19

I was wondering if it is possible to write a very simple game in TikZ; e.g. (very simple) Super Mario Bros. Here is Mario:

\documentclass[border=5pt,tikz]{standalone}
\begin{document}
    \begin{tikzpicture}
        \draw[step=.5cm] (0,0) grid (8,8);
        \foreach \x in {2.5,3,...,4.5}
        {
            \fill[red] (\x,7.5) rectangle +(.5,.5);
        }
        \foreach \x in {2,2.5,...,6}
        {
            \fill[red] (\x,7) rectangle +(.5,.5);
        }
        \foreach \x in {2,2.5,3}
        {
            \fill[brown!70!black] (\x,6.5) rectangle +(.5,.5);
        }
        \foreach \x in {3.5,4}
        {
            \fill[yellow!50!orange] (\x,6.5) rectangle +(.5,.5);
        }
        \fill[black] (4.5,6.5) rectangle +(.5,.5);
        \fill[yellow!50!orange] (5,6.5) rectangle +(.5,.5);
            \fill[brown!70!black] (1.5,6) rectangle +(.5,.5);
            \fill[yellow!50!orange] (2,6) rectangle +(.5,.5);
            \fill[brown!70!black] (2.5,6) rectangle +(.5,.5);
            \fill[yellow!50!orange] (3,6) rectangle +(.5,.5);
            \fill[yellow!50!orange] (3.5,6) rectangle +(.5,.5);
            \fill[yellow!50!orange] (4,6) rectangle +(.5,.5);
                \fill[black] (4.5,6) rectangle +(.5,.5);
                \fill[yellow!50!orange] (5,6) rectangle +(.5,.5);
                \fill[yellow!50!orange] (5.5,6) rectangle +(.5,.5);
                \fill[yellow!50!orange] (6,6) rectangle +(.5,.5);
        \begin{scope}[shift={(0,-.5)}]
            \fill[brown!70!black] (1.5,6) rectangle +(.5,.5);
            \fill[yellow!50!orange] (2,6) rectangle +(.5,.5);
            \fill[brown!70!black] (2.5,6) rectangle +(.5,.5);
            \fill[brown!70!black] (3,6) rectangle +(.5,.5);
            \fill[yellow!50!orange] (3.5,6) rectangle +(.5,.5);
            \fill[yellow!50!orange] (4,6) rectangle +(.5,.5);
                \fill[yellow!50!orange] (4.5,6) rectangle +(.5,.5);
                \fill[black] (5,6) rectangle +(.5,.5);
                \fill[yellow!50!orange] (5.5,6) rectangle +(.5,.5);
                \fill[yellow!50!orange] (6,6) rectangle +(.5,.5);
                \fill[yellow!50!orange] (6.5,6) rectangle +(.5,.5);
        \end{scope}
        \fill[brown!70!black] (2,5) rectangle +(.5,.5);
        \foreach \x in {2.5,3,...,4}
        {
            \fill[yellow!50!orange] (\x,5) rectangle +(.5,.5);
        }
        \foreach \x in {4.5,5,...,6}
        {
            \fill[black] (\x,5) rectangle +(.5,.5);
        }
        \foreach \x in {2.5,3,...,5}
        {
            \fill[yellow!50!orange] (\x,4.5) rectangle +(.5,.5);
        }
        \foreach \x in {2,2.5,...,5.5}
        {
            \fill[red] (\x,4) rectangle +(.5,.5);
        }
        \fill[blue] (3,4) rectangle +(.5,.5);
        \fill[blue] (4.5,4) rectangle +(.5,.5);
        \begin{scope}[shift={(0,-.5)}]
            \foreach \x in {1.5,2,...,6}
            {
                \fill[red] (\x,4) rectangle +(.5,.5);
            }
            \fill[blue] (3,4) rectangle +(.5,.5);
            \fill[blue] (4.5,4) rectangle +(.5,.5);
        \end{scope}
        \begin{scope}[shift={(0,-1)}]
            \foreach \x in {1,1.5,2,...,6.5}
            {
                \fill[red] (\x,4) rectangle +(.5,.5);
            }
            \fill[blue] (3,4) rectangle +(.5,.5);
                \fill[blue] (3.5,4) rectangle +(.5,.5);
