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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