Talk:Abelian sandpile model
optimizations ?
the figures are 8xtimes symmetric like drawing a circle. But how to calculate the generating triangle. See 16->
00100 02120 11011 02120 00100 the origin is the starting point of the sandpile. is rotated and mirrored of X20 011 mirrored at main diagonal 100 120 011 rotated 3 times by 90° around the origin.
A more readable Python solution.
This solution is more interactive, as it can also be used in imports on a Python console; plus, it is more readable. <lang python> from os import system, name from time import sleep
def clear(): if name == 'nt': _ = system('cls') else: _ = system('clear')
def exit(): import sys clear() sys.exit()
def make_area(x, y): area = [[0]*x for _ in range(y)] return area
def make_sandpile(area, loc, height): loc=list(n-1 for n in loc) x, y = loc
try: area[y][x]+=height except IndexError: pass
def run(area, by_frame=False): def run_frame(): for y_index, group in enumerate(area): y = y_index+1
for x_index, height in enumerate(group): x = x_index+1
if height < 4: continue
else: make_sandpile(area, (x+1, y), 1) make_sandpile(area, (x, y+1), 1)
if x_index-1 >= 0: make_sandpile(area, (x-1, y), 1) if y_index-1 >= 0: make_sandpile(area, (x, y-1), 1)
make_sandpile(area, (x, y), -4)
while any([any([pile>=4 for pile in group]) for group in area]): if by_frame: clear() run_frame() if by_frame: show_area(area); sleep(.05)
def show_area(area): display = [' '.join([str(item) if item else ' ' for item in group]) for group in area] [print(i) for i in display]
clear() if __name__ == '__main__': area = make_area(10, 10) print('\nBefore:') show_area(area) make_sandpile(area, (5, 5), 64); run(area) print('\nAfter:') show_area(area) </lang>
Output: <lang> Before: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
After: 0 0 0 0 0 0 0 0 0 0 0 0 0 1 2 1 0 0 0 0 0 0 2 2 2 2 2 0 0 0 0 1 2 2 2 2 2 1 0 0 0 2 2 2 0 2 2 2 0 0 0 1 2 2 2 2 2 1 0 0 0 0 2 2 2 2 2 0 0 0 0 0 0 1 2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 </lang>