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1

With package bidi-atbegshi, this even becomes much easier (you only have to change \AtBeginShipoutUpperLeft in @egreg's code to \AtBeginShipoutUpperRight: \documentclass{article} \usepackage{bidi-atbegshi,picture,xcolor} \usepackage[RTLdocument]{bidi} \AtBeginShipout{% \AtBeginShipoutUpperRight{% {\color{red}% \put(\dimexpr 1in+\oddsidemargin, ...

0

Just for fun solution with PSTricks. \documentclass[pstricks,border=15mm,12pt]{standalone} \psset{unit=6mm,dimen=monkey,linewidth=2pt} \begin{document} \begin{pspicture}[showgrid](31,31) \psframe[linecolor=red](8,8)(23,23) \psframe[linecolor=blue](12,12)(19,19) \rput(0,-1.5){0}\rput(31,-1.5){1}\psline{<->}(1,-1.5)(30,-1.5) ...

4

Here a short MWE. See the pgfmanual for descriptions. Adjust the scale-option to your needs (to get the image fit to your page-size) \documentclass[tikz, border=5mm]{standalone} \begin{document} \begin{tikzpicture}[scale=.5] % Grid \draw [step=1, dotted] (0,0) grid (31,31); % Grid units \foreach \x in {0,...,31} { \node [below] at (\x,0) {\x}; ...

4

Jesse's comment is probably easiest, but something along the lines of the code below works, also \documentclass{article} \usepackage[margin=1in]{geometry} \usepackage{tikz} \begin{document} \begin{center} \begin{tikzpicture}[scale=0.5] \foreach\x in {0,...,31} \draw(\x,31)--(\x,0) node[anchor=north]{\x} (31,\x)--(0,\x) node[anchor=east]{\x}; ...

3

And a Metapost approach... prologues := 3; outputtemplate := "%j%c.eps"; beginfig(1); % r = side of hexagon, n = repetitions of the grid (- and +) r = 5mm; n=10; % make a shape to draw path tri; tri = for t=0 step 120 until 359: origin -- (r,0) rotated t -- endfor cycle; % save the pattern as a picture centered on the origin picture grid; grid = ...

3

Another way could be to draw hexagonal nodes over an adjusted coordinate system. The idea came adapting Paul Gaborit's Pascal triangle for How can I draw Pascal's triangle with some its properties?. shapes.geometric library helps to draw hexagon where the minimum size is the diameter of the circumcircle. Therefore, selecting adjusted values for x ...

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