Bigl and \Bigr
  % (i.e. the next larger size after that of \bigl and \bigr)

\(\abs{M} + \abs{M^2} + \bigl\lvert \abs{M^2}^2 \bigr\rvert\)

\end{document}

How can I achieve this? For the above code, this would amount to defining \abs in a way that makes the two lines look the same.

(Here the default behavior of \left and \right (i.e. scaling to \bigl/\bigr) looks better, but I'm trying to keep the example short; I'd have to load a lot of packages to demonstrate my personal need. Also, a simple reason is that I need to embed certain formulas in paragraph text.)

1 Answer 1

9

The following example measures the formula inside \abs with \left/\right and with \bigl/\bigr. If the version \left/\right is larger, then the version with \bigl/\bigr is used.

\documentclass{article}
\usepackage{amsmath}
\usepackage{mleftright}

\makeatletter
\newcommand*{\abs}[1]{%
  \mathpalette\@abs{#1}%
}
% Version with the horizontal spacing of `\left` and `\right` in all cases:
% \newcommand*{\abs}[1]{%
%   \mathinner{\mathpalette\@abs{#1}}%
% }
\newcommand*{\@abs}[2]{%
  \sbox0{$\m@th#1\mleft\lvert#2\mright\rvert$}%
  \sbox2{$\m@th#1\bigl\lvert#2\bigr\rvert$}%   
  \ifdim\wd0>\wd2 %
    \bigl\lvert#2\bigr\rvert
  \else
    \ifdim\ht0>\ht2 %
      \bigl\lvert#2\bigr\rvert
    \else
      \ifdim\dp0>\dp2 %
        \bigl\lvert#2\bigr\rvert
      \else
        \mleft\lvert#2\mright\rvert
      \fi
    \fi
  \fi
}
\makeatother

\begin{document}

\(\abs{M} + \abs{M^2} + \abs{\abs{M^2}^2}\)

\(\abs{M} + \abs{M^2} + \bigl\lvert \abs{M^2}^2 \bigr\rvert\)

\end{document}

Result

Remarks:

  • Instead of \left/\right macros \mleft/\mright of package mleftright are used to avoid the additional spacing of \left/\right.

  • The method is not "bulletproof". It makes the following assumption. If the delimiters, picked by \mleft and \mright, are larger than the version with \bigl and \bigr, then this case can only be detected, if at least either the width, height or depth increases. For the case that the width remains the same, then the larger delimiters need to be slightly larger than the formula, otherwise the formula would hide the height/depth of the delimiters. I haven't analyzed all resizeable delimiters, but I think, usually the width of the delimiter will become larger that is easily detected.

Second approach

This approach implements the algorithm of minimal delimiter size according to "The TeXbook" by D. E. Knuth, "Appendix G: Generating Boxes from Formulas", item "19.":

If the math list begins and ends with boundary items, compute the maximum height h and depth d of the boxes in the translation of the math list [...]. Let a = σ22 be the axis height, and let δ = max(h-a, d+a) be the amount by which the formula extends away from the axis. Replace the boundary items by delimiters whose height plus depth is at least max(⎣δ/500⎦f, 2δ-l), where f is the \delimiterfactor and l is the \delimitershortfall. [...]

It can deal correctly with an expression like M^{2^{2^2}}_{f_{f_f}}, which trips up the first approach by violating its assumption about delimiter dimensions.

