Consider a circle with n points on circumference of it where n is even. Count number of ways we can connect these points such that no two connecting lines to cross each other and every point is connected with exactly one other point. Any point can be connected with any other point.
Consider a circle with 4 points.
1
2 3
4
In above diagram, there are two
non-crossing ways to connect
{{1, 2}, {3, 4}} and {{1, 3}, {2, 4}}.
Note that {{2, 3}, {1, 4}} is invalid
as it would cause a cross
Examples :
Input : n = 2 Output : 1 Input : n = 4 Output : 2 Input : n = 6 Output : 5 Input : n = 3 Output : Invalid n must be even.
We need to draw n/2 lines to connect n points. When we draw a line, we divide the points in two sets that need to connected. Each set needs to be connected within itself. Below is recurrence relation for the same.
Let m = n/2
// For each line we draw, we divide points
// into two sets such that one set is going
// to be connected with i lines and other
// with m-i-1 lines.
Count(m) = ∑ Count(i) * Count(m-i-1)
where 0 <= i < m
Count(0) = 1
Total number of ways with n points
= Count(m) = Count(n/2)
If we take a closer look at above recurrence, it is actually recurrence of Catalan Numbers. So the task reduces to finding n/2’th Catalan number.
Below is implementation based on above idea.
C++
// C++ program to count number of ways to connect n (where n// is even) points on a circle such that no two connecting// lines cross each other and every point is connected with// one other point.#include <iostream>using namespace std;// A dynamic programming based function to find nth// Catalan numberunsigned long int catalanDP(unsigned int n){ // Table to store results of subproblems unsigned long int catalan[n + 1]; // Initialize first two values in table catalan[0] = catalan[1] = 1; // Fill entries in catalan[] using recursive formula for (int i = 2; i <= n; i++) { catalan[i] = 0; for (int j = 0; j < i; j++) catalan[i] += catalan[j] * catalan[i - j - 1]; } // Return last entry return catalan[n];}// Returns count of ways to connect n points on a circle// such that no two connecting lines cross each other and// every point is connected with one other point.unsigned long int countWays(unsigned long int n){ // Throw error if n is odd if (n & 1) { cout << "Invalid"; return 0; } // Else return n/2'th Catalan number return catalanDP(n / 2);}// Driver program to test above functionint main(){ cout << countWays(6) << " "; return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
C
// C program to count number of ways to connect n (where n// is even) points on a circle such that no two connecting// lines cross each other and every point is connected with// one other point.#include <stdio.h>// A dynamic programming based function to find nth// Catalan numberunsigned long int catalanDP(unsigned int n){ // Table to store results of subproblems unsigned long int catalan[n + 1]; // Initialize first two values in table catalan[0] = catalan[1] = 1; // Fill entries in catalan[] using recursive formula for (int i = 2; i <= n; i++) { catalan[i] = 0; for (int j = 0; j < i; j++) catalan[i] += catalan[j] * catalan[i - j - 1]; } // Return last entry return catalan[n];}// Returns count of ways to connect n points on a circle// such that no two connecting lines cross each other and// every point is connected with one other point.unsigned long int countWays(unsigned long int n){ // Throw error if n is odd if (n & 1) { printf("Invalid"); return 0; } // Else return n/2'th Catalan number return catalanDP(n / 2);}// Driver program to test above functionint main(){ printf("%ld", countWays(6)); return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
Java
// Java program to count number of ways to connect n (where// n is even) points on a circle such that no two connecting// lines cross each other and every point is connected with// one other point.import java.io.*;class GFG { // A dynamic programming based function to find nth // Catalan number static int catalanDP(int n) { // Table to store results of subproblems int[] catalan = new int[n + 1]; // Initialize first two values in table catalan[0] = catalan[1] = 1; // Fill entries in catalan[] using recursive formula for (int i = 2; i <= n; i++) { catalan[i] = 0; for (int j = 0; j < i; j++) catalan[i] += catalan[j] * catalan[i - j - 1]; } // Return last entry return catalan[n]; } // Returns count of ways to connect n points on a circle // such that no two connecting lines cross each other // and every point is connected with one other point. static int countWays(int n) { // Throw error if n is odd if (n < 1) { System.out.println("Invalid"); return 0; } // Else return n/2'th Catalan number return catalanDP(n / 2); } // Driver Code public static void main(String[] args) { System.out.println(countWays(6) + " "); }}// This code is contributed by Aditya Kumar (adityakumar129) |
