Check if N is Strong Prime (original) (raw)

Last Updated : 27 Aug, 2022

Given a positive integer N, the task is to check if N is a strong prime or not.
In number theory, a strong prime is a prime number that is greater than the arithmetic mean of nearest prime numbers i.e next and previous prime numbers.
First few strong prime numbers are 11, 17, 29, 37, 41, 59, 67, 71, ...
A strong prime Pn can be represented as-

Strong prime

where n is its index in the ordered set of prime numbers.

Examples:

Input: N = 11
Output: Yes
11 is 5th prime number, the arithmetic mean of 4th and 6th prime number i.e. 7 and 13 is 10.
11 is greater than 10 so 11 is a strong prime.

Input: N = 13
Output: No
13 is 6th prime number, the arithmetic mean of 5th (11) and 7th (17) is (11 + 17) / 2 = 14.
13 is smaller than 14 so 13 is not a strong prime.

Approach:

If N is not a prime number or it is the first prime number i.e. 2 then print No.
Else find the primes closest to N (one on the left and one on the right) and store their arithmetic mean in mean.
- If N > mean then print Yes.
- Else print No.

Below is the implementation of the above approach:

C++ `

// C++ program to check if given number is strong prime #include <bits/stdc++.h> using namespace std;

// Utility function to check // if a number is prime or not bool isPrime(int n) { // Corner cases if (n <= 1) return false; if (n <= 3) return true;

// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
    return false;

for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0)
        return false;

return true;

}

// Function that returns true if n is a strong prime static bool isStrongPrime(int n) { // If n is not a prime number or // n is the first prime then return false if (!isPrime(n) || n == 2) return false;

// Initialize previous_prime to n - 1
// and next_prime to n + 1
int previous_prime = n - 1;
int next_prime = n + 1;

// Find next prime number
while (!isPrime(next_prime))
    next_prime++;

// Find previous prime number
while (!isPrime(previous_prime))
    previous_prime--;

// Arithmetic mean
int mean = (previous_prime + next_prime) / 2;

// If n is a strong prime
if (n > mean)
    return true;
else
    return false;

}

// Driver code int main() {

int n = 11;

if (isStrongPrime(n))
    cout << "Yes";
else
    cout << "No";

return 0;

}

Java

// Java program to check if given number is strong prime class GFG {

// Utility function to check
// if a number is prime or not
static boolean isPrime(int n)
{
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;

    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;

    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;

    return true;
}

// Function that returns true if n is a strong prime
static boolean isStrongPrime(int n)
{
    // If n is not a prime number or
    // n is the first prime then return false
    if (!isPrime(n) || n == 2)
        return false;

    // Initialize previous_prime to n - 1
    // and next_prime to n + 1
    int previous_prime = n - 1;
    int next_prime = n + 1;

    // Find next prime number
    while (!isPrime(next_prime))
        next_prime++;

    // Find previous prime number
    while (!isPrime(previous_prime))
        previous_prime--;

    // Arithmetic mean
    int mean = (previous_prime + next_prime) / 2;

    // If n is a strong prime
    if (n > mean)
        return true;
    else
        return false;
}

// Driver code
public static void main(String args[])
{

    int n = 11;

    if (isStrongPrime(n))
        System.out.println("Yes");

    else
        System.out.println("No");
}

}

Python3

Python 3 program to check if given

number is strong prime

from math import sqrt

Utility function to check if a

number is prime or not

def isPrime(n):

# Corner cases
if (n <= 1):
    return False
if (n <= 3):
    return True

# This is checked so that we can skip
# middle five numbers in below loop
if (n % 2 == 0 or n % 3 == 0):
    return False

k = int(sqrt(n)) + 1
for i in range(5, k, 6):
    if (n % i == 0 or n % (i + 2) == 0):
        return False

return True

Function that returns true if

n is a strong prime

def isStrongPrime(n):

# If n is not a prime number or
# n is the first prime then return false
if (isPrime(n) == False or n == 2):
    return False

# Initialize previous_prime to n - 1
# and next_prime to n + 1
previous_prime = n - 1
next_prime = n + 1

# Find next prime number
while (isPrime(next_prime) == False):
    next_prime += 1

# Find previous prime number
while (isPrime(previous_prime) == False):
    previous_prime -= 1

# Arithmetic mean
mean = (previous_prime + next_prime) / 2

# If n is a strong prime
if (n > mean):
    return True
else:
    return False

Driver code

if name == 'main': n = 11

if (isStrongPrime(n)):
    print("Yes")
else:
    print("No")

This code is contributed by

Sanjit_prasad

C#

// C# program to check if a given number is strong prime using System; class GFG {

// Utility function to check
// if a number is prime or not
static bool isPrime(int n)
{
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;

    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;

    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;

    return true;
}

// Function that returns true if n is a strong prime
static bool isStrongPrime(int n)
{
    // If n is not a prime number or
    // n is the first prime then return false
    if (!isPrime(n) || n == 2)
        return false;

    // Initialize previous_prime to n - 1
    // and next_prime to n + 1
    int previous_prime = n - 1;
    int next_prime = n + 1;

    // Find next prime number
    while (!isPrime(next_prime))
        next_prime++;

    // Find previous prime number
    while (!isPrime(previous_prime))
        previous_prime--;

    // Arithmetic mean
    int mean = (previous_prime + next_prime) / 2;

    // If n is a strong prime
    if (n > mean)
        return true;
    else
        return false;
}

// Driver code
public static void Main()
{
    int n = 11;

    if (isStrongPrime(n))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}

}

PHP

i∗i * i∗i <= $n; i=i = i=i + 6) if ($n % $i == 0 || nn % (ni + 2) == 0) return false; return true; } // Function that returns true // if n is a strong prime function isStrongPrime($n) { // If n is not a prime number or // n is the first prime then return false if (!isPrime($n) || $n == 2) return false; // Initialize previous_prime to n - 1 // and next_prime to n + 1 previousprime=previous_prime = previousprime=n - 1; nextprime=next_prime = nextprime=n + 1; // Find next prime number while (!isPrime($next_prime)) $next_prime++; // Find previous prime number while (!isPrime($previous_prime)) $previous_prime--; // Arithmetic mean mean=(mean = (mean=(previous_prime + $next_prime) / 2; // If n is a strong prime if ($n > $mean) return true; else return false; } // Driver code $n = 11; if (isStrongPrime($n)) echo ("Yes"); else echo("No"); // This code is contributed // by Shivi_Aggarwal ?>

JavaScript

Time Complexity: O(n1/2)

Auxiliary Space: O(1), since no extra space has been taken.