                \fill[blue] (4,4) rectangle +(.5,.5);
            \fill[blue] (4.5,4) rectangle +(.5,.5);
        \end{scope}
        \fill[yellow!50!orange] (1,2.5) rectangle +(.5,.5);
        \fill[yellow!50!orange] (1.5,2.5) rectangle +(.5,.5);
            \fill[red] (2,2.5) rectangle +(.5,.5);
            \fill[blue] (2.5,2.5) rectangle +(.5,.5);
                \fill[yellow!50!orange] (3,2.5) rectangle +(.5,.5);
            \fill[blue] (3.5,2.5) rectangle +(.5,.5);
            \fill[blue] (4,2.5) rectangle +(.5,.5);
                \fill[yellow!50!orange] (4.5,2.5) rectangle +(.5,.5);
            \fill[blue] (5,2.5) rectangle +(.5,.5);
            \fill[red] (5.5,2.5) rectangle +(.5,.5);
                \fill[yellow!50!orange] (6,2.5) rectangle +(.5,.5);
                \fill[yellow!50!orange] (6.5,2.5) rectangle +(.5,.5);
        \foreach \x in {1,1.5,2}
        {
            \fill[yellow!50!orange] (\x,2) rectangle +(.5,.5);
        }
        \foreach \x in {2.5,3,...,5}
        {
            \fill[blue] (\x,2) rectangle +(.5,.5);
        }
        \fill[yellow!50!orange] (5.5,2) rectangle +(.5,.5);
        \fill[yellow!50!orange] (6,2) rectangle +(.5,.5);
        \fill[yellow!50!orange] (6.5,2) rectangle +(.5,.5);
        \begin{scope}[shift={(0,-1)}]
            \fill[yellow!50!orange] (1,2.5) rectangle +(.5,.5);
            \fill[yellow!50!orange] (1.5,2.5) rectangle +(.5,.5);
                \fill[blue] (2,2.5) rectangle +(.5,.5);
                \fill[blue] (2.5,2.5) rectangle +(.5,.5);
                    \fill[blue] (3,2.5) rectangle +(.5,.5);
                \fill[blue] (3.5,2.5) rectangle +(.5,.5);
                \fill[blue] (4,2.5) rectangle +(.5,.5);
                    \fill[blue] (4.5,2.5) rectangle +(.5,.5);
                \fill[blue] (5,2.5) rectangle +(.5,.5);
                \fill[blue] (5.5,2.5) rectangle +(.5,.5);
                    \fill[yellow!50!orange] (6,2.5) rectangle +(.5,.5);
                    \fill[yellow!50!orange] (6.5,2.5) rectangle +(.5,.5);
        \end{scope}
            \foreach \x in {2,2.5,3}
            {
                \fill[blue] (\x,1) rectangle +(.5,.5);
            }
        \begin{scope}[shift={(2.5,0)}]
            \foreach \x in {2,2.5,3}
            {
                \fill[blue] (\x,1) rectangle +(.5,.5);
            }
        \end{scope}
        \foreach \x in {1.5,2,2.5}
        {
            \fill[brown!70!black] (\x,.5) rectangle +(.5,.5);
        }
        \begin{scope}[shift={(3.5,0)}]
            \foreach \x in {1.5,2,2.5}
            {
                \fill[brown!70!black] (\x,.5) rectangle +(.5,.5);
            }
        \end{scope}
        \begin{scope}[shift={(0,-.5)}]
            \foreach \x in {1,1.5,2,2.5}
            {
                \fill[brown!70!black] (\x,.5) rectangle +(.5,.5);
            }
            \begin{scope}[shift={(4,0)}]
                \foreach \x in {1,1.5,2,2.5}
                {
                    \fill[brown!70!black] (\x,.5) rectangle +(.5,.5);
                }
            \end{scope}
        \end{scope}
    \end{tikzpicture}
\end{document}

Here is the output:

Screenshot

And here is the original:

Screenshot

Because you can create simple animations and controls in TikZ (see the Rubiks thread), it is maybe possible to generate a world (e.g. from Mario) , which is refreshed by some buttons.

