# Draw a $\epsilon$ neighborhood

I would like how to draw a $\epsilon$ neighborhood of a set with smooth boundary in "tikz".

                \documentclass[tikz, border=5mm]{standalone}
\usepackage{tikz}
\usepackage{amsmath}
\begin{document}

\begin{tikzpicture}[>=latex]
\draw plot [smooth cycle, tension=0.6] coordinates {(4.4,0.4) (5,0.2) (5.8,0.6) (6.5773,0.5421)(6.4905,1.1074)  (5.9752,1.2828) (5.4,1.4) (4.6,1) };
\end{tikzpicture}
\end{document}

• geogebra might be of help to you! Oct 25, 2016 at 12:58
• This question and answer would probably be useful to you. Most of the code you probably would like is in there. Oct 25, 2016 at 13:14
• Based on comments to answer #1, I it sounds more like copying an object (eg tex.stackexchange.com/questions/12106/tikz-copy-command) and using scale would be the neatest solution. Oct 25, 2016 at 14:15
• Copying and scaling would not keep the same distance between the two outlines all around ... Oct 25, 2016 at 14:53
• You are looking for the Minkowski addition of your curve and a disk. Using such keywords, maybe these links can help: tex.stackexchange.com/questions/27171/…, tex.stackexchange.com/questions/316219/…. Oct 25, 2016 at 19:14

The first thing to realise is that this is impossible. PDF, and therefore TikZ, can only draw cubic bézier lines and enlargening a cubic bézier by the same distance all around can produce something that is not a cubic bézier any more.

So any solution is going to be something of a hack.

Having said that, here's a solution that does produce a path that is a set distance from the original curve. It exploits the fact that when a line is drawn then its thickness obeys exactly the constraint that you are trying to force: that the line is drawn so that at any point on the curve, the width of the line as measured orthogonal to the curve is the given thickness. So if only there were a way to draw only the outer edge of a thick line when drawing the curve ...

That's what this does. To get the edge, we draw the line twice with the second time being white and a little thinner than the first time (as you want it dashed, I decided not to use the double as that can lead to artifacts when viewing the document). To get only the outer edge, we clip against the original curve.

The odd dashing effect is because the dashes are the correct width along the original curve but then scale proportionally through the thickness of the line (taking out the white over-draw shows what's going on there).

\documentclass[tikz, border=5mm]{standalone}
%\url{http://tex.stackexchange.com/q/335826/86}
\usepackage{tikz}
\usepackage{amsmath}
\begin{document}

\begin{tikzpicture}[>=latex]
\begin{scope}[even odd rule]
\clip (3,-1) rectangle (8,3) plot [smooth cycle, tension=0.6] coordinates {(4.4,0.4) (5,0.2) (5.8,0.6) (6.5773,0.5421)(6.4905,1.1074)  (5.9752,1.2828) (5.4,1.4) (4.6,1) };
\draw[line width=1cm,dashed] plot [smooth cycle, tension=0.6] coordinates {(4.4,0.4) (5,0.2) (5.8,0.6) (6.5773,0.5421)(6.4905,1.1074)  (5.9752,1.2828) (5.4,1.4) (4.6,1) };
\draw[line width=.9cm,white] plot [smooth cycle, tension=0.6] coordinates {(4.4,0.4) (5,0.2) (5.8,0.6) (6.5773,0.5421)(6.4905,1.1074)  (5.9752,1.2828) (5.4,1.4) (4.6,1) };
\end{scope}
\draw[line width=.5mm] plot [smooth cycle, tension=0.6] coordinates {(4.4,0.4) (5,0.2) (5.8,0.6) (6.5773,0.5421)(6.4905,1.1074)  (5.9752,1.2828) (5.4,1.4) (4.6,1) };
\end{tikzpicture}
\end{document}


Just remember:

If there's something strange,
Who're you gonna call?
TeX-busters!

