Tron游戏碰撞乌龟Python

Question

我想做一个特隆游戏，它是一个本地2玩家。我的代码在下面，但它是一个未完成的产品。我想知道我怎样才能做到，这样，如果形状敌人和形状玩家彼此接触，或者他们创造的线，就会有结果。就像打印游戏和改变背景。 我一直把错误写在下面的评论里。在

def up():
    player.fd(15)
def right():
    player.fd(15)
def left():
    player.fd(15)
def down():
    player.fd(15)

playerPath = [[[100],[100]]]
enemyPath = [[[600],[600]]]
previousMove = "na"

#I keep getting this error: TclError: bad event type or keysym "up"

if previousMove != "up":
        #Check for other methods
    if previousMove == right():
        playerPath.append((playerPath[0][0][0] + 90, playerPath[0][0][1]))
    if previousMove == left():
        playerPath.append((playerPath[0][0][0] - 90, playerPath[0][0][1]))
    if previousMove == down():
        playerPath.append((playerPath[0][0][0] + 180, playerPath[0][0][1]))
        #repeat for other directions
if previousMove == "up":
    playerPath[0][0].append(playerPath[0][0][0] + 30)
    playerPath[0][1].append(playerPath[0][1][0] + 30)
previousMove = "up"


if previousMove != "right":
        #Check for other methods
    if previousMove == up():
        playerPath.append((playerPath[0][0][0] - 90, playerPath[0][0][1]))
    if previousMove == left():
        playerPath.append((playerPath[0][0][0] + 180, playerPath[0][0][1]))
    if  previousMove ==down():
        playerPath.append((playerPath[0][0][0] + 90, playerPath[0][0][1]))
       #repeat for other directions
if previousMove == "right":
    playerPath[0][0].append(playerPath[0][0][0] + 30)
    playerPath[0][1].append(playerPath[0][1][0] + 30)
previousMove = "right"


if previousMove != "left":
    #Check for other methods
    if previousMove == up():
        playerPath.append((playerPath[0][0][0] + 90, playerPath[0][0][1]))
    if previousMove == right():
        playerPath.append((playerPath[0][0][0] + 180, playerPath[0][0][1]))
    if previousMove == down():
        playerPath.append((playerPath[0][0][0] - 90, playerPath[0][0][1]))
        #repeat for other directions
if previousMove == "left":
    playerPath[0][0].append(playerPath[0][0][0] + 30)
    playerPath[0][1].append(playerPath[0][1][0] + 30)
previousMove = "left"


if previousMove != "down":
        #Check for other methods
    if previousMove == up():
        playerPath.append((playerPath[0][0][0] + 180, playerPath[0][0][1]))
    if previousMove == left():
        playerPath.append((playerPath[0][0][0] + 90, playerPath[0][0][1]))
    if previousMove == right():
        playerPath.append((playerPath[0][0][0] - 90, playerPath[0][0][1]))
        #repeat for other directions
if previousMove == "down":
    playerPath[0][0].append(playerPath[0][0][0] + 30)
    playerPath[0][1].append(playerPath[0][1][0] + 30)
previousMove = "down"

#This code gives me this error: IndexError: list index out of range    
#for subPath in enemyPath:
#    if player.position()[0] in range(subPath[0][0], subPath[0][1]) and player.position()[1] in range(subPath[1][0], subPath[1][1]):
#        print("Collision")


onkey(up, "up")
onkey(left, "left")
onkey(right, "right")
onkey(down, "down")

onkey(up1, "w")
onkey(left1, "a")
onkey(right1, "d")

listen()
mainloop()  

Answer 1

turtle“position”方法在这个例子中非常有用：http://interactivepython.org/runestone/static/IntroPythonTurtles/Summary/summary.html

具体来说，你可以做到

if player.position() == enemy.position():
      print  "Game Over"
      return

对于与一条线发生冲突的场景，您需要将两个玩家的移动路径存储在一个列表或数组中，然后在上面类似的if语句中确定玩家的移动是否进入其中一个空间。在

可以通过创建二维数组或列表来存储路径。如果玩家在同一个方向上移动，你可以在相应的轴上附加一个玩家移动距离范围的列表。如果玩家向新方向移动，则附加一个新列表，其中包含沿轴移动的距离范围。在

