Rune - 2 months ago 5x

PHP Question

We are looking for the Nth root in PHP. We need to do this with a very large number, and the windows calculator returns 2. With the following code we are getting 1. Does anybody have an idea how this works?

`echo bcpow(18446744073709551616, 1/64);`

Answer

Well it seems that PHP and the BC lib has some limits, and after searching on the internet i found this interesting article/code:

So you should use this function:

```
<?php
function NRoot($num, $n) {
if ($n<1) return 0; // we want positive exponents
if ($num<=0) return 0; // we want positive numbers
if ($num<2) return 1; // n-th root of 1 or 2 give 1
// g is our guess number
$g=2;
// while (g^n < num) g=g*2
while (bccomp(bcpow($g,$n),$num)==-1) {
$g=bcmul($g,"2");
}
// if (g^n==num) num is a power of 2, we're lucky, end of job
if (bccomp(bcpow($g,$n),$num)==0) {
return $g;
}
// if we're here num wasn't a power of 2 :(
$og=$g; // og means original guess and here is our upper bound
$g=bcdiv($g,"2"); // g is set to be our lower bound
$step=bcdiv(bcsub($og,$g),"2"); // step is the half of upper bound - lower bound
$g=bcadd($g,$step); // we start at lower bound + step , basically in the middle of our interval
// while step!=1
while (bccomp($step,"1")==1) {
$guess=bcpow($g,$n);
$step=bcdiv($step,"2");
$comp=bccomp($guess,$num); // compare our guess with real number
if ($comp==-1) { // if guess is lower we add the new step
$g=bcadd($g,$step);
} else if ($comp==1) { // if guess is higher we sub the new step
$g=bcsub($g,$step);
} else { // if guess is exactly the num we're done, we return the value
return $g;
}
}
// whatever happened, g is the closest guess we can make so return it
return $g;
}
echo NRoot("18446744073709551616","64");
?>
```

Hope this was helpful ...

Source (Stackoverflow)

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