For the beginn the pong game would be absolutely enough I think, because it is really hard to implement such things especially in TikZ.

As the title says, it would sound great if something like that would exist, so my question is: Does anybody have an idea how to do that? This is not a do-it-for-me – question, rather collecting ideas.

Henri Menke
  • 109,596
current_user
  • 5,235
  • LaTeX will always produce pdf files I think. So if you want to produce a game, you'd have to create pdfs for all possible outcomes beforehand. Not sure if that would be fun. (+1 for the Mario ;-) –  Aug 06 '18 at 23:44
  • @marmot: Yes, that's right, but I thought just like an interactive-animation-thing, you know? And thank you! ;) – current_user Aug 06 '18 at 23:45
  • The animate package has controls. I guess the user who could help you most is @AlexG, who was recently on the chat, meaning you could ping him there. He managed to do a really nice animation in which two ducks got exchanged when one clicked on them. –  Aug 06 '18 at 23:48
  • 3
    Interesting idea. Pkg animate, however, might not be the right tool here, as it does not allow branching. But PDF Layers (OCGs) could be an option (#pixels x #colours), to be enabled/hidden by JavaScript, possibly using random numbers, and/or by buttons. – AlexG Aug 07 '18 at 07:00
  • @marmot not, quite. Standard LaTeX output file is .dvi (and I think .ps), while pdflatex output file's .pdf. – naphaneal Aug 07 '18 at 07:44
  • 1
    @naphaneal Well, technically I agree but your probably know what I wanted to say, don't you? Even more explicitly: LaTeX will create a static document. –  Aug 07 '18 at 13:58
  • 1
    A non-trivial game could be a puzzle: reorder previously shuffled cells to show Mario again by moving sub-rows and sub-columns towards a whole (Sliding Puzzle). – AlexG Aug 07 '18 at 14:08
  • @AlexG: This is great! Is maybe Tetris possible? – current_user Aug 07 '18 at 15:39
  • See also https://tex.stackexchange.com/a/510755/128553 – CampanIgnis Apr 28 '20 at 14:49

2 Answers2

22

Sliding Puzzle

(Acrobat Reader required; pdflatex or lualatex)

This is a non-trivial game with user interaction, based on PDF Layers (OCGs, package ocgx2).

The picture is split into 15 tiles, stacked upon each other for every cell of the puzzle board, each on an individual PDF Layer. The total number of PDF Layers is thus 4 x 4 x 15 = 240. Their visibility is controlled by JavaScript, triggered by clicking on an array of transparent buttons overlaid on top of the puzzle board.

enter image description here

\documentclass[tikz,margin=1mm]{standalone}

\usepackage{xsavebox}
\usepackage[tikz]{ocgx2}
\usepackage{media9} % \mediabutton


\usepackage{tikzmarmots}
\usepackage{xcolor}\pagecolor{gray}
\usepackage{multido}
\usepackage{ifluatex}
\ifluatex\def\pdfpageattr{\pdfvariable pageattr}\fi


\begin{xlrbox}{Marmot}
  \begin{tikzpicture}[xscale=-1,transform shape]
    \useasboundingbox (0,0) rectangle (4.4,4.4);
    \node at (current bounding box.center) {\tikz\marmot[scale=2.1];};
  \end{tikzpicture}
\end{xlrbox}


%generating tiles Marmot.1, ..., Marmot.15
\multido{\nX=0.0+1.1,\iI=0+1}{4}{
  \multido{\nY=3.3+-1.1,\iJ=0+1}{4}{
    \ifnum\numexpr\iI+\iJ\relax>0
      \begin{xlrbox}{Marmot.\the\numexpr\iI+\iJ*4\relax}
        \begin{tikzpicture}
          \useasboundingbox (\nX,\nY) rectangle ++(1.1,1.1);
          \clip (\nX,\nY) rectangle ++(1.1,1.1);
          \node[inner sep=0pt, outer sep=0pt, anchor=south west,fill=white] at (0,0) {\theMarmot};
        \end{tikzpicture}
      \end{xlrbox}
    \fi
  }
}


