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Maximum number of line intersections formed through intersection of N planes
  • Last Updated : 01 Apr, 2021

Given N planes. The task is to find the maximum number of line intersections that can be formed through the intersections of N planes.
Examples: 
 

Input: N = 3 
Output: 3
Input: N = 5 
Output: 10
 

 

Approach:
Let there be N planes such that no 3 planes intersect in a single line of intersection and no 2 planes are parallel to each other. Adding ‘N+1’th plane to this space should be possible while retaining the above two conditions. In that case, this plane would intersect each of the N planes in ‘N’ distinct lines. 
Thus, the ‘N+1’th plane could atmost add ‘N’ new lines to the total count of lines of intersection. Similarly, the Nth plane could add at most “N-1′ distinct lines of intersection. It is easy therefore to see that, for ‘N’ planes, the maximum number of lines of intersection could be: 
 

(N-1) + (N-2) +...+ 1 = N*(N-1)/2

Below is the implementation of the above approach: 
 



C++




// C++ implementation of the above pproach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count maximum number of
// intersections possible
int countIntersections(int n)
{
    return n * (n - 1) / 2;
}
 
// Driver Code
int main()
{
    int n = 3;
 
    cout << countIntersections(n);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
     
    // Function to count maximum number of
    // intersections possible
    static int countIntersections(int n)
    {
        return n * (n - 1) / 2;
    }
     
    // Driver Code
    public static void main (String[] args)
    {
        int n = 3;
     
        System.out.println(countIntersections(n));
    }
}
 
// This code is contributed by AnkitRai01

Python3




# Python3 implementation of the above pproach
 
# Function to count maximum number of
# intersections possible
def countIntersections(n):
    return n * (n - 1) // 2
 
# Driver Code
n = 3
 
print(countIntersections(n))
 
# This code is contributed by mohit kumar

C#




// C# implementation of the approach
using System;
 
class GFG
{
     
    // Function to count maximum number of
    // intersections possible
    static int countIntersections(int n)
    {
        return n * (n - 1) / 2;
    }
     
    // Driver Code
    public static void Main (String[] args)
    {
        int n = 3;
     
        Console.WriteLine(countIntersections(n));
    }
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// Javascript implementation of the above approach
 
// Function to count maximum number of
// intersections possible
function countIntersections(n)
{
    return n * (n - 1) / 2;
}
 
// Driver Code
var n = 3;
document.write(countIntersections(n));
 
</script>
Output: 
3

 

Time Complexity: O(1)
 

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