Printing prime number from 1 to 100

Question

This program is supposed to output the prime numbers between 1 and 100. Will anyone please explain me the flow the programme below? I am having difficulty in writing the progra

Answer 1

The number which is only divisible by itself and 1 is known as prime number. Here is the simplest version of the code for finding prime numbers between 1 to 100.

import java.io.*;
import java.util.*;
class solution
{
    public static void main(String args[])
    {
        Scanner scan = new Scanner(System.in);
        int n = 100;
        /*
            A prime number is a whole number greater than 1, whose only two whole-number factors are 1 and itself.
        */
        for(int i=2;i<=n;i++) // 1.So we are starting with initialization i = 2
        {
            int flag = 1;
            for(int j=2;j<=i/2;j++)  // 2.Try dividing the number by half check whether it divisible
            {
                if(i%j==0) // 3. If the number is divisible by other number ->Not a prime Number
                {
                    flag = 0;
                    break;
                }

            }
            if(flag==1) // 4. If the number is not divisible by any other numbers but only by itself and 1 -> prime no
            {
                System.out.print(i+" ");
            }
        }
    }
}
/*
    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
*/

Answer 2

    public static List<Integer> getPrimeNumbers(int from , int to) {
    List<Integer> list = new ArrayList<>();
    for (int i = from;i <= to; i++) {
        int count = 0;
        for (int num = i; num>=1;num--) {
            if(i%num == 0){
                count++;
            }
        }
        if(count ==2) {
            list.add(i);
        }
    }
    return list;
}

Answer 3

If you split the various parts out into their own methods with appropriate names it becomes a bit easier to understand:

for (int n = 1; n < 100; n++)
    if (isPrime(n))
        System.out.println(n);

private boolean isPrime(int n) {
    for (int f = 2; f < n; f++) {
        if (isFactor(f, n))
            return false;
    }
    return true;
}

private boolean isFactor(int factor, int number) {
    return number % factor == 0;
}

This is also an area where Java 8 streams can make things a bit clearer:

List<Integer> primes = IntStream.range(1, 100)
    .filter(this::hasNoFactors)
    .collect(Collectors.toList());

private boolean hasNoFactors(int number) {
    return IntStream.range(2, number)
        .noneMatch(f -> number % f == 0);
}

Also note that this is a horribly inefficient algorithm. You don't need to check every possible factor from 2 to n, just the primes. You can also take advantage of multi-processor machines:

List<Integer> primes = new ArrayList<>();
IntStream.range(2, 100)
    .filter(n -> primes.parallelStream().noneMatch(p -> n % p == 0))
    .forEach(primes::add);

Answer 4

How would you find a prime number with plain vanilla solution? If number is prime. It will not be a multiple of any number other than itself. So assume number is x. This number will not be divisible by any number when starting from 2 till x-1. Why start from 2 and not 1 because every number is divisible by 1. The above code is trying to replicate the same behavior. To find all primes between 1 till 99 (as per the loop):

1. From 2 till number from (outer loop - 1)
2. Try dividing the number and check if it's divisible. (remainder should be zero).
3. If true number is not prime. Else number is prime.