Step by Step solution about How to resolve the algorithm Long primes step by step in the Java programming language

The given Java code is designed to find and count a specific type of prime numbers called "long primes" within a specified range. A long prime is a prime number for which the period of its reciprocal is equal to the prime number minus one. In other words, when you repeatedly divide 1 by the prime number and take the remainder, the sequence of remainders will eventually repeat itself after a certain number of steps. For a long prime, this number of steps is equal to the prime number minus one.

Here's a detailed breakdown of the code:

1. sieve Method (Sieve of Eratosthenes):

   • This method uses the Sieve of Eratosthenes algorithm to generate a list of prime numbers up to a specified limit.
   • It creates a boolean array c, where each element corresponds to a number from 0 to the limit. Initially, all elements are set to false.
   • The algorithm iterates through odd numbers starting from 3 and marks the multiples of each prime as true in the array.
   • The resulting list of primes is stored in a List called primes.

2. findPeriod Method:

   • This method calculates the period of the reciprocal of a given integer n.
   • It initializes r to 1 and period to 0.
   • It iterates through values from 1 to n-1, repeatedly multiplying r by 10 and taking the remainder when dividing by n.
   • The period is the number of iterations required for r to return to its initial value.

3. main Method:

   • Initializes an array numbers with various limits up to which long primes will be found.
   • Initializes an array totals to store the count of long primes for each limit in numbers.
   • Calls the sieve method to generate the list of primes up to 64000.
   • Iterates through the primes and checks if the period of the reciprocal of each prime is equal to the prime minus one. If so, it is considered a long prime and added to the longPrimes list.
   • Counts the number of long primes within each limit in numbers and stores the counts in the totals array.
   • Prints the long primes up to the first limit and the counts of long primes for all limits.

In summary, this code efficiently generates long primes within a specified range using the Sieve of Eratosthenes algorithm and calculates their periods to identify those that meet the definition of long primes. It then counts and reports the number of long primes found up to different limits.

import java.util.LinkedList;
import java.util.List;

public class LongPrimes
{
    private static void sieve(int limit, List<Integer> primes)
    {
        boolean[] c = new boolean[limit];
        for (int i = 0; i < limit; i++)
            c[i] = false;
        // No need to process even numbers
        int p = 3, n = 0;
        int p2 = p * p;
        while (p2 <= limit)
        {
            for (int i = p2; i <= limit; i += 2 * p)
                c[i] = true;
            do
                p += 2;
            while (c[p]);
            p2 = p * p;
        }
        for (int i = 3; i <= limit; i += 2)
            if (!c[i])
                primes.add(i);
    }

    // Finds the period of the reciprocal of n
    private static int findPeriod(int n)
    {
        int r = 1, period = 0;
        for (int i = 1; i < n; i++)
            r = (10 * r) % n;
        int rr = r;
        do
        {
            r = (10 * r) % n;
            ++period;
        }
        while (r != rr);
        return period;
    }
    
    public static void main(String[] args)
    {
        int[] numbers = new int[]{500, 1000, 2000, 4000, 8000, 16000, 32000, 64000};
        int[] totals = new int[numbers.length]; 
        List<Integer> primes = new LinkedList<Integer>();
        List<Integer> longPrimes = new LinkedList<Integer>();
        sieve(64000, primes);
        for (int prime : primes)
            if (findPeriod(prime) == prime - 1)
                longPrimes.add(prime);
        int count = 0, index = 0;
        for (int longPrime : longPrimes)
        {
            if (longPrime > numbers[index])
                totals[index++] = count;
            ++count;
        }
        totals[numbers.length - 1] = count;
        System.out.println("The long primes up to " + numbers[0] + " are:");
        System.out.println(longPrimes.subList(0, totals[0]));
        System.out.println();
        System.out.println("The number of long primes up to:");
        for (int i = 0; i <= 7; i++)
            System.out.printf("  %5d is %d\n", numbers[i], totals[i]);
    }
}