• Ahaha! Brilliant! Oct 25, 2016 at 19:16
• Cases to test this one: sharp corners, shapes with narrow regions (smaller than epsilon), shapes with something behind it. Note that the outline thickness code basically does the work of generating pseudo-beziers and removing cusps; doing it manually is going to be really annoying in a real(tm) programming language, let alone TeX.
– Yakk
Oct 25, 2016 at 19:29
• @Yakk Exactly. A solution to the full problem is going to be quite tricky to do as there are many edge cases as you describe. I'm hoping that the OP doesn't need a full solution and that there's enough here to work with. Oct 25, 2016 at 20:13

If the boundary of the set is parameterized counterclockwise as (x(t),y(t)) then the boundary of its epsilon-neighborhood is parametrized as: Now suppose we are given only a finite set of, say, N points on the curve (x(t), y(t)), so t assumes integer values modulo N. Then we can approximate the derivative x'(t) by (x(t+1)-x(t-1))/2, and analogously for y'(t).

Putting this into practice:

\documentclass[tikz, border=5mm]{standalone}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
%\draw[help lines,line width=.6pt,step=1] (4,0) grid (7,2);
%\draw[help lines,line width=.3pt,step=.1] (4,0) grid (7,2);
\draw plot [smooth cycle, tension=0.6] coordinates
{(4.4,0.4)(4.6,0.25)(5,0.2)(5.4,0.4)(5.8,0.6)(6.3,0.55)(6.55,0.55)
(6.6,0.8)(6.5,1.1)(6.35,1.2)(6,1.3)(5.8,1.35)(5.4,1.4)(5,1.25)
(4.6,1)(4.45,0.8)};
\draw[dashed] plot [smooth cycle, tension=0.6] coordinates
{(4.3,0.37)(4.56,0.14)(5.03,0.06)(5.47,0.26)(5.83,0.44)(6.29,0.42)
(6.59,0.5)(6.7,0.81)(6.57,1.14)(6.39,1.29)(6.03,1.4)(5.82,1.46)
(5.38,1.54)(4.93,1.39)(4.52,1.1)(4.34,0.84)};
\end{tikzpicture}
\end{document}


Here I used twice the number the points as originally given. The result is not perfect (especially on the regions of larger curvature), but increasing the number of points should of course improve quality.

I did the calulations using a Google spreadsheet. I'm sure there is a way of telling TikZ to do the calculations, but I was too lazy to find it out. If someone wants to append my answer and automate this part, please feel free to do so.

With Asymptote, I take points along the curve, say G, then get points that is $\epsilon$ distant in the direction of the normal vector, and finally smoothly connect them. So we can draw the boundary of the $\varepsilon$-neighbourhood with desired accuracy.

This is enough for usual cases of small $\epsilon$ and G has quite simple shape. For more complex shapes of G, $\epsilon$-boundary curve may self-intersect, and some improvement is needed.

unitsize(1cm);
real e=.1;  // size of $\epsilon$
pair[] epoints;
pair[] points={(4.4,0.4),(4.6,0.25),(5,0.2),(5.4,0.4),(5.8,0.6),(6.3,0.55),(6.55,0.55),
(6.6,0.8),(6.5,1.1),(6.35,1.2),(6,1.3),(5.8,1.35),(5.4,1.4),(5,1.25),
(4.6,1),(4.45,0.8)};
guide c=operator..(...points)..cycle;

for (real t=0; t<length(c); t=t+.1){
pair pt=point(c,t);
pair qt=pt+scale(e)*rotate(-90)*dir(c,t);
epoints.push(qt);
//draw(circle(pt,e),cyan+opacity(.3));  // to see rolling circles along the curve
}

draw(c);
draw(operator..(...epoints)..cycle,red);


With rolling circles along the curve:

Update In several situations, we need e-neighbbourhood of a (planar) domain D bounded by curve c with not small value of e, that is Minkowski sum of D and the e-radius circle. In case of convex polygon c, the above approach works well.

unitsize(1cm);
real e=.5;  // size of $\epsilon$

// for convex polygon >> OK!
//pair[] points={(0,0), (5,0), (3,4), (1,3.5)};

// concave polygon >> fill is OK, draw is not OK
pair[] points={(0,0), (5,0), (2,1), (1,3.5)};

path c=operator--(...points)--cycle;

pair[] epoints;  // points on $\epsilon$-boundary of c
real tstep=.01;
for (real t=0; t<length(c); t=t+tstep){
pair pt=point(c,t);
pair qt=pt+scale(e)*rotate(-90)*dir(c,t);
epoints.push(qt);
//draw(circle(pt,e),cyan+opacity(.5));  // to see rolling circles along the curve
}

fill(operator..(...epoints)..cycle,yellow+opacity(.5));
draw(operator..(...epoints)..cycle,red);
draw(c);


However, in cases of concave polygon, filling inside the e-boundary is ok but drawing is not what expected due to self-intersection. At present I have no idea to overcome this.

