[Tex/LaTex] tombstones and beer mugs

funsymbolstheorems

I'm trying to find an alternative to the standard tombstone qed symbol for more informal papers.
I initially thought of an empty square with Pub written inside, but what would actually be better would be a small beer mug symbol. (much like the one found in this set of icons http://dutchicon.com/iconsets/food-and-drinks-icons)

Unfortunately there doesn't seem to such a symbol in the comprehensive latex symbol list (although many funny symbols can be found there).
Has anyone thought about this? Is there any quick-and-dirty solution?

Best Answer

I personally don't like the idea of the beer mug, but you could redefine \qedsymbol to use a previously saved image of a beer mug:

\documentclass{article}
\usepackage{graphicx}
\usepackage{amsthm}

\renewcommand\qedsymbol{\raisebox{-4pt}{\includegraphics[height=10pt]{beer-mug}}}

\begin{document}

\begin{proof}
Test proof of a really basic theorem.
\end{proof}

\end{document}

enter image description here

1- Semi-Automatic Version

Here is a first attempt via markings. Board and ladders are drawing using integer coordinates of cells. But you have to find/specify manually the correct number of bars of each ladder.

enter image description here

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{decorations.markings,calc}
\def\ladderwidth{1.5mm}
\def\ladderstep{3mm}
\tikzset{
  board/.style={decorate,decoration={
      markings,
      mark=between positions 0 and 1 step 1cm with
      {\draw[fill=lime!20] (-.5,-.5) rectangle (.5,.5);}
    },
  },
  ladder/.style={decorate,decoration={
      markings,
      mark=between positions {1/#1/2} and {-1/#1/2} step {1/#1} with {
        \draw[thick,orange]
        (0,-\ladderwidth) -- (0,\ladderwidth)
        (\ladderstep,-\ladderwidth) -- (-\ladderstep,-\ladderwidth)
        (\ladderstep,\ladderwidth) -- (-\ladderstep,\ladderwidth);
      }
    },
  },
}
\begin{document}
\begin{tikzpicture}
  \draw[board]
  (1,0) -- (6,0) -- (6,2) -- (1,2) -- (1,4) -- (6,4) -- (6,8) -- (1,8) -- (1,6) -- (4,6);
  \draw[ladder=5] (1,4) -- (1,6);
  \draw[ladder=6] (1,0) -- (3,2);
  \draw[ladder=7] (5,4) -- (6,7);
  \draw[ladder=18] (5,0) -- (2,8);
\end{tikzpicture}
\end{document}

2- Automatic Version

Here is a second attempt with automatic calculation (via a to path) of number of bars for each ladder... The two images show the result with different values of \ladderstep (1.5mm on the left and 3mm on the right).

1.5mm enter image description here 3mm

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{decorations.markings,calc}
\def\ladderwidth{1.5mm}
\def\ladderstep{1.5mm}
\tikzset{
  board/.style={decorate,decoration={
      markings, mark=between positions 0 and 1 step 1cm with
      {\draw[fill=lime!20] (-.5,-.5) rectangle (.5,.5);}
    },
  },
  ladder/.style={decorate,decoration={
      markings, mark=between positions {1/#1/2} and {-1/#1/2} step {1/#1} with {
        \draw[thick,orange] (0,-\ladderwidth) -- (0,\ladderwidth)
        (\ladderstep,-\ladderwidth) -- (-\ladderstep,-\ladderwidth)
        (\ladderstep,\ladderwidth) -- (-\ladderstep,\ladderwidth);
      }
    },
  },
  ladder auto/.style={to path={
      let \p1=($(\tikztostart) - (\tikztotarget)$), \n1={veclen(\x1,\y1)} in
      \pgfextra{
        \pgfmathsetmacro{\bars}{int(\n1/\ladderstep/2)+1}
        \pgfinterruptpath
        \draw[ladder=\bars] (\tikztostart) -- (\tikztotarget);
        \endpgfinterruptpath
      }
    },
  },
}
\begin{document}
\begin{tikzpicture}
  \draw[board] (1,0) -- (6,0) -- (6,2) -- (1,2) -- (1,4) -- (6,4)
  -- (6,8) -- (1,8) -- (1,6) -- (4,6);
  \draw (1,4) to[ladder auto] (1,6) (1,0) to[ladder auto] (3,2)
  (5,4) to[ladder auto] (6,7) (5,0) to[ladder auto] (2,8);
\end{tikzpicture}
\end{document}

3- Complete Automatic Version

Here is a third attempt. I added the chutes (always via a to path).

