# [GIF] :christmas_tree: (Conformal map of upper half-plane to triangle)

Posted 2 months ago
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 :christmas_tree:The Schwarz–Christoffel mappings are conformal transformations from the upper half-plane (or unit disk) to convex polygons; the existence of such maps is guaranteed by the Riemann Mapping Theorem. In general, these are somewhat challenging to compute/estimate, but in some cases they can be written down explicitly. For example, the following is an explicit Schwarz–Christoffel mapping to an equilateral triangle: f[z_] := (E^((2 π I)/3) Gamma[2/3])/(2^(1/3) Gamma[1/3] Gamma[4/3]) (z - 1)^(1/3) Hypergeometric2F1[1/3, 2/3, 4/3, (1 - z)/2]; To make the animation, simply apply this map to the rectangular grid on the upper half-plane (and apply some cheats); the motion comes from sliding the vertical lines to the right.Here's the code: With[{r = 12., s = .67, cols = RGBColor /@ {"#f2f9f1", "#388e3c"}}, Manipulate[ Graphics[ {cols[[1]], Polygon[{{s Cos[π/3] - 1, s Sin[π/3]}, {s Cos[2 π/3], s Sin[π/3]}, {-1/2, Sqrt[3]/2}}], FaceForm[None], EdgeForm[Directive[cols[[1]], Thickness[.006]]], Polygon[{{0, 0}, {-1, 0}, {-1/2, Sqrt[3]/2}}], Thickness[.006], CapForm[None], Table[Line[Table[ReIm[f[t^3 + I y]], {t, -3., 3, 1/40}]], {y, 1, r}], Table[ Line[Table[ReIm[f[x + I Exp[t]]], {t, -13., 4, 1/6}]], {x, -r + smootheststep[u], r + smootheststep[u], 1}]}, ImageSize -> 540, Background -> cols[[2]], PlotRange -> {{-(6/5), 1/5}, {Sqrt[3]/4 - 7/10, Sqrt[3]/4 + 7/10}}], {u, 0, 1}] ] 
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Posted 2 months ago
 A very neat idea, @Clayton, have not seen that yet around. Sooo it is that time of the year again when we get showered in code- and math- Xmas trees, awesome! I will post here a few the WL code for which I wrote the past. The first one comes from a very simple formula: $$t * sin (t) ≈ Christmas Tree$$It is a simple expanding 3D spiral and it is made with the following code: PD = .5; s[t_, f_] := t^.6 - f; dt[cl_, ps_, sg_, hf_, dp_, f_] := {PointSize[ps], Hue[cl, 1, .6 + sg .4 Sin[hf s[t, f]]], Point[{-sg s[t, f] Sin[s[t, f]], -sg s[t, f] Cos[s[t, f]], dp + s[t, f]}]}; frames = ParallelTable[ Graphics3D[Table[{dt[1, .01, -1, 1, 0, f], dt[.45, .01, 1, 1, 0, f], dt[1, .005, -1, 4, .2, f], dt[.45, .005, 1, 4, .2, f]}, {t, 0, 200, PD}], ViewPoint -> Left, BoxRatios -> {1, 1, 1.3}, ViewVertical -> {0, 0, -1}, ViewCenter -> {{0.5, 0.5, 0.5}, {0.5, 0.55}}, Boxed -> False, Lighting -> "Neutral", PlotRange -> {{-20, 20}, {-20, 20}, {0, 20}}, Background -> Black], {f, 0, 1, .01}]; Export["tree.gif", frames] Another one is made as ASCII art, which I really enjoy. Watch closely - the tree is also changing - snow sticks to branches then falls ;-)To build it we start from simple observation that a bit of randomness builds a nice tree: Column[Table[Row[RandomChoice[{"+", ".", "*", "~", "^", "o"}, k]], {k, 1, 35, 2}], Alignment -> Center] And now with a bit more sophistication a scalable dynamic ASCII tree: DynamicModule[{atoms, tree, pos, snow, p = .8, sz = 15}, atoms = { Style["+", White], Style["*", White], Style["o", White], Style[".", Green], Style["~", Green], Style["^", Green], Style["^", Green] }; pos = Flatten[Table[{m, n}, {m, 18}, {n, 2 m - 1}], 1]; tree = Table[RandomChoice[atoms, k], {k, 1, 35, 2}]; snow = Table[ RotateLeft[ArrayPad[{RandomChoice[atoms[[1 ;; 2]]]}, {0, sz}, " "], RandomInteger[sz]], {sz + 1}]; Dynamic[Refresh[ Overlay[{ tree[[Sequence @@ RandomChoice[pos]]] = RandomChoice[atoms]; Column[Row /@ tree, Alignment -> Center, Background -> Black], Grid[ snow = RotateRight[ RotateLeft[#, RandomChoice[{(1 - p)/2, p, (1 - p)/2} -> {-1, 0, 1}]] & /@ snow , {1, 0}]] }, Alignment -> Center] , UpdateInterval -> 0, TrackedSymbols -> {}] ] ] And the last one is not mine, but just to make you smile, from xkcd:
Posted 2 months ago
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 Yes, Clayton forgot to include the definition for that function: smootheststep[t_] := -20 t^7 + 70 t^6 - 84 t^5 + 35 t^4 Alternatively, you could also try using ResourceFunction["SmootherStep"] from the Function Repository, or any of a number of other sigmoidal functions like ssteprat = #^3/(1 - 3 # (1 - #)) &;