Minimum Cost Polygon Triangulation - GeeksforGeeks

Minimum Cost Polygon Triangulation - GeeksforGeeks
A triangulation of a convex polygon is formed by drawing diagonals between non-adjacent vertices (corners) such that the diagonals never intersect. The problem is to find the cost of triangulation with the minimum cost. The cost of a triangulation is sum of the weights of its component triangles. Weight of each triangle is its perimeter (sum of lengths of all sides)

This problem has recursive substructure. The idea is to divide the polygon into three parts: a single triangle, the sub-polygon to the left, and the sub-polygon to the right. We try all possible divisions like this and find the one that minimizes the cost of the triangle plus the cost of the triangulation of the two sub-polygons.

Let Minimum Cost of triangulation of vertices from i to j be minCost(i, j)
If j <= i + 2 Then
  minCost(i, j) = 0
Else
  minCost(i, j) = Min { minCost(i, k) + minCost(k, j) + cost(i, k, j) }
                  Here k varies from 'i+1' to 'j-1'

Cost of a triangle formed by edges (i, j), (j, k) and (k, j) is 
  cost(i, j, k)  = dist(i, j) + dist(j, k) + dist(k, j) 

// A Dynamic programming based function to find minimum cost for convex

// polygon triangulation.

double mTCDP(Point points[], int n)

{

   // There must be at least 3 points to form a triangle

   if (n < 3)

      return 0;

   // table to store results of subproblems.  table[i][j] stores cost of

   // triangulation of points from i to j.  The entry table[0][n-1] stores

   // the final result.

   double table[n][n];

   // Fill table using above recursive formula. Note that the table

   // is filled in diagonal fashion i.e., from diagonal elements to

   // table[0][n-1] which is the result.

   for (int gap = 0; gap < n; gap++)

   {

      for (int i = 0, j = gap; j < n; i++, j++)

      {

          if (j < i+2)

             table[i][j] = 0.0;

          else

          {

              table[i][j] = MAX;

              for (int k = i+1; k < j; k++)

              {

                double val = table[i][k] + table[k][j] + cost(points,i,j,k);

                if (table[i][j] > val)

                     table[i][j] = val;

              }

          }

      }

   }

   return  table[0][n-1];

}

double dist(Point p1, Point p2)

{

    return sqrt((p1.x - p2.x)*(p1.x - p2.x) +

                (p1.y - p2.y)*(p1.y - p2.y));

}

// A utility function to find cost of a triangle. The cost is considered

// as perimeter (sum of lengths of all edges) of the triangle

double cost(Point points[], int i, int j, int k)

{

    Point p1 = points[i], p2 = points[j], p3 = points[k];

    return dist(p1, p2) + dist(p2, p3) + dist(p3, p1);

}

double mTC(Point points[], int i, int j)

{

   // There must be at least three points between i and j

   // (including i and j)

   if (j < i+2)

      return 0;

   // Initialize result as infinite

   double res = MAX;

   // Find minimum triangulation by considering all

   for (int k=i+1; k<j; k++)

        res = min(res, (mTC(points, i, k) + mTC(points, k, j) +

                        cost(points, i, k, j)));

   return  res;

}

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