Represent the given number as the sum of two composite numbers

Given an integer N, the task is to represent N as the sum of two composite integers. There can be multiple ways possible, print any one of them. If it is not possible to represent the number as the sum of two composite numbers then print -1.
Examples:

Input: N = 13 
Output: 4 9 
4 + 9 = 13 and both 4 and 9 are composite.
Input: N = 18 
Output: 4 14 

Approach: When N ? 11 then only 8 and 10 are the integers which can be represented as the sum of two composite integers i.e. 4 + 4 and 4 + 6 respectively. 
When N > 11 then there are two cases: 

1. When N is even: N can be represented as 4 + (N – 4) since both are composite.
2. When N is odd: N can be represented as 9 + (N – 9).

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to find two composite
// numbers which when added
// give sum as n
void findNums(int n)
{

    // Only 8 and 10 can be represented
    // as the sum of two composite integers
    if (n <= 11) {
        if (n == 8)
            cout << "4 4";
        if (n == 10)
            cout << "4 6";
        else
            cout << "-1";
        return;
    }

    // If n is even
    if (n % 2 == 0)
        cout << "4 " << (n - 4);

    // If n is odd
    else
        cout << "9 " << (n - 9);
}

// Driver code
int main()
{
    int n = 13;

    findNums(n);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

// Function to find two composite
// numbers which when added
// give sum as n
static void findNums(int n)
{

    // Only 8 and 10 can be represented
    // as the sum of two composite integers
    if (n <= 11)
    {
        if (n == 8)
            System.out.print("4 4");
        if (n == 10)
            System.out.print("4 6");
        else
            System.out.print("-1");
        return;
    }

    // If n is even
    if (n % 2 == 0)
        System.out.print("4 " + (n - 4));

    // If n is odd
    else
        System.out.print("9 " + (n - 9));
}

// Driver code
public static void main(String args[])
{
    int n = 13;

    findNums(n);
}
}

// This code is contributed by andrew1234

Python3

# Python3 implementation of the approach

# Function to find two composite
# numbers which when added
# give sum as n
def findNums(n):

    # Only 8 and 10 can be represented
    # as the sum of two composite integers
    if (n <= 11):
        if (n == 8):
            print("4 4", end = " ")
        if (n == 10):
            print("4 6", end = " ")
        else:
            print("-1", end = " ")

    # If n is even
    if (n % 2 == 0):
        print("4 ", (n - 4), end = " ")

    # If n is odd
    else:
        print("9 ", n - 9, end = " ")

# Driver code
n = 13

findNums(n)

# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;

class GFG
{

    // Function to find two composite
    // numbers which when added
    // give sum as n
    static void findNums(int n)
    {

        // Only 8 and 10 can be represented
        // as the sum of two composite integers
        if (n <= 11)
        {
            if (n == 8)
                Console.Write("4 4");
            if (n == 10)
                Console.Write("4 6");
            else
                Console.Write("-1");
            return;
        }

        // If n is even
        if (n % 2 == 0)
            Console.Write("4 " + (n - 4));

        // If n is odd
        else
            Console.Write("9 " + (n - 9));
    }

    // Driver code
    public static void Main()
    {
        int n = 13;

        findNums(n);
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>

// javascript implementation of the approach


// Function to find two composite
// numbers which when added
// give sum as n
function findNums(n)
{

    // Only 8 and 10 can be represented
    // as the sum of two composite integers
    if (n <= 11)
    {
        if (n == 8)
            document.write("4 4");
        if (n == 10)
            document.write("4 6");
        else
            document.write("-1");
        return;
    }

    // If n is even
    if (n % 2 == 0)
        document.write("4 " + (n - 4));

    // If n is odd
    else
        document.write("9 " + (n - 9));
}

// Driver code

var n = 13;

findNums(n);

// This code contributed by shikhasingrajput

</script>

Output:

9 4

Time Complexity: O(1)

Auxiliary Space: O(1)