Table of Contents

## What is a Prime Number?

A prime number is a natural number greater than 1 that is only divided by 1 and itself.

For example, 5 is a prime number because it can be divided by only 1 and 5.

Similarly, 6 is not a prime number because it can be divided by 1, 2, 3, and 6.

## Algorithm to Find Prime Numbers between 1 and N

There are many algorithms to find a prime number within the given range. The “Sieve of Eratosthenes” is one of the popular and simple algorithms to find prime numbers up to a given range.

The algorithm to find prime numbers between 1 and N has the following steps.

- Create a list of consecutive numbers from 2 to N i.e. (2,3,4…N).
- Start with the first and the smallest prime number 2. Let’s say variable p=2.
- Find the multiples of p i.e. 2p, 3p, 4p up to N and mark them in the list as not prime numbers. Note that the number p should not be marked itself.
- Find the next number in the list that is not marked and follow the earlier step with it.
- When the numbers in the list are exhausted, the algorithm is terminated. The list of numbers that are not marked are the prime numbers between 1 and N.

Below GIF gives a visual representation of the algorithm (Source: Wikipedia).

## Java Implementation for Finding Prime Numbers between 1 and N

```
Copy
package com.journaldev.examples;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
public class PrimeNumbers {
public static void main(String[] args) {
System.out.println(findPrimeNumbers(100));
}
// generating the list of prime numbers from 2 to the given number
// using the Sieve of Eratosthenes algorithm
private static List<Integer> findPrimeNumbers(int n) {
// initialize the array with "true", index denotes the numbers from 0 to n.
boolean[] primes = new boolean[n + 1];
Arrays.fill(primes, true);
//loop from 2 to x until x*x becomes greater than n
for (int i = 2; i * i < n; i++) {
// process if the number is not already marked
if (primes[i]) {
// find the multiples and mark them as false
for (int j = i * i; j <= n; j += i)
primes[j] = false;
}
}
// populate the list of prime numbers from the array and return it
List<Integer> primeNumbers = new ArrayList<>();
for (int i = 2; i <= n; i++) {
if (primes[i])
primeNumbers.add(i);
}
return primeNumbers;
}
}
```

**Output**: `[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]`

## Algorithm to Check if a Number is Prime?

The simplest method to check if a number is prime is using the trial division.

For a given number N, check if there is a prime number M between 2 to √N (square root) that evenly divides it i.e. remainder is 0.

If there is a number M that evenly divides N, then N is not a prime number. Otherwise, N is a prime number.

## Java Implementation to Check Prime Number

Here is the implementation to check if a number is prime or not. We will utilize the earlier defined method to find out the prime numbers between 2 and √N.

```
Copy
package com.journaldev.examples;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
public class PrimeNumbers {
public static void main(String[] args) {
int number = 123456789;
boolean isPrime = checkPrime(number);
System.out.printf("%d is Prime: %b.\n", number, isPrime);
System.out.printf("%d is Prime: %b.\n", 3511, checkPrime(3511));
System.out.printf("%d is Prime: %b.\n", 123, checkPrime(123));
}
// a number is not prime if there is a prime number between 2 to the square root of
// num that evenly divides it
private static boolean checkPrime(int num) {
// get the prime numbers from 2 to square root of num.
List<Integer> primes = findPrimeNumbers(Double.valueOf(Math.sqrt(num)).intValue());
System.out.println(primes);
for (int i : primes) {
if (num % i == 0) {
System.out.printf("%d is divided by %d.\n", num, i);
return false;
}
}
return true;
}
}
```

## Leave a Reply