N以内的素数计算(Java代码)

列出小于N的所有素数

普通计算方式, 校验每个数字

优化的几处:

• 判断是否整除时, 除数使用小于自身的平方根的素数
• 大于3的素数, 都在6的整数倍两侧, 即 6m - 1 和 6m + 1 

public class DemoPrime {
    private int[] primes;
    private int max;
    private int pos;
    private int total;

    public DemoPrime(int max) {
        this.max = max;
        int length = max / 3;
        primes = new int[length];
        pos = 0;
        total = 0;
    }

    private void put(int prime) {
        primes[pos] = prime;
        if (pos < primes.length - 1) {
            pos++;
        } else {
            throw new RuntimeException("Length exceed");
        }
    }

    private boolean isPrime(int num) {
        int limit = (int)Math.sqrt(num);
        for (int i = 0; i < pos; i++) {
            if (primes[i] > limit) {
                break;
            }
            total++;
            if (num % primes[i] == 0) return false;
        }
        return true;
    }

    public void calculate() {
        put(2);
        put(3);
        int val = 1;
        for (int i = 0; val <= max - 6;) {
            val += 4;
            if (isPrime(val)) {
                put(val);
            }
            val += 2;
            if (isPrime(val)) {
                put(val);
            }
        }
        System.out.println("Tried: " + total);
    }

    public void print() {
        System.out.println("Total: " + pos);
        /*for (int i = 0; i < pos; i++) {
            System.out.println(primes[i]);
        }*/
    }

    public static void main(String[] args) {
        DemoPrime dp = new DemoPrime(10000000);
        dp.calculate();
        dp.print();
    }
}

.使用数组填充的方式

一次性创建大小为N的int数组的方式, 在每得到一个素数时, 将其整数倍的下标(除自身以外)的元素都置位, 并且只需要遍历到N的平方根处. 最后未置位的元素即为素数, 在过程中可以统计素数个数. 这种方法比前一种效率高一个数量级.

public class DemoPrime2 {
    private int[] cells;
    private int max;
    private int total;

    public DemoPrime2(int max) {
        this.max = max;
        cells = new int[max];
        total = max;
    }

    private void put(int prime) {
        int i = prime + prime;
        while (i < max) {
            if (cells[i] == 0) {
                cells[i] = 1;
                total--;
            }
            i += prime;
        }
    }

    public void calculate() {
        total -= 2; // Exclude 0 and 1
        put(2);
        put(3);
        int limit = (int)Math.sqrt(max);
        for (int i = 4; i <= limit; i++) {
            if (cells[i] == 0) {
                put(i);
            }
        }
    }

    public void print() {
        System.out.println("Total: " + total);
        /*for (int i = 2; i < max; i++) {
            if (cells[i] == 0) {
                System.out.println(i);
            }
        }*/
    }

    public static void main(String[] args) throws InterruptedException {
        DemoPrime2 dp = new DemoPrime2(10000000);
        Thread.sleep(1000L);
        long ts = System.currentTimeMillis();
        dp.calculate();
        dp.print();
        long elapse = System.currentTimeMillis() - ts;
        System.out.println("Time: " + elapse);
    }
}

.

判断大数是否为素数

Miller-Rabin算法

对于大数的素性判断，目前Miller-Rabin算法应用最广泛。一般底数仍然是随机选取，但当待测数不太大时，选择测试底数就有一些技巧了。比如，如果被测数小于4 759 123 141，那么只需要测试三个底数2, 7和61就足够了。当然，你测试的越多，正确的范围肯定也越大。如果你每次都用前7个素数(2, 3, 5, 7, 11, 13和17)进行测试，所有不超过341 550 071 728 320的数都是正确的。如果选用2, 3, 7, 61和24251作为底数，那么10^16内唯一的强伪素数为46 856 248 255 981。这样的一些结论使得Miller-Rabin算法在OI中非常实用。通常认为，Miller-Rabin素性测试的正确率可以令人接受，随机选取k个底数进行测试算法的失误率大概为4^(-k)。

Miller-Rabin算法是一个RP算法。RP是时间复杂度的一种，主要针对判定性问题。一个算法是RP算法表明它可以在多项式的时间里完成，对于答案为否定的情形能够准确做出判断，但同时它也有可能把对的判成错的（错误概率不能超过1/2）。RP算法是基于随机化的，因此多次运行该算法可以降低错误率。还有其它的素性测试算法也是概率型的，比如Solovay-Strassen算法.

