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# Fastest algorithm for primality test

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## Tags :

Fastest algorithm for primality test , Fastest , algorithm , primality , test ,

## What would be the fastest method to test for primality in Java?

Tag : java , By : user157138
Date : March 29 2020, 07:55 AM
it helps some times I am trying to find the fastest way to check whether a given number is prime or not (in Java). Below are several primality testing methods I came up with. Is there any better way than the second implementation(isPrime2)? , Here's another way:
``````boolean isPrime(long n) {
if(n < 2) return false;
if(n == 2 || n == 3) return true;
if(n%2 == 0 || n%3 == 0) return false;
long sqrtN = (long)Math.sqrt(n)+1;
for(long i = 6L; i <= sqrtN; i += 6) {
if(n%(i-1) == 0 || n%(i+1) == 0) return false;
}
return true;
}
``````
``````import java.math.BigInteger;
import java.util.BitSet;

public class Main {

static BitSet primes;

static boolean isPrime(int p) {
return p > 0 && (p == 2 || (p%2 != 0 && primes.get(p/2)));
}

static void generatePrimesUpTo(int n) {
primes = new BitSet(n/2);

for(int i = 0; i < primes.size(); i++) {
primes.set(i, true);
}

primes.set(0, false);
int stop = (int)Math.sqrt(n) + 1;
int percentageDone = 0, previousPercentageDone = 0;
System.out.println("generating primes...");
long start = System.currentTimeMillis();

for(int i = 0; i <= stop; i++) {
previousPercentageDone = percentageDone;
percentageDone = (int)((i + 1.0) / (stop / 100.0));

if(percentageDone <= 100 && percentageDone != previousPercentageDone) {
System.out.println(percentageDone + "%");
}

if(primes.get(i)) {
int number = (i * 2) + 1;

for(int p = number * 2; p < n; p += number) {
if(p < 0) break; // overflow
if(p%2 == 0) continue;
primes.set(p/2, false);
}
}
}
long elapsed = System.currentTimeMillis() - start;
System.out.println("finished generating primes ~" + (elapsed/1000) + " seconds");
}

private static void test(final int certainty, final int n) {
int percentageDone = 0, previousPercentageDone = 0;
long start = System.currentTimeMillis();
System.out.println("testing isProbablePrime(" + certainty + ") from 1 to " + n);
for(int i = 1; i < n; i++) {
previousPercentageDone = percentageDone;
percentageDone = (int)((i + 1.0) / (n / 100.0));
if(percentageDone <= 100 && percentageDone != previousPercentageDone) {
System.out.println(percentageDone + "%");
}
BigInteger bigInt = new BigInteger(String.valueOf(i));
boolean bigIntSays = bigInt.isProbablePrime(certainty);
if(isPrime(i) != bigIntSays) {
System.out.println("ERROR: isProbablePrime(" + certainty + ") returns "
+ bigIntSays + " for i=" + i + " while it " + (isPrime(i) ? "is" : "isn't" ) +
" a prime");
return;
}
}
long elapsed = System.currentTimeMillis() - start;
System.out.println("finished testing in ~" + ((elapsed/1000)/60) +
" minutes, no false positive or false negative found for isProbablePrime(" + certainty + ")");
}

public static void main(String[] args) {
int certainty = Integer.parseInt(args[0]);
int n = Integer.MAX_VALUE;
generatePrimesUpTo(n);
test(certainty, n);
}
}
``````
``````java -Xmx1024m -cp . Main 15
``````

## Fastest primality test

Tag : algorithm , By : Thomas Plunkett
Date : March 29 2020, 07:55 AM
To fix the issue you can do The only deterministic, polynomial-time algorithm for primality testing I know of is the AKS primality test (http://en.wikipedia.org/wiki/AKS_primality_test). However, there are a lot of very good randomized primality tests that are fast and have extremely good probability of success. They usually work by finding whether the number is composite with exponentially good probability, so they'll either report that the number is composite or will require you to say "maybe" with very good confidence.

## What is the fastest deterministic primality test for numbers in the range 2^1024 to 2^4096?

Tag : algorithm , By : thatotherguy
Date : March 29 2020, 07:55 AM
PRIMALITY TESTING by Richard P. Brent: http://cs.anu.edu.au/student/comp4600/lectures/comp4600_primality.pdf

## Why naive primality test algorithm is not polynomial

Tag : algorithm , By : John Phipps
Date : March 29 2020, 07:55 AM
wish of those help The input size is typically measured in bits. To represent the number n the input size would be log2(n). The primitive primality test is linear in n, but exponential in log2(n).

## Primality Test algorithm fails

Tag : python , By : Giles
Date : March 29 2020, 07:55 AM
To fix the issue you can do You have your else condition inside the loop. At any point in time, it'll only check for one value...
Modifying your for loop to print out the number it is checking for:
``````for i in range (2, number):
print (i)
if number % i == 0:
print ("No")
break
else:
print ("Yes")
``````
``````2
Yes
3
No
``````
``````flag = False
for i in range (2, number):
if number % i == 0:
print ("No")
flag = True
break
if (!flag)
print("Yes")
``````