A much easier way to find perfect squares:
This method uses integer roots to find perfect squares instead of integer squares to find integer roots.
<?
# initial value
$i = 0;
#set any limit you want
$setSomeLimit = 10000;
# do it! .... do it! ... do it!
for ($n = 1; $n < sqrt($setSomeLimit; $n++)
{
// Find the square
$s= $n * $n;
print "<br>perfect_square_root is of $s is $n";
}
?>
sqrt
(PHP 4, PHP 5)
sqrt — 平方根
説明
float sqrt ( float $arg )arg の平方根を返します。
パラメータ
- arg
処理する引数。
返り値
arg の平方根を返します。
例
例 1115. sqrt() の例
<?php
// 小数点以下の精度は、precision ディレクティブの設定に依存します
echo sqrt(9); // 3
echo sqrt(10); // 3.16227766 ...
?>
参考
| pow() |
sqrt
30-Mar-2006 05:03
php NOSPAM at primezero dot com
13-Jul-2005 11:15
13-Jul-2005 11:15
creating perfect squares... (no decimals)
<?
# initial value
$i = 0;
#set any limit you want
$setSomeLimit = 10000;
# do it! .... do it! ... do it!
do {
# find the square root
$sqRoot = sqrt($i);
# not a decimal? that it is a perfect square! ^_^
if ( ereg ("^[0-9]{1,12}$", $sqRoot)) {
# suppose that the_square_root = teh_square_root
print "<br>perfect_square_root is of $i is $sqRoot";
}
$i++; # next!!
# keeps running until it reaches the limit
} while ($i < $setSomeLimit);
?>
btharper1221 at gmail dot com
16-May-2005 12:48
16-May-2005 12:48
this is something you can use if you need an exact answer to a root problem, this includes an echo outside of the function to show how it works, it will return it such as
$ret[0] is the number still part of the root (this is the exactness part of it)
$ret[1] is the number of roots and
$ret[2] is 1 or 0, 1 being that there should be an i following the number of roots (imaginary)
<?php
function newrt($val,$rt){
$i = 0;
if(($rt % 2 == 0) && ($val < 1)){
$i = 1;
$val = (-$val);
}
$c = 1;
if(($rt % 2 != 0) && ($val < 0)){
$c = -$c;
$val = -$val;
}
for($d = 2; pow($d,$rt) <= abs($val); $d++){
if($val % (pow($d,$rt)) == 0){
$val = ($val/(pow($d,$rt)));
$c = ($c*$d);
$d--;
}
}
$ret[0] = $val;
$ret[1] = $c;
$ret[2] = $i;
return $ret;
}//end function newrt
$ret = newrt($num,$num2);
if($ret[0] != 1){
if($ret[2] == 0) echo $ret[1]." roots of ".$ret[0];
if($ret[2] == 1) echo $ret[1]."<b><i>i</i></b> roots of ".$ret[0];
}elseif($ret[0] == 1){
if($ret[2] == 0) echo $ret[1];
if($ret[2] == 1) echo "$ret[1]<b><i>i</i></b>";
}
?>
chris DOT rutledge AT gmail DOT com
31-Mar-2005 07:24
31-Mar-2005 07:24
Just a note to say you can take the square root of a negative number - it returns an imaginary number.
One might do it like this
//take initial value of $x
$abs_x = abs($x);
$answer = sqrt($abs_x);
echo $answer;
if ($x < 0) echo"<b><i>i</i></b>";
jouhni at web dot de
18-Feb-2005 04:46
18-Feb-2005 04:46
To get any root of a number your can use the pow() function:
pow(8, 1/3)
which gives you the third root of eight.
Jouhni
php user
14-May-2004 04:25
14-May-2004 04:25
How to calculate triangle's area if you haven't height measure.
You need only sides' measure.
it would have to work.
(formula of erone)
$s1 = side 1
$s2 = side 2
$s3 = side 3
<?
function area($s1,$s2,$s3)
{
$calc1 = ($s1+$s2+$s3)/2;
$calc2 = $p*($p-$s1)*($p-$s2)*($p-$s3);
$result = round(sqrt($p),2);
return $result;
}
?>
example of I use:
<?
//This will make to return the area of a triangle with sides of 3,4,5
$area = area(3,4,5);
echo $area;
?>
bishop
18-Jul-2003 02:09
18-Jul-2003 02:09
Compute a triangle's area:
function triangleArea($a, $b /* ... */) {
if (func_num_args() === 3) {
$s = .5 * ($a + $b + ($c = func_get_arg(2)));
return sqrt($s * ($s-$a) * ($s-$b) * ($s-$c));
} else {
return (.5 * $a * $b);
}
}
With 2 args, parameters are triangle's base and height. With 3 args, parameters are the lengths of each side. Argument order is not important.
Either way, you get the area.
tcn at pandora dot be
14-Nov-2002 05:37
14-Nov-2002 05:37
Checking if the square you want to take from a number is valid or not (you can't take the root of a negative number)
if ($i<0)
{
echo "Error! The number you specified can't be rooted! It's negative!
}
else
{
echo "The square root of your number is ".sqrt($i);
}
morvandiumNOSPAM at yahoo dot com
05-Nov-2001 09:44
05-Nov-2001 09:44
I find this useful for any grid system where distances need to be computed (maps and graphs, mostly).
Just use the distance between points formula (pseudo-code): sqrt((x2-x1)^2+(y2-y1)^2) to find the distance between two points.