这是Python中的额外学分分配。我已经完成了大部分,直到最后一部分,我必须确定所选的tictactoe盒子的面积。在
编辑:更新,我只能检测到对角线框,用于合并下面的代码回复。 我可以检测到这3个框,但其余的仍然显示为无,而且大多数逻辑都与循环一起使用,所以我可以理解我哪里出错了。在
import turtle
from time import sleep
import sys
CURSOR_SIZE = 20
SQUARE_SIZE = 99
FONT_SIZE = 40
FONT = ('Arial', FONT_SIZE, 'bold')
BOXES = {}
# TRACK BOX
pen = turtle.Turtle()
pen.penup()
def mouse(x, y):
print('|--------------X={0} Y={1}--------------|'.format(x, y))
for key in BOXES:
minx, miny, maxx, maxy = BOXES[key]
print(key, BOXES[key])
if (minx <= x <= maxx) and (miny <= y <= maxy):
print("Found", key)
return key
print('None')
return None # Not found.
class TicTacToe:
global BOXES
def __init__(self):
# CREATES 2D LIST FOR INTERROGATION
self.board = [['?'] * 3 for i in range(3)]
def minmax(self, points):
""" Find extreme x and y values in a list of 2-D coordinates. """
minx, miny, maxx, maxy = points[0][0], points[0][1], points[0][0], points[0][1]
for x, y in points[1:]:
if x < minx:
minx = x
if y < minx:
miny = y
if x > maxx:
maxx = x
if y > maxy:
maxy = y
return minx, miny, maxx, maxy
def drawBoard(self):
##############################################
turtle.shape('square')
turtle.shapesize(SQUARE_SIZE * 3 / CURSOR_SIZE)
turtle.color('black')
turtle.stamp()
turtle.hideturtle()
##############################################
for j in range(3):
for i in range(3):
# CREATES SHAPE AND STORES IN PLACEHOLDER
turtle.shape('square')
box = turtle.shape('square')
# CREATES SHAPE SIZE AND STORES IN PLACEHOLDER
turtle.shapesize(SQUARE_SIZE / CURSOR_SIZE)
boxsize = turtle.shapesize()
# CREATES SHAPE COLOR
turtle.color('white')
turtle.penup()
# CREATES SHAPE POS AND STORES IN PLACEHOLDER
turtle.goto(i * (SQUARE_SIZE + 2) - (SQUARE_SIZE + 2), j * (SQUARE_SIZE + 2) - (SQUARE_SIZE + 2))
boxpos = turtle.pos()
mypos = []
pen.goto(boxpos[0]-50,boxpos[1]+50)
##############################################
for line in range(0, 4):
pen.forward(SQUARE_SIZE)
pen.right(90)
mypos.append(pen.pos())
turtle.showturtle()
turtle.stamp()
##############################################
a = mypos[0]
b = mypos[1]
c = mypos[2]
d = mypos[3]
self.board[j][i] = [a, b, c, d]
##############################################
BOXES['BOX01'] = self.minmax(self.board[0][0])
BOXES['BOX02'] = self.minmax(self.board[0][1])
BOXES['BOX03'] = self.minmax(self.board[0][2])
##############################################
BOXES['BOX11'] = self.minmax(self.board[1][0])
BOXES['BOX12'] = self.minmax(self.board[1][1])
BOXES['BOX13'] = self.minmax(self.board[1][2])
##############################################
BOXES['BOX21'] = self.minmax(self.board[2][0])
BOXES['BOX22'] = self.minmax(self.board[2][1])
BOXES['BOX23'] = self.minmax(self.board[2][2])
##############################################
turtle.onscreenclick(mouse)
turtle.setup(800, 600)
wn = turtle.Screen()
z = TicTacToe()
z.drawBoard()
turtle.mainloop()
我相信你没有充分利用pythonthurtle会让问题变得更加棘手。在点击屏幕时,不要试图在棋盘上找到一个方块,而是让棋盘上的方块变成海龟,对鼠标的点击做出反应。那就没什么好弄清楚的了,就位置而言。在
下面是一个重新实现,可以绘制一个板,允许您单击它,交替地将单击的部分设置为“X”或“O”:
您可以使用
board
来评分,或者实现一个智能计算机播放器,或者任何你想要的东西。在这将为您提供基本的思想,即计算每个框的最小和最大x和y值,并将它们存储在}的给定坐标是否在其中任何一个内。在
BOXES
中。这使得很容易确定传递给x
和{对于实际代码中的多个框,确保将新的
minmax()
函数应用到每个框的角上。在相关问题 更多 >
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