Is there an easy way to reproduce the following image in latex?
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5Welcome to TeX.SX! On this site, a question should typically revolve around an abstract issue (e.g. "How do I get a double horizontal line in a table?") rather than a concrete application (e.g. "How do I make this table?"). Questions that look like "Please do this complicated thing for me" tend to get closed because they are either "off topic", "too broad", or "unclear". Please try to make your question clear and simple by giving a minimal working example (MWE): you'll stand a greater chance of getting help. – Stefan Pinnow Oct 13 '19 at 7:23
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\documentclass[tikz, margin=3mm]{standalone}
\usetikzlibrary{chains,
positioning
}
\usepackage{amsmath}
\begin{document}
\begin{tikzpicture}[
node distance = 1pt and 0pt,
start chain = going right,
block/.style = {rectangle, minimum width=1.5em, outer sep=0pt,
on chain},
dot/.style = {circle, fill, node contents={}}
]
\foreach \i in {0,...,14}{
\node (n\i) [block] {\i};
}
\draw (n0.north west) -- (n14.north east);
\draw (n1.south west) -- ++ (0,-0.2) -| (n13.south east)
node[pos=0.25,below] {$\text{Range} = 13 - 1 = 12$};
\node [dot,above=of n1];
\node [dot,above=of n2];
\node (d4) [dot,above=of n4];
\node [dot,above=of d4];
\node [dot,above=of n6];
\node (d8) [dot,above=of n8];
\node (d8a) [dot,above=of d8];
\node [dot, above=of d8a];
\node [dot, above=of n9];
\node (d11) [dot, above=of n11];
\node [dot, above=of d11];
\node [dot, above=of n12];
\node [dot, above=of n13];
\end{tikzpicture}
\end{document}
Addedndum (1): you can define nodes styles for two and three dots. With them the code for picture is:
\documentclass[tikz, margin=3mm]{standalone}
\usetikzlibrary{chains,
positioning
}
\usepackage{amsmath}
\begin{document}
\begin{tikzpicture}[
node distance = 1pt and 0pt,
start chain = going right,
block/.style = {rectangle, minimum width=1.5em, outer sep=0pt,
on chain},
dot/.style = {circle, fill, node contents={}},
ddot/.style = {circle, fill,
append after command={node[dot,above=of \tikzlastnode]},
node contents={}},
dddot/.style = {circle, fill,
append after command={node (aux) [dot,above=of \tikzlastnode]
node [dot,above=of aux]},
node contents={}}
]
\foreach \i in {0,...,14}{
\node (n\i) [block] {\i};
}
\draw (n0.north west) -- (n14.north east);
\draw (n1.south west) -- ++ (0,-0.2) -| (n13.south east)
node[pos=0.25,below] {$\text{Range} = 13 - 1 = 12$};
%
\node [dot,above=of n1];
\node [dot,above=of n2];
\node [ddot,above=of n4]; % <---
\node [dot,above=of n6];
\node [dddot,above=of n8]; % <---
\node [dot, above=of n9];
\node [ddot, above=of n11];% <---
\node [dot, above=of n12];
\node [dot, above=of n13];
\end{tikzpicture}
\end{document}
Addedndum (2): version with dots as labels in the loop. A bit shorter code:
\documentclass[tikz, margin=3mm]{standalone}
\usetikzlibrary{chains,
positioning
}
\usepackage{amsmath}
\begin{document}
\begin{tikzpicture}[
node distance = 1pt and 0pt,
start chain = going right,
block/.style = {rectangle, minimum width=1.5em, outer sep=0pt,
on chain},
dot/.style = {circle, fill, node contents={}},
ddot/.style = {circle, fill,
append after command={node[dot,above=of \tikzlastnode]}},
dddot/.style = {circle, fill,
append after command={node (aux) [dot,above=of \tikzlastnode]
node [dot,above=of aux]},
node contents={}}
]
\foreach \i [count=\j from 0] in { , dot, dot, ,ddot, , dot,
,dddot,dot, ,ddot, dot, dot, }{
\node (n\j) [block, label={[yshift=3pt,\i]}] {\j};
}
\draw (n0.north west) -- (n14.north east);
\draw (n1.south) -- ++ (0,-0.1) -| (n13.south)
node[pos=0.25,below] {$\text{Range} = 13 - 1 = 12$};
\end{tikzpicture}
\end{document}
Result is the same as before.
