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As we all know our site's logo was completely generated by Mathematica. I suppose it is quite natural to make the next step -- to generate the animated version of this logo. There's a lot of space for creativity here, and I suggest to consider the following options.

  1. Animated process of construction from scratch, as it is described in Verbeia's blog post.
  2. Animated morphing of original pentagonal star to the current heptagonal one (J.M.'s idea in the comment)
  3. Some less fussy, a neutral animation of the logo itself, more suitable for placing on webpages.
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4
  • 4
    $\begingroup$ I'd sure like to see somebody automagically morph a hyperbolic pentagon to a heptagon... $\endgroup$ Oct 11, 2012 at 1:53
  • $\begingroup$ @J.M. You mean something like a flash shape tween between the two? $\endgroup$
    – VF1
    Oct 13, 2012 at 0:54
  • $\begingroup$ @VF1, yes, something like that... $\endgroup$ Oct 13, 2012 at 1:28
  • $\begingroup$ If the intention is to change the static logo at the top of each mathematica.se page to a dynamic logo, please don't: it needlessly wastes a bit of bandwidth and, more significantly, distracts from the content. (As a programming exercise, that's another matter.) $\endgroup$
    – murray
    Nov 13, 2012 at 16:15

9 Answers 9

59
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Let me join.

enter image description here

logo = Cases[
   p7 /. triangulate /. moretriangles /. shrink /. shrink /. shrink /. colour3[] /. colour4["SunsetColors", 1, 28/34] , {c__, Polygon[pts__]}, \[Infinity]];
logo = SortBy[logo, First];
p = Evaluate[InterpolatingPolynomial[{
     {0, {0, 0, 0, 0}}, {Pi, {Pi, 0, 0, 0, 0}}, {2 Pi, {2 Pi, 0, 0, 0}}},#]] &;
pp[a_] := If[Abs[a - Pi] < .6, Pi, p@a];(*to stabilize flickering*)
nf = 37;(*number of frames*)
frames = Table[
   Graphics[Thread[Rotate[logo, p@angle]],
    PlotRange -> 1.2, ImageSize -> 240
    ],
   {angle, 0, 2 Pi, (2 Pi)/(nf - 1)}];
ListAnimate@frames
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  • $\begingroup$ The only annoying issue is this polygon jumping. I think this is a numerical instability artefact in Rotate. $\endgroup$
    – faleichik
    Oct 18, 2012 at 11:26
  • $\begingroup$ Have you tried using RotationTransform instead? $\endgroup$
    – Mr.Wizard
    Oct 18, 2012 at 12:24
  • $\begingroup$ Not really, but I'm pretty sure that it won't help. This is possibly related to this problem: mathematica.stackexchange.com/q/9560/219 $\endgroup$
    – faleichik
    Oct 18, 2012 at 12:46
53
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Breathing with occluded borders, per Toad's request:

enter image description here

Run the following command to get the Mathematica code

NotebookPut@ImportString[Uncompress@FromCharacterCode@Flatten@ImageData[
               Import@ "http://i.stack.imgur.com/VqjJ9.png","Byte"],"NB"]
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43
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As per the blog:

Export["breathing.gif", Table[Graphics[
    p7 /. triangulate /. moretriangles /. shrink /. shrink /. shrink /. colour3[] /.
    colour4["SunsetColors", 1, 28/34] /. curve /. bolicsn[(1 - Cos[2 \[Pi] t])/2], 
  ImageSize -> 150], {t, 0, 1, 0.05}]];

breathing

Some good old fashioned colour cycling:

Clear[f];
f[c_] /; c > 2 := c - 2;
f[c_] /; c > 1 := 2 - c;
f[c_] := c;

colour4c[s_: "SunsetColors", a_?NumericQ, b_?NumericQ, c_?NumericQ] :=
  Polygon[v_] /; Length[v] == 4 :>
    {ColorData[s, f[c + a - b Norm[PolygonCentroid[v]]]], Polygon[v]}

Export["ColourCycleLogo.gif", Table[Graphics[
    p7 /. triangulate /. moretriangles /. shrink /. shrink /. shrink /. colour3[] /.
    colour4c["SunsetColors", 1, 28/34, t], 
  ImageSize -> 150], {t, 0, 2, 0.05}], "DisplayDurations" -> ConstantArray[0.05, 41]];

colours

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2
  • $\begingroup$ The first one is my favorite so far. +1! $\endgroup$
    – faleichik
    Oct 16, 2012 at 22:22
  • $\begingroup$ Alternatively: f[c_] := 1 - Abs[1 - c] $\endgroup$ Oct 20, 2012 at 14:31
43
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Who wanted the automagic? :)

