How can I invert a 'clip' selection within TikZ?

Question

I wish to draw an arc for a reflex angle on the outside of a triangle with coordinates (A), (B) and (C).

If I wanted to draw the arc inside the triangle I could use something like

\begin{scope}
\path[clip] (A) -- (B) -- (C) -- cycle;
\draw (B) circle (2mm);    % this is the little angle marker
\end{scope}

Is there a straightforward way to exclude the part of the circle that's in the triangle, as in the pseudo-code below?

\begin{scope}
\path[magicalinverseclipcommand] (A) -- (B) -- (C) -- cycle;
\draw (B) circle (2mm);    % this is the little angle marker
\end{scope}

EDIT: Although I included the code above as a motivating example of what an inverse clip would do, I would prefer a general solution that works systematically with any closed path, and not just in the particular example of a triangle with a circle clipped out of it. Looking at my question, I realise this isn't clear, and I will be happy with the best solution to my particular problem, but wonder if there is a less ad hoc solution than those listed so far.

Answer 1

What you can do is add a rectangle to your clipping path that's larger than the current bounding box, and clip with that. Andrew Stacey suggested using the current page as the clipping rectangle, because that will catch all elements that follow. By using the pgfinterruptboundingbox environment when defining the clipping rectangle, the actual size of the tikzpicture will not be influenced.

Note that, in order to use the current page, the remember picture,overlay options need to be passed to the tikzpicture, and you need two compile runs to get the positioning of all the elements right. Furthermore, this doesn't work with the minimal documentclass.

\documentclass{article} % Has to be a proper class, not minimal
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}[remember picture,overlay]

% A path that follows the edges of the current page
\tikzstyle{reverseclip}=[insert path={(current page.north east) --
  (current page.south east) --
  (current page.south west) --
  (current page.north west) --
  (current page.north east)}
]

\coordinate (A) at (0,0);
\coordinate (B) at (1,0);
\coordinate (C) at (1,1);

\begin{pgfinterruptboundingbox} % To make sure our clipping path does not mess up the placement of the picture
\path [clip] (A) -- (B) -- (C) -- cycle [reverseclip];
\end{pgfinterruptboundingbox}

\draw[thick] (A) circle (2mm);
\draw[thick] (B) circle (2mm);    
\draw[thick] (C) circle (2mm);   

\end{tikzpicture}
\end{document}

---

And just to show that it works for the general case:

\documentclass{article}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}[remember picture,overlay]

% A path that follows the edges of the current page
\tikzstyle{reverseclip}=[insert path={(current page.north east) --
  (current page.south east) --
  (current page.south west) --
  (current page.north west) --
  (current page.north east)}
]

\draw [step=0.1,red] (0,0) grid  (2,2);

\begin{pgfinterruptboundingbox} % To make sure our clipping path does not mess up the placement of the picture
\path [clip,rounded corners] (0,0) -- (.75,0) -- (1.2,.8) -- (2,1) -- (1.4,1) -- (1.2,2) -- (.3,.75) -- cycle [reverseclip];
\end{pgfinterruptboundingbox}

\draw [step=0.1,thick] (0,0) grid  (2,2);

\end{tikzpicture}
\end{document}

Answer 2

To avoid remember picture and overlay, I mix Jack's solution and Altermundus's solution using the bigger rectangle that TikZ/PGF (TeX?) can used (Edit: as suggested by Qrrbrbirlbel, I add [reset cm] to get a solution independent from any scale transformations).

First tikzpicture shows two (inv)clipping triangles.

Second tikzpicture shows the effect of nonzero rule (even odd rule can't be used directly in a clipping path, see note below).