enter image description here

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{decorations.markings,calc}
\def\ladderwidth{1.5mm}
\def\ladderstep{1.5mm}
\tikzset{
  board/.style={decorate,decoration={
      markings,
      mark=between positions 0 and 1 step 1cm with
      {\draw[fill=lime!20] (-.5,-.5) rectangle (.5,.5);}
    },
  },
  ladder/.style={decorate,decoration={
      markings,
      mark=between positions {1/#1/2} and {-1/#1/2} step {1/#1} with {
        \draw[thick,orange]
        (0,-\ladderwidth) -- (0,\ladderwidth)
        (\ladderstep,-\ladderwidth) -- (-\ladderstep,-\ladderwidth)
        (\ladderstep,\ladderwidth) -- (-\ladderstep,\ladderwidth);
      },
    },
  },
  ladder auto/.style={
    to path={
      let
      \p1=($(\tikztostart) - (\tikztotarget)$),
      \n1={veclen(\x1,\y1)}
      in
      \pgfextra{
        \pgfmathsetmacro{\bars}{int(\n1/\ladderstep/2)+1}
        \pgfinterruptpath
        \draw[ladder=\bars] (\tikztostart) -- (\tikztotarget);
        \endpgfinterruptpath
      }
    },
  },
  chute auto/.style={
    to path={
      let
      \p1=([xshift=\ladderwidth]\tikztostart),
      \p2=([xshift=-\ladderwidth]\tikztostart),
      \p3=([xshift=\ladderwidth]\tikztotarget),
      \p4=([xshift=-\ladderwidth]\tikztotarget),
      \p5=($(\p1)!.5!(\p3)$),
      \p6=($(\p2)!.5!(\p4)$)
      in
      \pgfextra{
        \pgfinterruptpath
        \path[thick,draw=orange]
        (\p1) sin (\p5) cos (\p3)
        (\p4) sin (\p6) cos (\p2);
        \fill[fill=orange!20]
        (\p1) sin (\p5) cos (\p3) --
        (\p4) sin (\p6) cos (\p2) -- cycle;
        \endpgfinterruptpath
      }
    },
  },
}
\begin{document}
\begin{tikzpicture}
  \draw[board] (1,0) -- (6,0) -- (6,2) -- (1,2) -- (1,4) -- (6,4)
  -- (6,8) -- (1,8) -- (1,6) -- (4,6);
  \draw[chute auto] (1,2) to (3,0) (2,4) to (5,2) (5,8) to (2,2);
  \draw[ladder auto] (1,4) to (1,6) (1,0) to (3,2)
  (5,4) to (6,7) (5,0) to (2,8);
\end{tikzpicture}
\end{document}

4- Complete Automatic Version with Styles, Scale and Variants

Here is a fourth attempt with scale feature as requested... but without scale option. Changes:

You can choose the global size via \cellsize: each ... auto style takes into account the \cellsize distance and changes the coordinate system to keep integer coordinates at center of cells.
There are three new styles to choose drawing and filling options: cell, ladder and chute.
There is a new ladder auto bis style to draw horizontal bars on ladders.