AKS算法

AKS最关键的重要性在于它是第一个被发表的一般的、多项式的、确定性的和无仰赖的素数判定算法。先前的算法至多达到了其中三点，但从未达到全部四个。

• AKS算法可以被用于检测任何一般的给定数字是否为素数。很多已知的高速判定算法只适用于满足特定条件的素数。例如，卢卡斯-莱默检验法仅对梅森素数适用，而Pépin测试仅对费马数适用。
• 算法的最长运行时间可以被表为一个目标数字长度的多项式。ECPP和APR能够判断一个给定数字是否为素数，但无法对所有输入给出多项式时间范围。
• 算法可以确定性地判断一个给定数字是否为素数。随机测试算法，例如米勒-拉宾检验和Baillie–PSW，可以在多项式时间内对给定数字进行校验，但只能给出概率性的结果。
• AKS算法并未“仰赖”任何未证明猜想。一个反例是确定性米勒检验：该算法可以在多项式时间内对所有输入给出确定性结果，但其正确性却基于尚未被证明的广义黎曼猜想。

AKS算法的时间复杂度是 O(log(n)), 比Miller-Rabin要慢

/***************************************************************************
 * Team
 **************
 * Arijit Banerjee
 * Suchit Maindola
 * Srikanth Manikarnike
 *
 **************
 * This is am implementation of Agrawal–Kayal–Saxena primality test in java.
 *
 **************
 * The algorithm is -
 * 1. l <- log n
 * 2. for i<-2 to l
 *      a. if an is a power fo l
 *              return COMPOSITE
 * 3. r <- 2
 * 4. while r < n
 *      a. if gcd( r, n) != 1
 *              return COMPSITE
 *      b. if sieve marked n as PRIME
 *              q <- largest factor (r-1)
 *              o < - r-1 / q
 *              k <- 4*sqrt(r) * l
 *              if q > k and n <= r
 *                      return PRIME
 *      c. x = 2
 *      d. for a <- 1 to k
 *              if (x + a) ^n !=  x^n + mod (x^r - 1, n)
 *                      return COMPOSITE
 *      e. return PRIME
 */

public class DemoAKS {
    private int log;
    private boolean sieveArray[];
    private int SIEVE_ERATOS_SIZE = 100000000;

    /* aks constructor */
    public DemoAKS(BigInteger input) {
        sieveEratos();

        boolean result = checkIsPrime(input);

        if (result) {
            System.out.println("1");
        } else {
            System.out.println("0");
        }
    }

    /* function to check if a given number is prime or not */
    public boolean checkIsPrime(BigInteger n) {
        BigInteger lowR, powOf, x, leftH, rightH, fm, aBigNum;
        int totR, quot, tm, aCounter, aLimit, divisor;
        log = (int) logBigNum(n);
        if (findPower(n, log)) {
            return false;
        }
        lowR = new BigInteger("2");
        x = lowR;
        totR = lowR.intValue();
        for (lowR = new BigInteger("2");
             lowR.compareTo(n) < 0;
             lowR = lowR.add(BigInteger.ONE)) {
            if ((lowR.gcd(n)).compareTo(BigInteger.ONE) != 0) {
                return false;
            }
            totR = lowR.intValue();
            if (checkIsSievePrime(totR)) {
                quot = largestFactor(totR - 1);
                divisor = (int) (totR - 1) / quot;
                tm = (int) (4 * (Math.sqrt(totR)) * log);
                powOf = mPower(n, new BigInteger("" + divisor), lowR);
                if (quot >= tm && (powOf.compareTo(BigInteger.ONE)) != 0) {
                    break;
                }
            }
        }
        fm = (mPower(x, lowR, n)).subtract(BigInteger.ONE);
        aLimit = (int) (2 * Math.sqrt(totR) * log);
        for (aCounter = 1; aCounter < aLimit; aCounter++) {
            aBigNum = new BigInteger("" + aCounter);
            leftH = (mPower(x.subtract(aBigNum), n, n)).mod(n);
            rightH = (mPower(x, n, n).subtract(aBigNum)).mod(n);
            if (leftH.compareTo(rightH) != 0) return false;
        }
        return true;
    }

