How can I draw a circle in a data array/map in python

Question

I have an array that will be 100 * 100, I can access any point like

map[x][y]

It kinda will look like this:

for i in map:
    for ii in i:
        print ii,
    print '\n',

output:

. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .

I want to make a circle in it like:

. . . . . # . . . . .
. . . # # . # # . . .
. . # . . . . . # . .
. # . . . . . . . # .
. # . . . . . . . # .
# . . . . . . . . . #
. # . . . . . . . # .
. # . . . . . . . # .
. . # . . . . . # . .
. . . # # . # # . . .
. . . . . # . . . . .

How can i do this?

I want to try and make a triangulation system where i will find the point that 3 circles will overlap. Is there any other way I can achieve this.

I just want to get the distance (dots from center) and the direction.

Answer 1

The basic formula for a circle is

(x - a)**2 + (y - b)**2 = r**2

Where (x, y) is a point, (a, b) is the center of the circle and r is the radius.

width, height = 11, 11
a, b = 5, 5
r = 5
EPSILON = 2.2

map_ = [['.' for x in range(width)] for y in range(height)]

# draw the circle
for y in range(height):
    for x in range(width):
        # see if we're close to (x-a)**2 + (y-b)**2 == r**2
        if abs((x-a)**2 + (y-b)**2 - r**2) < EPSILON**2:
            map_[y][x] = '#'

# print the map
for line in map_:
    print ' '.join(line)

This results in

. . . # # # # # . . .
. . # . . . . . # . .
. # . . . . . . . # .
# . . . . . . . . . #
# . . . . . . . . . #
# . . . . . . . . . #
# . . . . . . . . . #
# . . . . . . . . . #
. # . . . . . . . # .
. . # . . . . . # . .
. . . # # # # # . . .

You'll have to fiddle with the value for EPSILON with this method.

Alternatively, iterate by angle and calculate the (x,y) coordinate as you go

import math
# draw the circle
for angle in range(0, 360, 5):
    x = r * math.sin(math.radians(angle)) + a
    y = r * math.cos(math.radians(angle)) + b
    map_[int(round(y))][int(round(x))] = '#'

Gives:

. . . # # # # # . . .
. # # . . . . . # # .
. # . . . . . . . # .
# . . . . . . . . # #
# . . . . . . . . . #
# . . . . . . . . . #
# . . . . . . . . . #
# . . . . . . . . . #
. # . . . . . . . # .
. # # . . . . . # # .
. . . # # # # # . . .

Answer 2

The formula of circle is

(x-a)^2 + (y-b)^2 - r^2 = 0

where x,y are coodrinates of point, a,b are coordinates of center and r is the circle radius. Just find all points for which this equation is true. Since your field is integer, you will need to replace =0 with <1 or even <= 1 whatever looks best.