Random dweets | Dwitter

function u(t) {

c.width=1920 // clear the canvas
for(i=0;i<9;i++)
x.fillRect(400+i*100+S(t)*300,400,50,200) // draw 50x200 rects

} // 122/140

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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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Current theme challenge: #Food

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prplz

function u(t) {

for(c.width=-1,q=77,i=33;--i;)for(j=33;--j;)x.fillRect(q+q*C(i+t)*C(j+t),q+q*C(i)*S(j),1,1)

Sep 11 2018 5:01 AM UTC

}// 91/140

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u/donbright

cool donut
u/prplz

i donut know what you mean
u/fujimotos

Dough! Nut getting it?
u/lionleaf

Nice one! #donut

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katkip

function u(t) {

eval(unescape(escape`砮景湴㴧㜰灸∧㭰㴨搽湥眠䑡瑥⠩Ɽ⤮来瑔業攨⤯ㅥ㐻砮晩汬呥硴⠧猧ⰹ〰ⰵ㘰⤻景爨椽〻椫⬼㄰㬩砮晩汬呥硴⠧瀧ⰹ〰⭃⡰⩰⼱攱〫椪瀯ㅥ㌩⨹㤹⽩ⰵ㘰⭓⡰⩰⼱攱〪椪㤹⽰⭩⩰⼱攳⤪㤹㤯椩`.replace(/u(..)/g,"$1%"))); //lol

Oct 1 2018 1:14 PM UTC

}// 140/140

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Real time planets position (not really precise but I tried); s = sun, p = planet. Orininal code: x.font='70px"';p=(d=new Date(),d).getTime()/1e4 x.fillText('s',900,560);for(i=0;i++<10;)x.fillText('p', 900+C(p*p/1e10+i*p/1e3)*999/i, 560+S(p*p/1e10*i*99/p+i*p/1e3)*999/i)

u/donbright

awesome use of time

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Purpzie

remix of d/10165 by u/Purpzie

function u(t) {

r=_=>Math.random()-.5;c.width|=0;x.fillStyle=R(t*25);x.font=`${t*99}px"`;x.fillText('FRICK',300-t*25+r()*t*5,600+t*25+r()*t*5)

Aug 11 2019 2:25 AM UTC

}// 126/140

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when someone says not to swear

u/katkip

Welcome back to dwitter, female humanoid
u/katkip

#meme #text
u/13thptr

wb
u/joeytwiddle

🙊
u/lewdev

"FRICK" intensifies

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Niklas

function u(t) {

for(c.width|=0,e=960,f=540,r=260,i=r+1;i--;)a=.02417*i+t,g=S(t)*r,h=C(t)*r,x.lineTo(e+g+C(a)*r,f+h+S(a)*r),i%2&&x.lineTo(e+h,f+g);x.stroke()

Feb 24 2017 1:16 PM UTC

}// 140/140

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ciaccodavide

remix of d/24223 by u/ciaccodavide

function u(t) {

eval(unescape(escape`𩡯𬠨𬠽𣑡𭁨𛡲𨑮𩁯𫐬𪐽𜱥𝀻𪐭𛐻𚑻𦐽𬠨𦀽𬠨𨐽𨠽𪰽𜀩𚐊𭱨𪑬𩐨𨐫𨠼𜠦𪰫𚰼𞐰𚑥👡𛁡👡𚡡𛑢𚡢𚱘𚠮𜀲𛐮𜀱𛁢🐲𚡥𚡢𚱙𚠮𜀱𛐮𞀊𮀮𩡩𫁬𤱴𮑬𩐽𨁨𬱬𚀤𮱫𚡡𯐬𞐹𙐬𝐰𙑠𒡸𛡦𪑬𫁒𩑣𭀨𦀪𜡥𜰬𦐪𜑥𜰬𭀼𞐬𪰯𜰰𚑽`.replace(/u../g,'')))

Oct 25 2021 8:45 AM UTC

}// 130/140

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rainbow #mandelbrot forest

u/rep_movsd

Nice

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jasnh

function u(t) {

var F=function(b,m){return Math.abs(((t*60)%b)*510/b-255)};
x.fillStyle =R(F(91),0,F(63));
x.fillRect(19,0,F(200)*5,1000);

