我编写了一个使用Fortunes算法生成Voronoi图的程序。它适用于大多数情况，但我发现了一些我不知道如何解决的特殊情况。
其中之一是，如果共享相同y坐标和x坐标的2个站点接近，那么圆形事件（结束2个边缘并开始新边缘的事件）不会发生，或者在计算中的某处出错了。
其他是2个站点共享x坐标，y坐标接近，则不会发生弧交叉。我该如何处理？
using System.Collections;
using System.Collections.Generic;
using UnityEngine;
public class Fortune{
Voronoi diagram;
SortedQueue<AEvent> events;
Arc root;
float ly;
public Voronoi MakeDiagram()
{
// Broken seeds: 14, 15,
Random.InitState(16);
diagram = new Voronoi();
events = new SortedQueue<AEvent>(sortByY);
createSites(6);
constructDiagram();
return diagram;
}
private void createSites(int count)
{
for(int i = 0; i < count; i++)
{
events.Enqueue(new SiteEvent(new Vector2(Random.Range(0, 100), Random.Range(0, 100))));
}
}
private void constructDiagram()
{
while(events.Size() > 0)
{
events.Dequeue().handleEvent(this);
}
FinishEdges(root);
for(int i = 0; i < diagram.edges.Count; i++)
{
VEdge e = diagram.edges[i];
if(e.neighbour != null)
e.start = e.neighbour.end;
}
}
//Inser new parabol to beachline
public void AddParabol(Vector2 point)
{
ly = point.y;
Debug.DrawLine(point, point + Vector2.up, Color.blue, 100f);
if (root == null)
{
root = new Arc(point);
return;
}
if(root.isLeaf && root.site.y - point.y < 1)
{
root.isLeaf = false;
root.SetLeft(new Arc(root.site.point));
root.SetRight(new Arc(point));
VPoint mid2 = new VPoint((root.site.x + point.x) / 2, 100);
if(root.site.x > point.x)
{
root.edge = new VEdge(mid2, point, root.site.point);
}else
{
root.edge = new VEdge(mid2, root.site.point, point);
}
diagram.AddEdge(root.edge);
return;
}
Arc a = GetArcUnderPoint(root, point);
if(a.cevent != null)
{
events.Remove(a.cevent);
}
VPoint mid = new VPoint(point.x, GetY(a, point.x));
Debug.DrawLine(mid.point, mid.point + Vector2.up, Color.cyan, 100f);
VEdge l = new VEdge(mid, a.site.point, point);
VEdge r = new VEdge(mid, point, a.site.point);
diagram.AddEdge(l);
l.neighbour = r;
Arc lc = new Arc(a.site.point);
Arc m = new Arc(point);
Arc rc = new Arc(a.site.point);
a.SetLeft(lc);
Arc inter = new Arc();
inter.SetLeft(m);
inter.SetRight(rc);
a.SetRight(inter);
a.isLeaf = false;
a.edge = l;
inter.edge = r;
CheckCircleEvent(lc);
CheckCircleEvent(rc);
}
//Remove arc from tree
public void RemoveParabol(CircleEvent e)
{
ly = e.y;
Arc a = e.a;
Arc lp = a.leftParent;
Arc rp = a.rightParent;
Arc lc = lp.leftChild;
if (lc.cevent != null) events.Remove(lc.cevent);
Arc rc = rp.rightChild;
if (rc.cevent != null) events.Remove(rc.cevent);
VPoint end = new VPoint(e.x, GetY(a,e.x));
lp.edge.end = end;
rp.edge.end = end;
VEdge ne = new VEdge(end, lc.site.point, rc.site.point);
Arc higher = new Arc();
Arc cur = a;
while (cur != root)
{
cur = cur.parent;
if (cur == lp) higher = lp;
if (cur == rp) higher = rp;
}
higher.edge = ne;
diagram.AddEdge(ne);
Arc gp = a.parent.parent;
if(a.parent.left == a)
{
if (gp.left == a.parent) gp.SetLeft(a.parent.right);
if (gp.right == a.parent) gp.SetRight(a.parent.right);
}
else
{
if (gp.left == a.parent) gp.SetLeft(a.parent.left);
if (gp.right == a.parent) gp.SetRight(a.parent.left);
}
CheckCircleEvent(lc);
CheckCircleEvent(rc);
}
//Check if there is any circle events to come
public void CheckCircleEvent(Arc a)
{
Arc lp = a.leftParent;
Arc rp = a.rightParent;
if (lp == null || rp == null) return;
Arc lc = lp.leftChild;
Arc rc = rp.rightChild;
if (lc.site == rc.site) return;
VPoint s = Intersection(lp.edge, rp.edge);
if (s == null) return;
float r = Vector2.Distance(s.point, a.site.point);
if (s.y - r > ly) return;
CircleEvent e = new CircleEvent(a);
e.y = s.y - r;
e.x = s.x;
e.a = a;
a.cevent = e;
events.Enqueue(e);
}
//Find intersection of 2 edges and check if edge directions point to intersection point
public VPoint Intersection(VEdge left, VEdge right)
{
/*
* y = f1*x + g1
* y = f2*x + g2
* f1*x +g1 = f2*x+g2
* f1*x - f2*x = g2-g1
* x = (g2-g1)/(f1-f2)
*/
float x = (right.g - left.g) / (left.f - right.f);
float y = left.f * x + left.g;
