Diagonal strikeout starting too low and ending too high

Question

I want to strikeout an equation with diagonal line but I've not managed to get the diagonal line to draw as I want. The closest I've gotten is via the first example in Frédéric's answer of \cancel draws under thing being canceled:

\documentclass{minimal}

\usepackage{tikz}

\newcommand{\hcancel}[1]{%
    \tikz[baseline=(tocancel.base)]{
        \node[inner sep=0pt,outer sep=0pt] (tocancel) {#1};
        \draw[red] (tocancel.south west) -- (tocancel.north east);
    }%
}%

\begin{document}

\begin{equation}\label{eq:2}
\hcancel{$h_1 \land h_2 \land h_3 \land h_4 \land h_5 \land h_6 \land h_7 \land h_8 \land h_9 \land h_{10}$}
\end{equation}

\end{document}

This produces:

But the line starts too low and ends too high. The following is what I want:

I'd be happy for an option to make the line start slightly earlier and end slightly later too.

Answer 1

You can modify Frédéric's code so that \hcancel receives four more mandatory arguments controlling the vertical and horizontal shifting for the starting and ending points:

\documentclass{article}

\usepackage{tikz}
\usetikzlibrary{calc}
\newcommand{\hcancel}[5]{%
    \tikz[baseline=(tocancel.base)]{
        \node[inner sep=0pt,outer sep=0pt] (tocancel) {#1};
        \draw[red] ($(tocancel.south west)+(#2,#3)$) -- ($(tocancel.north east)+(#4,#5)$);
    }%
}%

\begin{document}

\begin{equation}\label{eq:1}
\hcancel{$h_1 \land h_2 \land h_3 \land h_4 \land h_5 \land h_6 \land h_7 \land h_8 \land h_9 \land h_{10}$}{0pt}{0pt}{0pt}{0pt}
\end{equation}

\begin{equation}\label{eq:2}
\hcancel{$h_1 \land h_2 \land h_3 \land h_4 \land h_5 \land h_6 \land h_7 \land h_8 \land h_9 \land h_{10}$}{-3pt}{3pt}{3pt}{-2pt}
\end{equation}

\end{document}

The new syntax:

\hcancel{<text>}{<start. point horiz. shifting>}{<start. point vertical shifting>}{<end. point horiz. shifting>}{<end. point vertical shifting>}

EDIT: using the xparse package, the definition of the new command is much more flexible; using something like

\usepackage{xparse}
\DeclareDocumentCommand{\hcancel}{mO{0pt}O{0pt}O{0pt}O{0pt}}{%
    \tikz[baseline=(tocancel.base)]{
        \node[inner sep=0pt,outer sep=0pt] (tocancel) {#1};
        \draw[red] ($(tocancel.south west)+(#2,#3)$) -- ($(tocancel.north east)+(#4,#5)$);
    }%
}%

allows the use of \hcancel{<text>} for the standard behaviour of the command as defined by Frédéric and to use the four (now optional) arguments to control the horizontal/vertical shifting:

\documentclass{article}
\usepackage{xparse}
\usepackage{tikz}
\usetikzlibrary{calc}

\DeclareDocumentCommand{\hcancel}{mO{0pt}O{0pt}O{0pt}O{0pt}}{%
    \tikz[baseline=(tocancel.base)]{
        \node[inner sep=0pt,outer sep=0pt] (tocancel) {#1};
        \draw[red] ($(tocancel.south west)+(#2,#3)$) -- ($(tocancel.north east)+(#4,#5)$);
    }%
}%

\begin{document}

\begin{equation}\label{eq:1}
\hcancel{$h_1 \land h_2 \land h_3 \land h_4 \land h_5 \land h_6 \land h_7 \land h_8 \land h_9 \land h_{10}$}
\end{equation}

\begin{equation}\label{eq:2}
\hcancel{$h_1 \land h_2 \land h_3 \land h_4 \land h_5 \land h_6 \land h_7 \land h_8 \land h_9 \land h_{10}$}[-3pt][3pt][3pt][-2pt]
\end{equation}

\end{document}

Answer 2

You can create nodes at the beginning and the end of the line, and shift them vertically:

\tikzstyle{nosep}=[inner sep=0pt, outer sep=0pt]
\newcommand{\hcancel}[1]{%
    \tikz[baseline=(tocancel.base)]{
        \node[nosep] (tocancel) {#1};
        \node[nosep, yshift=.5ex]  (from) at (tocancel.south west) {};
        \node[nosep, yshift=-.5ex] (to)   at (tocancel.north east) {};
        \draw[red] (from) -- (to);
    }%
}%

You can find the best shifts by trial and error. I chose .5ex and -.5ex arbitrarily.

Answer 3

Another possibility is to create a style.

\documentclass[]{scrartcl}
\usepackage{tikz}
\usetikzlibrary{calc}  


\begin{document}

\begin{tikzpicture}[cancel/.style={path picture={ \draw[#1]
($ (path picture bounding box.south west)+(-3pt,6pt)$) -- ($(path picture bounding box.north east)+(3pt,-6pt)$);
}}]  
\node   [inner sep=3pt,cancel=red] {$2x+3=y$};
\end{tikzpicture}

\end{document}

Answer 4

I use PSTricks in this answer. Please adjust the parameters until they suit your preference best. The parameters given in the code below are self-explanatory.

\documentclass{minimal}

\usepackage{pstricks-add}
\psset{linecolor=red}
\def\myeq{\psDefBoxNodes{A}{h_1 \land h_2 \land h_3 \land h_4 \land h_5 \land h_6 \land h_7 \land h_8 \land h_9 \land h_{10}}}
\begin{document}

\centering

Your settings:
\begin{equation}
\myeq
\ncline[nodesep=3pt,offsetA=-1pt,offsetB=3pt]{A:bl}{A:tr}
\end{equation}
\\[5mm]

Controlling the length:
\begin{equation}
\myeq
\ncline[nodesep=10pt,offsetA=-1pt,offsetB=3pt]{A:bl}{A:tr}
\end{equation}
\\[5mm]

Controlling the left node:
\begin{equation}
\myeq
\ncline[nodesep=3pt,offsetA=-1pt,offsetB=-10pt]{A:bl}{A:tr}
\end{equation}
\\[5mm]


Controlling the right node:
\begin{equation}
\myeq
\ncline[nodesep=3pt,offsetA=10pt,offsetB=3pt]{A:bl}{A:tr}
\end{equation}

\end{document}