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Count ways to divide circle using N non-intersecting chords

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  • Difficulty Level : Hard
  • Last Updated : 22 Jun, 2022
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Given a number N, find the number of ways you can draw N chords in a circle with 2*N points such that no 2 chords intersect. 
Two ways are different if there exists a chord which is present in one way and not in other.
Examples: 
 

Input : N = 2
Output : 2
Explanation: If points are numbered 1 to 4 in 
clockwise direction, then different ways to 
draw chords are:
{(1-2), (3-4)} and {(1-4), (2-3)}


Input : N = 1
Output : 1
Explanation: Draw a chord between points 1 and 2.

 

If we draw a chord between any two points, can you observe the current set of points getting broken into two smaller sets S_1 and S_2. If we draw a chord from a point in S_1 to a point in S_2, it will surely intersect the chord we’ve just drawn. 
So, we can arrive at a recurrence that Ways(n) = sum[i = 0 to n-1] { Ways(i)*Ways(n-i-1) }. 
Here we iterate over i, assuming that size of one of the sets is i and size of another set automatically is (n-i-1) since we’ve already used a pair of points and i pair of points in one set. 
 

C++




// cpp code to count ways
// to divide circle using
// N non-intersecting chords.
#include <bits/stdc++.h>
using namespace std;
 
int chordCnt( int A){
 
    // n = no of points required
    int n = 2 * A;
     
    // dp array containing the sum
    int dpArray[n + 1]={ 0 };
    dpArray[0] = 1;
    dpArray[2] = 1;
    for (int i=4;i<=n;i+=2){
        for (int j=0;j<i-1;j+=2){
             
          dpArray[i] +=
            (dpArray[j]*dpArray[i-2-j]);
        }
    }
 
    // returning the required number
    return dpArray[n];
}
// Driver function
int main()
{
 
    int N;
    N = 2;
cout<<chordCnt( N)<<'\n';
    N = 1;
cout<<chordCnt( N)<<'\n';
    N = 4;
cout<<chordCnt( N)<<'\n';
    return 0;
}
 
// This code is contributed by Gitanjali.

Java




// Java code to count ways
// to divide circle using
// N non-intersecting chords.
import java.io.*;
 
class GFG {
    static int chordCnt(int A)
    {
 
        // n = no of points required
        int n = 2 * A;
 
        // dp array containing the sum
        int[] dpArray = new int[n + 1];
        dpArray[0] = 1;
        dpArray[2] = 1;
        for (int i = 4; i <= n; i += 2) {
            for (int j = 0; j < i - 1; j += 2)
            {
                dpArray[i] += (dpArray[j] *
                              dpArray[i - 2 - j]);
            }
        }
 
        // returning the required number
        return dpArray[n];
    }
    public static void main(String[] args)
    {
        int N;
        N = 2;
        System.out.println(chordCnt(N));
        N = 1;
        System.out.println(chordCnt(N));
        N = 4;
        System.out.println(chordCnt(N));
    }
}
 
// This code is contributed by Gitanjali.

Python 3




# python code to count ways to divide
# circle using N non-intersecting chords.
def chordCnt( A):
 
    # n = no of points required
    n = 2 * A
 
    # dp array containing the sum
    dpArray = [0]*(n + 1)
    dpArray[0] = 1
    dpArray[2] = 1
    for i in range(4, n + 1, 2):
        for j in range(0, i-1, 2):
            dpArray[i] += (dpArray[j]*dpArray[i-2-j])
 
    # returning the required number
    return int(dpArray[n])
 
# driver code
N = 2
print(chordCnt( N))
N = 1
print(chordCnt( N))
N = 4
print(chordCnt( N))

C#




// C# code to count ways to divide
// circle using N non-intersecting chords.
using System;
 
class GFG {
     
    static int chordCnt(int A)
    {
        // n = no of points required
        int n = 2 * A;
 
        // dp array containing the sum
        int[] dpArray = new int[n + 1];
        dpArray[0] = 1;
        dpArray[2] = 1;
         
        for (int i = 4; i <= n; i += 2)
        {
            for (int j = 0; j < i - 1; j += 2)
            {
                dpArray[i] += (dpArray[j] * dpArray[i - 2 - j]);
            }
        }
 
        // returning the required number
        return dpArray[n];
    }
     
    // Driver code
    public static void Main()
    {
        int N;
        N = 2;
        Console.WriteLine(chordCnt(N));
        N = 1;
        Console.WriteLine(chordCnt(N));
        N = 4;
        Console.WriteLine(chordCnt(N));
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP code to count ways
// to divide circle using
// N non-intersecting chords.
function chordCnt( $A)
{
 
    // n = no of points required
    $n = 2 * $A;
     
    // dp array containing the sum
    $dpArray = array_fill(0, $n + 1, 0);
    $dpArray[0] = 1;
    $dpArray[2] = 1;
    for ($i = 4; $i <= $n; $i += 2)
    {
        for ($j = 0; $j < $i - 1; $j += 2)
        {
             
            $dpArray[$i] += ($dpArray[$j] *
                             $dpArray[$i - 2 - $j]);
        }
    }
 
    // returning the required number
    return $dpArray[$n];
}
 
// Driver Code
$N = 2;
echo chordCnt($N), "\n";
$N = 1;
echo chordCnt($N), "\n";
$N = 4;
echo chordCnt($N), "\n";
     
// This code is contributed by Ryuga
?>

Javascript




<script>
 
// JavaScript code to count ways
// to divide circle using
// N non-intersecting chords.
 
function chordCnt( A){
 
    // n = no of points required
    var n = 2 * A;
     
    // dp array containing the sum
    var dpArray = Array(n+1).fill(0);
    dpArray[0] = 1;
    dpArray[2] = 1;
    for (var i=4;i<=n;i+=2){
        for (var j=0;j<i-1;j+=2){
             
          dpArray[i] +=
            (dpArray[j]*dpArray[i-2-j]);
        }
    }
 
    // returning the required number
    return dpArray[n];
}
 
 
// Driver function
var N;
N = 2;
document.write( chordCnt( N) + '<br>');
N = 1;
document.write( chordCnt( N) + '<br>');
N = 4;
document.write( chordCnt( N) + '<br>');
 
 
</script>

Output:  

2
1
14

Time Complexity: O(n2
Auxiliary Space: O(n)

Please suggest if someone has a better solution which is more efficient in terms of space and time.
This article is contributed by Aarti_Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 


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