搜索素数的方法

Question

I need to build a method that calculates prime numbers.

I'm testing it and it's giving me the wrong answers and I don't know how to solve it! For example, it returns true to 98262679 instead of false . Where is my mistake?

public static boolean itsPrime(int nbTest){   // Tests for prime numbers method

        boolean prime = false;

        if (nbTest <= 1){
             return false;
        } else if (nbTest == 2){     // Two is prime   
            return true;
        } else if ((nbTest != 2) && (nbTest % 2 == 0)){       // Evens except number 2
            return false;                                     // are not prime
        } else if (nbTest % 2 != 0){    // For all remaining odds
            for(int i = 3; i <= Math.sqrt(nbTest); i = i+2){ 
                if (nbTest % i == 0){
                    prime = false;
                } else {
                    prime = true;
                }
            }   
        }

        return prime; 
    }

I'm learning Java and my professor asked us to construct the method itsPrime was based on this:

    For the first subtask, write a function `itsPrime` that takes an `int` 
            as argument and returns a `boolean`, `true` if the argument integer is prime 
        and `false` otherwise.

                To test whether an integer x is prime, one can proceed as follows:
                all integers less than or equal to 1 are not prime;
                2 is prime;
                all other even integers are not prime;
                for all remaining integers (obviously odd), search for a divisor:

                loop from 3 to the square root of the integer x (The square root of x can 
    be computed as ‘‘Math.sqrt(x)'' in Java.); if the remainder of the integer 
division of x by the loop index is zero, then x is not prime;
 if all remainders were non-zero at the end of the loop, then x is prime; 

Answer 1

You should stop once you here:

if (nbTest % i == 0){
   return false;
}

Answer 2

Delete the prime variable (it is an unnecessary step, see below).

Change the for loop:

for(int i = 3; i <= Math.sqrt(nbTest); i = i+2){ 
    if (nbTest % i == 0){
        prime = false;
    } else {
        prime = true;
    }
}

To this:

for(int i = 3; i <= Math.sqrt(nbTest); i = i+2)
    if (nbTest % i == 0)
        return false;

This simply breaks out of the function early if it isn't prime (if a factor has been found).

Answer 3

I know the answer is there somewhere in all the answers above, but I think that each require an explanation.

Here's a summary of all the enhancements you could make to your code:

1) Don't declare a boolean to return, since you are already returning true or false throughout your code. Remove this line from your code (call it [1]):

boolean prime = false;

You'll see why after you've fixed the rest of your function. Comment it out if desired, for now.

2) In your second else if , let's call it [2], you have:

else if ((nbTest != 2) && (nbTest % 2 == 0)){
    return false;
}

You already checked if nbTest is 2 in your first else if , so you don't need to check if it's not 2 again. If it entered the first if else , your function will return true . When a function returns, it is done. It returns the value to the caller and the rest of the code is not executed.

Thus, you may replace that second if else , [2], with:

else if (nbTest % 2 == 0) { // all other even integers are not prime
    return false;
}

3) If you enter third else if , this means that the rest of the code above already executed, and it either returned, or the program continued.

You may replace that third else if (nbTest % 2 != 0){ for:

else {

4) This is the one error that you really have to make your function return the wrong answer (call this snippet [4]):

if (nbTest % i == 0){
    prime = false;

If you find that the number you are testing is divisible (ie the remainder is zero), you are done. You definitely know that it is not prime.

You may replace this code, [4], with:

if(nbTest % counter == 0) {
    return false;
}

Thus, returning false. It is not a number. And the function does not keep executing. Your error was continuing execution after the function finds out that your input is not prime.

Finally, you may leave your return true at the end of the function body. If the function never returned from the previous tests, or from the loop, it has to be a prime number. Remember that first line I told you to remove? The boolean declaration? Since you never return a value from a variable, you just return true or false , you don't need that [1] line.

As an extra, a good read on finding prime numbers, which you might want to share with your professor:

https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes

Answer 4

You shouldn't keep testing for primality once prime has been set to false for a number.

Replace:

if (nbTest % i == 0){
    prime = false;

with:

if (nbTest % i == 0){
    return false;

And the last test is useless, you can just keep a basic else:

else if (nbTest % 2 != 0){ => else {

Answer 5

Your for loop is not correct.

A prime number is a number which is divisible by 1 and itself only. So, if a number A is divisible by any other number except 1 and itself, then A is a nonprime.

Just replace your for loop with the one below

for(int i = 3; i <= Math.sqrt(nbTest); i = i+2) {
    if (nbTest % i == 0)
      return false;
}

Once it is found that nbTest is divisible by some number, there is not point in continuing the loop. Return `nbTest is nonprime, then and there.