With operator--, things are bad ^^

fill(operator--(...epoints)--cycle,yellow+opacity(.5));
draw(operator--(...epoints)--cycle,red);


• Wish the same could be done with TikZ! Jan 28, 2021 at 22:09
• @SebGlav As I have just updated, for Asymptote, even for a concave polygon, drawing its e-boundary is not what expected. Such elaborate oparations on path is too difficult to handle with in TikZ. Jan 30, 2021 at 11:40

This is a simple solution with ellipse shape, but maybe you were looking for that specific set shape..

\documentclass[tikz, border=5mm]{standalone}
\usepackage{tikz}
\usepackage{amsmath}
\begin{document}

\begin{tikzpicture}[>=latex]
\draw[dashed] (0,0) ellipse (4cm and 2cm);
\draw (0,0) ellipse (3cm and 1.5cm);
\draw[<->] (0,1.5) -- (0,2) node[right,midway] {$\epsilon$};
\end{tikzpicture}

\end{document}


• I Draw the following set Oct 25, 2016 at 13:05
• \documentclass[tikz, border=5mm]{standalone} \usepackage{tikz} \usepackage{amsmath} \begin{document} \begin{tikzpicture}[>=latex] \draw plot [smooth,tension=0.8] coordinates {(-1.3697,3.7545) (1.5,4) (3.5,5) (5.5,3) (8,2) (8,-0.5)} ; \end{tikzpicture} \end{document} Oct 25, 2016 at 13:05
• @VictorHugo Mmm impressive, but I think it has nothing to do with the question or with a possible solution Oct 25, 2016 at 13:10
• I need a curved set, as mentioned in the comment. Is there a command to increase proportionally a drawing? Oct 25, 2016 at 13:18
• The epsilon-neighborhood of a (non-circular) ellipse is not an ellipse. Oct 28, 2016 at 18:09

The accepted answer says, “this is impossible. PDF, and therefore TikZ, can only draw cubic bézier lines,” but I’m afraid that’s not quite true. If you convert to a coordinate system where the solution falls out simply and naturally, pgfplots can graph parameterized polar curves. (Technically, it does approximate them in terms of other primitives.)

A Bézier curve is mathematically equivalent to a cubic spline, and polar coordinates are just another coordinate system, so we can convert the control points of a Bézier curve to a piecewise cubic function in polar coordinates. And if you express the problem as a polar function, with the origin in the interior of your figure, adding ε to the radius of your function gives you a close approximation to the boundary of its ε-neighborhood. (See below for a proof.)

If we convert the control points you gave to the new coordinate system, that is, (θ, ρ) = (atan2(y′,x′), √(x′²+y′²)), where (x′, y′) are the translated Cartesian coordinates relative to the new origin we chose, we can interpolate ρ(θ) as a cubic-spline approximation. To make it a smooth, closed loop, we add 2π to the angle of the first two points, add these duplicates to the end of our list of points, and use natural boundary conditions.

Here is a plot of the original data points you gave (after translation), their polar cubic-spline interpolation, and its ε-neighborhood for ε=0.2.

\documentclass{standalone}
\usepackage{pgfplots}

\pgfplotsset{compat=1.17}
\usepgfplotslibrary{polar}

\begin{document}
\begin{tikzpicture}
\begin{axis}
-1.1    -0.6
-0.5    -0.8
0.3    -0.4
1.0773 -0.4579
0.9905  0.1074
0.4752  0.2828
-0.1     0.4
-0.9     0.0
};
\end{axis}
\end{tikzpicture}
\end{document}


A somewhat computationally-easier approximation would be to do what Jairo Bochi did and generate a new set of points at the boundary of the ε-neighborhood and draw a new loop through them. However, instead of estimating the delta between pairs of points, we can convert to polar, and add ε to the polar radius of each original point.

