如何计算两个矩形之间的距离? (上下文:Lua中的游戏.)

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问题描述

给出两个x，y，宽度，高度(以像素为单位)和旋转值(以度为单位)的矩形-如何计算它们的轮廓彼此之间的最接近距离?

Given two rectangles with x, y, width, height in pixels and a rotation value in degrees -- how do I calculate the closest distance of their outlines toward each other?

背景:在用Lua编写的游戏中，我正在随机生成地图，但要确保某些矩形彼此之间的距离不是太近-这是必需的，因为如果矩形之间的距离接近一定距离，则地图将无法解决位置，因为球需要在它们之间传递.速度不是一个大问题，因为我没有太多矩形，并且每个级别只生成一次地图.我在StackOverflow上找到的先前链接是 this 和此

Background: In a game written in Lua I'm randomly generating maps, but want to ensure certain rectangles aren't too close to each other -- this is needed because maps become unsolvable if the rectangles get into certain close-distance position, as a ball needs to pass between them. Speed isn't a huge issue as I don't have many rectangles and the map is just generated once per level. Previous links I found on StackOverflow are this and this

非常感谢！

推荐答案

不是在Lua中，这是基于M Katz的建议的Python代码:

Not in Lua, a Python code based on M Katz's suggestion:

def rect_distance((x1, y1, x1b, y1b), (x2, y2, x2b, y2b)):
    left = x2b < x1
    right = x1b < x2
    bottom = y2b < y1
    top = y1b < y2
    if top and left:
        return dist((x1, y1b), (x2b, y2))
    elif left and bottom:
        return dist((x1, y1), (x2b, y2b))
    elif bottom and right:
        return dist((x1b, y1), (x2, y2b))
    elif right and top:
        return dist((x1b, y1b), (x2, y2))
    elif left:
        return x1 - x2b
    elif right:
        return x2 - x1b
    elif bottom:
        return y1 - y2b
    elif top:
        return y2 - y1b
    else:             # rectangles intersect
        return 0.

其中

dist是点之间的欧式距离
矩形. 1由点(x1, y1)和(x1b, y1b)
矩形. 2由点(x2, y2)和(x2b, y2b)

dist is the euclidean distance between points
rect. 1 is formed by points (x1, y1) and (x1b, y1b)
rect. 2 is formed by points (x2, y2) and (x2b, y2b)

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