Minimum Cost Polygon Triangulation

29 Jun, 2022
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)See following example taken from this source.  
Two triangulations of the same convex pentagon. The triangulation on the left has a cost of 8 + 2√2 + 2√5 (approximately 15.30), the one on the right has a cost of 4 + 2√2 + 4√5 (approximately 15.77).  
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, i) is 
  cost(i, j, k)  = dist(i, j) + dist(j, k) + dist(k, i)
Following is implementation of above naive recursive formula. 
C++
Java
Python3
C#
Javascript
// Recursive implementation for minimum cost convex polygon triangulation#include <iostream>#include <cmath>#define MAX 1000000.0using namespace std; // Structure of a point in 2D planestruct Point{    int x, y;}; // Utility function to find minimum of two double valuesdouble min(double x, double y){    return (x <= y)? x : y;} // A utility function to find distance between two points in a planedouble 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 triangledouble 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);} // A recursive function to find minimum cost of polygon triangulation// The polygon is represented by points[i..j].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;} // Driver program to test above functionsint main(){    Point points[] = {{0, 0}, {1, 0}, {2, 1}, {1, 2}, {0, 2}};    int n = sizeof(points)/sizeof(points[0]);    cout << mTC(points, 0, n-1);    return 0;}
// Class to store a point in the Euclidean planeclass Point{  int x, y;  public Point(int x, int y)  {    this.x = x;    this.y = y;  }   // Utility function to return the distance between two  // vertices in a 2-dimensional plane  public double dist(Point p)  {     // The distance between vertices `(x1, y1)` & `(x2,    // y2)` is `√((x2 − x1) ^ 2 + (y2 − y1) ^ 2)`    return Math.sqrt((this.x - p.x) * (this.x - p.x)                     + (this.y - p.y) * (this.y - p.y));  }} class GFG{   // Function to calculate the weight of optimal  // triangulation of a convex polygon represented by a  // given set of vertices `vertices[i..j]`  public static double MWT(Point[] vertices, int i, int j)  {     // If the polygon has less than 3 vertices,    // triangulation is not possible    if (j < i + 2)    {      return 0;    }     // keep track of the total weight of the minimum    // weight triangulation of `MWT(i,j)`    double cost = Double.MAX_VALUE;     // consider all possible triangles `ikj` within the    // polygon    for (int k = i + 1; k <= j - 1; k++)    {       // The weight of a triangulation is the length      // of perimeter of the triangle      double weight = vertices[i].dist(vertices[j])        + vertices[j].dist(vertices[k])        + vertices[k].dist(vertices[i]);       // choose the vertex `k` that leads to the      // minimum total weight      cost = Double.min(cost,                        weight + MWT(vertices, i, k)                        + MWT(vertices, k, j));    }    return cost;  }   // Driver code  public static void main(String[] args)  {     // vertices are given in clockwise order    Point[] vertices      = { new Point(0, 0), new Point(2, 0),         new Point(2, 1), new Point(1, 2),         new Point(0, 1) };     System.out.println(MWT(vertices,                           0, vertices.length - 1));  }} // This code is contributed by Priiyadarshini Kumari
# Recursive implementation for minimum# cost convex polygon triangulationfrom math import sqrtMAX = 1000000.0 # A utility function to find distance# between two points in a planedef dist(p1, p2):    return sqrt((p1[0] - p2[0])*(p1[0] - p2[0]) + \                (p1[1] - p2[1])*(p1[1] - p2[1])) # A utility function to find cost of# a triangle. The cost is considered# as perimeter (sum of lengths of all edges)# of the triangledef cost(points, i, j, k):    p1 = points[i]    p2 = points[j]    p3 = points[k]    return dist(p1, p2) + dist(p2, p3) + dist(p3, p1)  # A recursive function to find minimum# cost of polygon triangulation# The polygon is represented by points[i..j].def mTC(points, i, 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    res = MAX         # Find minimum triangulation by considering all    for k in range(i + 1, j):        res = min(res, (mTC(points, i, k) + \                        mTC(points, k, j) + \                        cost(points, i, k, j)))         return round(res, 4)  # Driver codepoints = [[0, 0], [1, 0], [2, 1], [1, 2], [0, 2]]n = len(points)print(mTC(points, 0, n-1)) # This code is contributed by SHUBHAMSINGH10
using System;using System.Collections.Generic; // Class to store a point in the Euclidean planepublic  class Point {  public int x, y;   public Point(int x, int y) {    this.x = x;    this.y = y;  }   // Utility function to return the distance between two  // vertices in a 2-dimensional plane  public double dist(Point p) {     // The distance between vertices `(x1, y1)` & `(x2,    // y2)` is `√((x2 − x1) ^ 2 + (y2 − y1) ^ 2)`    return Math.Sqrt((this.x - p.x) * (this.x - p.x) +                     (this.y - p.y) * (this.y - p.y));  }} public class GFG {   // Function to calculate the weight of optimal  // triangulation of a convex polygon represented by a  // given set of vertices `vertices[i..j]`  public static double MWT(Point[] vertices, int i, int j) {     // If the polygon has less than 3 vertices,    // triangulation is not possible    if (j < i + 2) {      return 0;    }     // keep track of the total weight of the minimum    // weight triangulation of `MWT(i,j)`    double cost = 9999999999999.09;     // consider all possible triangles `ikj` within the    // polygon    for (int k = i + 1; k <= j - 1; k++) {       // The weight of a triangulation is the length      // of perimeter of the triangle      double weight = vertices[i].dist(vertices[j]) +        vertices[j].dist(vertices[k])        + vertices[k].dist(vertices[i]);       // choose the vertex `k` that leads to the      // minimum total weight      cost = Math.Min(cost, weight +                      MWT(vertices, i, k) +                      MWT(vertices, k, j));    }    return Math.Round(cost,4);  }   // Driver code  public static void Main(String[] args) {     // vertices are given in clockwise order    Point[] vertices = { new Point(0, 0),                        new Point(2, 0),                        new Point(2, 1),                        new Point(1, 2),                        new Point(0, 1) };     Console.WriteLine(MWT(vertices, 0, vertices.Length - 1));  }} // This code is contributed by gauravrajput1
// A JavaScript program for a// Recursive implementation for minimum cost convex polygon triangulationconst MAX = 1.79769e+308; // Utility function to find minimum of two double valuesfunction min(x, y){    return (x <= y)? x : y;} // A utility function to find distance between two points in a planefunction dist(p1, p2){    return Math.sqrt((p1[0] - p2[0])*(p1[0] - p2[0]) + (p1[1] - p2[1])*(p1[1] - p2[1]));} // A utility function to find cost of a triangle. The cost is considered// as perimeter (sum of lengths of all edges) of the trianglefunction cost(points, i, j, k){    p1 = points[i], p2 = points[j], p3 = points[k];    return dist(p1, p2) + dist(p2, p3) + dist(p3, p1);} // A recursive function to find minimum cost of polygon triangulation// The polygon is represented by points[i..j].function mTC(points, i, 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   let res = MAX;    // Find minimum triangulation by considering all   for (let k=i+1; k<j; k++){       res = min(res, (mTC(points, i, k) + mTC(points, k, j) + cost(points, i, k, j)));   }        return  res;} // Driver program to test above functions{    let points =    [[0, 0], [1, 0], [2, 1], [1, 2],[0, 2]]    let n = points.length;    console.log(mTC(points, 0, n-1));} // The code is contributed by Nidhi Goel
Output: 
15.3006
The above problem is similar to Matrix Chain Multiplication. The following is recursion tree for mTC(points[], 0, 4). 
It can be easily seen in the above recursion tree that the problem has many overlapping subproblems. Since the problem has both properties: Optimal Substructure and Overlapping Subproblems, it can be efficiently solved using dynamic programming.Following is C++ implementation of dynamic programming solution. 