\documentclass{standalone}
\usepackage{tikz}
\begin{document}

\tikzset{invclip/.style={clip,insert path={{[reset cm]
      (-16383.99999pt,-16383.99999pt) rectangle (16383.99999pt,16383.99999pt)
    }}}}

\begin{tikzpicture}[outer sep=0mm]
  \coordinate (A) at (0,0);
  \coordinate (B) at (1,0);
  \coordinate (C) at (.5,1);
  \coordinate (Ap) at (0,1);
  \coordinate (Bp) at (1,1);
  \coordinate (Cp) at (.5,0);
  \begin{scope}
  \begin{pgfinterruptboundingbox} % useful to avoid the rectangle in the bounding box
    \path[invclip]
    (A) -- (B) -- (C) -- (A)
    (Ap) -- (Cp) -- (Bp) -- (Ap);
  \end{pgfinterruptboundingbox} 

  \fill[orange!50] (-1,-1) rectangle (2,2);

  \draw (A) circle (2mm);    % this is the little angle marker
  \draw (B) circle (2mm);    % this is the little angle marker
  \draw (C) circle (2mm);    % this is the little angle marker

  \draw (Ap) circle (2mm);    % this is the little angle marker
  \draw (Bp) circle (2mm);    % this is the little angle marker
  \draw (Cp) circle (2mm);    % this is the little angle marker

  \end{scope}

  \draw[red] (current bounding box.south west)
  rectangle (current bounding box.north east);
\end{tikzpicture}

\begin{tikzpicture}[outer sep=0mm]
  \coordinate (A) at (0,0);
  \coordinate (B) at (1,0);
  \coordinate (C) at (.5,1);
  \coordinate (Ap) at (0,1);
  \coordinate (Bp) at (1,1);
  \coordinate (Cp) at (.5,0);
  \begin{scope}
  \begin{pgfinterruptboundingbox} % useful to avoid the rectangle in the bounding box
    \path[invclip]
    (A) -- (B) -- (C) -- (A)
    (Ap) -- (Bp) -- (Cp) -- (Ap);
  \end{pgfinterruptboundingbox} 

  \fill[orange!50] (-1,-1) rectangle (2,2);

  \draw (A) circle (2mm);    % this is the little angle marker
  \draw (B) circle (2mm);    % this is the little angle marker
  \draw (C) circle (2mm);    % this is the little angle marker

  \draw (Ap) circle (2mm);    % this is the little angle marker
  \draw (Bp) circle (2mm);    % this is the little angle marker
  \draw (Cp) circle (2mm);    % this is the little angle marker

  \end{scope}

  \draw[red] (current bounding box.south west)
  rectangle (current bounding box.north east);
\end{tikzpicture}
\end{document}

Note about rules and clip:

It's impossible to combine clip and even odd rule in a path (it seems to me that it's almost a bug). But, if you add the even odd rule option to the enclosing scope, the clip operation uses it. Applied on the previous example, the clipping paths can use any direction of rotation:

\documentclass{standalone}
\usepackage{tikz}
\tikzset{invclip/.style={clip,insert path={{[reset cm]
        (-16383.99999pt,-16383.99999pt) rectangle (16383.99999pt,16383.99999pt)}}}}
\begin{document}
\begin{tikzpicture}[outer sep=0mm]


  \coordinate (A) at (0,0);
  \coordinate (B) at (1,0);
  \coordinate (C) at (.5,1);
  \coordinate (Ap) at (0,1);
  \coordinate (Bp) at (1,1);
  \coordinate (Cp) at (.5,0);
  \begin{scope}[even odd rule]

    \begin{pgfinterruptboundingbox} % useful to avoid the rectangle in the bounding box
      \path[invclip]
      (A) -- (B) -- (C) -- (A)
      (Ap) -- (Bp) -- (Cp) -- (Ap);
    \end{pgfinterruptboundingbox} 

    \fill[orange!50] (-1,-1) rectangle (2,2);

    \draw (A) circle (2mm);    % this is the little angle marker
    \draw (B) circle (2mm);    % this is the little angle marker
    \draw (C) circle (2mm);    % this is the little angle marker

    \draw (Ap) circle (2mm);    % this is the little angle marker
    \draw (Bp) circle (2mm);    % this is the little angle marker
    \draw (Cp) circle (2mm);    % this is the little angle marker

  \end{scope}

  \draw[red] (current bounding box.south west)
  rectangle (current bounding box.north east);
\end{tikzpicture}

Answer 3

@Jack send me here from a question regarding invclip: Reverse clipping in a 3D figure - Projection issue.

