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a/Master/texmf-dist/doc/latex/mh/xfrac.pdf +++ b/Master/texmf-dist/doc/latex/mh/xfrac.pdf diff --git a/Master/texmf-dist/source/latex/mh/breqn-technotes.tex b/Master/texmf-dist/source/latex/mh/breqn-technotes.tex index e8c6174cbbf..d87e2c538ba 100644 --- a/Master/texmf-dist/source/latex/mh/breqn-technotes.tex +++ b/Master/texmf-dist/source/latex/mh/breqn-technotes.tex @@ -1,10 +1,62 @@ -\documentclass{article} +\documentclass[twocolumn]{report} \usepackage{lmodern} \usepackage[T1]{fontenc} \usepackage{breqn} \def\eqnumside{L} \providecommand*\cs[1]{\texttt{\textbackslash#1}} +\providecommand{\qq}[1]{\textquotedblleft#1\textquotedblright} +\providecommand{\mdash}{\textemdash} +\providecommand{\ndash}{\textendash} + +\newcommand{\ititle}[1]{\textit{#1}} + +\newcommand{\LR}[2][.4]{% + \framebox[#1\displaywidth]{$\displaystyle{}#2$}% +} + +\newcommand{\LHS}[1]{\LR[\relifactor]{#1}} + +\newdimen\relindent \newdimen\rhswd + +\newcommand{\dwline}{% + \hbox to\curdw{\vrule height1ex + \leaders\hrule height.55ex depth-.45ex\hfil + \tiny \space display width + \leaders\hrule height.55ex depth-.45ex\hfil + \vrule height1ex}% +} + +\newenvironment{layout}[1][.15]{% + \noindent + $$\edef\curdw{\the\displaywidth}% + \def\relifactor{#1}% + \gdef\layoutcr{\cr}\def\\{\layoutcr}% + \binoppenalty 10000 \relpenalty 10000 + \setbox8\vbox\bgroup + \advance\baselineskip .35\baselineskip + \advance\lineskip .35\baselineskip \lineskiplimit\lineskip + \relindent=#1\displaywidth + \rhswd=\displaywidth \advance\rhswd-\relindent + \global\row 0 \gdef\rhsskew{}% + \halign\bgroup \global\advance\row 1 $\hfil\displaystyle{}##$&% + \ifnum\row>1 \rhsskew \fi $\displaystyle{}##\hfil$\cr +}{% + \crcr\egroup\egroup + \vcenter{\halign{\hfil##\hfil\cr + \hbox{\hss\dwline\hss}\cr\noalign{\vskip.6\baselineskip}\box8 \cr}}% + $$\relax + \ignorespacesafterend +} + +\newcommand{\stagger}{\omit\span\gdef\layoutcr{\cr\omit\span}} + +\newcount\row + +\newcommand{\rhsskew}{} +\newcommand{\skewleft}[1]{\gdef\rhsskew{\kern-#1\relax}} + + \usepackage{hyperref} \title{Technical notes on equation breaking} @@ -13,8 +65,9 @@ \maketitle +\onecolumn -\section{Tag placement} +\chapter{Tag placement} The method used by the breqn package to place the equation number is rather more complicated than you might think, and the whole reason is @@ -139,7 +192,271 @@ columnwidth and abovedisplayskip are used: \end{enumerate} +\twocolumn +\chapter{Equation Layouts} + +\section{Misc examples} + +Let us consider which of these have 50\% or more of wasted whitespace +\emph{within the bounding box of the visible material}. +\begin{layout}[.4] +\LHS{L}&=\LR[.35]{R_{1}}\\ +&=\LR[.25]{R_{1}} +\end{layout} + +\section{Ladder and step layouts} + +\subsection{Straight ladder layout} +This is distinguished by a relatively short LHS and one or more RHS's of +any length. +\begin{layout} +\LHS{L} &= \LR[.5]{R_{1}}\\ +&=\LR[.3]{R_{2}}\\ +&=\LR[.25]{R_{3}}\\ +&\qquad\ldots +\end{layout} +The simplest kind of equation that fits on one line and has only one RHS +may be viewed as a trivial subcase of the straight ladder layout: +\begin{layout} +\LHS{L} &= \LR[.5]{R} +\end{layout} +If some of the RHS's are too wide to fit on a single line they may be +broken at binary operator symbols such as plus or minus. This is still +classified as a straight ladder layout if none of the fragments intrude +into the LHS column, because the underlying parshape is the same. +\begin{layout} +\LHS{L} &= \LR[.4]{R_{1a}}\\ +&\quad +\LR[.6]{R_{1b}}\\ +&=\LR[.3]{R_{2}}\\ +&=\LR[.25]{R_{3a}}\\ +&\quad +\LR[.45]{R_{3b}}\\ +&\quad +\LR[.54]{R_{3c}}\\ +&\qquad\ldots +\end{layout} + +\subsection{Skew ladder layout} +\begin{layout}[.5] +\skewleft{.35\displaywidth} +\LHS{L}&= \LR[.3]{R_{1}}\\ +&=\LR[.6]{R_{2}}\\ +&=\LR[.25]{R_{3}}\\ +&\qquad\ldots +\end{layout} +In a skew ladder layout, the combined LHS width plus width of $R_{1}$ +does not exceed the available width, but one of the other RHS's is so +wide that aligning its relation symbol with the others cannot be done +without making it run over the right margin: $\mbox{width}(L) + +\mbox{width}_{\mathrm{max}}(R_{i})>\mbox{width}_{\mathrm{avail}}$. In +that case we next try aligning all but the first relation symbol, +allowing all the $R_{i}$ after $R_1$ to shift leftward. + +\subsection{Drop ladder layout} +\begin{layout}[.6] +\makebox[.15\displaywidth][l]{\LHS{L}}\\ +&= \LR[.6]{R_{1}}\\ +&=\LR[.3]{R_{2}}\\ +&=\LR[.25]{R_{3}}\\ +&\qquad\ldots +\end{layout} +The drop ladder layout is similar to the skew ladder layout but with the +width of $R_1$ too large for it to fit on the same line as the LHS. Then +we move $R_1$ down to a separate line and try again to align all the +relation symbols. Note that this layout consumes more vertical space +than the skew ladder layout. + +\subsection{Step layout} +\begin{layout}[.6] +\stagger +\LHS{R_{a}}\\ +\qquad + \LR[.7]{R_{b}}\\ +\qquad\qquad + \LR[.6]{R_{c}}\\ +\qquad\qquad\qquad + \LR[.45]{R_{d}}\\ +\qquad\qquad\qquad\qquad\ldots +\end{layout} +The chief characteristic of the step layout is that there is no relation +symbol, so that the available line breaks are (usually) all at binary +operator symbols. Let $w_1$ and $w_l$ be the widths of the first and +last fragments. We postulate that the ideal presentation is as follows: +Choose a small stairstep indent $I$ (let's say 1 or 2 em). We want the +last fragment to be offset at least $I$ from the start of the first +fragment, and to end at least $I$ past the end of the first fragment. If +there are only two lines these requirements determine a target width +$w_T=\max(w_1+I,w_l+I)$. If there are more than two lines ($l>2$) then +use $w_T = \max(w_1 + (l-1)I, w_l+I, w_{\mathrm{avail}}$ and reset $I$ +to $w_T/(l-1)$ if $w_T = w_{\mathrm{avail}}$. + +Furthermore, we would like the material to be distributed as evenly as +possible over all the lines rather than leave the last line exceedingly +short. If the total width is $1.1(\mbox{width}_{\mathrm{avail}})$, we +don't want to have .9 of that on line 1 and .2 of it on line 2: +\begin{layout}[.9] +\stagger +\LHS{R_{a}\hfil+\hfil R_{b}\hfil+\hfil R_{c}}\\ +\qquad + \LR[.1]{R_{d}} +\end{layout} +Better to split it as evenly as possible, if the available breakpoints +permit. +\begin{layout}[.5] +\stagger +\LHS{R_{a}\hfil+\hfil R_{b}}\\ +\qquad + \LR[.5]{R_{c}\hfil+\hfil R_d} +\end{layout} +A degenerate step layout may arise if an unbreakable fragment of +the equation is so wide that indenting it to its appointed starting +point would cause it to run over the right margin. In that case, we want +to shift the fragment leftward just enough to bring it within the right +margin: +\begin{layout}[.6] +\stagger +\LHS{L_{a}}\\ +\quad + \LR[.8]{L_{b}}\\ +\qquad + \LR[.7]{L_{c}}\\ +\; + \LR[.87]{L_{d}}\\ +\qquad\ldots +\end{layout} +And then we may want to regularize the indents as in the drop ladder +layout. Let's call this a dropped step layout: +\begin{layout}[.6] +\stagger +\LHS{L_{a}}\\ +\quad + \LR[.8]{L_{b}}\\ +\quad + \LR[.7]{L_{c}}\\ +\quad + \LR[.87]{L_{d}}\\ +\qquad\ldots +\end{layout} + +\section{Strategy} + +Here is the basic procedure for deciding which equation layout to use, +before complications like equation numbers and delimiter clearance come +into the picture. Let $A$ be the available width, $w_{\mathrm{total}}$ +the total width of the equation contents, $w(L)$ the width of the +left-hand side, $w_{\max}(R)$ the max width of the right-hand sides, $I$ +the standard indent for step layout, and $O$ the standard offset for +binary operators if a break occurs in the middle of an RHS. Also let +$t_L$ and $t_R$ represent certain thresholds for the width of the LHS or +the RHS at which a layout decision may change, as explained below. + +\newpage +\begin{small} +\begin{enumerate} +\renewcommand{\labelenumi}{(\theenumi)} +\item \ititle{Does everything fit on one line?}\label{i:LR} + $w_{\mathrm{total}}\leq A$? + +Yes: print the equation on a single line (done). + +No: Check whether the equation has both LHS and RHS (\ref{i:lhs-check}). + +\item \ititle{Is there a left-hand side?}\label{i:lhs-check} +Are there any relation symbols in the equation? + +Yes: Try a ladder layout (\ref{i:ladder}). + +No: Try a step layout (\ref{i:step}). + +\item\ititle{Does the LHS leave room to fit the widest RHS?