C++
Java
Python3
C#
// A Dynamic Programming based program to find minimum cost of convex// polygon triangulation#include <iostream>#include <cmath>#define MAX 1000000.0using namespace std; // Structure of a point in 2D planestruct Point{    int x, y;}; // Utility function to find minimum of two double valuesdouble min(double x, double y){    return (x <= y)? x : y;} // A utility function to find distance between two points in a planedouble 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 triangledouble 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);} // 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];} // Driver program to test above functionsint main(){    Point points[] = {{0, 0}, {1, 0}, {2, 1}, {1, 2}, {0, 2}};    int n = sizeof(points)/sizeof(points[0]);    cout << mTCDP(points, n);    return 0;}
// A Dynamic Programming based program to find minimum cost// of convex polygon triangulationimport java.util.*; class GFG{     // Structure of a point in 2D plane  static class Point {    int x, y;    Point(int x, int y)    {      this.x = x;      this.y = y;    }  }   // Utility function to find minimum of two double values  static double min(double x, double y)  {    return (x <= y) ? x : y;  }   // A utility function to find distance between two  // points in a plane  static double dist(Point p1, Point p2)  {    return Math.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  static 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);  }   // A Dynamic programming based function to find minimum  // cost for convex polygon triangulation.  static 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 = new double[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] = 1000000.0;          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];  }   // Driver program to test above functions  public static void main(String[] args)  {    Point[] points = { new Point(0, 0), new Point(1, 0),                      new Point(2, 1), new Point(1, 2),                      new Point(0, 2) };    int n = points.length;    System.out.println(mTCDP(points, n));  }} // This code is contributed by Karandeep Singh
# A Dynamic Programming based program to find minimum cost# of convex polygon triangulation import math  class GFG:    # Structure of a point in 2D plane    class Point:        x = 0        y = 0         def __init__(self, x,  y):            self.x = x            self.y = y    # Utility function to find minimum of two double values     @staticmethod    def min(x,  y):        return x if (x <= y) else y    # A utility function to find distance between two    # points in a plane     @staticmethod    def dist(p1,  p2):        return math.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     @staticmethod    def cost(points,  i,  j,  k):        p1 = points[i]        p2 = points[j]        p3 = points[k]        return GFG.dist(p1, p2) + GFG.dist(p2, p3) + GFG.dist(p3, p1)    # A Dynamic programming based function to find minimum    # cost for convex polygon triangulation.     @staticmethod    def mTCDP(points,  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.        table = [[0.0] * (n) for _ in range(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.        gap = 0        while (gap < n):            i = 0            j = gap            while (j < n):                if (j < i + 2):                    table[i][j] = 0.0                else:                    table[i][j] = 1000000.0                    k = i + 1                    while (k < j):                        val = table[i][k] + table[k][j] + \                            GFG.cost(points, i, j, k)                        if (table[i][j] > val):                            table[i][j] = val                        k += 1                i += 1                j += 1            gap += 1        return table[0][n - 1]  # Driver program to test above functionsif __name__ == "__main__":    points = [GFG.Point(0, 0), GFG.Point(1, 0), GFG.Point(        2, 1), GFG.Point(1, 2), GFG.Point(0, 2)]    n = len(points)    print(GFG.mTCDP(points, n)) # This code is contributed by Aarti_Rathi
using System; // A Dynamic Programming based program to find minimum cost// of convex polygon triangulation // Structure of a point in 2D planepublic class Point {  public int x;  public int y;} public static class Globals {  public const double MAX = 1000000.0;  // Utility function to find minimum of two double values  public static double min(double x, double y)  {    return (x <= y) ? x : y;  }   // A utility function to find distance between two  // points in a plane  public static double dist(Point p1, Point p2)  {    return Math.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  public static double cost(Point[] points, int i, int j,                            int k)  {    Point p1 = points[i];    Point p2 = points[j];    Point p3 = points[k];     return (dist(p1, p2) + dist(p2, p3) + dist(p3, p1));  }   // A Dynamic programming based function to find minimum  // cost for convex polygon triangulation.  public static 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 = new double[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];  }   // Driver program to test above functions  public static void Main()  {    Point[] points = { new Point(){ x = 0, y = 0 },                      new Point(){ x = 1, y = 0 },                      new Point(){ x = 2, y = 1 },                      new Point(){ x = 1, y = 2 },                      new Point(){ x = 0, y = 2 } };     int n = points.Length;    Console.Write(mTCDP(points, n));  }} // This code is contributed by Aarti_Rathi
Output: 
15.3006
Time complexity of the above dynamic programming solution is O(n3). Please note that the above implementations assume that the points of convex polygon are given in order (either clockwise or anticlockwise)Exercise: Extend the above solution to print triangulation also. For the above example, the optimal triangulation is 0 3 4, 0 1 3, and 1 2 3.Sources: http://www.cs.utexas.edu/users/djimenez/utsa/cs3343/lecture12.html http://www.cs.utoronto.ca/~heap/Courses/270F02/A4/chains/node2.htmlPlease write comments if you find anything incorrect, or you want to share more information about the topic discussed above 
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