答案1
由于你似乎并不要求曲线平滑（因为如何绘制周长相同的随机简单封闭平滑曲线？decoration=penciline确实如此），您可以尝试调整或的参数\freedraw：
参考：
代码：
\documentclass{article}
\pagestyle{empty}
\usepackage{tikz}
\usetikzlibrary{calc,decorations.pathmorphing,patterns,shapes}
%% https://tex.stackexchange.com/questions/39296/simulating-hand-drawn-lines: percusse
\pgfdeclaredecoration{penciline}{initial}{
\state{initial}[width=+\pgfdecoratedinputsegmentremainingdistance,auto corner on length=1mm,]{
\pgfpathcurveto%
{% From
\pgfqpoint{\pgfdecoratedinputsegmentremainingdistance}
{\pgfdecorationsegmentamplitude}
}
{% Control 1
\pgfmathrand
\pgfpointadd{\pgfqpoint{\pgfdecoratedinputsegmentremainingdistance}{0pt}}
{\pgfqpoint{-\pgfdecorationsegmentaspect\pgfdecoratedinputsegmentremainingdistance}%
{\pgfmathresult\pgfdecorationsegmentamplitude}
}
}
{%TO
\pgfpointadd{\pgfpointdecoratedinputsegmentlast}{\pgfpoint{0.5pt}{1.5pt}}
}
}
\state{final}{}
}
%% https://tex.stackexchange.com/questions/39296/simulating-hand-drawn-lines: Alain Matthes
\pgfdeclaredecoration{free hand}{start}
{
\state{start}[width = +0pt,
next state=step,
persistent precomputation = \pgfdecoratepathhascornerstrue]{}
\state{step}[auto end on length = 3pt,
auto corner on length = 3pt,
width=+2pt]
{
\pgfpathlineto{
\pgfpointadd
{\pgfpoint{2pt}{0pt}}
{\pgfpoint{rand*0.15pt}{rand*0.15pt}}
}
}
\state{final}
{}
}
\tikzset{free hand/.style={
decorate,
decoration={free hand}
}
}
\def\freedraw#1;{\draw[free hand] #1;}
\begin{document}
\textbf{decoration=penciline}
\par
\begin{tikzpicture}
\coordinate (A) at (0,0);
\coordinate (B) at (4,0);
\coordinate (C) at (7,0);
\begin{scope}[decoration=penciline,scale=1]
\draw[thick, fill=blue!25, fill opacity=.25, draw=red, decorate] (A) rectangle (2,2);
\draw[thick, fill=green!25, draw=brown, radius=1cm, decorate] (B) circle ;
\draw[thick, fill=red!20, draw=blue, x radius=1cm, y radius=1.5cm, rotate=30, shape=circle, decorate,] (C) circle ;
\end{scope}
\end{tikzpicture}
\textbf{\textbackslash freedraw}
\par
\begin{tikzpicture}
\coordinate (A) at (0,0);
\coordinate (B) at (4,0);
\coordinate (C) at (7,0);
\freedraw[thick, fill=brown!25, draw=blue] (A) rectangle (2,2);
\freedraw[thick, fill=violet!25, fill opacity=.25, draw=red] (B) circle [radius=1cm];
\freedraw[thick, fill=orange!25, fill opacity=.25, draw=brown, x radius=0.15cm, y radius=1.5cm, rotate=30, shape=circle,] (C) circle {};
\end{tikzpicture}
\end{document}
答案2
这里有一个分形解决方案流畅的线条。
两个圆形和两个三角形的示例：
代码：
\documentclass[convert={size=480},margin=1mm]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\usetikzlibrary{decorations.pathreplacing}
\tikzset{
fractal lineto/.style n args={2}{%
% #1 is a ratio of length to move the middle of each segment
% #2 is the mininum length to apply the recurrence
to path={
let
\p1=(\tikztostart), % start point
\p2=(\tikztotarget), % end point
\n1={veclen(\x1-\x2,\y1-\y2)}, % distance
\p3=($(\p1)!.5!(\p2)$), % middle point
\p4=(rand*#1*\n1,rand*#1*\n1), % random vector
\p5=(\x3+\x4,\y3+\y4) % random moved middle point
in \pgfextra{
