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Delaunay triangulation : too many triangles

I'm trying to implement the Delaunay triangulation in C++. Currently it's working, but I'm not getting the correct amount of triangles. I try it with 4 points in a square pattern : (0,0), (1,0), (0,1), (1,1).

Here's the algorithm I use :

std::vector<Triangle> Delaunay::triangulate(std::vector<Vec2f> &vertices) {

// Determinate the super triangle
float minX = vertices[0].getX();
float minY = vertices[0].getY();
float maxX = minX;
float maxY = minY;

for(std::size_t i = 0; i < vertices.size(); ++i) {
    if (vertices[i].getX() < minX) minX = vertices[i].getX();
    if (vertices[i].getY() < minY) minY = vertices[i].getY();
    if (vertices[i].getX() > maxX) maxX = vertices[i].getX();
    if (vertices[i].getY() > maxY) maxY = vertices[i].getY();
}

float dx = maxX - minX;
float dy = maxY - minY;
float deltaMax = std::max(dx, dy);
float midx = (minX + maxX) / 2.f;
float midy = (minY + maxY) / 2.f;

Vec2f p1(midx - 20 * deltaMax, midy - deltaMax);
Vec2f p2(midx, midy + 20 * deltaMax);
Vec2f p3(midx + 20 * deltaMax, midy - deltaMax);    

// Add the super triangle vertices to the end of the vertex list
vertices.push_back(p1);
vertices.push_back(p2);
vertices.push_back(p3);

// Add the super triangle to the triangle list
std::vector<Triangle> triangleList = {Triangle(p1, p2, p3)};

// For each point in the vertex list
for(auto point = begin(vertices); point != end(vertices); point++) 
{
    // Initialize the edges buffer
    std::vector<Edge> edgesBuff;

    // For each triangles currently in the triangle list    
    for(auto triangle = begin(triangleList); triangle != end(triangleList);) 
    {
        if(triangle->inCircumCircle(*point))
        {
            Edge tmp[3] = {triangle->getE1(), triangle->getE2(), triangle->getE3()};
            edgesBuff.insert(end(edgesBuff), tmp, tmp + 3); 
            triangle = triangleList.erase(triangle);
        }
        else
        {
            triangle++;
        }
    }

    // Delete all doubly specified edges from the edge buffer
    // Black magic by https://github.com/MechaRage 
    auto ite = begin(edgesBuff), last = end(edgesBuff);

    while(ite != last) {
        // Search for at least one duplicate of the current element
        auto twin = std::find(ite + 1, last, *ite);
        if(twin != last)
            // If one is found, push them all to the end.
            last = std::partition(ite, last, [&ite](auto const &o){ return !(o == *ite); });
        else
            ++ite;
    }

    // Remove all the duplicates, which have been shoved past "last".
    edgesBuff.erase(last, end(edgesBuff));

    // Add the triangle to the list
    for(auto edge = begin(edgesBuff); edge != end(edgesBuff); edge++)
        triangleList.push_back(Triangle(edge->getP1(), edge->getP2(), *point));


}

// Remove any triangles from the triangle list that use the supertriangle vertices
triangleList.erase(std::remove_if(begin(triangleList), end(triangleList), [p1, p2, p3](auto t){
    return t.containsVertex(p1) || t.containsVertex(p2) || t.containsVertex(p3);
}), end(triangleList));

return triangleList;

}

And here's what I obtain :

Triangle:
 Point x: 1 y: 0
 Point x: 0 y: 0
 Point x: 1 y: 1

Triangle:
 Point x: 1 y: 0
 Point x: 1 y: 1
 Point x: 0 y: 1

Triangle:
 Point x: 0 y: 0
 Point x: 1 y: 1
 Point x: 0 y: 1

While this would be the correct output :

Triangle:
 Point x: 1 y: 0
 Point x: 0 y: 0
 Point x: 0 y: 1

Triangle:
 Point x: 1 y: 0
 Point x: 1 y: 1
 Point x: 0 y: 1

I have no idea why there is a triangle with the (0, 0) and the (1, 1). I need an outside eye to review the code and find out what's going wrong.

All the sources are on my Github repo . Feel free to fork it and to PR your code.