^{pr2}$

当检查是否发生了碰撞时，遍历子路径，检查玩家位置X和Y值在已经遍历的X和Y路径中。在

for subPath in enemyPath:
    if player.position()[0] in range(subPath[0][0], subPath[0][1]) and
    player.position()[1] in range(subPath[1][0], subPath[1][1]):
    print "Collision"
    return

Answer 2

你的代码不可能像给定的那样工作，比如每次击键都有大量的代码块，如果它只运行一次，那么它就处于顶层。下面是您的代码的一个完整的重做。在

你需要测试你是否越过了敌人或你自己的防线。这段代码跟踪片段，并在每次移动时测试它们，完全消除了不小心踩到一条线的失败者：

from turtle import Turtle, Screen

screen = Screen()
screen.bgcolor('black')

def up(who):
    global previousMove

    turtle, path = players[who]
    turtle.setheading(90)

    if previousMove != 'up':
        path.append(turtle.position())
    previousMove = 'up'

    turtle.fd(15)

    if checkCollision(turtle.position(), path, players[1 - who][PATH]):
        collision(turtle)

def right(who):
    global previousMove

    turtle, path = players[who]
    turtle.setheading(0)

    if previousMove != 'right':
        path.append(turtle.position())
    previousMove = 'right'

    turtle.fd(15)

    if checkCollision(turtle.position(), path, players[1 - who][PATH]):
        collision(turtle)

def left(who):
    global previousMove

    turtle, path = players[who]
    turtle.setheading(180)

    if previousMove != 'left':
        path.append(turtle.position())
    previousMove = 'left'

    turtle.fd(15)

    if checkCollision(turtle.position(), path, players[1 - who][PATH]):
        collision(turtle)

def down(who):
    global previousMove

    turtle, path = players[who]
    turtle.setheading(270)

    if previousMove != 'down':
        path.append(turtle.position())
    previousMove = 'down'

    turtle.fd(15)

    if checkCollision(turtle.position(), path, players[1 - who][PATH]):
        collision(turtle)

def collision(turtle):
    for key in ('Up', 'Left', 'Right', 'Down', 'w', 'a', 'd', 'x'):
        screen.onkey(None, key)  # disable game
    turtle.clear()  # remove the loser from the board!

def checkCollision(position, path1, path2):
    if len(path1) > 1:

        A, B = position, path1[-1]  # only check most recent line segment

        if len(path1) > 3:  # check for self intersection
            for i in range(len(path1) - 3):
                C, D = path1[i:i + 2]

                if intersect(A, B, C, D):
                    return True

        if len(path2) > 1:  # check for intersection with other turtle's path
            for i in range(len(path2) - 1):
                C, D = path2[i:i + 2]

                if intersect(A, B, C, D):
                    return True
    return False

X, Y = 0, 1

def ccw(A, B, C):
    """ https://stackoverflow.com/a/9997374/5771269 """
    return (C[Y] - A[Y]) * (B[X] - A[X]) > (B[Y] - A[Y]) * (C[X] - A[X])

def intersect(A, B, C, D):
    """ Return true if line segments AB and CD intersect """
    return ccw(A, C, D) != ccw(B, C, D) and ccw(A, B, C) != ccw(A, B, D)

player = Turtle('circle')
player.shapesize(6 / 20)
player.color('red')
player.pensize(6)
player.speed('fastest')
player.penup()
player.setposition(100, 100)
player.pendown()

enemy = Turtle('circle')
enemy.shapesize(6 / 20)
enemy.color('blue')
enemy.pensize(6)
enemy.speed('fastest')
enemy.penup()
enemy.setposition(-300, -300)
enemy.pendown()

players = [[player, [player.position()]], [enemy, [enemy.position()]]]
PLAYER, ENEMY = 0, 1
TURTLE, PATH = 0, 1

previousMove = None  # consolidate moves in same direction into single line segment

screen.onkey(lambda: up(PLAYER), 'Up')
screen.onkey(lambda: left(PLAYER), 'Left')
screen.onkey(lambda: right(PLAYER), 'Right')
screen.onkey(lambda: down(PLAYER), 'Down')

screen.onkey(lambda: up(ENEMY), 'w')
screen.onkey(lambda: left(ENEMY), 'a')
screen.onkey(lambda: right(ENEMY), 'd')
screen.onkey(lambda: down(ENEMY), 'x')

screen.listen()

screen.mainloop()

线段交叉代码取自How can I check if two segments intersect?

以上代码是不完整的，它需要额外的工作，成为一个完整的游戏。但它基本上可以让你尝试一些想法：