}\label{i:ladder} + $w(L)+w_{\max}(R)<A$? + +Yes: Use a straight ladder layout (\ref{i:straight-ladder}). +No: Check the width of the LHS (\ref{i:check-lhs}). + +\item\ititle{Is the LHS relatively short?}\label{i:check-lhs} +$w(L)\leq t_L$? (where $t_L$ is typically $0.4A$). + +Yes: Subdividing one or more of the RHS's may permit us to use a +straight ladder layout (\ref{i:straight-ladder}). + +No: The straight ladder layout is unlikely to work. +Try a skew or drop ladder layout (\ref{i:skew-drop}). + +\item\ititle{Straight ladder layout}\label{i:straight-ladder} +Set up a straight ladder parshape [0pt $A$ $w(L)$ $A-w(L)$] and run +a trial break. If the combined width of the LHS plus the longest RHS is +no greater than $A$ then we should get a satisfactory layout with all +line breaks occurring at major division points (relation symbols). +Otherwise, we hope, some additional line breaks at minor division points +will allow everything to fit within the text column. + +\ititle{Line breaks OK?} + +\begingroup \hbadness=9999 +Yes: The straight ladder layout succeeded + (done).\par\endgroup + +No: Try a skew or drop ladder layout (\ref{i:skew-drop}). + +\item\ititle{Do the LHS and the first RHS fit on one + line?}\label{i:skew-drop} $w(L)+w(R_1) \leq A$? + +Yes: Try a skew ladder layout (\ref{i:skew}). + +No: Try a drop ladder layout (\ref{i:drop}). + +\item\ititle{Skew ladder layout}\label{i:skew} +Set up a parshape [0pt $A$ $I$ $A-I$] and run a trial break. + +\ititle{Line breaks OK?} + +Yes: Skew ladder layout succeeded (done). + +No: One of the unbreakable fragments of the $R_i$ ($i>1$) is wider than +$A-I$; try an almost-columnar layout (\ref{i:almost-columnar}). + +\item\ititle{Drop ladder layout}\label{i:drop} +Set up a parshape [0pt $w(L)$ $I$ $A-I$] and run a trial break. +This is the same parshape as for a skew ladder layout except that the +width of the first line is limited to the LHS width, so that the RHS is +forced to drop down to the next line. + +\ititle{Line breaks OK?} + +Yes: Drop ladder layout succeeded (done). + +No: One of the unbreakable fragments of the $R_i$ ($i>1$) is wider than +$A-I$; try an almost-columnar layout (\ref{i:almost-columnar}). + +\item\ititle{Almost-columnar layout}\label{i:almost-columnar} +This presupposes a trial break that yielded a series of expressions or +fragments, one per line. Let $w(F)$ denote the width of the first +fragment and $w(R_i)$ the widths of the remaining fragments. +Set up a parshape [0pt $w(F)$ $A-w_{\max}(R_i)$ $w_{\max}(R_i)$]: in other +words, set the first line flush left and the longest line flush right +and all other lines indented to the same position as the longest line. +But as a matter of fact there is one other refinement for extreme cases: +if $w_{\max}(R_i)>A$ then the parshape can be simplified without loss to +[0pt $w(F)$ 0pt $A$]\mdash for that is the net effect of substituting +$\min(A,w_{\max})$ in stead of $w_{\max}$. +(Done.) + +\item\ititle{Step layout}\label{i:step} Set target width $w_T$ to $A - + 2I$. Set parshape to [0pt $w_T$ $I$ $w_T -I$ $2I$ $w_T -2I$ \ldots\ + $(l-1)I$ $w_T - (l-1)I$], where $l=\lceil w_{\mathrm{total}}/A\rceil$ + is the expected number of lines that will be required. Trial break + with that parshape in order to find out the width of the last line. + +\ititle{Indents OK?} + +Yes: Step layout succeeded (done). + +No: One of the fragments is too wide to fit in +the allotted line width, after subtracting the indent specified by the +parshape. Try a dropped step layout (\ref{i:drop-step}) + +\item\ititle{Dropped step layout}\label{i:drop-step} Set up a parshape + [0pt $A$ $I$ $A-I$] and run a trial break. Note that this is actually + the same parshape as for a skew ladder layout. + +\ititle{Line breaks OK?} + +Yes: Dropped step layout succeeded (done). + +No: One of the unbreakable fragments of the $R_i$ ($i>1$) is wider than +$A-I$; as a last resort try an almost-columnar layout (\ref{i:almost-columnar}). + +\end{enumerate} +\par\end{small} \end{document}
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