Python3
# Python3 program to count number # of ways to connect n (where n # is even) points on a circle such # that no two connecting lines# cross each other and every point # is connected with one other point. # A dynamic programming based # function to find nth Catalan numberdef catalanDP(n): # Table to store results # of subproblems catalan = [1 for i in range(n + 1)] # Fill entries in catalan[] # using recursive formula for i in range(2, n + 1): catalan[i] = 0 for j in range(i): catalan[i] += (catalan[j] * catalan[i - j - 1]) # Return last entry return catalan[n]# Returns count of ways to connect# n points on a circle such that # no two connecting lines cross # each other and every point is# connected with one other point.def countWays(n): # Throw error if n is odd if (n & 1): print("Invalid") return 0 # Else return n/2'th Catalan number return catalanDP(n // 2)# Driver Code print(countWays(6))# This code is contributed # by sahilshelangia |
C#
// C# program to count number // of ways to connect n (where // n is even) points on a circle// such that no two connecting// lines cross each other and // every point is connected with// one other point.using System;class GFG{ // A dynamic programming // based function to find // nth Catalan numberstatic int catalanDP(int n){ // Table to store // results of subproblems int []catalan = new int [n + 1]; // Initialize first // two values in table catalan[0] = catalan[1] = 1; // Fill entries in catalan[] // using recursive formula for (int i = 2; i <= n; i++) { catalan[i] = 0; for (int j = 0; j < i; j++) catalan[i] += catalan[j] * catalan[i - j - 1]; } // Return last entry return catalan[n];}// Returns count of ways to // connect n points on a circle// such that no two connecting// lines cross each other and// every point is connected // with one other point.static int countWays(int n){ // Throw error if n is odd if (n < 1) { Console.WriteLine("Invalid"); return 0; } // Else return n/2'th // Catalan number return catalanDP(n / 2);}// Driver Codestatic public void Main (){ Console.WriteLine(countWays(6) + " ");}}// This code is contributed // by M_kit |
PHP
<?php// PHP program to count number of// ways to connect n (where n is // even) points on a circle such // that no two connecting lines // cross each other and every // point is connected with one // other point.// A dynamic programming based // function to find nth Catalan numberfunction catalanDP($n){ // Table to store results // of subproblems Initialize // first two values in table $catalan[0] = $catalan[1] = 1; // Fill entries in catalan[] // using recursive formula for ($i = 2; $i <= $n; $i++) { $catalan[$i] = 0; for ($j = 0; $j < $i; $j++) $catalan[$i] += $catalan[$j] * $catalan[$i - $j - 1]; } // Return last entry return $catalan[$n];}// Returns count of ways to connect // n points on a circle such that // no two connecting lines cross // each other and every point is // connected with one other point.function countWays($n){// Throw error if n is oddif ($n & 1){ echo "Invalid"; return 0;}// Else return n/2'th// Catalan numberreturn catalanDP($n / 2);}// Driver Codeecho countWays(6) , " ";// This code is contributed by aj_36 ?> |
Javascript
<script>// javascript program to count number of// ways to connect n (where n is // even) points on a circle such // that no two connecting lines // cross each other and every // point is connected with one // other point.// A dynamic programming based // function to find nth Catalan numberfunction catalanDP(n){ // Table to store results // of subproblems Initialize // first two values in table let catalan = [] catalan[0] = catalan[1] = 1; // Fill entries in catalan[] // using recursive formula for (let i = 2; i <= n; i++) { catalan[i] = 0; for (let j = 0; j < i; j++) catalan[i] += catalan[j] * catalan[i - j - 1]; } // Return last entry return catalan[n];}// Returns count of ways to connect // n points on a circle such that // no two connecting lines cross // each other and every point is // connected with one other point.function countWays(n){// Throw error if n is oddif (n & 1){ document.write("Invalid"); return 0;}// Else return n/2'th// Catalan numberreturn catalanDP(n / 2);}// Driver Codedocument.write(countWays(6) + " ");// This code is contributed by _saurabh_jaiswal</script> |
Output :
5
Time Complexity : O(n2)
Auxiliary Space : O(n)
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