Equally spaced circle in UI

AZOM

@mikael
Yeah, thanks for all.

mikael

@adomanim, here you go. You need the vector.py from here.

import ui
import vector

chars = 'ABCDEFGH'
start_angle = 0 # First character on the right

circle_color = 'red'
char_color = 'white'
char_font = ('Apple SD Gothic Neo', 32)

diameter = min(ui.get_screen_size())/2

root = ui.View()
root.present()

pointer = vector.Vector()
pointer.magnitude = diameter/2
pointer.degrees = start_angle

for c in chars:
    label = ui.Label(
        text=c,
        text_color=char_color,
        alignment=ui.ALIGN_CENTER,
        font=char_font)
    label.center = root.bounds.center() + tuple(pointer)
    pointer.degrees += 360/len(chars)
    root.add_subview(label)
    
    
class CircleView(ui.View):
    
    def layout(self):
        self.corner_radius = self.width/2
        
circle = CircleView(
    width=diameter, height=diameter,
    border_width=1, border_color=circle_color,
    center = root.bounds.center()
)

root.add_subview(circle)
circle.send_to_back()

AZOM

This is for my friend: can someone make a 3 points generator on the circumference of a circle that has a diameter of 2 units. And with these three x, y coordinates, look if the point (0,5:0) is in the triangle made with the three points?

mikael

@AZOM, sounds like a school exercise. Wouldn’t it be more useful for your friend to spend the time on cracking it?

AZOM

@mikael

No, we are in high school, and this is not a homework or something like that, he just like maths and wants help to “verify” his answer to a question he gets somewhere on the web.

AZOM

@mikael

The only thing I am doing wrong is with rounding up numbers (I think it is my only problem) beau a se if 2 values that I compare are like at 0,000001 appart, they are not equal. Also, I am using cos and sin to try to find the three points. And my friend doesn’t program btw.

mikael

@AZOM, try

0.1 + 0.2 == 0.3

If you are not satisfied with the result, you might try to use the Decimal class, where you can also

getcontext().prec == 6

mikael

@AZOM, meanwhile, for your problem, I would use ui.Path with the three points, then hit_test with the given point. I do not know whether this would be any more accurate than the method you are using, though.

AZOM

@mikael

Right, now, I don’t have my program, but I’m going to send it at ~5h pm.

mikael

@AZOM, going to get my beauty sleep now, but here’s what I meant by the hit test:

import random
import ui
import vector

random_angle = lambda: random.random() * 360

p = ui.Path()
point = vector.Vector(0, 1)

point.degrees = random_angle()
p.move_to(*point)
print(point)
for _ in range(2):
    point.degrees = random_angle()
    p.line_to(*point)
    print(point)
p.close()
    
print('IN' if p.hit_test(0.5,0) else 'OUT')

JonB

So, are you trying to input 3 points (not all colinear) and find the circle (radius and x,y). Then want to determine if a fourth point is on the edge of the circle?

A simple method is to just subtract the center, then compute the length, then compare to the circle radius. But, as you found, you want to use abs(point_to_center_dist-circle_radius)< threshold, where you'll have to decide what is an acceptable threshold.