Perfect square if,else

Question

having trouble with perfectsquare code

import java.util.*;


public class perfectsquare {

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub

Scanner s= new Scanner(System.in); 
double number; 

System.out.println("Enter number >");

number= s.nextDouble();

System.out.println(Math.sqrt(number));





    }

}

Using if,else how could you make it return a perfect square like sqrt of 9=3 but false if the square root of that number is a decimal for example squareroot of 10 would be 3.122222.....

If you want us to do your homework, at least know what return type you want. — jma127, Feb 02 '13 at 02:14
possible duplicate of : http://stackoverflow.com/questions/343852/whats-a-good-algorithm-to-determine-if-an-input-is-a-perfect-square — Sednus, Feb 02 '13 at 02:25

score 3 · Answer 1 · answered Feb 02 '13 at 02:18

3

if(Math.round(Math.sqrt(number))==Math.sqrt(number)){
  //it is a perfect square
}

answered Feb 02 '13 at 02:18

Neil Locketz

4,228
1
23
34

1

`if (Math.sqrt(number)==Math.sqrt(number)%1)` – tckmn Feb 02 '13 at 02:24
You can also use `Math.ceil` or `Math.floor` or multiply the sqrt's truncation to `int` by itself and see if it was the original number (after a cast to `double`). – eboix Feb 02 '13 at 02:24
^ This would work also, there is many ways to solve this problem this is just the first one that came to my head :P – Neil Locketz Feb 02 '13 at 02:28

score 2 · Answer 2 · 2013-02-02T02:25:36.493

2

You can change the return type to Object(But you will have to check if it's a number using instanceof when using the function) or return -1 instead of false

To see if it's a perfect square, use this:

if(Math.sqrt(number) % 1 == 0) {
    return number;
} else {
    return -1;
}

edited Feb 02 '13 at 02:25

answered Feb 02 '13 at 02:17

2

This doesn't return what the above poster asked for. – Aidanc Feb 02 '13 at 02:17
34 doesn't have decimals, and it is not a perfect square. – eboix Feb 02 '13 at 02:21
I saw the edit. It still does not answer the OP's question. – eboix Feb 02 '13 at 02:23
This is what he means by a perfect square: `a perfect square like sqrt of 9=3 but false if the square root of that number is a decimal for example squareroot of 10 would be 3.122222.....` – Feb 02 '13 at 02:27
You just added that. Anyway, the OP specified `true` if it is a perfect square, and `false` if it is not. – eboix Feb 02 '13 at 02:31
No he didn't: `return a perfect square like sqrt of 9=3 but false if` – Feb 02 '13 at 02:34
just a little project i was working on and thnxs everyone – NBera Feb 02 '13 at 02:44
@Runemoro Thanks Your answer helped me – Zohra Khan Mar 28 '16 at 15:10

score 0 · Accepted Answer · edited Nov 10 '15 at 03:37

0

import java.util.*;

public class perfectsquare {
    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub

        Scanner s= new Scanner(System.in); 
        double number; 

        System.out.println("Enter number >");

        number= s.nextDouble();
        double sqr= Math.sqrt(number);
        //System.out.println(sqr);
        double num=sqr;
        double x;
        x=num%1;
        x=num-Math.floor(num);
        if(x>0){
            System.out.println("False");
        }
        else{
            System.out.println(sqr);
        }
    }
}

This will work.

edited Nov 10 '15 at 03:37

andrewsi

10,807
132
35
51

answered Feb 02 '13 at 02:21

user1658435

574
1
9
28

Your logic seems faulty, I would do `% 1`. – eboix Feb 02 '13 at 02:27
In java it is not working. I'm new to java – user1658435 Feb 02 '13 at 02:31
Sorry, I wasn't clear enough. Earlier, I said that Java double mod operations were not well defined, but I take that back. They're not well defined in C, but they are well defined in Java. The reason this is not working is that you are doing `% 10` instead of `% 1`. Your statement will return whether the square root has a one's digit. `% 1` will get you whether or not it has decimals, which is what you want to know. Also, you should change your `double num=sqr; if(num ....` to just `if(sqr ....`. There isn't a reason to have an auxiliary variable `num`, there. – eboix Feb 02 '13 at 02:33
yaa i edited it, this program will work now :) – user1658435 Feb 02 '13 at 02:39
This would be more helpful if you formatted the code. Thanks. – Kurtymckurt Feb 02 '13 at 06:04

Perfect square if,else

3 Answers3