RSA encryption algorithm which is commonly used in secure commerce web sites, is based on the fact that it is easy to take two big prime numbers and multiply them, while it is computationaly difficult to do the opposite (factorize a big number into two prime numbers) on the current hardware.
Check a number is Prime: Java Coding Problem
Carvia Tech  May 24, 2019  3 min read  31 views  Java Coding Challenges
Prime numbers have great importance in computer science world. They are used in Cryptography & Security (RSA for example), Generating hashcodes, random number generators, etc. In this article we will learn if a given number is prime or not. The solution presented here may not be best optimized, but it should be sufficient from interview point of view.
Definition of Prime
A prime number is a whole number greater than 1 whose only factors are 1 and itself. The only even prime number is 2.
The first few prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.
Trivial implementation
In very first naive attempt, we can check every integer from 2 to itself (exclusive) and test whether it divides evenly or not.
boolean isPrime(long n) {
for (int i = 2; i < n; i++) {
if (n % i == 0)
return false;
}
return true;
}
Better implementation (check till square root)
There is lot of scope for the optimization in the previous code sample. Instead of checking till n (exclusive), we can check till square root of n.
\$i <= sqrt(n)\$
This is because if we list out all the factors of a number, the square root will always be in the middle.
Second optimization that we can do is to check only odd numbers, since every even number is divisible by 2 itself. We can check once if number is divisible by 2, if so it is not a prime number.
So here is the improved version:
boolean isPrimeV2(long n) {
if (n % 2 == 0) return false; (1)
for (int i = 3; i * i <= n; i += 2) { (2)
if (n % i == 0)
return false;
}
return true;
}
1  check if n is a multiple of 2 
2  just check the odds till sqrt of n 
Similar implementation using Java 8 version may look like below:
boolean isPrime(long n) {
if (n % 2 == 0) return false;
return n > 1 && IntStream.rangeClosed(2, (int) Math.sqrt(n))
.noneMatch(divisor > n % divisor == 0);
}
That’s all for this article.
Other considerations

There are many efficient algorithms for finding a prime number which are out of scope for this article, as we want to focus more on the coding skills rather than pure mathematics.

The implementations mentioned above does not utilize multicore hardware, we can use parallel processing to speedup the program.
Using Java 8 Parallel Streams for performance improvement
Using parallel streams can significantly improve the performance of Prime number implementation on multicore hardware.
public class PrimeUtils {
public static void main(String[] args) throws ExecutionException, InterruptedException {
ForkJoinPool forkJoinPool = new ForkJoinPool();
List<Integer> primeList = forkJoinPool.submit(() > new PrimeUtils().collectPrimes(1000000)).get();
System.out.println("primeList = " + primeList);
forkJoinPool.shutdown();
}
List<Integer> collectPrimes(int max) {
return IntStream.range(2, max).parallel().filter(this::isPrime).boxed().collect(Collectors.toList());
}
boolean isPrime(long n) {
if (n % 2 == 0) return false;
return n > 1 && IntStream.rangeClosed(2, (int) Math.sqrt(n)).noneMatch(divisor > n % divisor == 0);
}
}
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