Find three prime numbers with given sum

Given an integer N, the task is to find three prime numbers X, Y and Z such that the sum of these three numbers is equal to N i.e. X + Y + Z = N.

Examples:

Input: N = 20
Output: 2 5 13

Input: N = 34
Output: 2 3 29

Approach:



  • Generate prime numbers using Sieve of Eratosthenes
  • Start from the first prime number.
  • Take another number from the generated list.
  • Subtract first number and second number from the original number to obtain the third number.
  • Check if the third number is a prime number.
  • If the third number is a prime number then output the three numbers.
  • Otherwise, repeat the process for the second number and consequently the first number
  • If the answer does not exist then print -1.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
  
#include <bits/stdc++.h>
using namespace std;
  
const int MAX = 100001;
  
// The vector primes holds
// the prime numbers
vector<int> primes;
  
// Function to generate prime numbers
void initialize()
{
  
    // Initialize the array elements to 0s
    bool numbers[MAX] = {};
    int n = MAX;
    for (int i = 2; i * i <= n; i++)
        if (!numbers[i])
            for (int j = i * i; j <= n; j += i)
  
                // Set the non-primes to true
                numbers[j] = true;
  
    // Fill the vector primes with prime
    // numbers which are marked as false
    // in the numbers array
    for (int i = 2; i <= n; i++)
        if (numbers[i] == false)
            primes.push_back(i);
}
  
// Function to print three prime numbers
// which sum up to the number N
void findNums(int num)
{
  
    bool ans = false;
    int first = -1, second = -1, third = -1;
    for (int i = 0; i < num; i++) {
  
        // Take the first prime number
        first = primes[i];
        for (int j = 0; j < num; j++) {
  
            // Take the second prime number
            second = primes[j];
  
            // Subtract the two prime numbers
            // from the N to obtain the third number
            third = num - first - second;
  
            // If the third number is prime
            if (binary_search(primes.begin(),
                              primes.end(), third)) {
                ans = true;
                break;
            }
        }
        if (ans)
            break;
    }
    // Print the three prime numbers
    // if the solution exists
    if (ans)
        cout << first << " "
             << second << " " << third << endl;
    else
        cout << -1 << endl;
}
  
// Driver code
int main()
{
    int n = 101;
  
    // Function for generating prime numbers
    // using Sieve of Eratosthenes
    initialize();
  
    // Function to print the three prime
    // numbers whose sum is equal to N
    findNums(n);
  
    return 0;
}

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# Python3 implementation of the approach  


from math import sqrt 


  


MAX = 100001;  


  


# The vector primes holds  


# the prime numbers  


primes = [];  


  


# Function to generate prime numbers  


def initialize() : 


  


    # Initialize the array elements to 0s  


    numbers = [0]*(MAX + 1); 


    n = MAX;  


    for i in range(2, int(sqrt(n)) + 1) :  


        if (not numbers[i]) : 


            for j in range( i * i , n + 1, i) :  


  


                # Set the non-primes to true  


                numbers[j] = True;  


  


    # Fill the vector primes with prime  


    # numbers which are marked as false  


    # in the numbers array  


    for i in range(2, n + 1) : 


        if (numbers[i] == False) : 


            primes.append(i);  


  


# Function to print three prime numbers  


# which sum up to the number N  


def findNums(num) :  


  


    ans = False;  


    first = -1;  


    second = -1; 


    third = -1;  


    for i in range(num) : 


  


        # Take the first prime number  


        first = primes[i];  


        for j in range(num) : 


  


            # Take the second prime number  


            second = primes[j];  


  


            # Subtract the two prime numbers  


            # from the N to obtain the third number  


            third = num - first - second;  


  


            # If the third number is prime  


            if (third in primes) : 


                ans = True;  


                break;  


      


        if (ans) : 


            break;  


      


    # Print the three prime numbers  


    # if the solution exists  


    if (ans) : 


        print(first , second , third);  


    else : 


        print(-1);  


  


# Driver code  


if __name__ == "__main__" :  


  


    n = 101;  


  


    # Function for generating prime numbers  


    # using Sieve of Eratosthenes  


    initialize();  


  


    # Function to print the three prime  


    # numbers whose sum is equal to N  


    findNums(n);  


  


# This code is contributed by AnkitRai01 



















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Output:


2 2 97







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