\newcommand\MyStatus[2]{\ifnum#1=#2 visible\else invisible\fi}


%initialisations (JavaScript) on page-open (/O <<...>>)
\begingroup
\edef\x{\endgroup
  \pdfpageattr{
    \the\pdfpageattr
    /AA << /O << /S/JavaScript /JS (%
      var tile=new Array(4);
      var oldTile=new Array(4);
      for(var i=0;i<4;i++){
        tile[i]=new Array(4);
        oldTile[i]=new Array(4);
        for(var j=0;j<4;j++){
          tile[i][j]=new Array(15);
          oldTile[i][j]=i+j*4-1;
        }
      }
      ocg=this.getOCGs(this.pageNum);
      for(var i in ocg){
        ocgName=ocg[i].name.split('.');
        tile[ocgName[0]][ocgName[1]][ocgName[2]-1]=ocg[i];
      }%
      var swap_i=0; var swap_j=0;
      var onButtonClick=function(i,j){
        if(swap_i==i && swap_j!=j){
          for(var jj=swap_j; jj!=j; swap_j<j ? jj++ : jj--){
            if(jj!=swap_j) tile[i][jj][oldTile[i][jj]].state=false;
            newTile=oldTile[i][swap_j<j ? jj+1 : jj-1];
            oldTile[i][jj]=newTile;
            tile[i][jj][newTile].state=true;
          }
        } else if(swap_j==j && swap_i!=i){
          for(var ii=swap_i; ii!=i; swap_i<i ? ii++ : ii--){
            if(ii!=swap_i) tile[ii][j][oldTile[ii][j]].state=false;
            newTile=oldTile[swap_i<i ? ii+1 : ii-1][j];
            oldTile[ii][j]=newTile;
            tile[ii][j][newTile].state=true;
          }
        }
        if(swap_i==i || swap_j==j){
          if(oldTile[i][j]>-1)
            tile[i][j][oldTile[i][j]].state=false;
          oldTile[i][j]=-1;
          swap_i=i; swap_j=j;
        }
      };
    ) >> >>
  }
}
\x


\begin{document}


\begin{tikzpicture}
  \multido{\nX=0.0+1.1,\iI=0+1}{4}{%
    \multido{\nY=3.3+-1.1,\iJ=0+1}{4}{%
      %tiles
      \multido{\iK=1+1}{15}{%
        \begin{scope}[ocg={ref=\iI.\iJ.\iK,status=\MyStatus{\the\numexpr\iI+\iJ*4\relax}{\iK}}]%
          \node[inner sep=0pt, outer sep=0pt, anchor=south west] at (\nX,\nY) {\xusebox{Marmot.\iK}};
        \end{scope}
      }
      %button array
      \draw[
        line width=0,
        postaction={
          path picture={
            \path (path picture bounding box.south west) coordinate (p1)
              (path picture bounding box.north east) coordinate (p2)
              (p1) node[inner sep=0pt,anchor=south west,outer sep=0pt] {%
                \mediabutton[jsaction={try{onButtonClick(\iI,\iJ);}catch(e){}}]{%
                  \tikz \useasboundingbox (p1) rectangle (p2);%
                }%
              };
          }
        }
      ] (\nX,\nY) rectangle ++(1.1,1.1);
    }
  }
\end{tikzpicture}


\end{document}

Original Answer

For this kind of games based on a pixel array using a given set of pre-defined colours, the TikZ-part is straightforward to be implemented; making use of package ocgx2, we simply put each pixel colour on an individual PDF Layer.

The whole logic of the game can then be implemented in JavaScript. It programmatically controls the visibility of each pixel colour.

