Python 中的多边形

Question

我一直在尝试找出如何使用 Polygons 的方法找到 PI 的近似值。并画出近似圆周率的多边形。

我见过多种近似 PI 的方法，但是 none 其中 none 使用 Turtle 在屏幕上写出多边形（因此您可以看到近似的视觉方面）有人可以向我解释如何这样做吗？

如有任何帮助，我们将不胜感激。这更像是一个学习问题，然后是错误。

Answer 1

采用新方法，我使用海龟图形为 Craig Wood's Pi - Archimedes code from his Fun with Maths and Python 系列制作动画。

此代码仅说明内接的偶边多边形。它不使用 pi，而是从每个多边形估计它，并将越来越准确的结果输出到控制台：

from math import sqrt
from turtle import Turtle, Screen

SCALE = 300
ITERATIONS = 7

def draw_circle(turtle):
    turtle.goto(0, -SCALE)
    turtle.pendown()
    turtle.circle(SCALE)
    turtle.penup()

def inscribe_circle(turtle, sides, edge_length):
    turtle.goto(0, -SCALE)
    turtle.setheading(0)
    turtle.left(180 / sides)
    turtle.pendown()
    for _ in range(sides):
        turtle.forward(edge_length * SCALE)
        turtle.left(360 / sides)
    turtle.penup()

# based on code and analysis from http://www.craig-wood.com/nick/articles/pi-archimedes/
def pi_archimedes(turtle, n):
    """ Calculate n iterations of Archimedes PI recurrence relation """

    polygon_edge_length_squared = 2.0
    polygon_edge_length = sqrt(polygon_edge_length_squared)
    polygon_sides = 4

    inscribe_circle(turtle, polygon_sides, polygon_edge_length)
    print(polygon_sides * polygon_edge_length / 2)

    for _ in range(n - 1):
        polygon_sides *= 2
        polygon_edge_length_squared = 2 - 2 * sqrt(1 - polygon_edge_length_squared / 4)
        polygon_edge_length = sqrt(polygon_edge_length_squared)
        inscribe_circle(turtle, polygon_sides, polygon_edge_length)
        print(polygon_sides * polygon_edge_length / 2)

yertle = Turtle()
yertle.penup()

draw_circle(yertle)

pi_archimedes(yertle, ITERATIONS)

yertle.hideturtle()

Screen().exitonclick()

希望这能让您快速开始自己的 Archimedes Pi 插图。

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