Look for the GCD (greatest common divisor) of more than 2 integers?

问题

I already have a function that finds the GCD of 2 numbers.

function getGCDBetween($a, $b)
{
    while ($b != 0)
    {
        $m = $a % $b;
        $a = $b;
        $b = $m;
    }
    return $a;
}

But now, I would like to extend this function to find the GCD of N points. Any suggestion ?

回答1:

There is a more elegant way to do this :

// Recursive function to compute gcd (euclidian method)
function gcd ($a, $b) {
    return $b ? gcd($b, $a % $b) : $a;
}
// Then reduce any list of integer
echo array_reduce(array(42, 56, 28), 'gcd'); // === 14

If you want to work with floating points, use approximation :

function fgcd ($a, $b) {
    return $b > .01 ? fgcd($b, fmod($a, $b)) : $a; // using fmod
}
echo array_reduce(array(2.468, 3.7, 6.1699), 'fgcd'); // ~= 1.232

You can use a closure in PHP 5.3 :

$gcd = function ($a, $b) use (&$gcd) { return $b ? $gcd($b, $a % $b) : $a; };

回答2:

Had to do a bit of digging, but this is what I found.

The gcd of three numbers can be computed as gcd(a, b, c) = gcd(gcd(a, b), c), or in some different way by applying commutativity and associativity. This can be extended to any number of numbers.

You could use something like the following:

function multiGCD($nums)
{
    $gcd = getGCDBetween($nums[0], $nums[1]);

    for ($i = 2; $i < count($nums); $i++) { $gcd = getGCDBetween($gcd, $nums[$i]); }

    return $gcd;
}

回答3:

Take the GCD of numbers 1 and 2, and then the GCD of that and number 3, and so on.

回答4:

You can try

function gcd($a, $b) {
    if ($a == 0 || $b == 0)
        return abs(max(abs($a), abs($b)));
    $r = $a % $b;
    return ($r != 0) ? gcd($b, $r) : abs($b);
}

function gcd_array($array, $a = 0) {
    $b = array_pop($array);
    return ($b === null) ? (int) $a : gcd_array($array, gcd($a, $b));
}

echo gcd_array(array(50, 100, 150, 200, 400, 800, 1000)); // output 50

回答5:

I found a solution but it looks a bit ugly :

1) checking for every divisor of each integer

2) find the greater integer in every arrays

function getAllDivisorsOf($n)
{
    $sqrt = sqrt($n);
    $divisors = array (1, $n);
    for ($i = 2; ($i < $sqrt); $i++)
    {
        if (($n % $i) == 0)
        {
            $divisors[] = $i;
            $divisors[] = ($n / $i);
        }
    }
    if (($i * $i) == $n)
    {
        $divisors[] = $i;
    }
    sort($divisors);
    return $divisors;
}

function getGCDFromNumberSet(array $nArray)
{
    $allDivisors = array ();
    foreach ($nArray as $n)
    {
        $allDivisors[] = getAllDivisorsOf($n);
    }
    $allValues = array_unique(call_user_func_array('array_merge', $allDivisors));
    array_unshift($allDivisors, $allValues);
    $commons = call_user_func_array('array_intersect', $allDivisors);
    sort($commons);
    return end($commons);
}

echo getGCDFromNumberSet(array(50, 100, 150, 200, 400, 800, 1000)); // 50

Any better idea ?

回答6:

You can also use the gmp library:

<?php
    $gcd = gmp_gcd( '12', '21' );
    echo gmp_strval( $gcd );
?>

回答7:

You can store the numbers in an array and/or database and read from there. And then within a loop you can modular divide the array elements.

回答8:

I found this somewhere calculates gcd by recursion

function gcd(...$numbers) {
   if (count($numbers) > 2) {
        return array_reduce($numbers, 'gcd'); // use php's array reduce
  }

  $r = $numbers[0] % $numbers[1];
    return $r === 0 ? abs($numbers[1]) : gcd($numbers[1], $r);
}

来源：https://stackoverflow.com/questions/13828011/look-for-the-gcd-greatest-common-divisor-of-more-than-2-integers

标签

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greatest-common-divisor