Number of ways of Triangulation for a Polygon
Last Updated : 06 Nov, 2024
Given a convex polygon with n sides. The task is to calculate the number of ways in which triangles can be formed by connecting vertices with non-crossing line segments.
Examples:
Input: n = 3
Output: 1
It is already a triangle so it can only be formed in 1 way.
Input: n = 4
Output: 2
It can be cut into 2 triangles by using either pair of opposite vertices.
Input: n = 6
Output: 14 (Below, There are all 14 polygon).
It can be cut into 2 triangles by using either pair of opposite vertices.
The above problem is an application of a catalan numbers. The task is to only find the (n-2)’th Catalan Number. First few catalan numbers for n = 0, 1, 2, 3, 4, 5 are 1 1 2 5 14 42, … (considered from 0th number)
C++ // C++ code of finding Number of ways a convex polygon of n+2 // sides can split into triangles by connecting vertices #include <iostream> using namespace std; // Function to calculate the binomial coefficient C(n, k) int binomialCoeff(int n, int k) { // C(n, k) is the same as C(n, n-k) if (k > n - k) { k = n - k; } int res = 1; // Calculate the value of n! / (k! * (n-k)!) for (int i = 0; i < k; ++i) { res *= (n - i); res /= (i + 1); } return res; } // Function to find the nth Catalan number int countWays(int n) { n = n - 2; // Calculate C(2n, n) int c = binomialCoeff(2 * n, n); // Return the nth Catalan number return c / (n + 1); } int main() { int n = 6; cout << countWays(n) << endl; return 0; } // C code of finding Number of ways a convex polygon of n+2 // sides can split into triangles by connecting vertices #include <stdio.h> // Function to calculate the binomial coefficient C(n, k) int binomialCoeff(int n, int k) { // C(n, k) is the same as C(n, n-k) if (k > n - k) { k = n - k; } int res = 1; // Calculate the value of n! / (k! * (n-k)!) for (int i = 0; i < k; ++i) { res *= (n - i); res /= (i + 1); } return res; } // Function to find the nth Catalan number int countWays(int n) { n = n - 2; // Calculate C(2n, n) int c = binomialCoeff(2 * n, n); // Return the nth Catalan number return c / (n + 1); } int main() { int n = 6; printf("%d\n", countWays(n)); return 0; } // Java code of finding Number of ways a convex polygon of n+2 // sides can split into triangles by connecting vertices public class PolygonTriangulation { // Function to calculate the binomial coefficient C(n, k) static int binomialCoeff(int n, int k) { // C(n, k) is the same as C(n, n-k) if (k > n - k) { k = n - k; } int res = 1; // Calculate the value of n! / (k! * (n-k)!) for (int i = 0; i < k; ++i) { res *= (n - i); res /= (i + 1); } return res; } // Function to find the nth Catalan number static int countWays(int n) { n = n - 2; // Calculate C(2n, n) int c = binomialCoeff(2 * n, n); // Return the nth Catalan number return c / (n + 1); } public static void main(String[] args) { int n = 6; System.out.println(countWays(n)); } } # Python code of finding Number of ways a convex polygon of n+2 # sides can split into triangles by connecting vertices def binomialCoeff(n, k): # C(n, k) is the same as C(n, n-k) if k > n - k: k = n - k res = 1 # Calculate the value of n! / (k! * (n-k)!) for i in range(k): res *= (n - i) res //= (i + 1) return res # Function to find the nth Catalan number def countWays(n): n = n - 2 # Calculate C(2n, n) c = binomialCoeff(2 * n, n) # Return the nth Catalan number return c // (n + 1) n = 6 print(countWays(n))
// JavaScript code of finding Number of ways a convex polygon of n+2 // sides can split into triangles by connecting vertices function binomialCoeff(n, k) { // C(n, k) is the same as C(n, n-k) if (k > n - k) { k = n - k; } let res = 1; // Calculate the value of n! / (k! * (n-k)!) for (let i = 0; i < k; ++i) { res *= (n - i); res /= (i + 1); } return res; } // Function to find the nth Catalan number function countWays(n) { n = n - 2; // Calculate C(2n, n) let c = binomialCoeff(2 * n, n); // Return the nth Catalan number return c / (n + 1); } let n = 6; console.log(countWays(n)); Time Complexity: O(n), where n is nth catalan number.
Auxiliary Space: O(1)
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