Python 绘制嵌套正方形

Question

我正在编写一个递归函数，使用 turtle 在中心点周围绘制嵌套正方形。 .到目前为止，我所绘制的是嵌套的正方形，但围绕一个角点而不是正方形的中心。

def drawCentSq(t, center,side):

    xPt=center[0]
    yPt=center[1]
    t.up()
    t.goto(xPt,yPt)
    xPt-=20
    yPt+=20
    t.up()
    t.goto(xPt,yPt)
    t.down()
    for i in range(4):
        t.forward(side)
        t.right(90)


def drawNestSqCent(t, center, side):
    if side<1:
        return
    else:
        drawCentSq(t,center,side)
        drawNestSqCent(t,center,side-10)

def main():
    import turtle
    import random
    mad=turtle.Turtle()
    wn=mad.getscreen()
    print(drawNestSqCent(mad,(0,0),100))  
main()

有什么建议或更正吗？

Answer 1

你必须每次从中心点计算左上角(起点)和正方形的大小(我不确定Turtle中的+x，-x，+y和-y是什么方向但你应该明白这一点)。

import turtle 
import random

def drawCentSq(t, center,side):
    ## calculate top left corner
    xPt=center[0]-side/2
    yPt=center[1]+side/2
    t.up()
    t.goto(xPt, yPt)
    t.down()
    for i in range(4):
        t.forward(side)
        t.right(90)

def drawNestSqCent(t, center, side):
    if side<1:
        return
    ## else:  not necessary as long as the return comes first
    drawCentSq(t,center,side)
    drawNestSqCent(t,center,side-10)

mad=turtle.Turtle() 
wn=mad.getscreen() 
drawNestSqCent(mad,[0,0],100)