Vector2 s = new Vector2(x, y);
if ((x - left.start.x) / left.dir.x < 0) return null;
if ((y - left.start.y) / left.dir.y < 0) return null;
if ((x - right.start.x) / right.dir.x < 0) return null;
if ((y - right.start.y) / right.dir.y < 0) return null;
return new VPoint(s);
}
// What if no intersection
public Arc GetArcUnderPoint(Arc root, Vector2 point)
{
// drawTree(root);
Arc curr = root;
while (!curr.isLeaf)
{
float x = ArcIntersectionX(curr, point.x);
Debug.Log("Returned =" + x);
if (float.IsNaN(x) || float.IsInfinity(x))
{
// drawTree(curr);
}
if (x > point.x) curr = curr.left;
else curr = curr.right;
}
return curr;
}
//Draws parabols to see where issueas are
public void drawTree(Arc a)
{
if(a.isLeaf)
{
for (int i = 0; i < 100; i++)
{
float y = GetY(a, i);
Vector2 t = new Vector2(i, y);
Debug.DrawLine(t, t + Vector2.up, Color.yellow, 100f);
}
}
if (!a.isLeaf)
{
drawTree(a.left);
drawTree(a.right);
}
}
// Find x coordinate of 2 arc intersection. Arcs are thel eft and right child of given node(Arc in tree)
public float ArcIntersectionX(Arc a, float x)
{
Arc lc = a.leftChild;
Arc rc = a.rightChild;
VPoint ls = lc.site;
VPoint rs = rc.site;
/*Intersection of 2 parabols
*
* y1 = 1 / 4f1 x ^ 2 - v11 / 2f1 x + v11 ^ 2 / 4f1 + v12
* y2 = 1 / 4f2 x ^ 2 - v21 / 2f2 x + v21 ^ 2 / 4f2 + v22
*
* 1 / 4f1 x ^ 2 - 1 / 4f2 x ^ 2 - v11 / 2f1 x + v21 / 2f2 x + v11 ^ 2 / 4f1 - v21 ^ 2 / 4f2 + v12 - v22 =
* (1 / 4f1 - 1 / 4f2) x ^ 2 + (-v11 / 2f1 + v21 / 2f2) x + (v11 ^ 2 / 4f1 - v21 ^ 2 / 4f2 + c12 - v22) = 0
*
* v11 = ls.x
* v21 = rs.x
*
* v12 = ls.y - f1
* v22 = rs.y - f2
*
* f1 = (ls.y - ly) / 2
* f2 = (rs.y - ly) / 2
*/
float f1 = (ls.y - ly) / 2;
float f2 = (rs.y - ly) / 2;
float a1 = (1/(4*f1) - 1/(4*f2));
float b1 = -ls.x / (2 * f1) + rs.x / (2 * f2);
float c1 = ls.x * ls.x / (4 * f1) - rs.x * rs.x / (4 * f2) + (ls.y - f1) - (rs.y - f2);
float D = Mathf.Sqrt(b1 * b1 - 4 * a1 * c1);
float x1 = (-b1 + D) / (2 * a1);
float x2 = (-b1 - D) / (2 * a1);
Vector2 p1 = new Vector2(x1, GetY(lc, x1));
Vector2 p2 = new Vector2(x2, GetY(lc, x2));
if(ls.y > rs.y)
{
return Mathf.Min(x1, x2);
}
else
{
return Mathf.Max(x1, x2);
}
}
public bool HasValue(float val)
{
return (!float.IsNaN(val) && !float.IsInfinity(val));
}
// Find where x = cons and arc intersect
public float GetY(Arc a, float x)
{
/*
* y = 1 / 4f (x - v1)^2 + v2
*
* f = (a.site.y - ly) / 2
* v1 = a.site.x
* v2 = a.site.y - f
*/
float f = (a.site.y - ly) / 2;
return 1 / (4 * f) * (x - a.site.x) * (x - a.site.x) + (a.site.y-f);
}
public void FinishEdges(Arc root)
{
if (root.isLeaf) return;
VEdge e = root.edge;
if(e.end == null)
{
float mx;
if (e.dir.x > 0) mx = Mathf.Max(100, e.start.x + 10);
else mx = Mathf.Min(0, e.start.x - 10);
float my = mx * e.f + e.g;
/*
if(my > 100)
{
my = 100;
mx = (100 - e.g) / e.f;
}
else if(my < 0)
{
my = 0;
mx = -e.g / e.f;
}
*/
VPoint end = new VPoint(mx, my);
e.end = end;
}
FinishEdges(root.left);
FinishEdges(root.right);
}
public static int sortByY(AEvent a, AEvent b)
{
if (a.y > b.y) return -1;
if (a.y < b.y) return 1;
if (a.x < b.x) return -1;
if (a.x > b.x) return 1;
return 0;
}
}
解决
好的，所以我调试了几个小时，发现问题来自有时值为NaN或Infinity的事实。所以我只需要检查哪里可能会给我这两个中的一个。
答案 0 :(得分：0)
我遇到了和您一样的问题。在实施https://github.com/fewlinesofcode/FortunesAlgorithm时。您可以检查代码并在那里找到答案，但是我也记录了代码，所以这里有一些摘录：
@i !== null当断点恰好落在两个弧之间时，也会出现简并的情况。
这是我的处理方式（抱歉，我使用了Swift语言）
Case: There is only one site in the **Beachline**. Exisiting site shares Y coordinate with new site
In this case two parabolas are degenerating into Rays origining at their event point and pointing up.
They have no intersection point but we know two things about the future of this case:
- The *x* coordinate of the intersection point point will be exactly in the middle
- *y* coordinate will lay somewhere far above the points.
We replace *y* with an arbitrary value big enough to cover our case. (`yVal`)