As a trivial example, we implement a "game" whose sole purpose is to shuffle the pixel colours by pressing a button and to enjoy the result (Acrobat Reader or Foxit required for viewing):

enter image description here

\documentclass[tikz,margin=1mm]{standalone}
\usepackage[tikz]{ocgx2}
\usepackage{media9} % \mediabutton

\newcommand{\mycolour}[1]{\ifcase#1
white%0
\or
black%1
\or
red%2
\or
orange%3
\or
blue%4
\or
brown%5
\or
yellow%6
\fi}


\begin{document}


%set up canvas (16x16 pixels^2) x (7 colours)
\begin{tikzpicture}
  \foreach \X in {0,1,...,15} {
    \foreach \Y in {0,1,...,15} {
      \foreach \Z in {0,1,...,6} {
        \begin{scope}[ocg={ref=\X.\Y.\Z,status=invisible}]
          \node[minimum size=0.5cm,draw,fill=\mycolour{\Z}] at (\X/2,-\Y/2){};
        \end{scope}
      }
    }
  }
  \node at (3.75,-8) {%
    \mediabutton[
      jsaction={
        %initialisation
        if(typeof pixelcolour==='undefined'){
          var pixelcolour=new Array(16);
          var oldcolour=new Array(16);
          for(var i=0;i<16;i++){
            pixelcolour[i]=new Array(16);
            oldcolour[i]=new Array(16);
            for(var j=0;j<16;j++){
              pixelcolour[i][j]=new Array(7);
              oldcolour[i][j]=0;
            }
          }
          ocg=this.getOCGs(this.pageNum);
          for(i in ocg){
            ocgName=ocg[i].name.split('.');
            pixelcolour[ocgName[0]][ocgName[1]][ocgName[2]]=ocg[i];
          }

        }
        %randomise colours
        for(i=0;i<16;i++){
          for(j=0;j<16;j++){
            newcolour=Math.floor(Math.random()*7);
            pixelcolour[i][j][oldcolour[i][j]].state=false;
            pixelcolour[i][j][newcolour].state=true;
            oldcolour[i][j]=newcolour;
          }
        }
      }
    ]{\fbox{Random}}
  };
\end{tikzpicture}
\end{document}
Community
  • 1
AlexG
  • 54,894
  • Great! But how long does it take to compile? I'm compiling since ten minutes … – current_user Aug 07 '18 at 11:31
  • 1
    Are you on a RaspberryPi :-)? Yes, it takes a while. 20 seconds on my PC. – AlexG Aug 07 '18 at 11:33
  • Yes, I am, but it doesn't work. Could you post a little .gif? – current_user Aug 07 '18 at 11:35
  • 1
    Trying to upload a static PNG, but imgur rejects it for some reason. – AlexG Aug 07 '18 at 11:48
  • You could actually do a small screen recording and then convert the .mp4 (or .avi) to a .gif. – current_user Aug 07 '18 at 11:51
  • @AlexG Can you add a picture, please? – Sebastiano Aug 09 '18 at 20:37
  • 2
    @Sebastiano Pictures and a new game added :-). – AlexG Aug 10 '18 at 09:21
  • @AlexG Great! Very gooood :-) – Sebastiano Aug 10 '18 at 19:40
  • This is incredible work. How could one take out the black lines between the tiles? I want to hide a puzzle like this in one of the figures in my dissertation. :-) – jvriesem Oct 23 '20 at 01:13
  • 1
    @jvriesem That's easy. Look for the single \draw in the code and replace it with \path. Though I personally would keep the lines for this kind of game. – AlexG Oct 23 '20 at 07:09
  • Thanks! I actually embedded this as an Easter egg in one of the figures in my dissertation (rather than the marmot image here). As in this example, clicking on the figure in the right place would start the game. I want to get it past the proofreaders, so I need to keep it hidden. The figure has a white background, so I made the game above have a white background, and I adjusted a label or two so that the "missing piece" is not visible by default. It is not visible when printing it, either. Only problem is the 9x4 grid makes the figure look blocky when scrolling past it quickly. Dunno why! – jvriesem Oct 25 '20 at 19:19
  • @jvriesem Funny idea. Hope the referees take it with humor if they happen to discover the Easter egg. As for the loading speed, I think it is quite natural that the PDF reader wants to prerender the whole stack of 36*35=1260 layers of the image and also the OCG information when scrolling over the page, which may take some time. I just wonder whether it needs to do so on every re-visiting the page, or whether it uses some kind of cache for the pages seen. – AlexG Oct 25 '20 at 20:43
  • @AlexG: Eeep — so it IS that many layers! That makes sense. It's also a high resolution figure (7000x3000 pixels, <1 MB PNG file), so that probably doesn't help.... Fortunately, the "game" doesn't hurt the PDF size...almost at all! (I can't imagine the university proofreaders will notice. I plan to publish part of the dissertation content but not include the game for the subsequent publication, where somebody would definitely notice!) – jvriesem Oct 28 '20 at 03:15
  • 1
    @AlexG: Update: the game made it through the final published version! It doesn't seem to have any visual glitches or performance issues now, so perhaps the PDF reader updated how it handles these things in the past year. – jvriesem Oct 23 '21 at 17:40
12

Four steps with Super Mario.

\documentclass{article}
\usepackage{tikz}
\usepackage{animate}

\newcommand{\MarioStop}{
\fill[red] (2.5,8) -- (5,8) -- (5,7.5) -- (6.5,7.5) -- (6.5,7) -- (2,7) -- (2,7.5) -- (2.5,7.5);
\fill[brown!70!black] (2,7) -- (3.5,7) -- (3.5,6.5) -- (3,6.5) -- (3,6) -- (3.5,6) -- (3.5,5.5) -- (2.5,5.5) -- (2.5,6.5) -- (2,6.5);
\fill[brown!70!black] (4.5,7) -- (4.5,6) -- (5,6) -- (5,7);
\fill[brown!70!black](5,6) -- (5.5,6) -- (5.5,5.5) -- (6.5,5.5) -- (6.5,5) -- (4.5,5) -- (4.5,5.5) -- (5,5.5);
\fill[brown!70!black] (1.5,6.5) -- (2,6.5) -- (2,5.5) -- (2.5,5.5) -- (2.5,5) -- (1.5,5);
\fill[yellow!50!orange] (3.5,7) -- (4.5,7) -- (4.5,6) -- (5,6) -- (5,5.5) -- (4.5,5.5) -- (4.5,5) -- (6,5) -- (6,4.5) -- (2.5,4.5) -- (2.5,5.5) -- (3.5,5.5) -- (3.5,6) -- (3,6) -- (3,6.5) -- (3.5,6.5);
\fill[yellow!50!orange] (2,6.5) -- (2.5,6.5) -- (2.5,5.5) -- (2,5.5);
\fill[yellow!50!orange] (5,7) -- (5.5,7) -- (5.5,6.5) -- (6.5,6.5) -- (6.5,6) -- (7,6) -- (7,5.5) -- (5.5,5.5) -- (5.5,6) -- (5,6);
\fill[red] (2,4.5) -- (3,4.5) -- (3,3) -- (2.5,3) -- (2.5,2.5) -- (2,2.5) -- (2,3) -- (1,3) -- (1,3.5) -- (1.5,3.5) -- (1.5,4) -- (2,4);
\fill[red] (3.5,4.5) -- (5.5,4.5) -- (5.5,4) -- (6.5,4) -- (6.5,3.5) -- (7,3.5) -- (7,3) -- (6,3) -- (6,2.5) -- (5.5,2.5) -- (5.5,3) -- (5,3) -- (5,4) -- (4.5,4) -- (4.5,3.5) -- (3.5,3.5);
\fill[blue] (3,4.5) -- (3.5,4.5) -- (3.5,3.5) -- (4.5,3.5) -- (4.5,4) -- (5,4) -- (5,3) -- (5.5,3) -- (5.5,2) -- (6,2) -- (6,1) -- (4.5,1) -- (4.5,1.5) -- (3.5,1.5) -- (3.5,1) -- (2,1) -- (2,2) -- (2.5,2) -- (2.5,3) -- (3,3);
\fill[yellow!50!orange] (3,3) -- (3.5,3) -- (3.5,2.5) -- (3,2.5);
\fill[yellow!50!orange] (4.5,3) -- (5,3) -- (5,2.5) -- (4.5,2.5);
\fill[yellow!50!orange] (6,3) -- (7,3) -- (7,1.5) -- (6,1.5) -- (6,2) -- (5.5,2) -- (5.5,2.5) -- (6,2.5);
\fill[yellow!50!orange] (1,3) -- (2,3) -- (2,2.5) -- (2.5,2.5) -- (2.5,2) -- (2,2) -- (2,1.5) -- (1,1.5);
\fill[brown!70!black] (1.5,1) -- (3,1) -- (3,0) -- (1,0) -- (1,0.5) -- (1.5,0.5);
