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function u(t) {
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u(t) is called 60 times per second. t: Elapsed time in seconds. S: Shorthand for Math.sin. C: Shorthand for Math.cos. T: Shorthand for Math.tan. R: Function that generates rgba-strings, usage ex.: R(255, 255, 255, 0.5) c: A 1920x1080 canvas. x: A 2D context for that canvas.
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  • Same as the remixed dweet, except the HSL works on iOS Safari.

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Xen
remix of d/14887 by u/Blacklink

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  • had to have a play with this one

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remix of d/246 by u/PhiLho

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  • u/iverjo
    To make it more clear: I like the slow pace, but not the performance hog ;)
  • u/iverjo
    Here's a less performance-intensive version: c.width=96;M=255;for(h=96;h--;)for(v=54;v--;x.fillStyle=R(M*(1+S(t))/2,M*S(v/34.4),M*C(h/61.1)),x.fillRect(h,v,1,1))
  • u/iverjo
    It should still be smooth and slow-paced
  • u/iverjo
    The previous code might not work.. here's an alternative version that probably works: c.width=96;M=255;for(h=96;h--;)for(v=54;v--;){x.fillStyle=R(M*(1+S(t))/2,M*S(v/34.4),M*C(h/61.1)),x.fillRect(h,v,1,1)}

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function u(t) {

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  • As promised, this version prevents cheating, although there is still one "special move" you can use ;-)

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  • New Dwitter Compressor - Proof of Concept
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  • u/tomxor
    Ah, I don't mean input even, i mean output even... which I guess is easier to do with a condition that checks for the lower surrogate being zero and replaces with 0x20
  • u/tomxor
    argh no i'm getting confused I was right first time, the condition is on i%7
  • u/tomxor
    no that was a terrible idea, I can see why you chose !V, since you don't have to check length and even if you did the minimum number of extraneous bits is 1, which in the 7th position is 0x40 (@) which doesn't eval...
  • u/KilledByAPixel
    tomxor, good ideas! keep em comming. ?0: to || saves 1 byte!

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  • Smoler Isometric Divisions
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  • u/KilledByAPixel
    nice!
  • u/pavel
    Weirdly the numeric error from approximating pi/3 seems to matter a lot.
  • u/tomxor
    Yeah, I think it's because it causes a deviation in the sequence at j=0. Although the output of 0|3*C(r) generally gives the expected low period of +1,+2,+1,-1,-2,-1 as j decrements, it ends on a 3, but only when using lower precision PI/3. full precision PI/3 pushes all of the 2s into 3s, which can be corrected by lowering the coefficient to 2.9, but this also makes j=0 output 2 which then seems to destroy the complexity.
  • u/pavel
    I tried getting rid of the sin/cos by using hex coordinates but I was unable to make a complex picture. Usually just one of the six slices drew-in.

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function u(t) {

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  • Enjoy the ride but watch out for the missing track pieces! :)

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