在Java中测试素数

Question

The goal of my program is to write a program that will prompt the user to input an integer. The program will read the integer and decide whether it is a prime or not. And if it is not a prime, it will tell the user all the divisors.

This is how far I have got, but when i tested it with numbers like 169 or 289, the program said that they were prime number. I understand that the problem is lying on this line :

int[] div = new int[] { 2,3,4,5,6,7,8,9};

I tried to do something like this :

for (int s = nr; s != 0 ; s--) { 
      if (nr%s == 0) {
      int[] div = new int[]{s}; }

But it didn't work. A little help or a right direction would help a lot. Thank You!

public static void main(String[] args){
    System.out.println("enter a number:");
    Scanner scanner = new Scanner (System.in);
    int nr = scanner.nextInt();
    int[] div = new int[] { 2,3,4,5,6,7,8,9}; 
    boolean prime = nr >= 2;
    int i = nr;
    for(int j = 0; j< div.length && prime && i> div[j]; j++)
        if(i % div[j] == 0)
            prime = false;

    if(prime){
        System.out.println(i + " is a prime");
    }else{
        System.out.print(i + " is divisible");
        for(int k = 2; k < i; k++)
            if(i % k == 0){
                System.out.print( k);
                System.out.print(",");}

                }

            } }

Answer 1

You're only trying numbers up to 9 for divisors. So anything that would have factors of all 10 or higher, such as 169 being 13 by 13, you will not find the 13s.

Instead of storing the divisors in an array, consider using an integer and counting upwards. That is to say, instead of div[j] just use j as your divisor, and don't let it stop at 10. Let it stop at the highest possible divisor (which is the square root of the number you're trying to find the prime of).

Answer 2

<edit> I should have looked a little closer... I guess it IS homework, so this answer is not really any good to you, eh? Well, I'm gonna leave it up in case the information here becomes useful to someone somewhere... </edit>

If you don't actually need to do the calculations yourself (if this isn't homework, that is), you could always opt to use Java's BigInteger class. For example:

public class Prime { 
    public static void main(String[] args){
        long start = System.nanoTime();
        System.out.println("enter a number:");
        Scanner scanner = new Scanner(System.in);
        BigInteger bigInt = scanner.nextBigInteger();
        boolean prime = bigInt.isProbablePrime(10);
        System.out.println(prime);
    }
}

As with most things in life, there are at least a couple caveats:

1. the probability of reporting a number's primality is not 100%, nor can it be (but we'll talk about that in a moment); and

2. the more accurate you want the test to be, the longer it will take.

This is detailed in Oracle's Official Documentation :

isProbablePrime

public boolean isProbablePrime (int certainty)

Returns true if this BigInteger is probably prime, false if it's definitely composite. If certainty is <= 0 , true is returned.

Parameters:

certainty - a measure of the uncertainty that the caller is willing to tolerate: if the call returns true the probability that this BigInteger is prime exceeds (1 - 1/(2^certainty)) . The execution time of this method is proportional to the value of this parameter.

Returns:

true if this BigInteger is probably prime, false if it's definitely composite.

I got curious about how accurate it could be and how much time a highly-accurate primality test would take, so I made a quick benchmark and ran some numbers.

I calculated the primality of every odd number under a million for each certainty value between 1 and 50 (0 or less always returns true).

Times are in milliseconds (though the times were obtained via a call to System.nanoTime() , and then rounded to the nearest millisecond).

Here are my results:

Certainty       Time(ms)        Chance of Correctness
1               1417            50.0%
2               932             75.0%
3               1064            87.5%
4               1063            93.75%
5               1183            96.875%
6               1195            98.4375%
7               1313            99.21875%
8               1308            99.609375%
9               1441            99.8046875%
10              1443            99.90234375%
11              1567            99.951171875%
12              1571            99.9755859375%
13              1690            99.98779296875%
14              1691            99.993896484375%
15              1817            99.9969482421875%
16              1822            99.99847412109375%
17              1944            99.99923706054688%
18              1941            99.99961853027344%
19              2069            99.99980926513672%
20              2073            99.99990463256836%
21              2197            99.99995231628418%
22              2200            99.99997615814209%
23              2324            99.99998807907104%
24              2340            99.99999403953552%
25              2453            99.99999701976776%
26              2465            99.99999850988388%
27              2647            99.99999925494194%
28              2626            99.99999962747097%
29              2715            99.99999981373549%
30              2710            99.99999990686774%
31              2844            99.99999995343387%
32              2818            99.99999997671694%
33              2950            99.99999998835847%
34              3074            99.99999999417923%
35              3121            99.99999999708962%
36              3167            99.99999999854481%
37              3295            99.9999999992724%
38              3302            99.9999999996362%
39              3334            99.9999999998181%
40              3360            99.99999999990905%
41              3493            99.99999999995453%
42              3583            99.99999999997726%
43              3663            99.99999999998863%
44              3599            99.99999999999432%
45              3783            99.99999999999716%
46              3816            99.99999999999858%
47              3964            99.99999999999929%
48              3898            99.99999999999964%
49              4029            99.99999999999982%
50              3969            99.99999999999991%
total: 124312

As you can see, even at the highest certainty-value that I tested, evaluating the primality of 500,000 numbers only took 4 seconds and each evaluation had a 99.99999999999991% chance of being correct

So if your intended application isn't incredibly performance-critical, you can use a (relatively) high number, like 25. The algorithm will correctly report a number as prime 99.99999701976776% of the time. Not recommended if you're a rocket scientist. And if you are , a rocket scientist, then I hope you find your solution elsewhere :).

Answer 3

The reason you are getting 169 as a prime is because you are only dividing by the values 2-9 and 169 is 13 squared so it isn't divisible by any number 2-9.

you can use this method http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes to find all the prime numbers between 1 and n and use that to decide if your number is prime.

Answer 4

A very easy and basic way to find if a number is prime is this:

Continue dividing n by each number between 2 and n^1/2 inclusive. If any of them divide evenly, then n is not prime because you found a factor. If n has no factors less than its square root, then n is prime. It is sufficient to check only for divisors less than or equal to n^1/2 because if n = a*b, then a and b can't both exceed the square root of n.

for(int i=2;i<=n^1/2;i++) 
  {
   if(num%i)==0 //number not prime
    else continue;
  }

  if none of the if statements in the loop were true, number is prime

Answer 5

只需使用Java内置的BigInteger.isProbablePrime（）库函数即可。虽然不是100％确定如何使用确定性参数。

BigInteger.valueOf(13).isProbablePrime(10000)