Is cube number (?)

[m1tch37]

I need to function that finds if a number is a cube number. I have this at the moment:

 

function is_cube($n) {
$root = pow($n,1/3);
if ( intval($root) == $root )
	return true;
return false;
}

 

The problem, that i cant work out, is:

 

is_cube( returns true (As it should)

 

However is_cube(343) returns false. 7³ is 343, so why on earth is it false?

 

What is even more confusing is in the case of 343, $root is float(7), but intval($root) is int(6). WHAT?

 

I dont get it...

 

Thanks

Mitch

[Ken2k7]

What if you try to use the is_int() function?

 

function is_cube($n) {
     return is_int(pow($n, 1/3));
}

[sasa]

try

<?php
function is_cube($n) {
$root = round(pow($n,1/3));
$root = (int) ($root * $root * $root);
return $root == $n;
}

echo is_cube(343);
?>

[Ken2k7]

try

<?php
function is_cube($n) {
$root = round(pow($n,1/3));
$root = (int) ($root * $root * $root);
return $root == $n;
}

echo is_cube(343);
?>

That MAY not be accurate.

[LemonInflux]

If speed's not a concern...

 

function is_cube($number) {
    if(!is_int($number)) {
        return;
    }

    for($i=0;$i<$number;$i++) {
        if(pow($i, 2) == sqrt($number)) {
            return true;
        } // Or just pow($i, 3) == $number. Whatever you want.
        return;
    }
}

 

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[LemonInflux]

function is_cube($number) {
    return is_int(pow($number, 1/3));
}

 

?

 

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[corbin]

pow() always returns a float, so is_int is useless .

 

 

You could do

 

function is_cube($number) {
    $p = pow($number, 1/3);
    return $p == ((int) $p);
}

 

 

It would be better to do it the way sasa did it though.  With this way, there's a tiny, tiny chance of floating point numbers screwing everything up.  (I don't think it would actually ever happen, but there's a tiny chance.)

 

 

On sasa's, contrary to what Ken said, it would always be accurate.  (Well, I can't think of when it wouldn't be accurate....  Could be wrong.)

[Ken2k7]

If speed's not a concern...

 

function is_cube($number) {
    if(!is_int($number)) {
        return;
    }

    for($i=0;$i<$number;$i++) {
        if(pow($i, 2) == sqrt($number)) {
            return true;
        } // Or just pow($i, 3) == $number. Whatever you want.
        return;
    }
}

 

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via FoxyTunes

Can you use a return statement like that in PHP?

[corbin]

Yeah, it just returns null.  I personally would return false there though, since the number would not be cubic.  False would make more sense to me, but....

[Ken2k7]

Yeah, it just returns null.  I personally would return false there though, since the number would not be cubic.  False would make more sense to me, but....

Ah cool. Thanks for the info mate.

 

m1tch37: also consider that cubic numbers have to be positive integers. I'm not sure if is_int() would cover that, but I don't think it does.

[corbin]

The cube root of -27 is -3.

 

 

For a = x^1/b where b is odd, if x < 0, a < 0.

 

 

You may be thinking of even numbered exponents.  If an exponent is even, and the number is negative, it will result in an unreal answer.

[DarkWater]

Not exactly, corbin.

 

-3 ^ 2 = 9

 

You might be thinking of square roots of negative numbers, which is an imaginary number.

 

sqrt(-36) = 6i

[Ken2k7]

Not exactly, corbin.

 

-3 ^ 2 = 9

 

You might be thinking of square roots of negative numbers, which is an imaginary number.

 

sqrt(-36) = 6i

No, he's right. My bad.

 

His equation was: a = b ^ 1 / c.

 

So your equation is the same as: 6i = -36 ^ 1 / 2.

[corbin]

Ah DarkWater you're right.  I worded that horribly.  I meant if the fraction was odd (as in odd denominator) and the number was negative, and the numerator wouldn't make the number positive, the answer would be unreal.

 

 

Essentially xroot(n) where x is even.

 

 

Or, in fraction form, which is what I was trying to get at:

 

a^(b/c)

 

Where a is negative, b is odd, and c is even.

[jordanwb]

What about this:

 

function isCubed ($value)
{
$third_root = (int)pow ($value, 1/3);
return ((int)pow ($third_root, 3)) == $value;
}

[corbin]

sasa's example earlier was essentially the same, except his(/her's?) used rounding instead of flooring.