查找给定范围内的最高质数

Question

I need to find the highest prime number in a given range.
Here is my code which works for 0-100 but if I give 0-125 it is showing prime number as 125.

<?php
    $flag=0;
    $b=125;
    for($i=$b;$i>=0;$i--)
    {
        if($i%2!=0)
        {
            for($b=3;$b<10;$b++)
            {
                if($flag==0)
                {
                    echo('<br>');
                    if($i%$b!=0)
                    {
                        echo('highest prime number is'.$i);
                        $flag=1;
                        break;
                    }
                    elseif ($i%$b==0)
                    {
                        break;
                    }
                }
            }
        }
    }
?>

In the above code I have taken the range from 0-125

Answer 1

gmp_nextprime()

<?php
$start = 125;
$stop = 0;
for($x=$start;$x>=$stop;$x--){
    if(($prime = gmp_intval(gmp_nextprime($x)))<$start){
        echo 'The highest prime is '.$prime;
        break;
    }
}?>

Answer 2

thanks for your reply Samuel Cook ..but i need without using 'gmp_nextprime()' function.. iamgetting message 'Call to undefined function gmp_intval() in C:\\wamp\\www\\highprime.php on line 5' – user1659450 16 mins ago

Since you are having difficultly getting gmp functions to work then you can use

Example 1 : None GMP Function

$range = range(125, 0); // create the range
foreach ( $range as $v ) {
    if (isPrime($v)) {
        printf('highest prime number is %d', $v);
        break;
    }
}

If you are able to get gmp working then you can use gmp_prob_prime Function Used

Example 1 : Using gmp function

foreach ( $range as $v ) {
    if (gmp_prob_prime($v) == 2) {
        printf('highest prime number is %d', $v);
        break;
    }
}

Output

highest prime number is 113

Functions Used

function isPrime($num) {
    if ($num == 1)
        return false;

    if ($num == 2)
        return true;

    if ($num % 2 == 0) {
        return false;
    }

    for($i = 3; $i <= ceil(sqrt($num)); $i = $i + 2) {
        if ($num % $i == 0)
            return false;
    }
    return true;
}

Answer 3

Yup the problem is algorithmic...

1) You'll need to check up till sqrt($b) ie 11 in this case

2) The $flag logic is kinda messed up, no use changing the flag then breaking out right after...

<?php
        $flag=0;
        $b=125;
        $sq=sqrt($b); 

        for($i=$b;$i>=0;$i--)
        {

            if($i%2!=0)
            {
                    for($b=3;$b<=$sq;$b++)
                    {
                            if($i%$b!=0)
                            {
                                $flag=1;
                            }
                            elseif($i%$b==0)
                            {
                                $flag=0;
                                break;
                            }
                    }
            if($flag==1){ 
                echo('highest prime number is '.$i);  
                break;
            }
        }
    }
?>

Answer 4

The simplest method you can use is dividing a number starting from 3 with an Array filled by a 2, if modulus (%) gives you an an answer else than 0, it's prime. Then after you got a working code think what can I make better ?

• You could say I want to ignore even numbers! so why iterate over than at all ? just use i = i + 2 (and start by using 3 as seed prime)
• I divide through too many numbers! do you really need to check 21%11, 21%13, 21%17, 21%19 ? They are always smaller than 2 so lets ignore primes > half of it
• Optimize more ? how would limit using a root as maximum in your code affect it ? and Why ?
• Even more ? why couldn't I eliminate all non-primes beforehand and check what's left ?

Try to apply these on your code, or just google a working version.