Open In App

N expressed as sum of 4 prime numbers

Improve
Improve
Like Article
Like
Save
Share
Report

Express a given number as a summation of 4 positive primes. If it is not possible to express then print “-1”.

Examples: 

Input: 24
Output: 3 11 3 7  
Explanation : 3+11+3+7 = 24 and 3, 11, 7 are all prime. 

Input: 46
Output: 11 11 17 7 
explanation : 11+11+17+7 = 46 and 11, 7, 17 are all prime.

Approach : Every even integer greater than 2 can be expressed as the sum of two numbers by Goldbach’s conjecture.
Below are some facts for expressing a number as sum of 4 primes.  

  • Number must be greater than or equal to 8 as 2 is the smallest prime
  • If given number is even, we can break it as (2 + 2) + x so that x remains even and can broken into two primes.
  • If given number is odd, we can break it as (2 + 3) + x so that x remains even and can broken into two primes.

Now we can easily express n as sum of two primes using link 

C++




// CPP program to express n as sum of 4 primes.
#include <bits/stdc++.h>
using namespace std;
  
// function to check if a number is prime or not
int isPrime(int x)
{
    // does square root of the number
    int s = sqrt(x);
  
    // traverse from 2 to sqrt(n)
    for (int i = 2; i <= s; i++)
  
        // if any divisor found then non prime
        if (x % i == 0)
            return 0;
  
    // if no divisor is found then it is a prime
    return 1;
}
  
void Num(int x, int& a, int& b)
{
    // iterates to check prime or not
    for (int i = 2; i <= x / 2; i++) {
  
        // calls function to check if i and x-i
        // is prime or not
        if (isPrime(i) && isPrime(x - i)) {
  
            a = i;
            b = x - i;
  
            // if two prime numbers are found,
            // then return
            return;
        }
    }
}
  
// function to generate 4 prime numbers adding upto n
void generate(int n)
{
    // if n<=7 then 4 numbers cannot sum to
    // get that number
    if (n <= 7)
        cout << "Impossible to form" << endl;
  
    // a and b stores the last two numbers
    int a, b;
  
    // if it is not even then 2 and 3 are first
    // two of sequence
    if (n % 2 != 0) {
  
        // calls the function to get the other
        // two prime numbers considering first two
        // primes as 2 and 3 (Note 2 + 3 = 5)
        Num(n - 5, a, b);
  
        // print 2 and 3 as the firsts two prime
        // and a and b as the last two.
        cout << "2 3 " << a << " " << b << endl;
    }
  
    // if it is even then 2 and 2 are first two
    // of sequence
    else {
  
        /// calls the function to get the other
        // two prime numbers considering first two
        // primes as 2 and 2 (Note 2 + 2 = 4)
        Num(n - 4, a, b);
  
        // print 2 and 2 as the firsts two prime
        // and a and b as the last two.
        cout << "2 2 " << a << " " << b << endl;
    }
}
  
// driver program to test the above function
int main()
{
    int n = 28;
    generate(n);
    return 0;
}


Java




// Java program to express n as sum of
// 4 primes.
class GFG {
  
    static int a = 0, b = 0;
  
    // function to check if a number
    // is prime or not
    static int isPrime(int x)
    {
  
        // does square root of the
        // number
        int s = (int)Math.sqrt(x);
  
        // traverse from 2 to sqrt(n)
        for (int i = 2; i <= s; i++)
  
            // if any divisor found
            // then non prime
            if (x % i == 0)
                return 0;
  
        // if no divisor is found
        // then it is a prime
        return 1;
    }
  
    static void Num(int x)
    {
  
        // iterates to check prime
        // or not
        for (int i = 2; i <= x / 2; i++) {
  
            // calls function to check
            // if i and x-i is prime
            // or not
            if (isPrime(i) != 0 && isPrime(x - i) != 0) {
  
                a = i;
                b = x - i;
  
                // if two prime numbers
                // are found, then return
                return;
            }
        }
    }
  
    // function to generate 4 prime
    // numbers adding upto n
    static void generate(int n)
    {
  
        // if n<=7 then 4 numbers cannot
        // sum to get that number
        if (n <= 7)
            System.out.println("Impossible"
                               + " to form");
  
        // if it is not even then 2 and 3
        // are first two of sequence
        if (n % 2 != 0) {
  
            // calls the function to get the
            // other two prime numbers
            // considering first two primes
            // as 2 and 3 (Note 2 + 3 = 5)
            Num(n - 5);
  
            // print 2 and 3 as the firsts
            // two prime and a and b as the
            // last two.
            System.out.println("2 3 " + a + " " + b);
        }
  
        // if it is even then 2 and 2 are
        // first two of sequence
        else {
  
            /// calls the function to get the
            // other two prime numbers
            // considering first two primes as
            // 2 and 2 (Note 2 + 2 = 4)
            Num(n - 4);
  
            // print 2 and 2 as the firsts
            // two prime and a and b as the
            // last two.
            System.out.println("2 2 " + a + " " + b);
        }
    }
  
    // Driver function to test the above
    // function
    public static void main(String[] args)
    {
        int n = 28;
  
        generate(n);
    }
}
  
// This code is contributed by Anant Agarwal.