enter image description here

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{decorations.markings,calc}
\pgfmathsetmacro{\cellsize}{1cm}
\pgfmathsetmacro{\ladderstep}{\cellsize/6}
\pgfmathsetmacro{\ladderwidth}{\cellsize/6}
\tikzset{
  % styles to draw and to fill
  cell/.style={
    draw=green!50!black,top color=lime!10,bottom color=lime,
    line width=1pt,
    rounded corners=\cellsize/5
  },
  ladder/.style={
    draw=orange,line cap=round,line join=round,line width=2pt,
  },
  chute/.style={
    top color=blue!30,
    bottom color=blue!30,
    middle color=blue!5,
    draw=blue!50!black,line cap=round,
    line join=round,line width=1pt,
  },
  % styles to draw board, ladders and chutes
  board auto/.style={
    x=\cellsize pt,
    y=\cellsize pt,
    decorate,decoration={
      markings,
      mark=between positions 0 and 1 step \cellsize pt with
      {\path[cell] (-.5*\cellsize pt,-.5*\cellsize pt)
        rectangle (.5*\cellsize pt,.5*\cellsize pt);}
    },
  },
  ladder draw/.style={
    decorate,decoration={
      markings,
      mark=between positions {1/#1/2} and {-1/#1/2} step {1/#1} with {
        \path[ladder]
        (0,-\ladderwidth pt) -- (0,\ladderwidth pt)
        (\ladderstep pt,-\ladderwidth pt) -- (-\ladderstep pt,-\ladderwidth pt)
        (\ladderstep pt,\ladderwidth pt) -- (-\ladderstep pt,\ladderwidth pt);
      },
    },
  },
  ladder auto/.style={
    x=\cellsize pt,
    y=\cellsize pt,
    to path={
      let
      \p1=($(\tikztostart) - (\tikztotarget)$),
      \n1={veclen(\x1,\y1)}
      in
      \pgfextra{
        \pgfmathsetmacro{\bars}{int(\n1/\ladderstep/2)+1}
        \pgfinterruptpath
        \draw[ladder draw=\bars] (\tikztostart) -- (\tikztotarget);
        \endpgfinterruptpath
      }
    },
  },
  ladder auto bis/.style={
    x=\cellsize pt,
    y=\cellsize pt,
    to path={
      let
      \p1=(\tikztostart),
      \p2=(\tikztotarget),
      \p3=($(\tikztostart) - (\tikztotarget)$),
      \p4=([xshift=-\ladderwidth pt]\p1),
      \p5=([xshift=-\ladderwidth pt]\p2),
      \p6=([xshift=\ladderwidth pt]\p1),
      \p7=([xshift=\ladderwidth pt]\p2),
      \n1={veclen(\x3,\y3)}
      in
      \pgfextra{
        \pgfmathtruncatemacro{\bars}{int(\n1/\ladderstep/2)+1}
        \pgfinterruptpath
        \path[ladder] (\p4) -- (\p5);
        \path[ladder] (\p6) -- (\p7)
        \foreach \bar in {1,...,\bars}{
          \pgfextra{\pgfmathsetmacro{\pos}{1/(\bars+1)*\bar}}
          coordinate[pos=\pos] (p\bar)
        };
        \foreach \bar in {1,...,\bars}{
          \path[ladder] (p\bar) -- ++(-2*\ladderwidth pt,0);
        }
        \endpgfinterruptpath
      }
    },
  },
  chute auto/.style={
    x=\cellsize pt,
    y=\cellsize pt,
    to path={
      let
      \p1=([xshift=\ladderwidth pt]\tikztostart),
      \p2=([xshift=-\ladderwidth pt]\tikztostart),
      \p3=([xshift=\ladderwidth pt]\tikztotarget),
      \p4=([xshift=-\ladderwidth pt]\tikztotarget),
      \p5=($(\p1)!.5!(\p3)$),
      \p6=($(\p2)!.5!(\p4)$)
      in
      \pgfextra{
        \pgfinterruptpath
        \path[chute]
        (\p1) sin (\p5) cos (\p3) --
        (\p4) sin (\p6) cos (\p2) -- cycle;
        \endpgfinterruptpath
      }
    },
  },
}
\begin{document}
\begin{tikzpicture}
  \draw[board auto] (1,0) -- (6,0) -- (6,2) -- (1,2) -- (1,4) -- (6,4)
  -- (6,8) -- (1,8) -- (1,6) -- (4,6);
  \draw[chute auto] (1,2) to (3,0) (2,4) to (5,2) (5,8) to (2,2);
  \draw[ladder auto bis] (1,4) to (1,6) (1,0) to (3,2)
  (5,4) to (6,7) (5,0) to (2,8);
\end{tikzpicture}
\end{document}

[Tex/LaTex] Cite a theorem by its name and number

Here are two approaches. First, you can combine \ref and \nameref to format as desired. Second, you can use \mylabel to save preformatted text. The advantage of the latter is that it is implemented as a single hyperlink.

\documentclass{article}
\usepackage[colorlinks=true]{hyperref}

\newtheorem{theorem}{Theorem}

\makeatletter
\@ifpackageloaded{hyperref}%
  {\newcommand{\mylabel}[2]% #1=name, #2 = contents
    {\protected@write\@auxout{}{\string\newlabel{#1}{{#2}{\thepage}%
      {\@currentlabelname}{\@currentHref}{}}}}}%
  {\newcommand{\mylabel}[2]% #1=name, #2 = contents
    {\protected@write\@auxout{}{\string\newlabel{#1}{{#2}{\thepage}}}}}
\makeatother

\begin{document}
\begin{theorem}[Triangle inequality]\label{thm:triangle}
\mylabel{text:triangle}{Triangle inequality (Theorem \thetheorem)}
\end{theorem}
This is easily proved using the \nameref{thm:triangle} (Theorem \ref{thm:triangle}).

\medskip\noindent
This is easily proved using the \ref{text:triangle}.
\end{document}

Best Answer

Related Solutions

[Tex/LaTex] Chutes and Ladders

1- Semi-Automatic Version

2- Automatic Version

3- Complete Automatic Version

4- Complete Automatic Version with Styles, Scale and Variants

[Tex/LaTex] Cite a theorem by its name and number

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