मैंने पायथन कोड को सी # पर पोर्ट किया। ऐसा लगता है कि काम करता है।
using System;
using System.Collections.Generic;
using System.Drawing;
// Based on code here:
// http://code.activestate.com/recipes/117225/
// Jared Updike ported it to C# 3 December 2008
public class Convexhull
{
// given a polygon formed by pts, return the subset of those points
// that form the convex hull of the polygon
// for integer Point structs, not float/PointF
public static Point[] ConvexHull(Point[] pts)
{
PointF[] mpts = FromPoints(pts);
PointF[] result = ConvexHull(mpts);
int n = result.Length;
Point[] ret = new Point[n];
for (int i = 0; i < n; i++)
ret[i] = new Point((int)result[i].X, (int)result[i].Y);
return ret;
}
// given a polygon formed by pts, return the subset of those points
// that form the convex hull of the polygon
public static PointF[] ConvexHull(PointF[] pts)
{
PointF[][] l_u = ConvexHull_LU(pts);
PointF[] lower = l_u[0];
PointF[] upper = l_u[1];
// Join the lower and upper hull
int nl = lower.Length;
int nu = upper.Length;
PointF[] result = new PointF[nl + nu];
for (int i = 0; i < nl; i++)
result[i] = lower[i];
for (int i = 0; i < nu; i++)
result[i + nl] = upper[i];
return result;
}
// returns the two points that form the diameter of the polygon formed by points pts
// takes and returns integer Point structs, not PointF
public static Point[] Diameter(Point[] pts)
{
PointF[] fpts = FromPoints(pts);
PointF[] maxPair = Diameter(fpts);
return new Point[] { new Point((int)maxPair[0].X, (int)maxPair[0].Y), new Point((int)maxPair[1].X, (int)maxPair[1].Y) };
}
// returns the two points that form the diameter of the polygon formed by points pts
public static PointF[] Diameter(PointF[] pts)
{
IEnumerable<Pair> pairs = RotatingCalipers(pts);
double max2 = Double.NegativeInfinity;
Pair maxPair = null;
foreach (Pair pair in pairs)
{
PointF p = pair.a;
PointF q = pair.b;
double dx = p.X - q.X;
double dy = p.Y - q.Y;
double dist2 = dx * dx + dy * dy;
if (dist2 > max2)
{
maxPair = pair;
max2 = dist2;
}
}
// return Math.Sqrt(max2);
return new PointF[] { maxPair.a, maxPair.b };
}
private static PointF[] FromPoints(Point[] pts)
{
int n = pts.Length;
PointF[] mpts = new PointF[n];
for (int i = 0; i < n; i++)
mpts[i] = new PointF(pts[i].X, pts[i].Y);
return mpts;
}
private static double Orientation(PointF p, PointF q, PointF r)
{
return (q.Y - p.Y) * (r.X - p.X) - (q.X - p.X) * (r.Y - p.Y);
}
private static void Pop<T>(List<T> l)
{
int n = l.Count;
l.RemoveAt(n - 1);
}
private static T At<T>(List<T> l, int index)
{
int n = l.Count;
if (index < 0)
return l[n + index];
return l[index];
}
private static PointF[][] ConvexHull_LU(PointF[] arr_pts)
{
List<PointF> u = new List<PointF>();
List<PointF> l = new List<PointF>();
List<PointF> pts = new List<PointF>(arr_pts.Length);
pts.AddRange(arr_pts);
pts.Sort(Compare);
foreach (PointF p in pts)
{
while (u.Count > 1 && Orientation(At(u, -2), At(u, -1), p) <= 0) Pop(u);
while (l.Count > 1 && Orientation(At(l, -2), At(l, -1), p) >= 0) Pop(l);
u.Add(p);
l.Add(p);
}
return new PointF[][] { l.ToArray(), u.ToArray() };
}
private class Pair
{
public PointF a, b;
public Pair(PointF a, PointF b)
{
this.a = a;
this.b = b;
}
}
private static IEnumerable<Pair> RotatingCalipers(PointF[] pts)
{
PointF[][] l_u = ConvexHull_LU(pts);
PointF[] lower = l_u[0];
PointF[] upper = l_u[1];
int i = 0;
int j = lower.Length - 1;
while (i < upper.Length - 1 || j > 0)
{
yield return new Pair(upper[i], lower[j]);
if (i == upper.Length - 1) j--;
else if (j == 0) i += 1;
else if ((upper[i + 1].Y - upper[i].Y) * (lower[j].X - lower[j - 1].X) >
(lower[j].Y - lower[j - 1].Y) * (upper[i + 1].X - upper[i].X))
i++;
else
j--;
}
}
private static int Compare(PointF a, PointF b)
{
if (a.X < b.X)
{
return -1;
}
else if (a.X == b.X)
{
if (a.Y < b.Y)
return -1;
else if (a.Y == b.Y)
return 0;
}
return 1;
}
}
स्रोत
2008-12-03 20:20:24
आप केंद्र कहां रखते हैं? शायद सेंट्रॉइड के पास लेकिन मैं शर्त लगाता हूं कि मैं उन परिस्थितियों के साथ आ सकता हूं जहां उस सर्कल का केंद्र उस पर महत्वपूर्ण प्रभाव डालता है कि आपको सही जीएलडी मिलती है या नहीं। –
यह अवधारणात्मक रूप से त्रुटिपूर्ण है - एक समतुल्य त्रिकोण का circumcircle व्यास जीएलडी sqrt (3) बार है, जो पक्ष की लंबाई के बराबर है, – Jimmy