Python3




# Python3 program to express 
# n as sum of 4 primes.
import math;
# function to check if a 
# number is prime or not
def isPrime(x):
    # does square root
    # of the number
    s = int(math.sqrt(x))
      
    # traverse from 2 to sqrt(n)
    for i in range(2,s+1):
        # if any divisor found
        # then non prime
        if (x % i == 0):
            return 0
    # if no divisor is found
    # then it is a prime
    return 1
  
def Num(x):
    # iterates to check
    # prime or not
    ab=[0]*2
    for i in range(2,int(x / 2)+1):
        # calls function to check
        # if i and x-i is prime
        # or not
        if (isPrime(i) != 0 and isPrime(x - i) != 0):
            ab[0] = i
            ab[1] = x - i
            # if two prime numbers
            # are found, then return
            return ab
  
# function to generate 4 prime
# numbers adding upto n
def generate(n):
    # if n<=7 then 4 numbers cannot
    # sum to get that number
    if(n <= 7):
        print("Impossible to form")
      
    # if it is not even then 2 and
    # 3 are first two of sequence
      
    if (n % 2 != 0):
        # calls the function to get
        # the other two prime numbers
        # considering first two primes
        # as 2 and 3 (Note 2 + 3 = 5)
        ab=Num(n - 5)
          
        # print 2 and 3 as the firsts
        # two prime and a and b as the
        # last two.
        print("2 3",ab[0],ab[1])
          
        # if it is even then 2 and 2 are
        # first two of sequence
    else:
        # calls the function to get
        # the other two prime numbers
        # considering first two primes
        # as 2 and 2 (Note 2 + 2 = 4)
        ab=Num(n - 4)
          
        # print 2 and 2 as the firsts
        # two prime and a and b as the
        # last two.
        print("2 2",ab[0],ab[1]) 
  
# Driver Code
if __name__=='__main__':
    n = 28
    generate(n)
  
# This code is contributed by mits.


C#




// C# program to express n as sum of
// 4 primes.
using System;
  
class GFG {
  
    static int a = 0, b = 0;
  
    // function to check if a number
    // is prime or not
    static int isPrime(int x)
    {
  
        // does square root of the
        // number
        int s = (int)Math.Sqrt(x);
  
        // traverse from 2 to sqrt(n)
        for (int i = 2; i <= s; i++)
  
            // if any divisor found
            // then non prime
            if (x % i == 0)
                return 0;
  
        // if no divisor is found
        // then it is a prime
        return 1;
    }
  
    static void Num(int x)
    {
  
        // iterates to check prime
        // or not
        for (int i = 2; i <= x / 2; i++)
        {
  
            // calls function to check
            // if i and x-i is prime
            // or not
            if (isPrime(i) != 0 && 
                     isPrime(x - i) != 0)
            {
  
                a = i;
                b = x - i;
  
                // if two prime numbers
                // are found, then return
                return;
            }
        }
    }
  
    // function to generate 4 prime
    // numbers adding upto n
    static void generate(int n)
    {
  
        // if n<=7 then 4 numbers cannot
        // sum to get that number
        if (n <= 7)
            Console.Write("Impossible"
                        + " to form");
  
        // if it is not even then 2 and
        // 3 are first two of sequence
        if (n % 2 != 0) {
  
            // calls the function to get
            // the other two prime numbers
            // considering first two primes
            // as 2 and 3 (Note 2 + 3 = 5)
            Num(n - 5);
  
            // print 2 and 3 as the firsts
            // two prime and a and b as the
            // last two.
            Console.Write("2 3 " + a + " "
                                      + b);
        }
  
        // if it is even then 2 and 2 are
        // first two of sequence
        else {
  
            /// calls the function to get
            // the other two prime numbers
            // considering first two primes
            // as 2 and 2 (Note 2 + 2 = 4)
            Num(n - 4);
  
            // print 2 and 2 as the firsts
            // two prime and a and b as the
            // last two.
            Console.Write("2 2 " + a + " " 
                                      + b);
        }
    }
  
    // Driver function to test the above
    // function
    public static void Main()
    {
        int n = 28;
  
        generate(n);
    }
}
  
// This code is contributed by nitin mittal.


PHP





Javascript





Output: 

2 2 5 19

Time complexity: O(n sqrt(n)) 
Auxiliary space: O(1)

This article is contributed by Raja Vikramaditya

 



Last Updated : 18 Sep, 2023
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads