在 Java 中测试素性的最快方法是什么? [英] What would be the fastest method to test for primality in Java?

问题描述

我试图找到检查给定数字是否为质数的最快方法(在 Java 中).以下是我想出的几种素性测试方法.有没有比第二种实现(isPrime2)更好的方法?

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)?

public class Prime {
    public static boolean isPrime1(int n) {
        if (n <= 1) {
            return false;
        }
        if (n == 2) {
            return true;
        }
        for (int i = 2; i <= Math.sqrt(n) + 1; i++) {
            if (n % i == 0) {
                return false;
            }
        }
        return true;
    }
    public static boolean isPrime2(int n) {
        if (n <= 1) {
            return false;
        }
        if (n == 2) {
            return true;
        }
        if (n % 2 == 0) {
            return false;
        }
        for (int i = 3; i <= Math.sqrt(n) + 1; i = i + 2) {
            if (n % i == 0) {
                return false;
            }
        }
        return true;
    }
}

public class PrimeTest {
    public PrimeTest() {
    }
 
    @Test
    public void testIsPrime() throws IllegalArgumentException, IllegalAccessException, InvocationTargetException {
 
        Prime prime = new Prime();
        TreeMap<Long, String> methodMap = new TreeMap<Long, String>();
 
        for (Method method : Prime.class.getDeclaredMethods()) {
 
            long startTime = System.currentTimeMillis();
 
            int primeCount = 0;
            for (int i = 0; i < 1000000; i++) {
                if ((Boolean) method.invoke(prime, i)) {
                    primeCount++;
                }
            }
 
            long endTime = System.currentTimeMillis();
 
            Assert.assertEquals(method.getName() + " failed ", 78498, primeCount);
            methodMap.put(endTime - startTime, method.getName());
        }
 
 
        for (Entry<Long, String> entry : methodMap.entrySet()) {
            System.out.println(entry.getValue() + " " + entry.getKey() + " Milli seconds ");
        }
    }
}

推荐答案

这是另一种方式:

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;
}

和 BigInteger 的 isProbablePrime(...) 对所有 32 位 int 都有效.

编辑

请注意，isProbablePrime(certainty) 并不总是产生正确的答案.当确定性偏低时，就会产生误报，正如评论中提到的@dimo414.

Note that isProbablePrime(certainty) does not always produce the correct answer. When the certainty is on the low side, it produces false positives, as @dimo414 mentioned in the comments.

不幸的是，我找不到声称 isProbablePrime(certainty) 对所有(32 位)int 有效的来源(如果有足够的确定性！).

Unfortunately, I could not find the source that claimed isProbablePrime(certainty) is valid for all (32-bit) int's (given enough certainty!).

所以我进行了几次测试.我创建了一个大小为 Integer.MAX_VALUE/2 的 BitSet 表示所有奇数，并使用素数筛来查找 1..Integer.MAX_VALUE 范围内的所有素数.然后我从 i=1..Integer.MAX_VALUE 循环来测试每个 new BigInteger(String.valueOf(i)).isProbablePrime(certainty) == isPrime(i).

So I performed a couple of tests. I created a BitSet of size Integer.MAX_VALUE/2 representing all uneven numbers and used a prime sieve to find all primes in the range 1..Integer.MAX_VALUE. I then looped from i=1..Integer.MAX_VALUE to test that every new BigInteger(String.valueOf(i)).isProbablePrime(certainty) == isPrime(i).

对于确定性 5 和 10，isProbablePrime(...) 沿线产生了误报.但是使用 isProbablePrime(15)，没有测试失败.

For certainty 5 and 10, isProbablePrime(...) produced false positives along the line. But with isProbablePrime(15), no test failed.

这是我的测试设备:

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);
    }
}

我是这样做的:

java -Xmx1024m -cp . Main 15

在我的机器上生成素数需要大约 30 秒.而1..Integer.MAX_VALUE中所有i的实际测试耗时约2小时15分钟.

The generating of the primes took ~30 sec on my machine. And the actual test of all i in 1..Integer.MAX_VALUE took around 2 hours and 15 minutes.

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