enter image description here

mmastar[as_, nn_: 1] := Graphics[
   Scale[#, 1/max@#, {0, 0}] &[
    Polygon[pt /@ as] /. triangulate /. moretriangles /. shrink /. 
          shrink /. shrink /. colour3[] /. colour4[] /. curve /. 
     bolicsn[nn]], AspectRatio -> Automatic, PlotRange -> 0.025];
da = 0.0001;
max[zu_] := 
  Cases[zu, {_?NumericQ, _?NumericQ}, \[Infinity]] // Norm // Max;
pt[a_] := {Sin@a, Cos@a};
pts0 = Range[ 0, (2 - 2/5) Pi, 2 Pi/5] // N
pts1 = Append[Insert[pts0, pts0[[2]] - da, 2], pts0[[-1]] + da]
pts2 = Range[Pi/7, 2 Pi, 2 Pi/7] // N
ptsat[t_] := (1 - t) pts1 + t pts2;
nn0 = 1; nn1 = 0.0001;
nat[t_] := (1 - t) nn0 + t nn1;
frames = Table[mmastar[ptsat[t], nat[t]], {t, 0, 1, 1/16}] // Reverse;
ListAnimate[frames]
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2
  • $\begingroup$ Excellent, +1 - although you've shown the official logo upside down :) $\endgroup$
    – cormullion
    Oct 26, 2012 at 12:02
  • $\begingroup$ @cormillion, fixed! In previous version it was pts2 = Range[2 Pi/7, 2 Pi, 2 Pi/7] // N. $\endgroup$
    – faleichik
    Oct 26, 2012 at 15:07
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Load some images:

size = {200, 200};
foot = ImageResize[Import[
   "http://upload.wikimedia.org/wikipedia/commons/a/ab/Monty_python_foot.png"
  ], size];
spikey = ImageResize[Import[
   "http://upload.wikimedia.org/wikipedia/en/b/bf/MathematicaSpikeyVersion8.png"
  ], size];
mse = ImageResize[Import[
   "http://i.stack.imgur.com/yjrEY.png"
  ], size];

Crop them, squash them, transform them:

feet = Table[ImageCrop[foot, size {1, k}, Top], {k, 0.1, 0.9, 0.1}];
spoke = Table[ImageResize[spikey, size {1, k}], {k, 0.9, 0.1, -0.1}];
logos = Table[ImagePerspectiveTransformation[mse, 
  FindGeometricTransform[{{0, 0}, {1, 0},
    {0.5, 0.5} + {-(1/(-2 - 2 Cos[t])), (-4 - 3 Cos[t] + 8 Sin[t])/(8 + 8 Cos[t])},
    {0.5, 0.5} + {1/(-2 - 2 Cos[t]), (-4 - 3 Cos[t] + 8 Sin[t])/(8 + 8 Cos[t])}},
    {{0, 0}, {1, 0}, {1, 1}, {0, 1}}][[2]], Padding -> White], {t, 0, \[Pi]/2, \[Pi]/40}];
squish = Table[ImageCrop[logos[[1]], size {1, k}, Top], {k, 0.1, 0.9, 0.1}];

Assemble them together:

a = ImageAssemble[List /@ #] & /@ Thread[{feet, spoke}];
b = ImageAssemble[List /@ #] & /@ Thread[{Reverse@feet, squish}];
c = logos;
d = ConstantArray[Last@logos, 5];

Animate:

Export["logoanimate.gif", Join[a, b, c, d]]

1

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4
  • $\begingroup$ ahhahahahaha.... $\endgroup$
    – Mr.Wizard
    Oct 16, 2012 at 4:28
  • $\begingroup$ "too silly...."! $\endgroup$
    – cormullion
    Oct 16, 2012 at 13:00
  • 3
    $\begingroup$ Finally, something completely different! $\endgroup$ Oct 16, 2012 at 14:19
  • 1
    $\begingroup$ Ah, finally the foot. Where are the mints? $\endgroup$
    – Yves Klett
    Oct 16, 2012 at 20:21
33
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Here's a spinning "3D version" of the logo

enter image description here

Using the code from meta/blog to create the logo (assigned to the variable logo), continue with the following steps:

side[o_] := Block[{z, pts = Partition[
    Table[N[{Cos[t], Sin[t], z}], {t, Pi/14, 2 Pi, 2 Pi/7}], 2, 1, 1]},
    Composition[Polygon, Flatten[#, 1] &] /@ Thread[{pts /. z -> o/2, Reverse /@ pts /. z -> -o/2}]
]

