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# Ways to Remove Edges from a Complete Graph to make Odd Edges

Given a complete graph with N vertices, the task is to count the number of ways to remove edges such that the resulting graph has odd number of edges.

Examples:

```Input: N = 3
Output: 4 ```

The initial graph has 3 edges as it is a complete graph. We can remove edges (1, 2) and (1, 3) or (1, 2) and (2, 3) or (1, 3) and (2, 3) or we do not remove any of the edges.

```Input: N = 4
Output: 32 ```

Approach: As the graph is complete so the total number of edges will be E = N * (N – 1) / 2. Now there are two cases,

1. If E is even then you have to remove odd number of edges, so the total number of ways will be which is equivalent to .
2. If E is odd then you have to remove even number of edges, so the total number of ways will be which is equivalent to .

Note that if N = 1 then the answer will be 0.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Function to return the number of ways` `// to remove edges from the graph so that` `// odd number of edges are left in the graph` `int` `countWays(``int` `N)` `{` `    ``// Total number of edges` `    ``int` `E = (N * (N - 1)) / 2;`   `    ``if` `(N == 1)` `        ``return` `0;`   `    ``return` `pow``(2, E - 1);` `}`   `// Driver code` `int` `main()` `{` `    ``int` `N = 4;` `    ``cout << countWays(N);`   `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach ` `class` `GfG ` `{ `   `// Function to return the number of ways ` `// to remove edges from the graph so that ` `// odd number of edges are left in the graph ` `static` `int` `countWays(``int` `N) ` `{ ` `    ``// Total number of edges ` `    ``int` `E = (N * (N - ``1``)) / ``2``; `   `    ``if` `(N == ``1``) ` `        ``return` `0``; `   `    ``return` `(``int``)Math.pow(``2``, E - ``1``); ` `} `   `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `N = ``4``; ` `    ``System.out.println(countWays(N)); ` `}` `} `   `// This code is contributed by Prerna Saini`

## Python3

 `# Python3 implementation of the approach`   `# Function to return the number of ways` `# to remove edges from the graph so that` `# odd number of edges are left in the graph` `def` `countWays(N):` `    `  `    ``# Total number of edges` `    ``E ``=` `(N ``*` `(N ``-` `1``)) ``/` `2`   `    ``if` `(N ``=``=` `1``):` `        ``return` `0`   `    ``return` `int``(``pow``(``2``, E ``-` `1``))`   `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` `    ``N ``=` `4` `    ``print``(countWays(N))`   `# This code contributed by PrinciRaj1992 `

## C#

 `// C# implementation of the approach`   `using` `System;`   `public` `class` `GFG{` `    `  `// Function to return the number of ways ` `// to remove edges from the graph so that ` `// odd number of edges are left in the graph ` `static` `int` `countWays(``int` `N) ` `{ ` `    ``// Total number of edges ` `    ``int` `E = (N * (N - 1)) / 2; `   `    ``if` `(N == 1) ` `        ``return` `0; `   `    ``return` `(``int``)Math.Pow(2, E - 1); ` `} `   `// Driver code ` `    ``static` `public` `void` `Main (){` `    `  `    ``int` `N = 4; ` `    ``Console.WriteLine(countWays(N)); ` `    ``}` `} ` `// This code is contributed by ajit.`

## PHP

 ``

## Javascript

 `// JavaScript Code to demonstrate Ways to Remove Edges ` `// from a Complete Graph to make Odd Edges ` `function` `countWays(N) ` `{` `    ``// Total number of edges ` `    ``let E = (N * (N - 1)) / 2;`   `    ``if` `(N == 1)` `        ``return` `0;`   `    ``return` `2 ** (E - 1);` `}`   `// Driver code ` `let N = 4;` `console.log(countWays(N));` `//This code is contributed by chinmaya121221`

Output:

`32`

Time Complexity: O(log E), where E = (N * (N – 1)) / 2.

Auxiliary Space: O(1)

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