That question shows a problem of @PaulGaborit's answer that double braces in

\tikzset{invclip/.style={clip,insert path={{[reset cm]
      (-16383.99999pt,-16383.99999pt) rectangle (16383.99999pt,16383.99999pt)
    }}}}

does not limit the scope of reset cm. In fact, it took four in the case in the link above. Therefore, in the link above, I suggested the following code

\tikzset{
    clip even odd rule/.code={\pgfseteorule}, % Credit to Andrew Stacey 
    invclip/.style={
        clip,insert path=
            [clip even odd rule]{
                [reset cm](-\maxdimen,-\maxdimen)rectangle(\maxdimen,\maxdimen)
            }
    }
}

A complete example goes like this

\documentclass[border=9,tikz]{standalone}
\usetikzlibrary{graphs,graphs.standard}
\begin{document}

\tikzset{
    clip even odd rule/.code={\pgfseteorule}, % Credit to Andrew Stacey 
    invclip/.style={
        clip,insert path=
            [clip even odd rule]{
                [reset cm](-\maxdimen,-\maxdimen)rectangle(\maxdimen,\maxdimen)
            }
    }
}

\tikz{
    \begin{pgfinterruptboundingbox}
        \clip[invclip](1,0)--(0,1)--(0,-1)--(-1,0)--cycle;
    \end{pgfinterruptboundingbox}
    \graph{subgraph K_n[n=20,clockwise,radius=50]};
}

\end{document}

Answer 4

The solution to draw an arc outside the triangle is to use my new package tkz-euclide and with this way, you don' need to invert a clip selection

\documentclass{minimal}
\usepackage{tkz-euclide} 
\usetkzobj{all}
\usetikzlibrary{calc}

\begin{document}
\begin{tikzpicture}

\coordinate (A) at (0,0);
\coordinate (B) at (2,0);
\coordinate (C) at (2,2);

\draw (A) -- (B) -- (C) -- cycle;
\draw (B) circle (2mm);    % this is the little angle marker
\tkzDrawArc[R with nodes,color=red](B,1cm)(A,C)  
\tkzDrawArc[R with nodes,color=blue](B,1cm)(C,A) 
\end{tikzpicture}
\end{document}

Now if you want, I can extract the code from my package. You need to use an internal macro of pgf named \pgfmathanglebetweenpoints; in my package I use \tkzmathanglebetweenpoints because I want a good precision so I work with fp and not with pgfmath from some parts of the code. Then the problem it's to use angles with + or - because I want to draw from (B,A) towards (B,C) or from (B,C) towards (B,A). If you want all the details, I give you in some hours because now, I need to work with my students

I complete my answer without tkz-euclide

\documentclass{minimal}
\usepackage{tikz} 
\usetikzlibrary{calc}

\makeatletter
\def\drawarc{\pgfutil@ifnextchar[{\draw@arc}{\draw@arc[]}}  
\def\draw@arc[#1](#2,#3)(#4,#5){% 
 \begingroup
 \pgfmathanglebetweenpoints{\pgfpointanchor{#2}{center}}{%
                            \pgfpointanchor{#4}{center}} 
\global\let\FirstAngle\pgfmathresult 
 \pgfmathanglebetweenpoints{\pgfpointanchor{#2}{center}}{%
                            \pgfpointanchor{#5}{center}} 
\global\let\SecondAngle\pgfmathresult        
  \pgfmathgreaterthan{\FirstAngle}{0}   
  \ifdim\pgfmathresult pt=1 pt\relax%  
    \pgfmathgreaterthan{\FirstAngle}{\SecondAngle}
    \ifdim\pgfmathresult pt=1 pt\relax%
     \pgfmathsubtract{\FirstAngle}{360}
     \edef\FirstAngle{\pgfmathresult}%
 \fi 
 \else
     \pgfmathgreaterthan{\FirstAngle}{\SecondAngle}
    \ifdim\pgfmathresult pt=1 pt\relax%
        \pgfmathadd{\SecondAngle}{360}
        \edef\SecondAngle{\pgfmathresult}%
     \fi 
 \fi
     \draw[#1,shift = {(#2)}](\FirstAngle:#3) arc (\FirstAngle:\SecondAngle:#3);
\endgroup  
}  
\makeatother

