# All Primesby Isai Damier, Android Engineer @ Google

```/***************************************************************************
* Author: Isai Damier
* Title: All Primes
* Project: geekviewpoint
* Package: algorithms
*
* Statement:
*  Given a number n, return all prime numbers less than or equal to n.
*
* Sample Input: 35
* Sample Output: {2,3,5,7,11,13,17,19,23,29,31}
*
* Time Complexity of Solution:
*   Best = Average = Worst = O(Time) = O(n log log n),
*   where n is the max number to test. You can verify this as
*   O(Time) = n/p1+n/p2+... = n*sum(1/p) for all prime p < n^0.5
*
* Approach:
*   Arithmetic review:
*   0) 0 and 1 are not prime numbers.
*   1) The first prime number is 2.
*   2) A number is prime if it is only divisible by itself and by 1
*   3) If y is the square root of x (i.e. y^2 = x); then if x is not
*      divisible by any number between 2 and y inclusive, then x is prime.
*
*   This is an "innocent until proven guilty" algorithm that capitalizes on
*   the above arithmetic facts to find all primes between 2 and n. In the
*   innocent until proven guilty paradigm, all candidates start out as
*   winners. Then through a rigorous process, the algorithm winnows the
*   faulty candidates until the elite is left. Known as the sieve of
*   Eratosthenes, there are a number of ways to implement the set of
*   arithmetic rules above into an algorithm. The following is easy to
*   understand.
*
*   0] Declare a bit vector and initialize all the keys from 0 to
*      x with the value 'true' -- making them all primes.
*   1] For keys 0 and 1, set the values to false
*   2] For each number n between 2 and x, set all multiples of n to false.
*   3] Group together all keys whose values are still true and
*      return that group.
*
* There are a number of ways to improve the time complexity of this
* program, such as considering only odd numbers as opposed to all
* numbers between 2 and n.
**************************************************************************/
public List<Integer> allPrimes(int max) {
if (2 > max) {
return null;
}
boolean[] primes = new boolean[max + 1];
Arrays.fill(primes, true);
primes = primes = false;
for (int i = 2, sqrt = (int) Math.sqrt(max); i <= sqrt; i++) {
if (primes[i]) {
for (int k = 2; k * i <= max; k++) {
if (primes[k * i]) {
primes[k * i] = false;
}
}
}
}
List<Integer> result = new ArrayList<Integer>();
for (int i = 2; i <= max; i++) {
if (primes[i]) {
}
}
return result;
}```
```import java.util.Arrays;
import java.util.List;
import org.junit.Test;
import static org.junit.Assert.*;

public class NumbersTest {

/**
* Test of allPrimes method, of class Numbers.
*/
@Test
public void testAllPrimes() {
System.out.println("allPrimes");
int max = 0;
Numbers primes = new Numbers();
List<Integer> expected = Arrays.asList(
new Integer[] {2,3,5,7,11,13,17,19,23,29});
assertTrue(expected.equals(primes.allPrimes(30)));

expected = Arrays.asList(
new Integer[] {2,3,5,7,11,13,17,19,23,29,31});
assertTrue(expected.equals(primes.allPrimes(31)));

expected = Arrays.asList(
new Integer[] {2,3,5,7,11,13,17,19,23,29,31,37});
assertTrue(expected.equals(primes.allPrimes(40)));
}
}```