Non-crossing lines to connect points in a circle


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.
2        3
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


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 C++ implementation based on above idea.

// 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 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;

Output :


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

Asked in: Amazon

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