PHP - check if number is prime

Question

I\'m trying to create a function which checks whether the number is prime or not. BUT I want this function to echo to the user \'prime\' or \'NOT prime\' - and that\'s where my

Answer 1

in 54 59 bytes:

function is_prime($n){for($i=$n;--$i&&$n%$i;);return$i==1;}

loops $i down from $n-1 until it finds a divisor of $n; $n is prime if that divisor is 1.

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add 10 bytes for much better performance:

function is_prime($n){for($i=$n**.5|1;$i&&$n%$i--;);return!$i&&$n>1;}

loops $i from (approx.) sqrt($n) to 1 looking for a divisor with a post-decrement on $i.
If the divisor is 1, $i will be 0 at the end, and !$i gives true.

This solition uses a trick: For $n=2 or 3, $i will be initialized to 1 → loop exits in first iteration. For larger even square roots ($n**.5|0), |1 serves as +1. For odd square roots, +1 is not needed because: if $n is divisible by root+1, it is also divisible by 2. Unfortunately, this can cost a lot of iterations; so you better

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add another 7 bytes for even better performance:

function is_prime($n){for($i=~-$n**.5|0;$i&&$n%$i--;);return!$i&$n>2|$n==2;}

$n=2 needs a special case here: inital $i=2 divides $n=2 → final $i=1 → returns false.

Adding 1 to $n instead of the square root is enough to avoid failures; but:

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I did not count the iterations but only tested time consumption; and that only in TiO instead of a controlled environment. The difference between the last two versions was smaller than the deviation between several runs. Significant test results may be added later.

Answer 2

The break is only breaking the for loop, use return; instead. it will exit the function

Answer 3

First, don't make a mistake here:

for ($i = 2; $i < $num; $i++)

and then:

if ($num % $i == 0) return false;

2 % 2 equals 0 and then 2 will result as NOT prime.

Next, you don't have to check even numbers, so after you check if $num == 2, better performance is:

for ($i = 3; $i < $num/2; $i += 2)

Notice $num/2 - you don't have to check beyond that point. And even better is:

for ($i = 3; $i*$i <= $num; $i += 2)

This is because when you check for division with 2 and 3 all other NON prime numbers are product of two (or more) prime numbers (e.g. 49 = 7*7 or 55 = 5*11). In the worst case scenario your $i pointer would reach a square root of $num (like in 49 = 7*7). That's why you check until $i*$i <= $num.

Answer 4

You can find prime numbers and non prime numbers from 1 to your limit and its count.

public function prime_checker($count){
  $counter=0;
  $no_prime=0;
  for($i=2 ; $i<=$count; $i++ ){
    for($j=2 ; $j<$i ; $j++ ){
      if($i % $j == 0){
        echo $i.'is not prime<br/>';
        $no_prime=$i;
        break;
      }
    }
    if($i != $no_prime){
      $prime_numbers[$counter]=$i;
      $counter=$counter+1;
    }    
  }
  echo '<br/>'.$counter.'prime numbers<br/><br/>';
  for($i=0 ; $i<$counter ; $i++ ){
    echo $prime_numbers[$i].' is  prime<br/>';
  }
}

Answer 5

Put else part otherwise it will always return Prime

    for ($i = 2; $i < $num; $i++)
{
    if ($num % $i == 0)
    {
        echo 'NOT prime';
        break;
    }
    echo 'Prime';
}

Answer 6

try this

class IsPrime
{       
function check($num)
{
    for ($i = 2; $i < $num; $i++)
    {
        if ($num % $i == 0) 
        {
            echo 'NOT prime';
            break;
        }
        else
        {
            echo 'Prime';
        }
    }
    //echo 'Prime';           
}       
}
  $x = new IsPrime();
  $x->check(3);