\pgfmathsetmacro\mytest{(\n1<#2)?1:0}
\ifnum\mytest=1 %
\tikzset{fractal lineto/.style n args={2}{line to}}
\fi
} to[fractal lineto={#1}{#2}] (\p5) to[fractal lineto={#1}{#2}] (\p2)
},
},
%
fractal curveto/.style n args={4}{
to path={
% % #1 is ratio of length to move the middle of each segment
% % #2 is the mininum length to apply the recurrence
let
\p0=(\tikztostart),
\p1=(#3),
\p2=(#4),
\p3=(\tikztotarget),
\p4=($(\p0)!.5!(\p1)$),
\p5=($(\p1)!.5!(\p2)$),
\p6=($(\p2)!.5!(\p3)$),
\p7=($(\p4)!.5!(\p5)$),
\p8=($(\p5)!.5!(\p6)$),
\p9=($(\p7)!.5!(\p8)$),
\n1={veclen(\x0-\x0,\y0-\y9)+veclen(\x9-\x3,\y9-\y3)}, % distance
\p{rand}=(rand*#1*\n1,rand*#1*\n1), % random vector
\p{randang}=(rand*#1*\n1,rand*#1*\n1), % random vector
\p{new9}=(\x9+\x{rand},\y9+\y{rand}), % random moved middle point
\p{new7}=(\x7+\x{rand},\y7+\y{rand}), % random moved control point
\p{new8}=(\x8+\x{rand},\y8+\y{rand}) % random moved control point
in \pgfextra{
\pgfmathsetmacro\mytest{(\n1<#2)?1:0}
\ifnum\mytest=1 %
\tikzset{
fractal curveto/.style n args={4}{
curve to,controls=(####3) and (####4)
}
}
\fi
%\typeout{p9:\p9}
}
to[fractal curveto={#1}{#2}{\p4}{\p{new7}}] (\p{new9})
to[fractal curveto={#1}{#2}{\p{new8}}{\p{6}}] (\p3)
},
},
deformation/.style n args={3}{decorate,decoration={show path construction,
lineto code={
\path[#3]
(\tikzinputsegmentfirst)
to[fractal lineto={#1}{#2}]
(\tikzinputsegmentlast);
},
curveto code={
\path[#3]
(\tikzinputsegmentfirst)
to[fractal curveto=%
{#1}{#2}{\tikzinputsegmentsupporta}{\tikzinputsegmentsupportb}]
(\tikzinputsegmentlast);
},
closepath code={
\path[#3]
(\tikzinputsegmentfirst)
to[fractal lineto={#1}{#2}]
(\tikzinputsegmentlast);
},
},
}
}
\begin{document}
\begin{tikzpicture}
\pgfmathsetseed{\pdfuniformdeviate 10000000}
\def\ratio{.1}
\def\minlen{10mm}
\begin{scope}
\draw[deformation={\ratio}{\minlen}{draw=red,line width=1mm}] circle(5cm);
\draw[deformation={\ratio}{\minlen}{draw=blue,line width=1mm}] circle(5cm);
\end{scope}
\begin{scope}
\draw[deformation={\ratio}{\minlen}{draw=lime,line width=1mm}]
(0:4) -- (120:4) -- (-120:4) -- cycle;
\draw[deformation={\ratio}{\minlen}{draw=orange,line width=1mm}]
(0:4) -- (120:4) -- (-120:4) -- cycle;
\end{scope}
\end{tikzpicture}
\end{document}
答案3
这是我的贡献：
\documentclass[border=7mm]{standalone}
\usepackage{tikz}
% create some random points arround 0
% #1 is the number of points
% #2 is the minimal radius
% #3 is the maximal deviation (if =0 no randomness)
\newcommand{\rndpts}[3]{
\def\pts{}
\foreach[
evaluate=\x as \r using {#2+#3*rnd},
evaluate=\x as \a using {\la+720*rnd/#1},
remember=\a as \la (initially 0)]
\x in {0,...,#1}
{
\pgfmathparse{int(\a)}
\ifnum\pgfmathresult > 360\relax
\breakforeach
\else
\xdef\pts{\pts (\a:\r)}
\fi
}
}
\begin{document}
\begin{tikzpicture}
\foreach \npts/\rmin/\rdelta/\c in {10/1/2/red,20/1/3/green,30/1/4/blue,20/2/3/yellow} {
\rndpts{\npts}{\rmin}{\rdelta}
\draw[\c, ultra thick] plot[smooth cycle,tension=.7] coordinates {\pts};
}
\end{tikzpicture}
\end{document}