\begin{document}        
\begin{tikzpicture}

\coordinate (A) at (2,2);
\coordinate (B) at (1,0);
\coordinate (C) at (3,1);    
\draw (A) -- (B) -- (C) -- cycle;
\drawarc[red](B,1cm)(A,C)
\drawarc[blue](B,1cm)(C,A)     
\end{tikzpicture}  
\end{document}

Answer 5

To get the ball rolling on this one, here's a method that uses fadings. What I don't like about it is that to make this work, one has to specify a large rectangle in the fading that one hopes (!) is large enough - I don't know how to automate this. The issue is that when specifying a fading, everything outside the fading is assumed to be transparent, whereas for this one wants everything outside to be visible.

\documentclass{minimal}
\usepackage{tikz}
\usetikzlibrary{fadings}
\begin{document}

\begin{tikzfadingfrompicture}[name=fadeit]
\fill[white] (-10,-10) rectangle (10,10);
\path (0,0) coordinate (A) +(0:2) coordinate (B) +(50:2) coordinate (C);
\fill[black] (B) -- (A) -- (C) -- cycle;
\end{tikzfadingfrompicture}

\begin{tikzpicture}
\draw[path fading=fadeit,fit fading=false] (0,0) circle (1);
\end{tikzpicture}
\end{document}

It's also not so great because one has to split the definition of the cutout from the picture (I wonder if it's possible to fix this using remember picture ...).

Answer 6

this all can be done very easily using the even odd rule, see http://www.texample.net/tikz/examples/venn-diagram/

Answer 7

I search how to avoid the use of remember picture with the Jake's solution. I find this :

\documentclass{article}
\usepackage{tikz}

\begin{document}
\fbox{\begin{tikzpicture}[invclip/.style={insert path={(-1,-1) rectangle (2,2)}}]
 % a rectangle is necessary

\path coordinate (A) at (0,0)
      coordinate (B) at (1,0)
      coordinate (C) at (1,1);

\begin{pgfinterruptboundingbox} % useful to avoid the rectangle in the bounding box
\path[clip] (A) -- (B) -- (C) -- cycle [invclip];
\end{pgfinterruptboundingbox}  

\draw[thick] (A) circle (2mm)
             (B) circle (2mm)
             (C) circle (2mm);
\end{tikzpicture}}
\end{document}

( I added fbox to verify the bounding box)

Answer 8

As I needed a solution were I could reuse an existing path (which does not work in the otherwise best answer from @Symbol1), I’ve explored other related questions and found another solution from marmotton, also building on top of @Symbol1 and @paul-gaborit answers.

So basically it boils down to:

\documentclass[tikz]{standalone}

\makeatletter
\tikzset{
    reuse path/.code={\pgfsyssoftpath@setcurrentpath{#1}},
    even odd clip/.code={\pgfseteorule},
    reverseclip/.code={\clip[overlay,even odd clip,reuse path=#1] [reset cm] (-\maxdimen,-\maxdimen) rectangle (\maxdimen,\maxdimen);}
}
\makeatother

\begin{document}
\begin{tikzpicture}
    \coordinate (A) at (0,0);
    \coordinate (B) at (1,0);
    \coordinate (C) at (1,1);

    \draw [save path=\rectclip] (A) -- (B) -- (C) -- cycle;
    \begin{pgfinterruptboundingbox}
    \tikzset{reverseclip=\rectclip}
    \end{pgfinterruptboundingbox}
    \draw (A) circle (2mm)
          (B) circle (2mm)
          (C) circle (2mm);
\end{tikzpicture}
\end{document}

The use of overlay makes pgfinterruptboundingbox generally not necessary (in my actual use case I did not need it), but here if you don’t add that the figure is cut to the triangle (figure on the right).