\documentclass{standalone}
\usepackage{pgfplots}

\pgfplotsset{compat=1.17}
\usepgfplotslibrary{polar}

\begin{document}
\begin{tikzpicture}
\begin{axis}
\addplot+ [fill, smooth cycle, mark=*, mark options={black}, draw=gray, fill=lightgray] coordinates {
(-1.2755791145828763, -0.6957704261361151)
(-0.605999788000636, -0.9695996608010178)
(0.4199999999999998, -0.56)
(1.261363262096108, -0.5361350020549593)
(1.1893345582858206, 0.12895964821796777)
(0.6470676596962922, 0.38508151128390444)
(-0.14850712500726604, 0.5940285000290663)
(-1.1000000000000003, 0)
};
\addplot+ [fill, smooth cycle, mark=*, mark options={black}, draw=black, fill=darkgray] coordinates {
(-1.1, -0.6)
(-0.5, -0.8)
(0.3, -0.4)
(1.0773, -0.4579)
(0.9905, 0.1074)
(0.4752, 0.2828)
(-0.1, 0.4)
(-0.9, 0.0)
};
\end{axis}
\end{tikzpicture}
\end{document}


## Update

I received a comment claiming that this approach “dilates the function from the chosen origin, but that isn't the same as the epsilon neighbourhood.” This is of course an approximation of a cubic-spline interpolation that gives us a closed subset of the true neighborhood. All the other answers are approximations of interpolations, too. I think that’s fine: the purpose here is to visualize a diagram, not to find an exact analytical solution. But here is a proof that the approximation is mathematically justified.

For a circle, where r'(t) = 0 everywhere, the solution is exact. There might be others (the expression I took the limit of looks like you could multiply by a conjugate instead, for instance). I’m sure anything I came up with in a few minutes has been studied long, long before.

Of course, since our r(t) is a differentiable periodic piecewise function composed of cubic polynomials, we could calculate exact values of x(t), y(t) and their derivatives at every t.

• What you write doesn't contradict my statement about this being mathematically impossible since you admit that pgfplots essentially "joins the dots" with primitives (lines and cubic béziers). Moreover, what you end up with is not an epsilon-neighbourhood. You have shifted the control points radially from some origin but that is not the same as taking the Minkowski sum of the original region with an epsilon-disc, which is how the question was originally interpreted. Jan 28, 2021 at 20:46
• @AndrewStacey I agree with your first point, and had mentioned that PGF approximates parameterized polar curves in terms of other primitives. Jan 28, 2021 at 20:49
• @AndrewStacey Only the second figure radially perturbs the control points. I call this a “computationally-easier approximation.” I agree that this is not mathematically equivalent. Jan 28, 2021 at 20:51
• On the first point, may I suggest rephrasing it as "nevertheless, we can still approximate it" rather than "I'm afraid that's not quite true" since your comment suggests that you agree that it is true. On the second, you write that you "adding ε to the radius of your function". This dilates the function from the chosen origin, but that isn't the same as the epsilon neighbourhood. The epsilon neighbourhood is defined as the Minkowski sum of the original region and an epsilon disc, so every point that is at most epsilon from a point in the original region is included. Jan 28, 2021 at 21:02
• @AndrewStacey All the answers are approximations of interpolations of a set of points. But, I added the word “approximately.” You are correct that it should have been there. I also gave a proof that the approximation is mathematically-justified. I’m sure anything I came up with in minutes has been studied long, long before. Jan 29, 2021 at 0:12

This should work.

\documentclass[border=0pt]{standalone}
\usepackage{mathtools,tikz}\usetikzlibrary{decorations.markings,arrows}
\begin{document}\begin{tikzpicture}\tikzset{every node}=[font=\huge]
\tikzstyle{coarselydashed}=[dash pattern=on 7pt off 8pt]
\draw[coarselydashed,line width=1pt,rounded corners=52pt](-1,3)--(2,0)--(6,3)--(10,0)--(13,3)--(12,6)--(6,9)--(1,7)--cycle;
\draw[shade,top color=black!96,bottom color=black!16,fill opacity=0.64,line width=2pt,rounded corners=39pt](0,3)--(2,1)--(6,4)--(10,1)--(12,4)--(10,6.4)--(6,8)--(1,6)--cycle;
\begin{scope}[decoration={markings,mark=at position 1 with {\arrow[scale=1.2]{triangle 45},color=black!96}}]
\draw[line width=1pt,-,postaction={decorate}](6,7.8) to (6,8.5);\node (epsilon) at (6.5,8.1) {$\epsilon$};
\end{scope}
\end{tikzpicture}\end{document}

• It works, but it is clear that we do not have the same distance from the dotted line. I'm looking for a command to do this. Oct 25, 2016 at 14:38