Delaunay 2D算法

整个算法的流程如下所示：

Incremental Delaunay Triangulation.
    1 Construct an initial triangle, which is large.
    2 Randomly generate a point p in the unit square.
    3 InsertVertex(p).
    4 Repeat step 2 and 3

整个算法的关键在于insertVertex函数

InsertVertex(p)
    1 pFace = LocatePoint(p)
    2 FaceSplit( pFace )
    3 Legalize three edges of the original pFace.

inCircle函数

用于判断P是否位于三角形 $[v_i, v_j, v_k]$ 的外接圆中
若位于三角形 $[v_i, v_j, v_k]$ 中,则：

i n C i r c l e (p i, p j, p k, p) = d e t ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ p i x p k x p j x p x p i y p k y p j y p y p 2 i x + p 2 i y p 2 k x + p 2 k y p 2 j x + p 2 j y p 2 x + p 2 y 1111 ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ > 0

$inCircle(p_i, p_j, p_k, p) = det \left( \begin{array}\ p_{ix} & p_{iy} & p_{ix}^2+p_{iy}^2 & 1 \\ p_{kx} & p_{ky} & p_{kx}^2+p_{ky}^2 & 1\\ p_{jx} & p_{jy} & p_{jx}^2+p_{jy}^2 & 1\\ p_{x} & p_{y} & p_{x}^2+p_{y}^2 & 1\end{array}\right)>0$
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locatePoint函数

首先，三角形的面积计算如下：
Given a triangle [v0,v1,v2] with vi = (xi,yi), the area is given by

S (v 0, v 1, v 2) = 1 2 d e t ⎛ ⎝ ⎜ ⎜ ⎜ x 0 x 1 x 2 y 0 y 1 y 2 111 ⎞ ⎠ ⎟ ⎟ ⎟

$S(v_0, v_1, v_2) = \frac{1}{2}det \left( \begin{array}\ x_0 & y_0 & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{array} \right)$
对于一个三角形

[v0,v1,v2] $[v_0, v_1, v_2]$ ，p是同一平面上一点，重心坐标

α k = S ( p , v k + 1 , v k + 2 ) S ( v 0 , v 1 , v 2 )

$\alpha_k = \frac{S(p, v_{k+1}, v_{k+2})}{S(v_0, v_1, v_2)}$

(α0,α1,α2) $(\alpha_0, \alpha_1, \alpha_2)$ 称为P在三角形

[v0,v1,v2] $[v_0, v_1, v_2]$ 上的重心坐标
可见，每个

αk $\alpha_k$ 对应三角形的一条有向边

[vi,vj] $[v_i, v_j]$ ，若

αk $\alpha_k$ 为负数，则p在

[vi,vj] $[v_i, v_j]$ 外面

Face * LocatePoint( Point p)
    1 Arbitrarily choose the initial face [vi,vj,vk]
    2 Compute the barycentric coordinates of p with respect to current
    face.
    3 If αi,αj,αk are non-negative, then return the current face.
    4 Suppose αi is negative, get the face adjacent to the current face
    sharing edge [vj,vk], denote as F ˜
    5 If F ˜ is empty, return NULL. The point is outside the whole range.
    6 Set current face to be F ˜, repeat through step 2

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faceSplit函数

将p所在的face分割
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edgeSwap函数

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legalizeEdge函数

Suppose e = [v0,v1], the vertex against v is v2, compute the
    circum circle c through v,v0,v1.
    2 If v2 is outside c, then return false.
    3 EdgeSwap(e)
    4 Recursive call LegalizeEdge(v,[v1,v2]);
    5 Recursive call LegalizeEdge(v,[v0,v2]);
    6 return true.

运算结果：

调用函数：
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1000个随机点：