Very simple prime number test - I think I'm no

问题:

I am practicing past exam papers for a basic java exam, and I am finding it difficult to make a for loop work for testing whether a number is prime. I don't want to complicate it by adding efficiency measures for larger numbers, just something that would at least work for 2 digit numbers.

At the moment it always returns false even if n IS a prime number.

I think my problem is that I am getting something wrong with the for loop itself and where to put the "return true;" and "return false;"... I'm sure it's a really basic mistake I'm making...

public boolean isPrime(int n) {
    int i;
    for (i = 2; i <= n; i++) {
        if (n % i == 0) {
            return false;
        }
    }
    return true;
}

The reason I couldn't find help elsewhere on stackoverflow is because similar questions were asking for a more complicated implementation to have a more efficient way of doing it.

回答1:

Your for loop has a little problem. It should be: -

for (i = 2; i < n; i++)  // replace `i <= n` with `i < n`

Of course you don't want to check the remainder when n is divided by n. It will always give you 1.

In fact, you can even reduce the number of iterations by changing the condition to: - i <= n / 2. Since n can't be divided by a number greater than n / 2, except when we consider n, which we don't have to consider at all.

So, you can change your for loop to: -

for (i = 2; i <= n / 2; i++)  

回答2:

You can stop much earlier and skip through the loop faster with:

public boolean isPrime(long n) {
    // fast even test.
    if(n > 2 && (n & 1) == 0)
       return false;
    // only odd factors need to be tested up to n^0.5
    for(int i = 3; i * i <= n; i += 2)
        if (n % i == 0) 
            return false;
    return true;
}

回答3:

Error is i<=n

for (i = 2; i<n; i++){

回答4:

You should write i < n, because the last iteration step will give you true.

回答5:

public class PrimeNumberCheck {
  private static int maxNumberToCheck = 100;

  public PrimeNumberCheck() {
  }

    public static void main(String[] args) {
      PrimeNumberCheck primeNumberCheck = new PrimeNumberCheck();

      for(int ii=0;ii < maxNumberToCheck; ii++) {
        boolean isPrimeNumber = primeNumberCheck.isPrime(ii);

      System.out.println(ii + " is " + (isPrimeNumber == true ? "prime." : "not prime."));
    }
  }

  private boolean isPrime(int numberToCheck) {    
    boolean isPrime = true;

    if(numberToCheck < 2) {
      isPrime = false;
    }

    for(int ii=2;ii<numberToCheck;ii++) {
      if(numberToCheck%ii == 0) {
        isPrime = false;
        break;
      }
    }

    return isPrime;
  }
}

回答6:

With this code number divisible by 3 will be skipped the for loop code initialization.
For loop iteration will also skip multiples of 3.

private static boolean isPrime(int n) {

    if ((n > 2 && (n & 1) == 0) // check is it even
       || n <= 1  //check for -ve
       || (n > 3 && (n % 3 ==  0))) {  //check for 3 divisiable
            return false;
    }

    int maxLookup = (int) Math.sqrt(n);
    for (int i = 3; (i+2) <= maxLookup; i = i + 6) {
        if (n % (i+2) == 0 || n % (i+4) == 0) {
            return false;
        }
    }
    return true;
}

回答7:

You could also use some simple Math property for this in your for loop.

A number 'n' will be a prime number if and only if it is divisible by itself or 1. If a number is not a prime number it will have two factors:

n = a * b

you can use the for loop to check till sqrt of the number 'n' instead of going all the way to 'n'. As in if 'a' and 'b' both are greater than the sqrt of the number 'n', a*b would be greater than 'n'. So at least one of the factors must be less than or equal to the square root.

so your loop would be something like below:

for(int i=2; i<=Math.sqrt(n); i++)

By doing this you would drastically reduce the run time complexity of the code. I think it would come down to O(n/2).

回答8:

One of the fastest way is looping only till the square root of n.

  private static boolean isPrime(int n){
        int square = (int)Math.ceil((Math.sqrt(n)));//find the square root
        HashSet<Integer> nos = new HashSet<>(); 
        for(int i=1;i<=square;i++){
            if(n%i==0){
                if(n/i==i){
                    nos.add(i);
                }else{
                    nos.add(i);
                    int rem = n/i;
                    nos.add(rem);
                }
            }
        }
        return nos.size()==2;//if contains 1 and n then prime
    }

回答9:

You are checking i<=n.So when i==n, you will get 0 only and it will return false always.Try i<=(n/2).No need to check until i<n.

回答10:

The mentioned above algorithm treats 1 as prime though it is not. Hence here is the solution.

static boolean isPrime(int n) {
  int perfect_modulo = 0;
  boolean prime = false;

  for ( int i = 1; i <=  n; i++ ) {
    if ( n % i == 0 ) {
      perfect_modulo += 1;
    }
  }
  if ( perfect_modulo == 2 ) {
    prime = true;
  }

  return prime;
}

回答11:

Well, the for loop has some problem. Here's the code,

public static boolean checkPrimeNUmber(int number) 
{ 
   if(number <= 1) 
   { 
      return false; 
   } 
   for(int a = 2; a < Math.sqrt(number); a++) 
   { 
      if(number % a == 0) 
      { 
         return false; 
      } 
   } 
   return true;
}

回答12:

Doing it the Java 8 way is nicer and cleaner

    private static boolean isPrimeA(final int number) {
        return IntStream
               .rangeClosed(2, number/2)
               .noneMatch(i -> number%i == 0);
    }