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 = 2print(chordCnt( N)) N = 1print(chordCnt( N)) N = 4print(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 ?> |
Output:
2 1 14
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
Recommended Posts:
- Count the number of ways to divide N in k groups incrementally
- Minimum cuts required to divide the Circle into equal parts
- Program to calculate area of inner circle which passes through center of outer circle and touches its circumference
- Equation of circle when three points on the circle are given
- Check if a circle lies inside another circle or not
- Angle subtended by the chord to center of the circle when the angle subtended by the another equal chord of a congruent circle is given
- Find area of the larger circle when radius of the smaller circle and difference in the area is given
- Area of the circle that has a square and a circle inscribed in it
- Number of ways to divide an array into K equal sum sub-arrays
- Number of ways to divide string in sub-strings such to make them in lexicographically increasing sequence
- Count of arrays in which all adjacent elements are such that one of them divide the another
- Queries on count of points lie inside a circle
- Count of obtuse angles in a circle with 'k' equidistant points between 2 given points
- Closest Pair of Points using Divide and Conquer algorithm
- Maximum Subarray Sum using Divide and Conquer algorithm
- Convex Hull using Divide and Conquer Algorithm
- Product of all sorted subsets of size K using elements whose index divide K completely
- Count the number of ways to tile the floor of size n x m using 1 x m size tiles
- Count number of ways to fill a "n x 4" grid using "1 x 4" tiles
- Count ways to reach a score using 1 and 2 with no consecutive 2s
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
Improved By : AnkitRai01
