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3

An easy and flexible way is using plain TikZ. \documentclass{memoir} \usepackage{tikz} \usepackage{amsmath,amssymb} \begin{document} \begin{center} \begin{tikzpicture}[>=stealth] \path[left] (0,0) node (A) {\r{a}ben} ++(0,-1.6) node (L) {lukket} ++(0,-1.2) node (D) {dobbeltlukket}; \path[right] (A.east) +(1,.5) node (A1) {ka} +(1,-.5) node (A2) {...


6

One possibility with use of the tikz-cd package: \documentclass{memoir} \usepackage{tikz-cd} \begin{document} \begin{center} \begin{tabular}{c @{\hspace{4em}} l} \begin{tikzcd}[row sep=small, baseline=(current bounding box.center)] & \rlap{aaa} \\ \llap{x} \ar[ru]\ar[rd] & \\ &...


3

TikZ allows you to locate the point located at the intersection of the horizontal line passing through a and the vertical of foo1 that you are looking for is the syntax (foo1|-a) or both with the syntax (a-|foo1). I quote the TikZ 3.1.4 manual In general, the meaning of (p |- q) is the intersection of a vertical line through p and a horizontal line ...


1

Like this? \documentclass{article} \usepackage{tikz} \usetikzlibrary{positioning,intersections} \begin{document} \begin{tikzpicture}[node distance=-\pgflinewidth] \node[fill=black] (a) {}; \node[fill=black, xshift=\linewidth] (c) {}; \node[xshift=\linewidth*0.7] (b) {...}; \draw [name path=line1] (a) -- (b); \...


8

I don't know of a built-in way to do it, but here is a short function I wrote to do it. It uses the built-in length() and subpath() functions. unitsize(5cm); settings.outformat = "png"; void segmentArrows(path p) { for (int i = 0; i < length(p); ++i) { draw(subpath(p, i, i+1), arrow=Arrow); } } segmentArrows((0,0)--(1,0.2)--(0,1)--(2,...


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