Courbe de bézier de avec n points de contrôle (le premier est à l’origine, en haut à gauche): essayez render(5)

from math import *
from cmath import *
import kandinsky as k
import random as ran
dark=k.color(0,0,0)
red=k.color(255,0,0)
def factorial(x):
  x=int(x)
  if x>=0:
    s=1
    for loop in range(x):
      s*=(loop+1)
    return s

def binomial(n,k):
  return(factorial(n)/(factorial(k)*factorial(n-k)))     

def bernstein(m,i,u):
  return(binomial(m,i)*(u**i)*((1-u)**(m-i)))

def setpoint(x,y):
  for cx in range(7):
    k.set_pixel((x-3+cx),y,dark)
  for cy in range(7):
    k.set_pixel(x,y-3+cy,dark)

def color(x,y):
  k.set_pixel(x,y,red)

def P(t,list):
  s=0
  for loop in range(len(list)):
    s+=bernstein(len(list),loop,t)*list[loop]
  return s

def render(n):
  n-=1
  ptsx=[]
  ptsy=[]
  setpoint(0,0)
  for loop in range(n):
    ptsx.append(ran.randrange(0,320))
    ptsy.append(ran.randrange(0,222))
    setpoint(ptsx[loop],ptsy[loop])
  step=0
  while 1:
    t=(1/(2**step))
    for loop in range(1+2**step):
      k.fill_rect(int(P(t*loop,ptsx)),int(P(t*loop,ptsy)),1,1,(0,int(255-loop*t*255),0))
    step+=1