Два шарика в квадрате, Python 3

Language: Layout:

'''
В прямоугольнике атомы, случайно расположенные в нём.
Программа находит их взаимное положение, при котором атомы максимально
удалены друг от друга и максимально удалены и от ближайших стенок.
'''
import turtle as t
import random as rnd

l, r, d, u = -300, 300, -300, 300

t1 = t.Pen(shape='circle')
t2 = t.Pen(shape='square')
term = t.Pen()
term.pencolor('green')
term.pensize(3)
term.ht()
term.pu()
term.goto(l, u)
term.down()
term.goto(l, d)
term.goto(r, d)
term.goto(r, u)
term.goto(l, u)
term.goto(r, d)
term.goto(r, u)
term.goto(l, d)

term.forward(100)

t1.setpos(rnd.randint(l+10, r-10), rnd.randint(d+10, u-10))
t2.setpos(rnd.randint(l+10, r-10), rnd.randint(d+10, u-10))

while True:
    if abs (t1.xcor() - l) < abs(t1.xcor() - t2.xcor()):
        t1.goto(t1.xcor() + 1, t1.ycor())
    else:
        t1.goto(t1.xcor() - 1, t1.ycor())
        
    if abs (t2.xcor() - r) < abs(t1.xcor() - t2.xcor()):
        t2.goto(t2.xcor() - 1, t2.ycor())
    else:
        t2.goto(t2.xcor() + 1, t2.ycor())
        
    if abs (t1.ycor() - d) < abs(t1.ycor() - t2.ycor()):
        t1.goto(t1.xcor(), t1.ycor() + 1)
    else:
        t1.goto(t1.xcor(), t1.ycor() - 1)
        
    if abs (t2.ycor() - u) < abs(t1.ycor() - t2.ycor()):
        t2.goto(t2.xcor(), t2.ycor() - 1)
    else:
        t2.goto(t2.xcor(), t2.ycor() + 1)
    lx = min(abs(t1.xcor()-l), abs(t1.xcor()-r))
    ly = min(abs(t1.ycor()-u), abs(t1.ycor()-d)) 
    rx = min(abs(t2.xcor()-l), abs(t2.xcor()-r))
    ry = min(abs(t2.ycor()-u), abs(t2.ycor()-d)) 
    # print(t1.distance(t2) + lx + ly + rx + ry)

[ + ] Show input