# 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>

> You can just add your Python draft as an additional variant. As the Rosetta Code landing page puts it: 'to aid a person with a grounding in one approach to a problem in learning another'. The goal is to provide comparative insight rather than to construct any claim to exclusive canonicity Hout (talk) 21:53, 15 January 2020 (UTC)

Thanks for the information, I have added my code underneath the original.

## Incorrect .PPM File created by C code.

The current C code uses directly code directly copied from ppm Bitmap Writing page. This consequently leads to a useless .ppm image created.

The following lines of code need to be replaced:

for(i=0;i<sandPileEdge;i++){

for(j=0;j<sandPileEdge;j++){
colour[0] = (sandPile[i][j] + i)%256;
colour[1] = (sandPile[i][j] + j)%256;
colour[2] = (sandPile[i][j] + i*j)%256;
fwrite(colour,1,3,fp);
}

}

with these to get an image similar in style to the images on the wikipedia page, however any different combination of RGB values will serve.

for(i=0;i<sandPileEdge;i++){

for(j=0;j<sandPileEdge;j++){
if (sandPile[i][j] == 0){
colour[0] = 0;
colour[1] = 0;
colour[2] = 0;
} else if (sandPile[i][j] == 1){
colour[0] = 0;
colour[1] = 255;
colour[2] = 0;
} else if (sandPile[i][j] == 2){
colour[0] = 255;
colour[1] = 0;
colour[2] = 255;
} else if (sandPile[i][j] == 3){
colour[0] = 255;
colour[1] = 215;
colour[2] = 0;
};
fwrite(colour,1,3,fp);
};

};