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"Result is the same" except you corrected an error (the too-short bar below). Excellent! – user3445853 Oct 15 '19 at 12:35
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@user3445853, almost the same ... :-), recently I experiences some difficulties to upload images, so I didn't provide the recent one :-( (which not differ very much from showed) – Zarko Oct 15 '19 at 12:42
Use pics. You can define a pic tower list which draws the circles. Getting all the x coordinates and circles is then as simple as saying
\pic{tower list={0,1,1,0,2,0,1,0,3,1,0,2,1,1,0}};
where the integers indicate how many circles should be drawn at 0, 1, 2....
\documentclass[tikz,border=3mm]{standalone}
\begin{document}
\begin{tikzpicture}[font=\sffamily,pics/tower/.style={code={
\ifnum#1>0
\foreach \X in {1,...,#1}{\fill (0,\X*0.4) circle[radius=0.17cm];}
\fi}},
pics/tower list/.style={code={\path foreach \Y [count=\Z starting from 0] in {#1}
{ (\Z*0.5,0)node[below](tower-\Z){$\mathsf{\Z}$} (\Z*0.5,0) pic{tower=\Y}};}}]
\begin{scope}
\pic{tower list={0,1,1,0,2,0,1,0,3,1,0,2,1,1,0}};
\draw (tower-0.west|-0,0) -- (tower-13.east|-0,0);
\draw[blue] (tower-1.south) -- ++(0,-0.1)
-| (tower-13.south)
node[pos=0.25,below]{Range${}=\mathsf{13}-\mathsf{1}=\mathsf{12}$};
\end{scope}
\end{tikzpicture}
\end{document}
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Hello, then with such a technique you can built an Italian abacus with units, tens, hundreds? +1. Can I ask a question on LaTeX with reference to your code? – Sebastiano Oct 13 '19 at 8:32
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2@Sebastiano Sure you can define an abacus pic (will be back in a bit...) – user194703 Oct 13 '19 at 8:41
You don't really need TiKZ for this. Just marvosym, booktabs and stackengine:
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{marvosym}
\usepackage[svgnames, table]{xcolor}
\newcommand\MVTen{\MVOne\MVZero}
\newcommand\MVxi{\MVOne\MVOne}
\newcommand\MVxii{\MVOne\MVTwo}
\newcommand\MVxiii{\MVOne\MVThree}
\newcommand\MVxiv{\MVOne\MVFour}
\usepackage{array,booktabs}
\usepackage[usestackEOL]{stackengine}
\begin{document}
\setstackgap{S}{0pt}
\begin{tabular}{*{15}{c}}
& \CircSteel & \CircSteel & & \Shortstack{\CircSteel\\ \CircSteel} & & \CircSteel & & \Shortstack{\CircSteel\\ \CircSteel\\ \CircSteel} & \CircSteel & & \Shortstack{\CircSteel\\ \CircSteel} & \CircSteel & \CircSteel \\[-0.6ex]
\midrule[1pt]
\MVZero & \MVOne & \MVTwo & \MVThree & \MVFour & \MVFive & \MVSix & \MVSeven & \MVEight & \MVNine & \MVTen & \MVxi & \MVxii & \MVxiii & \MVxiv \\[-1ex]
\arrayrulecolor{LightSlateBlue!60} &\color{LightSlateBlue!60} \rule{1pt}{1.3ex} & \multicolumn{11}{c}{} & \color{LightSlateBlue!60}\rule{1pt}{1.3ex} \\[-1.46ex]
\cmidrule[1pt](l{0.88em}r{1.19em}){2-14}\addlinespace[-0.3ex]
& \multicolumn{13}{c}{\sffamily\bfseries Range = 13 – 1 = 12}
\end{tabular}
\end{document}
