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Fixes: #2724 Add description for Sieve of Eratosthenes algorithm. Refactor a little bit #2725

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74 changes: 53 additions & 21 deletions Others/SieveOfEratosthenes.java
Original file line number Diff line number Diff line change
@@ -1,44 +1,76 @@
package Others;

/** @author Varun Upadhyay (https://github.com/varunu28) */
import java.util.Arrays;

/**
* Sieve of Eratosthenes is an ancient algorithm for finding all prime numbers up to any given limit.
* It does so by iteratively marking as composite (i.e., not prime) the multiples of each prime,
* starting with the first prime number, 2.
* The multiples of a given prime are generated as a sequence of numbers starting from that prime,
* with constant difference between them that is equal to that prime.
* This is the sieve's key distinction from using trial division to sequentially test each
* candidate number for divisibility by each prime.
* Once all the multiples of each discovered prime have been marked as composites, the remaining
* unmarked numbers are primes.
* <p>
* Poetry about Sieve of Eratosthenes:
* <p><i>Sift the Two's and Sift the Three's:</i></p>
* <p><i>The Sieve of Eratosthenes.</i></p>
* <p><i>When the multiples sublime,</i></p>
* <p><i>The numbers that remain are Prime.</i></p>
*
* @see <a href="https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes">Wiki</a>
*/
public class SieveOfEratosthenes {

/**
* This method implements the Sieve of Eratosthenes Algorithm
*
* @param n The number till which we have to check for prime Prints all the prime numbers till n
* @param n The number till which we have to check for prime Prints all the prime numbers till n.
* Should be more than 1.
* @return array of all prime numbers between 0 to n
*/
public static void findPrimesTillN(int n) {
int[] arr = new int[n + 1];

for (int i = 0; i <= n; i++) {
arr[i] = 1;
}
public static int[] findPrimesTill(int n) {
// Create array where index is number and value is flag - is that number a prime or not.
// size of array is n + 1 cause in Java array indexes starts with 0
Type[] numbers = new Type[n + 1];

arr[0] = arr[1] = 0;
// Start with assumption that all numbers except 0 and 1 are primes.
Arrays.fill(numbers, Type.PRIME);
numbers[0] = numbers[1] = Type.NOT_PRIME;

for (int i = 2; i <= Math.sqrt(n); i++) {
if (arr[i] == 1) {
double cap = Math.sqrt(n);
// Main algorithm: mark all numbers which are multiples of some other values as not prime
for (int i = 2; i <= cap; i++) {
if (numbers[i] == Type.PRIME) {
for (int j = 2; i * j <= n; j++) {
arr[i * j] = 0;
numbers[i * j] = Type.NOT_PRIME;
}
}
}

//Write all primes to result array
int primesCount = (int) Arrays.stream(numbers)
.filter(element -> element == Type.PRIME)
.count();
int[] primes = new int[primesCount];

int primeIndex = 0;
for (int i = 0; i < n + 1; i++) {
if (arr[i] == 1) {
System.out.print(i + " ");
if(numbers[i] == Type.PRIME) {
primes[primeIndex++] = i;
}
}

System.out.println();
return primes;
}

private enum Type {
PRIME, NOT_PRIME
}

// Driver Program
public static void main(String[] args) {
int n = 100;

// Prints 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
findPrimesTillN(n);
System.out.println("Searching for all primes from zero to " + n);
int[] primes = findPrimesTill(n);
System.out.println("Found: " + Arrays.toString(primes));
}
}