PHP Get amount of square numbers up to number

Question

I have a function, that, given two integers A and B, return the number of whole squares within the interval [A..B] (both ends included)

For example, given $A = 4 and $B = 17, the function should return 3, because there are three squares of integers in the interval [4..17]. namely 4 = 2*, 9 = 3* and 14 = 4*

How would I get a number of square numbers up to the number?

Congratulations for having a function. Do you have a question? — Mark Baker, Apr 30 '17 at 11:12
http://stackoverflow.com/questions/295579/fastest-way-to-determine-if-an-integers-square-root-is-an-integer — u_mulder, Apr 30 '17 at 11:22
You should provide an approach including your relevant snippet of source code. Otherwise it smells like: 'Please do the coding job for me'. Consider a loop testing if the square root is integer. — Quasimodo's clone, Apr 30 '17 at 11:22
I managed to figure out the solution with Quasimodo's comment function solution($A, $B){ $P = 0; while ($A < $B){ if(gmp_perfect_square($A)){ $P++; } $A++; } return $P; } — Cleggy1012, Apr 30 '17 at 12:06
@Cleggy Click **EDIT** below your question if you need to update it. — Jocelyn, Apr 30 '17 at 13:11

Jocelyn · Answer 1 · 2017-05-03T17:30:11.157

This function loops through all integer numbers between $start and $end. If the square root of the number is equal to its integer part, then $count is increased by 1.

function countSquares($start, $end)
{
    $count = 0;
    for($n=$start;$n<=$end;$n++)
    {
        if(pow($n, 0.5) == intval(pow($n, 0.5)))
        {
            //echo "$n<br>";
            $count++;
        }
    }
    return $count;
}

echo countSquares(4, 17);

However this first function is quite slow when used with large numbers.
This other function does the job much more quickly, and the code is also much shorter.
The number of integer square roots between $start and $end is obtained by subtracting the number of integer square roots between 0 and $end to the number of integer square roots between 0 and $start-1. (I use $start-1 because your interval includes the starting number)

function countSquares2($start, $end)
{
    return floor(pow($end, 0.5)) - floor(pow($start-1, 0.5));
}

echo countSquares2(1000000, 10000000);

Really nice logic in `countSquares2`! To become an even better answer, you could explain more mathematical details, why `floor(pow($float_num, 0.5))` is the number of squares from 0. It is not necessarily obvious to people with a lower mathematical background. Explanation could even start at the point, that the power of 0.5 actually is the square root. — Quasimodo's clone, May 03 '17 at 08:01
PS: I didn't test it in PHP. What is the behavior when it is invoked with negative arguments? Consider to sanitize inputs - also when end is less than start. — Quasimodo's clone, May 03 '17 at 08:12

score 0 · Answer 2 · answered Mar 23 '18 at 21:48

0

Or you could do it by using sqrt function:

function getSquaresInRange($a, $b){
    return floor(sqrt($b)) - ceil(sqrt($a)) + 1;
}

answered Mar 23 '18 at 21:48

Rwakos

71
3

PHP Get amount of square numbers up to number

2 Answers2