Check if a number is a perfect square


How could I check if a number is a perfect square?

Speed is of no concern, for now, just working.

7/8/2019 2:37:59 AM

Accepted Answer

The problem with relying on any floating point computation (math.sqrt(x), or x**0.5) is that you can't really be sure it's exact (for sufficiently large integers x, it won't be, and might even overflow). Fortunately (if one's in no hurry;-) there are many pure integer approaches, such as the following...:

def is_square(apositiveint):
  x = apositiveint // 2
  seen = set([x])
  while x * x != apositiveint:
    x = (x + (apositiveint // x)) // 2
    if x in seen: return False
  return True

for i in range(110, 130):
   print i, is_square(i)

Hint: it's based on the "Babylonian algorithm" for square root, see wikipedia. It does work for any positive number for which you have enough memory for the computation to proceed to completion;-).

Edit: let's see an example...

x = 12345678987654321234567 ** 2

for i in range(x, x+2):
   print i, is_square(i)

this prints, as desired (and in a reasonable amount of time, too;-):

152415789666209426002111556165263283035677489 True
152415789666209426002111556165263283035677490 False

Please, before you propose solutions based on floating point intermediate results, make sure they work correctly on this simple example -- it's not that hard (you just need a few extra checks in case the sqrt computed is a little off), just takes a bit of care.

And then try with x**7 and find clever way to work around the problem you'll get,

OverflowError: long int too large to convert to float

you'll have to get more and more clever as the numbers keep growing, of course.

If I was in a hurry, of course, I'd use gmpy -- but then, I'm clearly biased;-).

>>> import gmpy
>>> gmpy.is_square(x**7)
>>> gmpy.is_square(x**7 + 1)

Yeah, I know, that's just so easy it feels like cheating (a bit the way I feel towards Python in general;-) -- no cleverness at all, just perfect directness and simplicity (and, in the case of gmpy, sheer speed;-)...

3/22/2010 1:30:00 AM

Use Newton's method to quickly zero in on the nearest integer square root, then square it and see if it's your number. See isqrt.

Python ≥ 3.8 has math.isqrt. If using an older version of Python, look for the "def isqrt(n)" implementation here.

import math

def is_square(i: int) -> bool:
    return i == math.isqrt(i) ** 2

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