在2D数组中创建圆的算法编辑：钻石会很好

Question

I have a 2D in python that represents a tile map, each element in the array is either a 1 or 0, 0 representing land and 1 representing water. I need an algorithm that takes 2 random coordinates to be the center of the circle, a variable for the radius (max 5) and replace the necessary elements in the array to form a full circle.

x = random.randint(0,MAPWIDTH)
y = random.randint(0,MAPHEIGHT)
rad = random.randint(0,5)

tileMap[x][y] = 1 #this creates the center of the circle

how would I do this?

Answer 1

You would have to set a coordinate to one if ((x – h)(x - h)) + ((y – k)(y - k)) = r * r is true. h is the centre x coordinate and k is the centre y coordinate.

Answer 2

As previously said, you can use the definition of a circle, like so:

import math

def dist(x1, y1, x2, y2):
    return math.sqrt((x1 - x2) ** 2 + (y1 - y2) ** 2)

def make_circle(tiles, cx, cy, r):
    for x in range(cx - r, cx + r):
        for y in range(cy - r, cy + r):
            if dist(cx, cy, x, y) <= r:
                tiles[x][y] = 1

width = 50
height = 50

cx = width // 2
cy = height // 2
r = 23

tiles = [[0 for _ in range(height)] for _ in range(width)]

make_circle(tiles, cx, cy, r)

print("\n".join("".join(map(str, i)) for i in tiles))

This outputs

00000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000
00000000000000000000000001000000000000000000000000
00000000000000000001111111111111000000000000000000
00000000000000001111111111111111111000000000000000
00000000000000111111111111111111111110000000000000
00000000000001111111111111111111111111000000000000
00000000000111111111111111111111111111110000000000
00000000001111111111111111111111111111111000000000
00000000011111111111111111111111111111111100000000
00000000111111111111111111111111111111111110000000
00000001111111111111111111111111111111111111000000
00000001111111111111111111111111111111111111000000
00000011111111111111111111111111111111111111100000
00000111111111111111111111111111111111111111110000
00000111111111111111111111111111111111111111110000
00001111111111111111111111111111111111111111111000
00001111111111111111111111111111111111111111111000
00001111111111111111111111111111111111111111111000
00011111111111111111111111111111111111111111111100
00011111111111111111111111111111111111111111111100
00011111111111111111111111111111111111111111111100
00011111111111111111111111111111111111111111111100
00011111111111111111111111111111111111111111111100
00011111111111111111111111111111111111111111111100
00111111111111111111111111111111111111111111111100
00011111111111111111111111111111111111111111111100
00011111111111111111111111111111111111111111111100
00011111111111111111111111111111111111111111111100
00011111111111111111111111111111111111111111111100
00011111111111111111111111111111111111111111111100
00011111111111111111111111111111111111111111111100
00001111111111111111111111111111111111111111111000
00001111111111111111111111111111111111111111111000
00001111111111111111111111111111111111111111111000
00000111111111111111111111111111111111111111110000
00000111111111111111111111111111111111111111110000
00000011111111111111111111111111111111111111100000
00000001111111111111111111111111111111111111000000
00000001111111111111111111111111111111111111000000
00000000111111111111111111111111111111111110000000
00000000011111111111111111111111111111111100000000
00000000001111111111111111111111111111111000000000
00000000000111111111111111111111111111110000000000
00000000000001111111111111111111111111000000000000
00000000000000111111111111111111111110000000000000
00000000000000001111111111111111111000000000000000
00000000000000000001111111111111000000000000000000
00000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000

Note that I deliberately used a rather large array and radius - this results in being able to actually see the circle a bit better. For some radius around 5, it would probably be pixelated beyond belief.

Answer 3

Inspired by Izaak van Dongen, just re-worked a bit:

from pylab import imshow, show, get_cmap
from numpy import random
import math

def dist(x1, y1, x2, y2):
    return math.sqrt((x1 - x2) ** 2 + (y1 - y2) ** 2)

def make_circle(tiles, cx, cy, r):
    for x in range(cx - r, cx + r):
        for y in range(cy - r, cy + r):
            if dist(cx, cy, x, y) < r:
                tiles[x][y] = 1
    return tiles

def generate_image_mask(iw,ih,cx,cy,cr):

    mask = [[0 for _ in range(ih)] for _ in range(iw)]
    mask = make_circle(mask, cx, cy, cr)
    #print("\n".join("".join(map(str, i)) for i in mask))
    imshow(mask, cmap=get_cmap("Spectral"), interpolation='nearest')
    show()

if __name__ == '__main__':

    image_w = 60
    image_h = 60
    circle_x = image_w/2
    circle_y = image_h/2
    circle_r = 15

    generate_image_mask(image_w,image_h,circle_x,circle_y,circle_r)