Thanks!

what about this implementation of Paul Bourke 's Delaunay triangulation algorithm. Take a look at Triangulate() I have used this source many times without any complains

#include <iostream>
#include <stdlib.h> // for C qsort 
#include <cmath>
#include <time.h> // for random

const int MaxVertices = 500;
const int MaxTriangles = 1000;
//const int n_MaxPoints = 10; // for the test programm
const double EPSILON = 0.000001;

struct ITRIANGLE{
  int p1, p2, p3;
};

struct IEDGE{
  int p1, p2;
};

struct XYZ{
  double x, y, z;
};

int XYZCompare(const void *v1, const void *v2);
int Triangulate(int nv, XYZ pxyz[], ITRIANGLE v[], int &ntri);
int CircumCircle(double, double, double, double, double, double, double, double, double&, double&, double&);
using namespace std; 

////////////////////////////////////////////////////////////////////////
// CircumCircle() :
//   Return true if a point (xp,yp) is inside the circumcircle made up
//   of the points (x1,y1), (x2,y2), (x3,y3)
//   The circumcircle centre is returned in (xc,yc) and the radius r
//   Note : A point on the edge is inside the circumcircle
////////////////////////////////////////////////////////////////////////

int CircumCircle(double xp, double yp, double x1, double y1, double x2, 
double y2, double x3, double y3, double &xc, double &yc, double &r){
  double m1, m2, mx1, mx2, my1, my2;
  double dx, dy, rsqr, drsqr;

/* Check for coincident points */
if(abs(y1 - y2) < EPSILON && abs(y2 - y3) < EPSILON)
  return(false);
if(abs(y2-y1) < EPSILON){ 
  m2 = - (x3 - x2) / (y3 - y2);
  mx2 = (x2 + x3) / 2.0;
  my2 = (y2 + y3) / 2.0;
  xc = (x2 + x1) / 2.0;
  yc = m2 * (xc - mx2) + my2;
}else if(abs(y3 - y2) < EPSILON){ 
        m1 = - (x2 - x1) / (y2 - y1);
        mx1 = (x1 + x2) / 2.0;
        my1 = (y1 + y2) / 2.0;
        xc = (x3 + x2) / 2.0;
        yc = m1 * (xc - mx1) + my1;
      }else{
         m1 = - (x2 - x1) / (y2 - y1); 
         m2 = - (x3 - x2) / (y3 - y2); 
         mx1 = (x1 + x2) / 2.0; 
         mx2 = (x2 + x3) / 2.0;
         my1 = (y1 + y2) / 2.0;
         my2 = (y2 + y3) / 2.0;
         xc = (m1 * mx1 - m2 * mx2 + my2 - my1) / (m1 - m2); 
         yc = m1 * (xc - mx1) + my1; 
       }
  dx = x2 - xc;
  dy = y2 - yc;
  rsqr = dx * dx + dy * dy;
  r = sqrt(rsqr); 
  dx = xp - xc;
  dy = yp - yc;
  drsqr = dx * dx + dy * dy;
  return((drsqr <= rsqr) ? true : false);
}
///////////////////////////////////////////////////////////////////////////////
// Triangulate() :
//   Triangulation subroutine
//   Takes as input NV vertices in array pxyz
//   Returned is a list of ntri triangular faces in the array v
//   These triangles are arranged in a consistent clockwise order.
//   The triangle array 'v' should be malloced to 3 * nv
//   The vertex array pxyz must be big enough to hold 3 more points
//   The vertex array must be sorted in increasing x values say
//
//   qsort(p,nv,sizeof(XYZ),XYZCompare);
///////////////////////////////////////////////////////////////////////////////

int Triangulate(int nv, XYZ pxyz[], ITRIANGLE v[], int &ntri){
  int *complete = NULL;
  IEDGE *edges = NULL; 
  IEDGE *p_EdgeTemp;
  int nedge = 0;
  int trimax, emax = 200;
  int status = 0;
  int inside;
  int i, j, k;
  double xp, yp, x1, y1, x2, y2, x3, y3, xc, yc, r;
  double xmin, xmax, ymin, ymax, xmid, ymid;
  double dx, dy, dmax; 

