# Tag Info

6

There is tikzmark library for this. You can make one handmade macro yourself too. \documentclass{article} \usepackage{tikz} \newcommand{\tikzmark}[3][]{\tikz[remember picture,baseline] \node [inner xsep=0pt,anchor=base,#1](#2) {#3};} \begin{document} $\tikzmark{a}{a_0} \to a_1 \to a_2 \to \dots \to \tikzmark{b}{a_k}$ ...

5

This is simply because your are using rounded corners rectangles. So, you should compensate for the trim effect (if you want the arrows to touch the rectangles) by issuing the option shorten >= -2pt, shorten <= -2pt. You can manually control the shorten distance. Note that you need only to compensate the two edges to node e3. Here is the modified ...

4

The following solution with TikZ works without \tikzmark and additional LaTeX run for remembering positions. The arrow head of the curved arrow matches the default appearance of LaTeX's \rightarrow or its synonym \to. Also the start of the error uses a round line cap. Parameter looseness is used to flatten the curve a bit to reduce the vertical space ...

4

You are trying to solve a problem in 3d geometry using 2d (screen) coordinates. It can't be done. There is no way to determine what plane (b) is supposed to be in just given its screen coordinates. The following is my best approximation to the mwe using 3d coordinates. My guesses are in blue. \documentclass{standalone} \usepackage{tikz} ...

4


3

I just combined the two answers that you cite, using David's technique applied to Heiko's code. The credit goes to them. Here was my addition to Heiko's code: \newcommand\xdashmapsto[2][]{\mathrel{\mapstochar\xdashrightarrow[#1]{#2}}} The only twist was to remember to employ arguments in the same manner as employed by \xdashrightarrow. ...

3

Second solution shows how to do this without usage of TikZ. Only pdfTeX primitives are expected. {\lccode\?=\p \lccode\!=\t \lowercase{\gdef\ignorept#1?!{#1}}} \def\usedim#1 {\expandafter\ignorept\the#1 \space} \def\cyclicseq#1{\setbox0=\hbox{\kern-.7em$#1$\kern-.7em}% \dimen0=.3\wd0 \dimen1=.7\wd0 \leavevmode \kern.7em ...

2

Use ,baseline=(current bounding box.center) in the options of circuitikz \documentclass[11pt]{article} \usepackage[margin=1in]{geometry} \usepackage{graphicx} \usepackage[english]{babel} \usepackage{circuitikz} \usepackage{tikz} \usetikzlibrary{shapes.geometric, arrows} \begin{document} \begin{center} \begin{circuitikz}[american ...

2

You could use mathtools. \documentclass{article} \usepackage{mathtools} \begin{document} $\xrightarrow{print(5)}$ \end{document}

2

For this kind of drawings I usually use the tikz-3dplot package that offers good 3d/perspective capabilities. It makes it easy to draw the projection of a point both in the main reference frame and in the rotated reference frame. The useful command in this case is \tdplotsetcoord, used as \tdplotsetcoord{<name>}{<r>}{<theta>}{<phi>} ...

1

Use \overleftrightarrow and \underleftrightarrow. \documentclass[a4paper]{article} \usepackage{amsmath} \begin{document} \begin{align} &\hphantom{(2\times 3)( }% \overleftrightarrow{\hphantom{1 \times 3) = (2 \times 1} } \nonumber \\[-07pt]% &( \underleftrightarrow{ 2\times \underleftrightarrow{ 3) ( 1 \times ...

1

Here is something to start with. The only thing you will have to do is to smash the diagonal dots in order to get them on the same height as the other stuff: % arara: pdflatex \documentclass{standalone} \usepackage{tikz-cd} \newcommand{\placeholder}{\phantom{(0,n)}} \begin{document} \begin{tikzcd}[% ,row sep={1cm,between origins} ,column ...

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