Finding perfect square

Question

I have this python code:

def sqrt(x):
    ans = 0
    if x >= 0:
        while ans*ans < x:
            ans = ans + 1
            if ans*ans != x:
                print x, 'is not a perfect square.'
                return None
            else:
                print x, ' is a perfect square.'
                return ans
    else:
        print x, ' is not a positive number.'
        return None

y = 16      
sqrt(y)

the output is:

16 is not a perfect square.

Whereas this works perfectly:

x = 16
ans = 0
if x >= 0:
    while ans*ans < x:
        ans = ans + 1
        #print 'ans =', ans
    if ans*ans != x:
        print x, 'is not a perfect square'  
    else: print ans, 'is a perfect square'
else: print x, 'is not a positive number'

What am I doing wrong?

Here is a related question on finding whether a number is a perfect square: http://stackoverflow.com/questions/295579/fastest-way-to-determine-if-an-integers-square-root-is-an-integer — Nick Craig-Wood, Oct 10 '09 at 10:30

Sunjay Varma · Answer 1 · 2015-10-29T02:19:09.997

10

Just thought I'd contribute a simpler solution:

def is_square(n):
    return sqrt(n).is_integer()

This is valid for n < 2**52 + 2**27 = 4503599761588224.

Examples:

>>> is_square(4)
True
>>> is_square(123)
False
>>> is_square(123123123432**2)
True

edited Oct 29 '15 at 02:19

answered Apr 23 '11 at 21:49

Sunjay Varma

5,007
6
34
51

While a good answer for *small* values, this has precision problems. `is_square(123456789123456789**7-1)` gives `True`. There's no particularly easy solution, though. [I don't think `123123123432156156165**7` is square anyway.](http://www.wolframalpha.com/input/?i=is+123456789123456789**7+square). – Veedrac May 26 '14 at 07:17
and 123123123432**7 is not a perfect square: its sqrt lies between 654920651322780642131484946101910838253 and 654920651322780642131484946101910838254 – blabla999 Oct 27 '15 at 20:15

score 7 · Accepted Answer · answered Oct 10 '09 at 06:30

7

Indent your code correctly to let the while statement execute until ans*ans < x:

def sqrt(x):
    ans = 0
    if x >= 0:
        while ans*ans < x:
            ans = ans + 1

        if ans*ans != x:  # this if statement was nested inside the while
            print x, 'is not a perfect square.'
            return None
        else:
            print x, ' is a perfect square.'
            return ans
    else:
        print x, ' is not a positive number.'
        return None

y = 16          
print sqrt(y)

Try it out here.

answered Oct 10 '09 at 06:30

Christian C. Salvadó

807,428
183
922
838

Aah! I was going crazy over this, indentation while really helps in reading the code can be really difficult to debug! :) – eozzy Oct 10 '09 at 06:49
2

Give me Python's indentation rules any day over the possibility my indentation may not match my braces in C et al. It's a lot easier if you make sure your editor is configured for spaces only. – paxdiablo Oct 10 '09 at 07:04

score 1 · Answer 3 · answered Oct 10 '09 at 06:20

1

Your while loop only executes once. No matter which branch the if statement inside it takes, the whole function will return immediately.

answered Oct 10 '09 at 06:20

Greg Hewgill

951,095
183
1,149
1,285

Strange! The same code works outside the function, why is that? – eozzy Oct 10 '09 at 06:23
Because the loop outside the function is not being exited prematurely with a return. – paxdiablo Oct 10 '09 at 06:33

ManicMailman · Answer 4 · 2009-10-10T06:39:29.213

0

EDIT I modified it, tried it out, and it works. You just need this piece of code

As soon as ans = 4, ans * ans is no longer smaller than x. Try while ans*ans <= x: instead of just <

def sqrt(x):
ans = 0
if x >= 0:
        while ans*ans <= x:                     
                if ans*ans == x:
                            print x, ' is a perfect square.'
                            return ans
        else:
            ans = ans + 1

edited Oct 10 '09 at 06:39

answered Oct 10 '09 at 06:18

ManicMailman

75
2
13

You can't return in your loop in the case where ans*ans is not x. def sqrt(x): ans = 0 if x >= 0: while ans*ans <= x: ans = ans + 1 if ans*ans == x: print x, ' is a perfect square.' return ans else: print x, ' is not a positive number.' return None print x, 'is not a perfect square.' return None y = 16 sqrt(y) – ManicMailman Oct 10 '09 at 06:27

score 0 · Answer 5 · answered Oct 10 '09 at 06:23

0

Change your code so it displays the value of ans as well as x, so you can tell how many times the loop is executed.

answered Oct 10 '09 at 06:23

pavium

14,808
4
33
50

I'm not a python programmer, but I could recognise the same problem as Greg Hewgill. – pavium Oct 10 '09 at 06:27

score 0 · Answer 6 · answered Oct 10 '09 at 06:37

If your code sample is actually correctly indentet the first round of the while will return on it's first round - always. So any positive value of x>1 will fullfil the ans*ans=1*1=1!=x, giving "x is not a perfect square".

You basically needs to get your indentation right - like you do in your other example. Again - if your code sample here actually is correctly indented. Try this:

def sqrt(x):
    ans = 0
    if x >= 0:
        while ans*ans < x:
            ans = ans + 1

        if ans*ans != x:
            print x, 'is not a perfect square.'
            return None
        else:
            print x, ' is a perfect square.'
            return ans
    else:
        print x, ' is not a positive number.'
        return None

score 0 · Answer 7 · answered Jul 10 '14 at 15:29

0

def isPerfectSquare(number):
    return len(str(math.sqrt(number)).split('.')[1]) == 1

answered Jul 10 '14 at 15:29

trycatch22

307
4
14

Suman Michael · Answer 8 · 2017-08-05T05:02:34.073

0

I think this is probably short.

 def issquare():
      return (m**.5 - int(m**.5)==0)

edited Aug 05 '17 at 05:02

answered Aug 04 '17 at 16:00

Suman Michael

11
3

score -1 · Answer 9 · answered Apr 21 '11 at 21:50

-1

If the goal is to determine whether a number is a perfect square, I would think it would be simpler (and perhaps more efficient) to use math builtins, e.g.:

def is_perfect_square(n):
  if not ( ( isinstance(n, int) or isinstance(n, long) ) and ( n >= 0 ) ):
    return False 
  else:
    return math.sqrt(n) == math.trunc(math.sqrt(n))

answered Apr 21 '11 at 21:50

gumption

1,108
13
10

1

This returns True for `(2**64) - 1` -- you're losing precision – bstpierre Apr 22 '11 at 13:45
1

Not relevant criticism, despite being true. This isnt a pure math page, this is a programming site. You cannot even store irrational numbers in computers. Precision is never infinite; thats the point of computers and anyone whose done any programming knows this ahead of time. Dont see a reason for criticizing. Numbers that large wont be tested except by mathematicians who probably have better notions and wont be on this site. – CogitoErgoCogitoSum Aug 17 '17 at 07:03

Finding perfect square

9 Answers9

Linked