Mar 27 2022 10:43 PM UTC

}// 122/140

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beginners first dweet

u/Xen

Welcome to the dwitter family!
u/jasnh

With unused m parameter to function

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joeytwiddle

remix of d/26971 by u/joeytwiddle

function u(t) {

x.fillRect(0,0,2e3,1080)
for(j=9;j--;)for(i=99;i--;)W=50-i/2,x.clearRect(550+j*99+S(t)*300+i/4*S((5+j)*i*i-t*3)-W/2,i?610+2*i:400,W,i?1:200)

Jan 29 2023 1:44 PM UTC

}// 140/140

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#default reflected in rippling #water (night, redux)

u/joeytwiddle

This is more efficient than fillRect() but it's too long! c.width|=c.style.filter='invert(100%)' (This way, we also need to change clearRect into fillRect)
u/magna

c.style.filter='invert(' suffices since browsers are notoriously forgiving

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cantelope

function u(t) {

eval(unescape(escape`𩡯𬠨𨰮𭱩𩁴𪁼👩🐱𝐻𪐭𛐻𚑦𫱲𚁪🐳𚱩𙐴𛁱🐶𛠳𛱪𞱪𛐭𞰩𩡯𬠨𪰽𫰽𜐷𜀻𪰭𛐻𮀮𩡩𫁬𤡥𨱴𚁯𚱩𙐵𚠴𜀰𚰨𨐫𚁓𚁳👱𚠨𪠫𜐩𚐪𫰭𨐩𛱯𚡫𚐪𬠬𫰫𚁩𛰵𯀰𚐪𜰶𜀫𚀨𨠽𠰨𬀩𚡯𚐫𚁃𚁳𚐪𫰭𨠩𛱯𚡫𚐪𬠬𝀬𝀩𚑲👍𨑴𪀮𬡡𫡤𫱭𚁡👓𚁰👱𚡪𚐪𫰩𚠪𛠵`.replace(/u../g,'')))

Aug 21 2024 7:55 PM UTC

}// 140/140

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place points randomly inside polygons

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Tesseract

function u(t) {

n=c.width=133
d=(a,b)=>[a+(o=a-7-C(t))*(k=.1/(o*o+b*b)),b-b*k]
for(l=1e4;l--;)x.fillRect(j=l%n,i=l/n,1,(C[l]=t?d(...C[l]):[i/5,j/5])[1])

Oct 30 2020 12:38 PM UTC

}// 136/140

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Schramm–Loewner evolution

u/Tesseract

If anyone knows how to get something that looks like brownian motion as a function of t in around 10 bytes, I'd like to know (it should replace C(t)).
u/rodrigo.siqueira

maybe C(t*i**8) is more random
u/rodrigo.siqueira

I see a very beautiful orange SLE (Schramm-Loewner Evolutions) here: http://www.statslab.cam.ac.uk/~jpmiller/teac… I wish this could be reproduced in javascript.
u/rodrigo.siqueira

I asked jpmiller about how he did that incredible beautiful orange SLE and got his answer, if you want to reproduce it The way that the image was made is using the following steps: "1) Sample an instance of the Gaussian free field (GFF) — see Page 24 of this paper for Mathematica code which produces it https://arxiv.org/pdf/math/0312099.pdf 2) The GFF is a function h defined on the grid {1,...,n}^2. If you consider its exponential exp(i c h), where i = sqrt(-1), c > 0 is a constant, then you get a vector field. 3) Then you can consider flow lines of this vector field by solving the equation eta’(t) = exp(i c h(eta (t)). The image essentially shows flow lines with angles +/- pi/2 starting from every point in the square where c = 1 / (2/Sqrt(128) + Sqrt(128)/2) — this is the value of c so that what you will see will be related to SLE with kappa = 128. Best wishes, Jason"

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poetaster

function u(t) {

x.font=`${99+90*S(t)}px a`
for(i=0;i<21;i++)
x.fillText('~',i*90+(i-9)*35-669,(i-9)*90+S(t+(i-9))*9*(i-9))
x.fillStyle=R(i,i,i,0.01)

Oct 19 2023 8:54 PM UTC

}// 135/140

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drawing birds, throwing birds