/* Allocate memory for the completeness list, flag for each triangle */
  trimax = 4 * nv;
  complete = new int[trimax];
/* Allocate memory for the edge list */
  edges = new IEDGE[emax];
/*
      Find the maximum and minimum vertex bounds.
      This is to allow calculation of the bounding triangle
*/
  xmin = pxyz[0].x;
  ymin = pxyz[0].y;
  xmax = xmin;
  ymax = ymin;
  for(i = 1; i < nv; i++){
    if (pxyz[i].x < xmin) xmin = pxyz[i].x;
    if (pxyz[i].x > xmax) xmax = pxyz[i].x;
    if (pxyz[i].y < ymin) ymin = pxyz[i].y;
    if (pxyz[i].y > ymax) ymax = pxyz[i].y;
  }
  dx = xmax - xmin;
  dy = ymax - ymin;
  dmax = (dx > dy) ? dx : dy;
  xmid = (xmax + xmin) / 2.0;
  ymid = (ymax + ymin) / 2.0;
/*
   Set up the supertriangle
   his is a triangle which encompasses all the sample points.
   The supertriangle coordinates are added to the end of the
   vertex list. The supertriangle is the first triangle in
   the triangle list.
*/
  pxyz[nv+0].x = xmid - 20 * dmax;
  pxyz[nv+0].y = ymid - dmax;
  pxyz[nv+1].x = xmid;
  pxyz[nv+1].y = ymid + 20 * dmax;
  pxyz[nv+2].x = xmid + 20 * dmax;
  pxyz[nv+2].y = ymid - dmax;
  v[0].p1 = nv;
  v[0].p2 = nv+1;
  v[0].p3 = nv+2;
  complete[0] = false;
  ntri = 1;
/*
   Include each point one at a time into the existing mesh
*/
  for(i = 0; i < nv; i++){
    xp = pxyz[i].x;
    yp = pxyz[i].y;
    nedge = 0;
/*
     Set up the edge buffer.
     If the point (xp,yp) lies inside the circumcircle then the
     three edges of that triangle are added to the edge buffer
     and that triangle is removed.
*/
  for(j = 0; j < ntri; j++){
    if(complete[j])
      continue;
    x1 = pxyz[v[j].p1].x;
    y1 = pxyz[v[j].p1].y;
    x2 = pxyz[v[j].p2].x;
    y2 = pxyz[v[j].p2].y;
    x3 = pxyz[v[j].p3].x;
    y3 = pxyz[v[j].p3].y;
    inside = CircumCircle(xp, yp, x1, y1, x2, y2, x3, y3, xc, yc, r);
    if (xc + r < xp)
// Suggested
// if (xc + r + EPSILON < xp)
      complete[j] = true;
    if(inside){
/* Check that we haven't exceeded the edge list size */
      if(nedge + 3 >= emax){
        emax += 100;
        p_EdgeTemp = new IEDGE[emax];
        for (int i = 0; i < nedge; i++) { // Fix by John Bowman
          p_EdgeTemp[i] = edges[i];   
        }
        delete []edges;
        edges = p_EdgeTemp;
      }
      edges[nedge+0].p1 = v[j].p1;
      edges[nedge+0].p2 = v[j].p2;
      edges[nedge+1].p1 = v[j].p2;
      edges[nedge+1].p2 = v[j].p3;
      edges[nedge+2].p1 = v[j].p3;
      edges[nedge+2].p2 = v[j].p1;
      nedge += 3;
      v[j] = v[ntri-1];
      complete[j] = complete[ntri-1];
      ntri--;
      j--;
    }
  }
/*
  Tag multiple edges
  Note: if all triangles are specified anticlockwise then all
  interior edges are opposite pointing in direction.
*/
  for(j = 0; j < nedge - 1; j++){
    for(k = j + 1; k < nedge; k++){
      if((edges[j].p1 == edges[k].p2) && (edges[j].p2 == edges[k].p1)){
        edges[j].p1 = -1;
        edges[j].p2 = -1;
        edges[k].p1 = -1;
        edges[k].p2 = -1;
      }
       /* Shouldn't need the following, see note above */
      if((edges[j].p1 == edges[k].p1) && (edges[j].p2 == edges[k].p2)){
        edges[j].p1 = -1;
        edges[j].p2 = -1;
        edges[k].p1 = -1;
        edges[k].p2 = -1;
      }
    }
  }
/*
     Form new triangles for the current point
     Skipping over any tagged edges.
     All edges are arranged in clockwise order.
*/
  for(j = 0; j < nedge; j++) {
    if(edges[j].p1 < 0 || edges[j].p2 < 0)
      continue;
    v[ntri].p1 = edges[j].p1;
    v[ntri].p2 = edges[j].p2;
    v[ntri].p3 = i;
    complete[ntri] = false;
    ntri++;
  }
}
/*
      Remove triangles with supertriangle vertices
      These are triangles which have a vertex number greater than nv
*/
  for(i = 0; i < ntri; i++) {
    if(v[i].p1 >= nv || v[i].p2 >= nv || v[i].p3 >= nv) {
      v[i] = v[ntri-1];
      ntri--;
      i--;
    }
  }
  delete[] edges;
  delete[] complete;
  return 0;
} 

int XYZCompare(const void *v1, const void *v2){
  XYZ *p1, *p2;

  p1 = (XYZ*)v1;
  p2 = (XYZ*)v2;
  if(p1->x < p2->x)
    return(-1);
  else if(p1->x > p2->x)
         return(1);
       else
         return(0);
}

I didn't go with a debugger, but from the resulting triangles it seems that this is an accuracy/ambiguity problem.

When you are triangulating a square there are two ways to split it into triangles and both are OK from Delaunay criteria (circumscribed circle center is on border of triangle).

So if you evaluate every triangle independently you may sometimes get even 4 triangles (depending on implementation).

Normally in such cases I recommend to build algorithm as a series of questions which cannot produce contradicting answers. In this case the question is "to which point goes triangle based on edge (1,0)-(1,1)". But often this requires significant changes to the algorithm.

A quick fix usually involves adding some tolerances for comparisons and extra checks (like non-intersecting triangles). But usually it just makes problems rarer.

Most likely you didn't delete all the double edges, especially not the edges from same triangles but with vertices only in another order. The correct function is in the answer from @cMinor.

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