    /* function that computes the log of a big number*/
    public double logBigNum(BigInteger bNum) {
        String str;
        int len;
        double num1, num2;
        str = "." + bNum.toString();
        len = str.length() - 1;
        num1 = Double.parseDouble(str);
        num2 = Math.log10(num1) + len;
        return num2;
    }

    /*function that computes the log of a big number input in string format*/
    public double logBigNum(String str) {
        String s;
        int len;
        double num1, num2;
        len = str.length();
        s = "." + str;
        num1 = Double.parseDouble(s);
        num2 = Math.log10(num1) + len;
        return num2;
    }

    /* function to compute the largest factor of a number */
    public int largestFactor(int num) {
        int i;
        i = num;
        if (i == 1) return i;
        while (i > 1) {
            while (sieveArray[i] == true) {
                i--;
            }
            if (num % i == 0) {
                return i;
            }
            i--;
        }
        return num;
    }

    /*function given a and b, computes if a is power of b */
    public boolean findPowerOf(BigInteger bNum, int val) {
        int l;
        double len;
        BigInteger low, high, mid, res;
        low = new BigInteger("10");
        high = new BigInteger("10");
        len = (bNum.toString().length()) / val;
        l = (int) Math.ceil(len);
        low = low.pow(l - 1);
        high = high.pow(l).subtract(BigInteger.ONE);
        while (low.compareTo(high) <= 0) {
            mid = low.add(high);
            mid = mid.divide(new BigInteger("2"));
            res = mid.pow(val);
            if (res.compareTo(bNum) < 0) {
                low = mid.add(BigInteger.ONE);
            } else if (res.compareTo(bNum) > 0) {
                high = mid.subtract(BigInteger.ONE);
            } else if (res.compareTo(bNum) == 0) {
                return true;
            }
        }
        return false;
    }

    /* creates a sieve array that maintains a table for COMPOSITE-ness
     * or possibly PRIME state for all values less than SIEVE_ERATOS_SIZE
     */
    public boolean checkIsSievePrime(int val) {
        if (sieveArray[val] == false) {
            return true;
        } else {
            return false;
        }
    }

    long mPower(long x, long y, long n) {
        long m, p, z;
        m = y;
        p = 1;
        z = x;
        while (m > 0) {
            while (m % 2 == 0) {
                m = (long) m / 2;
                z = (z * z) % n;
            }
            m = m - 1;
            p = (p * z) % n;
        }
        return p;
    }

    /* function, given a and b computes if a is a power of b */
    boolean findPower(BigInteger n, int l) {
        int i;
        for (i = 2; i < l; i++) {
            if (findPowerOf(n, i)) {
                return true;
            }
        }
        return false;
    }

    BigInteger mPower(BigInteger x, BigInteger y, BigInteger n) {
        BigInteger m, p, z, two;
        m = y;
        p = BigInteger.ONE;
        z = x;
        two = new BigInteger("2");
        while (m.compareTo(BigInteger.ZERO) > 0) {
            while (((m.mod(two)).compareTo(BigInteger.ZERO)) == 0) {
                m = m.divide(two);
                z = (z.multiply(z)).mod(n);
            }
            m = m.subtract(BigInteger.ONE);
            p = (p.multiply(z)).mod(n);
        }
        return p;
    }

    /* array to populate sieve array
     * the sieve array looks like this
     *
     *  y index -> 0 1 2 3 4 5 6 ... n
     *  x index    1
     *     |       2   T - T - T ...
     *     /      3     T - - T ...
     *             4       T - - ...
     *             .         T - ...
     *             .           T ...
     *             n
     *
     *
     *
     *
     */
    public void sieveEratos() {
        int i, j;
        sieveArray = new boolean[SIEVE_ERATOS_SIZE + 1];
        sieveArray[1] = true;
        for (i = 2; i * i <= SIEVE_ERATOS_SIZE; i++) {
            if (!sieveArray[i]) {
                for (j = i * i; j <= SIEVE_ERATOS_SIZE; j += i) {
                    sieveArray[j] = true;
                }
            }
        }
    }

    public static void main(String[] args) {
        new DemoAKS(new BigInteger("100000217"));
    }
}