Check if an integer can be expressed as a sum of two semi-primes
Given a positive integer N, check if it can be expressed as a sum of two semi-primes or not.
Semi-primes A number is said to be a semi-prime if it can be expressed as product of two primes number ( not necessarily distinct ).
Semi-primes in the range 1 -100 are:
4, 6, 9, 10, 14, 15, 21, 22, 25, 26, 33, 34, 35, 38, 39, 46, 49, 51, 55, 57, 58, 62, 65, 69, 74, 77, 82, 85, 86, 87, 91, 93, 94, 95.
Examples:
Input : N = 30 Output: YES Explanation: 30 can be expressed as '15 + 15' 15 is semi-primes as 15 is a product of two primes 3 and 5. Input : N = 45 Output : YES Explanation: 45 can be expressed as '35 + 10' 35 and 10 are also semi-primes as it can a be expressed as product of two primes: 35 = 5 * 7 10 = 2 * 5
Prerequisite:
A Simple Solution is to traverse form i=1 to and check if i and N-i are semi-primes or not. If yes, print i and n-i.
An Efficient Solution is to pre-compute semi-primes in an array up to the given range and then traverse the semi-prime array and check if n-arr[i] is semi-prime or not. As, arr[i] is already a semi-prime if n-arr[i] is also a semi-prime, then n can be expressed as sum of two semi-primes.
Below is the implementation of above approach:
C++
// CPP Code to check if an integer // can be expressed as sum of // two semi-primes #include <bits/stdc++.h> using namespace std; #define MAX 1000000 vector< int > arr; bool sprime[MAX]; // Utility function to compute // semi-primes in a range void computeSemiPrime() { memset (sprime, false , sizeof (sprime)); for ( int i = 2; i < MAX; i++) { int cnt = 0; int num = i; for ( int j = 2; cnt < 2 && j * j <= num; ++j) { while (num % j == 0) { num /= j, ++cnt; // Increment count // of prime numbers } } // If number is greater than 1, add it to // the count variable as it indicates the // number remain is prime number if (num > 1) ++cnt; // if count is equal to '2' then // number is semi-prime if (cnt == 2) { sprime[i] = true ; arr.push_back(i); } } } // Utility function to check // if a number sum of two // semi-primes bool checkSemiPrime( int n) { int i = 0; while (arr[i] <= n / 2) { // arr[i] is already a semi-prime // if n-arr[i] is also a semi-prime // then we a number can be expressed as // sum of two semi-primes if (sprime[n - arr[i]]) { return true ; } i++; } return false ; } // Driver code int main() { computeSemiPrime(); int n = 30; if (checkSemiPrime(n)) cout << "YES" ; else cout << "NO" ; return 0; } |
Java
// Java Code to check if an integer // can be expressed as sum of // two semi-primes import java.util.*; class GFG { static final int MAX = 1000000 ; static Vector<Integer> arr = new Vector<>(); static boolean [] sprime = new boolean [MAX]; // Utility function to compute // semi-primes in a range static void computeSemiPrime() { for ( int i = 0 ; i < MAX; i++) sprime[i] = false ; for ( int i = 2 ; i < MAX; i++) { int cnt = 0 ; int num = i; for ( int j = 2 ; cnt < 2 && j * j <= num; ++j) { while (num % j == 0 ) { num /= j; ++cnt; // Increment count // of prime numbers } } // If number is greater than 1, add it to // the count variable as it indicates the // number remain is prime number if (num > 1 ) ++cnt; // if count is equal to '2' then // number is semi-prime if (cnt == 2 ) { sprime[i] = true ; arr.add(i); } } } // Utility function to check // if a number is sum of two // semi-primes static boolean checkSemiPrime( int n) { int i = 0 ; while (arr.get(i) <= n / 2 ) { // arr[i] is already a semi-prime // if n-arr[i] is also a semi-prime // then we a number can be expressed as // sum of two semi-primes if (sprime[n - arr.get(i)]) { return true ; } i++; } return false ; } // Driver code public static void main(String[] args) { computeSemiPrime(); int n = 30 ; if (checkSemiPrime(n)) System.out.println( "YES" ); else System.out.println( "NO" ); } } |
Python3
# Python3 Code to check if an integer can # be expressed as sum of two semi-primes MAX = 10000 arr = [] sprime = [ False ] * ( MAX ) # Utility function to compute # semi-primes in a range def computeSemiPrime(): for i in range ( 2 , MAX ): cnt, num, j = 0 , i, 2 while cnt < 2 and j * j < = num: while num % j = = 0 : num / = j # Increment count of prime numbers cnt + = 1 j + = 1 # If number is greater than 1, add it # to the count variable as it indicates # the number remain is prime number if num > 1 : cnt + = 1 # if count is equal to '2' then # number is semi-prime if cnt = = 2 : sprime[i] = True arr.append(i) # Utility function to check # if a number sum of two # semi-primes def checkSemiPrime(n): i = 0 while arr[i] < = n / / 2 : # arr[i] is already a semi-prime # if n-arr[i] is also a semi-prime # then we a number can be expressed as # sum of two semi-primes if sprime[n - arr[i]] = = True : return True i + = 1 return False # Driver code if __name__ = = "__main__" : computeSemiPrime() n = 30 if checkSemiPrime(n) = = True : print ( "YES" ) else : print ( "NO" ) # This code is contributed by # Rituraj Jain |
C#
// C# Code to check if an integer // can be expressed as sum of // two semi-primes using System.Collections.Generic; class GFG { static int MAX = 1000000; static List< int > arr = new List< int >(); static bool [] sprime = new bool [MAX]; // Utility function to compute // semi-primes in a range static void computeSemiPrime() { for ( int i = 0; i < MAX; i++) sprime[i] = false ; for ( int i = 2; i < MAX; i++) { int cnt = 0; int num = i; for ( int j = 2; cnt < 2 && j * j <= num; ++j) { while (num % j == 0) { num /= j; ++cnt; // Increment count // of prime numbers } } // If number is greater than 1, add it to // the count variable as it indicates the // number remain is prime number if (num > 1) ++cnt; // if count is equal to '2' then // number is semi-prime if (cnt == 2) { sprime[i] = true ; arr.Add(i); } } } // Utility function to check // if a number is sum of two // semi-primes static bool checkSemiPrime( int n) { int i = 0; while (arr[i] <= n / 2) { // arr[i] is already a semi-prime // if n-arr[i] is also a semi-prime // then we a number can be expressed as // sum of two semi-primes if (sprime[n - arr[i]]) { return true ; } i++; } return false ; } // Driver code public static void Main() { computeSemiPrime(); int n = 30; if (checkSemiPrime(n)) System.Console.WriteLine( "YES" ); else System.Console.WriteLine( "NO" ); } } // This code is contributed by mits |
YES
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