\fill[brown!70!black] (5,1) -- (6.5,1) -- (6.5,0.5) -- (7,0.5) -- (7,0) -- (5,0);
}
\newcommand{\MarioFStep}{
\fill[red] (3,7.5) -- (5.5,7.5) -- (5.5,7) -- (7,7) -- (7,6.5) -- (2.5,6.5) -- (2.5,7) -- (3,7);
\fill[brown!70!black] (2.5,6.5) -- (4,6.5) -- (4,6) -- (3.5,6) -- (3.5,5.5) -- (4,5.5) -- (4,5) -- (3,5) -- (3,6) -- (2.5,6);
\fill[brown!70!black] (2,6) -- (2.5,6) -- (2.5,5) -- (3,5) -- (3,4.5) -- (2,4.5);
\fill[brown!70!black] (5,6.5) -- (5,5.5) -- (5.5,5.5) -- (5.5,6.5);
\fill[brown!70!black] (5.5,5.5) -- (6,5.5) -- (6,5) -- (7,5) -- (7,4.5) -- (5,4.5) -- (5,5) -- (5.5,5);
\fill[brown!70!black] (1.5,2.5) -- (2.5,2.5) -- (2.5,2) -- (2,2) -- (2,1) -- (1.5,1) -- (1.5,0.5) -- (1,0.5) -- (1,1.5) -- (1.5,1.5);
\fill[brown!70!black] (3.5,1) -- (5,1) -- (5,0.5) -- (5.5,0.5) -- (5.5,0) -- (3.5,0);
\fill[yellow!50!orange]  (2.5,6) -- (3,6) -- (3,5) -- (2.5,5);
\fill[yellow!50!orange] (4,6.5) -- (5,6.5) -- (5,5.5) -- (5.5,5.5) -- (5.5,5) -- (5,5) -- (5,4.5) -- (6.5,4.5) -- (6.5,3.5) -- (7,3.5) -- (7,3) -- (6.5,3) -- (6.5,2.5) -- (5.5,2.5) -- (5.5,3.5) -- (6,3.5) -- (6,4) -- (3,4) -- (3,5) -- (4,5) -- (4,5.5) -- (3.5,5.5) -- (3.5,6) -- (4,6);
\fill[yellow!50!orange]  (5.5,6.5) -- (6,6.5) -- (6,6) -- (7,6) -- (7,5.5) -- (7.5,5.5) -- (7.5,5) -- (6,5) -- (6,5.5) -- (5.5,5.5);
\fill[yellow!50!orange] (2,3.5) -- (2.5,3.5) -- (2.5,2.5) -- (1.5,2.5) -- (1.5,3) -- (2,3);
\fill[red] (2.5,4) -- (4.5,4) -- (4.5,3.5) -- (5,3.5) -- (5,4) -- (5.5,4) -- (5.5,2.5) -- (3,2.5) -- (3,3) -- (2.5,3);
\fill[blue] (4.5,4) -- (4.5,3.5) -- (5,3.5) -- (5,4);
\fill[blue] (2.5,3) -- (3,3) -- (3,2.5) -- (6,2.5) -- (6,1.5) -- (5.5,1.5) -- (5.5,1) -- (4,1) -- (4,1.5) -- (3.5,1.5) -- (3.5,1) -- (2,1) -- (2,2) -- (2.5,2);
}
\newcommand{\MarioSStep}{
\fill[red] (3,8) -- (5.5,8) -- (5.5,7.5) -- (7,7.5) -- (7,7) -- (2.5,7) -- (2.5,7.5) -- (3,7.5);
\fill[red] (1.5,4.5) -- (3.5,4.5) -- (3.5,3) -- (2.5,3) -- (2.5,3.5) -- (1.5,3.5);
\fill[red] (4.5,4.5) -- (5.5,4.5) -- (5.5,4) -- (6.5,4) -- (6.5,3.5) -- (7,3.5) -- (7,3) -- (6,3) -- (6,3.5) -- (5,3.5) -- (5,4) -- (4.5,4);
