• Subject: rotating ASCII donut
• From: Andrew Trevorrow <andrewtrevorrow@...>
• Date: Wed, 23 Feb 2022 10:38:49 +1100

Below is a Lua version of Andy Sloane's rotating ASCII donut
(see http://www.a1k0n.net/2011/07/20/donut-math.html).
Only tested in Terminal.app on macOS but should work in any
terminal that supports standard ANSI escape sequences.
Note that I aimed for an aesthetically pleasing donut shape
rather than the shortest possible code.

local io=io.write
local i,j,k local z,b={},{}
local a=".,-~:;=!*#\$@"local A=0.0
local sin=math.sin local cos=math.cos
local int=math.floor io("\27[2J")local B=
0 while true do for i= 0,1759 do b[i]=" " z
[i]=0.0 end j=0.0 while 6.28 > j do i=0 while
6.28>i do local c=sin(i)local d=cos(j) local e=
sin(A)local f=sin(j)         local g=cos(A) local
h=d+2 local D= 1/(             c*h*e+f*g+5) local
l=cos(i) local m=               cos(B)local n=sin
(B) t=c*h*g - f*e               local x=int(40+30
* D*(l*h*m - t*n))             local y= int(12+15
* D*(l*h*n + t*m))o=         int(x+80*y) local N=
int(8 *((f*e-c*d*g)*m - c*d*e - f*g - l*d*n))+1
if 22>y and y>0 and x>0 and 80>x and D > z[o]
then z[o]= D if N > 1 then b[o]= a:sub(N,N)
else b[o]=a:sub(1,1) end end i=i+0.02 end
j= j + 0.07 end io("\27[H")for k = 0,
1760 do if k%80>0.0 then io(b[k])
else io("\n") end end A= A+
0.04 B=B+0.02 end