Example file:

\documentclass{article}
\usepackage{amsmath}
\usepackage{mleftright}

\makeatletter
\newcommand*{\abs}[1]{%
  \mathpalette\@abs{#1}%
}
% Version with the horizontal spacing of `\left` and `\right` in all cases:
% \newcommand*{\abs}[1]{%
%   \mathinner{\mathpalette\@abs{#1}}%
% }
\newcommand*{\@abs}[2]{%
  % a := math axis height -> \dimen0
  \sbox0{$#1\vcenter{}$}%
  \dimen0=\ht0 %
  % formula without delimiters -> \box0
  % h := height of formula -> \ht0
  % d := depth of formula -> \dp0
  \sbox0{$#1#2$}%
  % delta := max(h-a, d+a) -> \dimen2
  \dimen2=\dimexpr\ht0-\dimen0\relax
  \dimen4=\dimexpr\dp0+\dimen0\relax
  \ifdim\dimen4>\dimen2 %
    \dimen2=\dimen4 %
  \fi
  % delimiter's total height >= max(floor(delta/500)*f, 2*delta - l)
  %   -> \dimen4
  % f := \delimiterfactor
  % l := \delimitershortfall
  \dimen4 = \dimen2 %
  \divide\dimen4 by 500 %
  \dimen4=\delimiterfactor\dimen4 %
  \dimen6=\dimexpr 2\dimen2 - \delimitershortfall\relax
  \ifdim\dimen6>\dimen4 %
    \dimen4=\dimen6 %
  \fi
  % formula with \bigl/\bigr
  \sbox0{$#1\bigl\lvert\bigr\rvert$}%
  % comparison: Use \mleft/\mright, if \bigl/\bigr is larger than needed
  \ifdim\dimexpr\ht0+\dp0\relax>\dimen4 %
    % Case 1: \bigl/\bigr is larger than the minimum needed height.
    %   This means that either (\lvert,\rvert) or (\bigl\lvert,\bigr\rvert)
    %   provide the smallest delimiter size that meet the minimum height requirement.
    %   Consequently we can have \mleft/\mright choose (they won't pick anything
    %   larger than \bigl/\bigr).
    \mleft\lvert#2\mright\rvert
  \else
    % Case 2: \bigl/\bigr is exactly right or smaller than the required minimum height
    %   => use \bigl\bigr
    \bigl\lvert#2\bigr\rvert
  \fi
}
\makeatother

\begin{document}

\(\abs{M} + \abs{M^2} + \abs{\abs{M^2}^2}\)

\(\abs{M} + \abs{M^2} + \bigl\lvert \abs{M^2}^2 \bigr\rvert\)

\( \abs{M^{2^{2^2}}_{f_{f_f}}} \)

\end{document}

Result

14
  • Can I replace (\mleft, \mright) by (\left, \right) in your code, or are there caveats I need to know about if I do this? Perhaps I should ask a different question: I understand that mleftright makes sure to not generate the mathinner character class (or should I write "mathinner spacing"?). I would like any "size-limited \left and \right" functionality to behave like what I ordinarily do in my documents. When exactly will (La)TeX normally generate mathinner? Do you have a good pointer so that I can decide about \mleft/\mright vs \left/\right? Commented Jun 8, 2013 at 4:21
  • 1
    @LoverofStructure: A formula with \left and \right is a \mathinner subformula that might be surrounded by additional space depending on the context. However, \bigl and \bigr do not have these special treatment. Therefore I have used \mleft and \mright so that \abs has a consistent spacing that does not change if the height/depth of the formula is larger or smaller. But feel free to add \mathinner if you want to have the additional spacing. Commented Jun 8, 2013 at 4:40
  • Thanks, this is a lucid explanation. Do I understand it correctly that achieving consistent mathinner classification would be achieved by doing the following two things? (1) replacing the 4 occurrences of \bigl\lvert#2\bigr\rvert by \mathinner{\bigl\lvert#2\bigr\rvert} and (2) replacing the 2 occurrences of \mleft\lvert#2\mright\rvert by \left\lvert#2\right\rvert. Commented Jun 8, 2013 at 12:02
  • @LoverofStructure: Yes, this should do it and you won't need package mleftright. Commented Jun 8, 2013 at 16:42
  • 1
    @LoverofStructure: Yes, you always need both \left and \right in this order. But a delimiter can be invisible by using a dot, e.g. \right.. In this case TeX inserts the horizontal space \nulldelimiterspace. Default is 1.2pt in LaTeX. Commented Jun 9, 2013 at 21:35

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