\fill[brown!70!black] (2.5,7) -- (4,7) -- (4,6.5) -- (3.5,6.5) -- (3.5,6) -- (4,6) -- (4,5.5) -- (3,5.5) -- (3,6.5) -- (2.5,6.5);
\fill[brown!70!black]  (2,6.5) -- (2.5,6.5) -- (2.5,5.5) -- (3,5.5) -- (3,5) -- (2,5);
\fill[brown!70!black]  (5,7) -- (5,6) -- (5.5,6) -- (5.5,7);
\fill[brown!70!black]  (5.5,6) -- (6,6) -- (6,5.5) -- (7,5.5) -- (7,5) -- (5,5) -- (5,5.5) -- (5.5,5.5);
\fill[brown!70!black]  (1,1.5) -- (2,1.5) -- (2,1) -- (2.5,1) -- (2.5,0.5) -- (3,0.5) -- (3,0) -- (1.5,0) -- (1.5,0.5) -- (1,0.5);
\fill[brown!70!black]  (6.5,2.5) -- (7,2.5) -- (7,3) -- (7.5,3) -- (7.5,1) -- (6.5,1);
\fill[yellow!50!orange] (4,7) -- (5,7) -- (5,6) -- (5.5,6) -- (5.5,5.5) -- (5,5.5) -- (5,5) -- (6.5,5) -- (6.5,4.5) -- (3,4.5) -- (3,5.5) -- (4,5.5) -- (4,6) -- (3.5,6) -- (3.5,6.5) -- (4,6.5);
\fill[yellow!50!orange] (2.5,6.5) -- (3,6.5) -- (3,5.5) -- (2.5,5.5);
\fill[yellow!50!orange] (5.5,7) -- (6,7) -- (6,6.5) -- (7,6.5) -- (7,6) -- (7.5,6) -- (7.5,5.5) -- (6,5.5) -- (6,6) -- (5.5,6);
\fill[yellow!50!orange] (0.5,4) -- (1.5,4) -- (1.5,3.5) -- (2,3.5) -- (2,3) -- (1.5,3) -- (1.5,2.5) -- (0.5,2.5);
\fill[yellow!50!orange] (6.5,4) -- (8,4) -- (8,3) -- (7,3) -- (7,3.5) -- (6.5,3.5);
\fill[blue] (3.5,4.5) -- (4.5,4.5) -- (4.5,4) -- (5,4) -- (5,3.5) -- (6,3.5) -- (6,2.5) -- (6.5,2.5) -- (6.5,1) -- (5,1) -- (5,1.5) -- (3.5,1.5) -- (3.5,1) -- (2,1) -- (2,1.5) -- (1.5,1.5) -- (1.5,2) -- (2,2) -- (2,2.5) -- (2.5,2.5) -- (2.5,3) -- (3.5,3);
\fill[yellow!50!orange] (4,3.5) -- (4.5,3.5) -- (4.5,3) -- (4,3);
}
\begin{document}

\begin{animateinline}[autoplay,loop,
    begin={\begin{tikzpicture}},
    end={\end{tikzpicture}}]{2}
\MarioStop\newframe
\MarioFStep\newframe
\MarioSStep\newframe
\MarioFStep\newframe
\MarioSStep\newframe
\MarioFStep\newframe
\MarioSStep\newframe
\MarioFStep\newframe
\MarioSStep\newframe
\MarioStop
\end{animateinline}
\end{document}

Super Mario Four Steps

vi pa
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