3

I want to be able to connect the white rectangles with lines as shown in the picture below (this example consists only of two rectangles, the original code consists of more than 10 of such rectangles). Right now I am working with coordinates. But this is a rather complicated approach and error-prone. It would be ideal if I could define a connecting point by "north-west, south-east" instead of using numeric coordinates. It would also be very helpful if I could additionally connect the lines coming from the left rectangle to the center of the right rectangle.

Is there an easier way to do it? Here is the code and the output it produces:

    \documentclass[a5paper]{article}

    \usepackage{tikz}
    \usetikzlibrary{decorations.pathreplacing}
    \usetikzlibrary{fadings}

    \usepackage{etoolbox}

    \begin{document}

    \newcommand{\sizein}{1.0}
    \newcommand{\distin}{0.15}
    \newcommand{\opacity}{0.8}
    \newcommand{\xDist}{2.0}

    \begin{figure}[htb!]
        \centering
        \begin{tikzpicture}
            \foreach \i[count=\xi] in {2,...,0}{
                \draw [xshift=\xDist*0cm, fill=gray, opacity=\opacity, draw=black] 
                (\distin*\i,\distin*\i) rectangle (\distin*\i+\sizein,\distin*\i+\sizein); 
            }

            \foreach \i[count=\xi] in {2,...,0}{
                \draw [xshift=\xDist*1cm, fill=gray, opacity=\opacity, draw=black] 
                (\distin*\i,\distin*\i) rectangle (\distin*\i+\sizein,\distin*\i+\sizein);
            }

            \draw [fill=white, draw=black] (0.1,0.1) rectangle (0.3,0.3); 
            \draw [fill=white, draw=black] (2.5,0.4) rectangle (2.7,0.6); 

            \draw (0.3,0.1) -- (2.7,0.4); 
            \draw (0.3,0.3) -- (2.5,0.6); 
    \end{tikzpicture}
\end{figure}

\end{document}

enter image description here

3

edit:

  • instead of rectangles i suggest to use nodes with rectangle shape
  • in mwe below the style of those nodes are defined in options of the tikzpicture:

    % for transparent gray "rectangles" 
    box/.style = {draw, fill=gray, opacity=\opacity, minimum size=\sizein cm},
    % for white "rectangles" 
    wbx/.style = {draw, fill=white, minimum size=\distin cm, outer sep=0pt}
    
  • nodes has defined anchors which you can exploit in connection between their corners

  • using nodes you can make code much shorter (compare your solution wth mwe below)
  • in cases, when you like to have other shapes, accordingly change style definition to their nodes: for example define it as circle. however in such cases defining of nodes anchors can become cumbersome (complex: for example using tangent coordinates)

\documentclass[a5paper]{article}
\usepackage{tikz}

\begin{document}
\newcommand{\sizein}{1.0}
\newcommand{\distin}{0.15}
\newcommand{\opacity}{0.8}
\newcommand{\xDist}{2.0}

    \begin{tikzpicture}[
box/.style = {draw, fill=gray, opacity=\opacity, minimum size=\sizein cm},  
wbx/.style = {draw, fill=white, minimum size=\distin cm, outer sep=0pt}
                        ]
\foreach \i in {2,1,0}%
{
    \node   [box,above right] at (\i*\distin,\i*\distin) {};
    \node   [box,above right] at (\xDist+\i*\distin,\i*\distin) {};
}
\node (w1) [wbx] at (0.3,0.3) {};
\node (w2) [wbx] at (\xDist+0.5,0.5) {};
\draw[very thin]   (w1.north west) -- (w2.north west)
        (w1.south east) -- (w2.south east);
    \end{tikzpicture}
\end{document}            

enter image description here

actual size and position of white boxes you can change according to your needs.

addendum: regarding to comment below @marmot answer, you need to replase existed line drawig command with:

\draw[very thin]
        (w1.north west) -- (w2.center)
        (w1.north east) -- (w2.center)
        (w1.south west) -- (w2.center)
        (w1.south east) -- (w2.center);

and you will obtain:

enter image description here

  • I just wondered, where in your code the squares are defined. What do I have to change if I want to have rectangles or circles? – Samuel Dec 30 '18 at 16:14
  • @Samuel, i will add some explanation to answer asap. – Zarko Dec 30 '18 at 16:24
1

I couldn't resist making Zarko's answer more concise. This comes with a style connect that picks and connects the right corners for you.

