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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 ``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

 ``
Output:
`3`

Time Complexity: O(1)

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