使用坐标计算随机三角形的边界框

Question

我正在寻找一种合适的方法来计算三角形的矩形边界框，该三角形具有三个角作为点 - (x1,y1)、(x2,y2)、(x3,y3)。

这是我正在使用的数据类型（按照建议，我添加了更多构造函数）：

data Shape =
     Circle Point Double |
     Rectangle Point Point |
     Triangle Point Point Point

边界框函数的格式应该是“bounding :: Shape -> Shape”。 我也试过矩形和圆形的边界框：

bounding :: Shape -> Shape
bounding (Rectangle (Point x y) (Point z z1)) = (Rectangle (Point x y) (Point z z1))
bounding (Circle (Point p w) r) = (Rectangle (Point (p-r) (w-r)) (Point (p+r) (w+r)))

如果应该使用图形坐标系，这些是否正确（其中 (x,y) 坐标应该被视为 (x,-y)）？

有人可以帮帮我吗？ 附言不需要图形库。

Answer 1

你试过什么？你有代码吗？

你需要：

 data Shape = Triangle Point Point Point | Rectangle Point Point

假设 x 向右增加，y 向上增加。左上角将是 (min{x1, x2, x3}, max{y1, y2, y3})，你需要右下角。

---

编辑

x 应该向右增加，y 向下增加。我添加了三角形一，删除了不必要的括号并派生（显示）以便能够打印形状数据类型。完整代码：

data Point = Point Double Double deriving (Show)

data Shape =
        Circle Point Double |
        Rectangle Point Point |
        Triangle Point Point Point deriving (Show)

bounding :: Shape -> Shape
bounding (Rectangle (Point x y) (Point z z1)) = Rectangle (Point x y) (Point z z1)
bounding (Circle (Point p w) r) = Rectangle (Point (p-r) (w-r)) (Point (p+r) (w+r))
bounding (Triangle (Point x1 y1) (Point x2 y2) (Point x3 y3)) = Rectangle 
        (Point (minimum [x1, x2, x3]) (minimum [y1, y2, y3])) 
        (Point (maximum [x1, x2, x3]) (maximum [y1, y2, y3]))

Answer 2

根据 @dave4420 的出色观察，我假设 Rectangles 与 x/y 轴对齐。

根据您对Circle 的规则，在Rectangle 中，x,y 最小的点排在第一位，x,y 最大的点排在第二位。

您可能应该坚持使用带有数字后缀的x 和y，而不是发明像p、w、z 和z1 这样非常令人困惑的新名称。

Rectangle 的边界框是 Rectangle 本身。

Triangle 的边界框将是一个Rectangle，从任意点来看，x,y 的最小和最大。

bounding :: Shape -> Shape
bounding rect@(Rectangle _ _) = rect
bounding (Circle (Point x y) r) = Rectangle p1 p2 where
  p1 = (Point (x-r) (y-r))
  p2 = (Point (x+r) (y+r))
bounding (Triangle (Point x1 y1) (Point x2 y2) (Point x3 y3)) = Rectangle p1 p2 where
  p1 = Point (minimum [x1 x2 x3]) (minimum [y1 y2 y3])
  p2 = Point (maximum [x1 x2 x3]) (maximum [y1 y2 y3])

Answer 3

我发布这个答案只是为了说明边界框和矩形不是一回事。

data Point = Point Double Double   -- Point x y

data BoundingBox = BoundingBox Double Double Double Double
                            -- top    left   bottom right

data Shape
    = Circle Point Double
    | Rectangle Point Point Double
        -- yes, you need two points and a scalar to specify arbitrary rectangles
    | Triangle Point Point Point

boundingBox :: Shape -> BoundingBox
boundingBox (Circle (Point x y) r) = BoundingBox (y-r) (x-r) (y+r) (x+r)
boundingBox (Rectangle (Point x0 y0) (Point x1 y1) d)
        = BoundingBox (minimum ys) (minimum xs) (maximum ys) (maximum xs) where
    xs = [x0, x1, x1+dx, x0+dx]
    ys = [y0, y1, y1+dy, y0+dy]
    d' = d / ((x0-x1)^^2 + ((y0-y1)^^2)
    dx = d' * (y0-y1)
    dy = d' * (x1-x0)
boundingBox (Triangle (Point x0 y0) (Point x1 y1) (Point x2 y2))
        = BoundingBox (minimum ys) (minimum xs) (maximum ys) (maximum xs) where
    xs = [x0, x1, x2]
    ys = [y0, y1, y2]

作为练习，从Rectangle 和Triangle 案例中提取公共代码。 （查找和修复我在Rectangle 案例中放置的任何错误的奖励积分。）