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Non-crossing lines to connect points in a circle

  • Difficulty Level : Hard
  • Last Updated : 01 Apr, 2021

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 : 
 

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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 number
unsigned 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 function
int main()
{
    cout << countWays(6) << " ";
    return 0;
}

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 akt_mit

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 number
def 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 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)
    {
        Console.WriteLine("Invalid");
        return 0;
    }
 
    // Else return n/2'th
    // Catalan number
    return catalanDP(n / 2);
}
 
// Driver Code
static 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 number
function 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($n)
{
// Throw error if n is odd
if ($n & 1)
{
    echo "Invalid";
    return 0;
}
 
// Else return n/2'th
// Catalan number
return catalanDP($n / 2);
}
 
// Driver Code
echo 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 number
function 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 odd
if (n & 1)
{
    document.write("Invalid");
    return 0;
}
 
// Else return n/2'th
// Catalan number
return catalanDP(n / 2);
}
 
// Driver Code
document.write(countWays(6) + " ");
 
// This code is contributed by _saurabh_jaiswal
</script>

Output : 

5

Time Complexity : O(n2
Auxiliary Space : O(n)
This article is contributed by Shivam Agrawal. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 




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