logo3D = With[{d = 0.1}, 
    Graphics3D[{
        {EdgeForm@None, #}, 
        {EdgeForm@None, FaceForm@RGBColor[0.5995136280878135`, 0.20347121886943803`, 0.37787606421753417`], side@d}
        }, Boxed -> False, Lighting -> "Neutral"
    ] & @@ (logo /. Polygon[x__] :> Polygon[{x /. {a_, b_} :> {a, b, d/2}, 
        x /. {a_, b_} :> {a, b, -d/2}}])
]

frames = Table[Graphics3D[
        {Rotate[First@logo3D, x, {0, 1, 0}]},
        Lighting -> "Neutral",
        ViewAngle -> 35 Degree, ViewVector -> {0, 0, 3.5},
        ViewCenter -> {1, 1, 1}/2, ViewRange -> All, ViewVertical -> {0, 1, 0},
        Axes -> False, Boxed -> False, ImageSize -> 400
    ], {x, 0, 2 Pi, Pi/20}
];

Export["spin.gif", frames, "DisplayDurations" -> 0.05];

A "true 3D version" of the logo would involve raised and beveled profiles for the various inner decorations, but that's considerably harder.

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4
  • $\begingroup$ Cool! Perhaps it will look more spectacular with removed gray polygons. The holes will make this "more three-dimensional". $\endgroup$
    – faleichik
    Oct 16, 2012 at 22:26
  • $\begingroup$ and that is considerably harder $\endgroup$ Oct 16, 2012 at 22:32
  • $\begingroup$ @faleichik I agree, and not just that, one would also need to bevel the various polygons "nicely" to give it depth and structure. As I mentioned in the last line, it's a much harder problem :) $\endgroup$
    – rm -rf
    Oct 16, 2012 at 22:49
  • $\begingroup$ OK, now I see why it is more involved than I thought. $\endgroup$
    – faleichik
    Oct 16, 2012 at 23:02
26
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A very rough interpretation, which I hope might at least give some ideas:

(* Final image *)
fin = (p7 /. triangulate /. moretriangles /. shrink
      /. shrink /. shrink /. colour3[] /. colour4["SunsetColors", 1, 28/34]);
icycle[ j_, k_] := 
      Table[Graphics[fin[[1 ;; i, j, k]], PlotRange -> 1], {i, 7}] 
kcycle[i_, j_] := 
      Table[Graphics[fin[[i, j, 1 ;; k]], PlotRange -> 1], {k, 4}]
raster = Rasterize/@
    Prepend[Drop[
       Module[{c}, 
        Flatten@
         {Table[(c = icycle[1, 1 ;; m])~Join~Reverse[c], {m, 4}], 
          Table[(c = kcycle[1 ;; 7, 1 ;; m])~Join~Reverse[c], {m, 4}]}], -4],
   Graphics[{White, Rectangle[]}]];
Export["logo.gif", raster]

image

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    $\begingroup$ nice! On the other hand I wouldn't want to have this amount of flickering on my webpage! $\endgroup$
    – chris
    Oct 11, 2012 at 6:51
  • $\begingroup$ @chris thanks. I was just having fun with this, and this is the end result. I'd love to play with it some more though and maybe get something smother if I have time. $\endgroup$
    – VF1
    Oct 11, 2012 at 14:01
  • 1
    $\begingroup$ aaah! Motion sickness! $\endgroup$
    – dearN
    Oct 14, 2012 at 22:04
  • $\begingroup$ Good for the tip of a nerd's Christmas tree $\endgroup$
    – Rojo
    Oct 17, 2012 at 12:40
25
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Somewhat belatedly, here is a version that starts from random points and slowly coalesces into the logo.

Begin with the logo from the blog entry which is here called img, and apply a jitter filter which randomizes the position of each pixel within a region of specified size. By starting with a large region (100 pixels by 100 pixels) and shrinking down to 1 by 1, the image changes from a point cloud into a geometric object.

video = Table[
   ImageFilter[RandomChoice[Flatten[#, 1]] &, img, i, Interleaving -> True], 
     {i, {100, 90, 80, 70, 60, 50, 40, 30, 25, 20, 15, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1}}];

enter image description here

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Not very interesting, but I learnt a few things...

tab = Table[Show[
    Graphics[Rectangle[{-1, -1}, {1, 1}]],
    i (* where i is the final graphic produced by Verbeia's blog post *)  
     /. 
     {GrayLevel[0.85] -> Opacity[0],
      Polygon[{a_, b_, c_, d_}] -> 
       {
        Scale[Rotate[Polygon[{a, b, c, d}], 2 Pi t, {0, 0}], t, {0, 0}]
       }
      }
    ],
   {t, 0, 1, 0.02}]; 

Export["stack-logo.gif", Flatten[Join[tab, Reverse[tab]]]]

stack overrun

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