We know that all primes above 3 can be generated using:
6 * k + 1
6 * k - 1
However we all numbers generated from the above formulas are not prime.
For Example:
6 * 6 - 1 = 35 which is clearly divisible by 5.
To Eliminate such conditions, I used a Sieve Method and removing the numbers which are factors of the numbers generated from the above formula.
Using the facts:
A number is said to be prime if it has no prime factors.
To generate primes below 1000.
ArrayList<Integer> primes = new ArrayList<>();
primes.add(2);//explicitly add
primes.add(3);//2 and 3
int n = 1000;
for (int i = 1; i <= (n / 6) ; i++) {
//get all the numbers which can be generated by the formula
int prod6k = 6 * i;
primes.add(prod6k - 1);
primes.add(prod6k + 1);
}
for (int i = 0; i < primes.size(); i++) {
int k = primes.get(i);
//remove all the factors of the numbers generated by the formula
for(int j = k * k; j <= n; j += k)//changed to k * k from 2 * k, Thanks to DTing
{
int index = primes.indexOf(j);
if(index != -1)
primes.remove(index);
}
}
System.out.println(primes);
However, this method does generate the prime numbers correctly. This runs in a much faster way as we need not check for all the numbers which we do check in a Sieve.
My question is that am I missing any edge case? This would be a lot better but I never saw someone using this. Am I doing something wrong?
Can this approach be much more optimized?
Taking a boolean[] instead of an ArrayList is much faster.
int n = 100000000;
boolean[] primes = new boolean[n + 1];
for (int i = 0; i <= n; i++)
primes[i] = false;
primes[2] = primes[3] = true;
for (int i = 1; i <= n / 6; i++) {
int prod6k = 6 * i;
primes[prod6k + 1] = true;
primes[prod6k - 1] = true;
}
for (int i = 0; i <= n; i++) {
if (primes[i]) {
int k = i;
for (int j = k * k; j <= n && j > 0; j += k) {
primes[j] = false;
}
}
}
for (int i = 0; i <= n; i++)
if (primes[i])
System.out.print(i + " ");
You don't need to add all possible candidates to the array. You can create a Set to store all non primes.
Also you can start checking at k * k, rather than 2 * k
public void primesTo1000() {
Set<Integer> notPrimes = new HashSet<>();
ArrayList<Integer> primes = new ArrayList<>();
primes.add(2);//explicitly add
primes.add(3);//2 and 3
for (int i = 1; i < (1000 / 6); i++) {
handlePossiblePrime(6 * i - 1, primes, notPrimes);
handlePossiblePrime(6 * i + 1, primes, notPrimes);
}
System.out.println(primes);
}
public void handlePossiblePrime(
int k, List<Integer> primes, Set<Integer> notPrimes) {
if (!notPrimes.contains(k)) {
primes.add(k);
for (int j = k * k; j <= 1000; j += k) {
notPrimes.add(j);
}
}
}
untested code, check corners
Here is a bit packing version of the sieve as suggested in the answer referenced by @Will Ness. Rather than return the nth prime, this version returns a list of primes to n:
public List<Integer> primesTo(int n) {
List<Integer> primes = new ArrayList<>();
if (n > 1) {
int limit = (n - 3) >> 1;
int[] sieve = new int[(limit >> 5) + 1];
for (int i = 0; i <= (int) (Math.sqrt(n) - 3) >> 1; i++)
if ((sieve[i >> 5] & (1 << (i & 31))) == 0) {
int p = i + i + 3;
for (int j = (p * p - 3) >> 1; j <= limit; j += p)
sieve[j >> 5] |= 1 << (j & 31);
}
primes.add(2);
for (int i = 0; i <= limit; i++)
if ((sieve[i >> 5] & (1 << (i & 31))) == 0)
primes.add(i + i + 3);
}
return primes;
}
There seems to be a bug in your updated code that uses a boolean array (it is not returning all the primes).
public static List<Integer> booleanSieve(int n) {
boolean[] primes = new boolean[n + 5];
for (int i = 0; i <= n; i++)
primes[i] = false;
primes[2] = primes[3] = true;
for (int i = 1; i <= n / 6; i++) {
int prod6k = 6 * i;
primes[prod6k + 1] = true;
primes[prod6k - 1] = true;
}
for (int i = 0; i <= n; i++) {
if (primes[i]) {
int k = i;
for (int j = k * k; j <= n && j > 0; j += k) {
primes[j] = false;
}
}
}
List<Integer> primesList = new ArrayList<>();
for (int i = 0; i <= n; i++)
if (primes[i])
primesList.add(i);
return primesList;
}
public static List<Integer> bitPacking(int n) {
List<Integer> primes = new ArrayList<>();
if (n > 1) {
int limit = (n - 3) >> 1;
int[] sieve = new int[(limit >> 5) + 1];
for (int i = 0; i <= (int) (Math.sqrt(n) - 3) >> 1; i++)
if ((sieve[i >> 5] & (1 << (i & 31))) == 0) {
int p = i + i + 3;
for (int j = (p * p - 3) >> 1; j <= limit; j += p)
sieve[j >> 5] |= 1 << (j & 31);
}
primes.add(2);
for (int i = 0; i <= limit; i++)
if ((sieve[i >> 5] & (1 << (i & 31))) == 0)
primes.add(i + i + 3);
}
return primes;
}
public static void main(String... args) {
Executor executor = Executors.newSingleThreadExecutor();
executor.execute(() -> {
for (int i = 0; i < 10; i++) {
int n = (int) Math.pow(10, i);
Stopwatch timer = Stopwatch.createUnstarted();
timer.start();
List<Integer> result = booleanSieve(n);
timer.stop();
System.out.println(result.size() + "\tBoolean: " + timer);
}
for (int i = 0; i < 10; i++) {
int n = (int) Math.pow(10, i);
Stopwatch timer = Stopwatch.createUnstarted();
timer.start();
List<Integer> result = bitPacking(n);
timer.stop();
System.out.println(result.size() + "\tBitPacking: " + timer);
}
});
}
0 Boolean: 38.51 μs
4 Boolean: 45.77 μs
25 Boolean: 31.56 μs
168 Boolean: 227.1 μs
1229 Boolean: 1.395 ms
9592 Boolean: 4.289 ms
78491 Boolean: 25.96 ms
664116 Boolean: 133.5 ms
5717622 Boolean: 3.216 s
46707218 Boolean: 32.18 s
0 BitPacking: 117.0 μs
4 BitPacking: 11.25 μs
25 BitPacking: 11.53 μs
168 BitPacking: 70.03 μs
1229 BitPacking: 471.8 μs
9592 BitPacking: 3.701 ms
78498 BitPacking: 9.651 ms
664579 BitPacking: 43.43 ms
5761455 BitPacking: 1.483 s
50847534 BitPacking: 17.71 s