How do I find perfect square from a set of numbers in Java?

Question

So I'm trying to find out the perfect squares within a set of numbers. I declared the necessary variables, added a for loop, added a sqroot = Math.sqrt(num) and a print method to list the numbers. What I can't figure out is how can I make the program pick out the perfect squares within the range of numbers, and find the average of them?

This is for an assignment I'm working on for a class and I've been stuck on this for a while now. I'm also rather new to Java so sorry if this is a stupid question. The code is below:


public class Test {

    public static void main(String[] args) {
        int num;
        double sqroot = 0;
        int sumPsq = 0; //sum variable
        int psq = 0; //counter for perfect squares
        double avg = 0;

        for(num = 101; num <= 149; num += 2){
           sqroot = Math.sqrt(num);

            if(sqroot*sqroot == num){ //Condition to find perfect squares
                psq += 1; //counting perfect squares
                sumPsq = sumPsq + num; //Find the sum of all the perfect squares

                System.out.println(num); //Print out the perfect squares
            }
        }
        avg = 1.0 * sumPsq/psq;
        System.out.println(avg);
    }

}

This is just a piece of the code from the entire assignment, so if you need more of it then I'm more than willing to provide it. Thanks!

You should probably either put the perfect squares you find in a list, or just sum them and keep a counter of how many there are. — chrylis -cautiouslyoptimistic-, Sep 24 '19 at 19:06
Your code seems to be doing exactly what you said you needed it to be doing, though that's really not the best way of testing for perfect-squaredness.... — Roddy of the Frozen Peas, Sep 24 '19 at 19:13
https://stackoverflow.com/questions/34056609/how-to-check-if-an-integer-is-a-perfect-square — Roddy of the Frozen Peas, Sep 24 '19 at 19:17
lets start analyzing the problem: what is the exact definition of a perfect square? — user85421, Sep 24 '19 at 19:17
@RoddyoftheFrozenPeas If this code gets the answer right, it's only by accident, because OP is never rounding the square root. — Dawood ibn Kareem, Sep 24 '19 at 19:24

Liviu Stirb · Accepted Answer · 2019-09-25T08:17:04.273

3

A perfect square is a number that can be expressed as the product of two equal integers so it must be an int after the sqrt. If you do a sqroot*sqroot == num you are just checking that sqrt is working correctly. Some number won't pass the check because of this but usually, you will get way more numbers than you want.

So what you need to do is just checking that after the sqrt the result is an int:

if (sqrootd % 1 == 0) { ... }

An optimization you can do is to check the number is an integer before sqrt. Other than that, your code looks fine to me

edited Sep 25 '19 at 08:17

answered Sep 24 '19 at 19:13

Liviu Stirb

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what is `variable`? – Roddy of the Frozen Peas Sep 24 '19 at 19:14
@RoddyoftheFrozenPeas sorry, is sqroot – Liviu Stirb Sep 24 '19 at 19:15
@CarlosHeuberger I got a downvote and an upvote, so there must be something wrong I guess – Liviu Stirb Sep 24 '19 at 19:20
no upvote, no downvote - probably it was revoked, that is, it must have **been** something wrong (the smiley at my comment should suggest it was not that serious [:-)) – user85421 Sep 24 '19 at 19:21
must be that then – Liviu Stirb Sep 24 '19 at 19:22
You can also use `if ((int) sqroot == sqroot)` ! Is it a good idea? Probably not. (Especially for doubles higher than int max) – Nexevis Sep 24 '19 at 19:23
The posted link (rounding, precision lost, ...) is hardly the reason for it not working: OP is using `double` **and** not checking that it is an integer (reason why I asked him/her for exact definition) – user85421 Sep 24 '19 at 19:31
@CarlosHeuberger I thought of that reason too but this will also fix it. and in theory, this might be an issue – Liviu Stirb Sep 24 '19 at 19:33
one of many many solutions.... (despite the result (in this case) is almost obvious straight-forward `132.5` [:-) ) – user85421 Sep 24 '19 at 19:36
The output is now 121 so I think this pretty much did it, but could you clarify what you mean by checking the number is an integer, for future reference? – Anonymous Sep 24 '19 at 20:32

score 1 · Answer 2 · answered Oct 15 '21 at 05:19

I used the following mathematical formula to find the perfect square:

1 + 3 + 5 + 7 .... = n ^ 2

for example: 1 + 3 + 5 = 9 = 3 ^ 2

and sample code:

   int i = 1;

    while (num > 0) {
        num = num - i;
        i = i + 2;
    }

    return num == 0;

Jignesh · Answer 3 · 2019-09-24T19:51:21.737

0

The best way to check if square root is an integer you will need below condition

if ((sqroot - Math.floor(sqroot)) == 0){

instead of

if(sqroot*sqroot == num){

The Math.sqrt() method finds the square root of the given number and the floor() method finds the largest (closest to positive infinity) floating-point value that less than or equal to the argument (here square root value returned by sqrt() method) and is equal to a mathematical integer.

Then we calculate the difference between these two to check if the difference is zero. For a perfect square number this difference is always zero. The reason is: square root of perfect square number is integer.

reference

edited Sep 24 '19 at 19:51

answered Sep 24 '19 at 19:15

Jignesh

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I think you explained the functionality of `round()` not `floor()` - it is not the "closest integer", more the largest integer that is less or equal to the given number – user85421 Sep 24 '19 at 19:43
@CarlosHeuberger thanks for pointing it out. you are right. I will update the answer – Jignesh Sep 24 '19 at 19:47

How do I find perfect square from a set of numbers in Java?

3 Answers3