\documentclass[a5paper]{article}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\newcommand{\sizein}{1.0}
\newcommand{\distin}{0.15}
\newcommand{\opacity}{0.8}
\newcommand{\xDist}{2.0}

    \begin{tikzpicture}[
box/.style = {draw, fill=gray, opacity=\opacity, minimum size=\sizein cm},  
wbx/.style = {draw, fill=white, minimum size=\distin cm, outer sep=0pt},
connect/.style args={#1 with #2}{insert path={
let \p1=($(#2.center)-(#1.center)$),\n1={int(mod(8+atan2(\y1,\x1)/90,2))}
in \pgfextra{\pgfmathtruncatemacro{\itest}{\n1}}
\ifcase\itest
(#1.north west) -- (#2.north west) (#1.south east) -- (#2.south east)
\or
(#1.south west) -- (#2.south west) (#1.north east) -- (#2.north east)
\fi
}}
                        ]
\foreach \i in {2,1,0}%
{
    \node   [box,above right] at (\i*\distin,\i*\distin) {};
    \node   [box,above right] at (\xDist+\i*\distin,\i*\distin) {};
}
\node (w1) [wbx] at (0.3,0.3) {};
\node (w2) [wbx] at (\xDist+0.5,0.5) {};
\node (w3) [wbx] at (0.1,0.9) {};
\node (w4) [wbx] at (0.5,-0.3) {};
\node (w5) [wbx] at (-0.5,0) {};

\draw[very thin] [connect=w1 with w2,connect=w1 with w3,connect=w1 with w4,connect=w1 with w5];
\end{tikzpicture}
\end{document}            

enter image description here

Thanks for the clarification! I added a second style cconnect that connects the outer corners of one rectangle with the center of another whereby the style figures out what the outer corners are.

\documentclass[a5paper]{article}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\newcommand{\sizein}{1.0}
\newcommand{\distin}{0.15}
\newcommand{\opacity}{0.8}
\newcommand{\xDist}{2.0}

    \begin{tikzpicture}[
box/.style = {draw, fill=gray, opacity=\opacity, minimum size=\sizein cm},  
wbx/.style = {draw, fill=white, minimum size=\distin cm, outer sep=0pt},
connect/.style args={#1 with #2}{insert path={
let \p1=($(#2.center)-(#1.center)$),\n1={int(mod(8+atan2(\y1,\x1)/90,2))}
in \pgfextra{\pgfmathtruncatemacro{\itest}{\n1}}
\ifcase\itest
(#1.north west) -- (#2.north west) (#1.south east) -- (#2.south east)
\or
(#1.south west) -- (#2.south west) (#1.north east) -- (#2.north east)
\fi
}},cconnect/.style args={#1 with center of #2}{insert path={
let \p1=($(#2.center)-(#1.north west)$),\p2=($(#2.center)-(#1.north east)$),
\p3=($(#2.center)-(#1.south west)$),\p4=($(#2.center)-(#1.south east)$),
\p5=($(#2.center)-(#1.center)$),
\n1={abs(atan2(\y1,\x1)-atan2(\y5,\x5))},
\n2={abs(atan2(\y2,\x2)-atan2(\y5,\x5))},
\n3={abs(atan2(\y3,\x3)-atan2(\y5,\x5))},
\n4={abs(atan2(\y4,\x4)-atan2(\y5,\x5))}
in
\ifdim\n1>\n2
(#1.north west) -- (#2.center) 
\else
(#1.north east) -- (#2.center) 
\fi
\ifdim\n3>\n4
(#1.south west) -- (#2.center)
\else
(#1.south east) -- (#2.center)
\fi
}}
                        ]
\foreach \i in {2,1,0}%
{
    \node   [box,above right] at (\i*\distin,\i*\distin) {};
    \node   [box,above right] at (\xDist+\i*\distin,\i*\distin) {};
}
\node (w1) [wbx] at (0.3,0.3) {};
\node (w2) [wbx] at (\xDist+0.5,0.5) {};
\node (w3) [wbx] at (0.1,0.9) {};
\node (w4) [wbx] at (0.5,-0.3) {};
\node (w5) [wbx] at (-0.5,0) {};

%\draw[very thin] [connect=w1 with w2,connect=w1 with w3,connect=w1 with w4,connect=w1 with w5];

\draw[very thin,blue] [cconnect=w1 with center of w2,cconnect=w1 with center of w3,cconnect=w1 with center of w4,cconnect=w1 with center of w5];
\end{tikzpicture}
\end{document}            

enter image description here

Of course, you can use both styles in the same picture. To verify this, just uncomment \draw[very thin] [connect=w1 with w2,connect=w1 with w3,connect=w1 with w4,connect=w1 with w5];. (I did not do this because it makes the plot a bit busy.) If you need to draw chains of these, I'd recommend using show path construction and adjusting the lineto code for this. (If you want this spelled out, could you perhaps consider asking a separate question?)

  • Great! I would like to have lines from the corners of the white left rectangle to the center of the white right rectangle. – Samuel Dec 30 '18 at 15:46
  • @Samuel I added a proposal for that. The square comes from minimum size=\sizein cm. If you choose minimum width=\sizein cm,minimum height=2*\sizein cm instead, you get a rectangle. – marmot Dec 30 '18 at 16:14

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