Find next perfect square not working in python

Thread Starter

terabaaphoonmein

Joined Jul 19, 2020
56
Code:
def find_next_square(sq):
    # Return the next square if sq is a square, -1 otherwise
    sq2=(sq**1/2)
    xyz=isinstance(sq2, int)
    if (xyz==True):
        print("Is perfect square")
        nextsq=sq+1
        print("Next perfect square=",nextsq**2)
    else:
        print("Not perfect square")
        return -1

n=int(input("Enter an integer"))
find_next_square(n)
Output-:
Enter an integer25
Not perfect square

Expected output-:
Enter an integer25
Next perfect square=36

My logic is that xyz checks if $(sq)^0.5$ is integer or not. If it is integer we find next perfect square, else we return -1
 
Last edited:

ApacheKid

Joined Jan 12, 2015
590
Code:
def find_next_square(sq):
    # Return the next square if sq is a square, -1 otherwise
    sq2=(sq**1/2)
    xyz=isinstance(sq2, int)
    if (xyz==True):
        print("Is perfect square")
        nextsq=sq+1
        print("Next perfect square=",nextsq**2)
    else:
        print("Not perfect square")
        return -1

n=int(input("Enter an integer"))
find_next_square(n)
Output-:
Enter an integer25
Not perfect square

Expected output-:
Enter an integer25
Next perfect square=36

My logic is that xyz checks if $(sq)^0.5$ is integer or not. If it is integer we find next perfect square, else we return -1
Yet 25 is a perfect square...
 

Thread Starter

terabaaphoonmein

Joined Jul 19, 2020
56
This works tbh.
Code:
import math

def find_next_square(sq):
    # Return the next square if sq is a square, -1 otherwise
    sq2=math.sqrt(sq)
    sq2=(int(sq2) if sq2.is_integer() else sq2)  # convert answer to int if we can do it
    xyz=isinstance(sq2, int)
    if (xyz==True):
        print("Is perfect square")
        nextsq=sq2+1
        print("Next perfect square=",nextsq**2)
    else:
        print("Not perfect square")
        return -1

n=int(input("Enter an integer"))
find_next_square(n)
 

Ramussons

Joined May 3, 2013
1,154
Code:
def find_next_square(sq):
    # Return the next square if sq is a square, -1 otherwise
    sq2=(sq**1/2)
    xyz=isinstance(sq2, int)
    if (xyz==True):
        print("Is perfect square")
        nextsq=sq+1
        print("Next perfect square=",nextsq**2)
    else:
        print("Not perfect square")
        return -1

n=int(input("Enter an integer"))
find_next_square(n)
Output-:
Enter an integer25
Not perfect square

Expected output-:
Enter an integer25
Next perfect square=36

My logic is that xyz checks if $(sq)^0.5$ is integer or not. If it is integer we find next perfect square, else we return -1
I did this change in line 3
sq2=int((sq**1/2))
and it runs fine.
 

ZCochran98

Joined Jul 24, 2018
192
You may want to check those parentheses.
sq**1/2 is not the same as sq**(1/2). The first one is the same as \(sq^1/2\), and the second one is \(sq^{1/2}\)
Using the int() is also a good idea, like what you had.
 

Ramussons

Joined May 3, 2013
1,154
You may want to check those parentheses.
sq**1/2 is not the same as sq**(1/2). The first one is the same as \(sq^1/2\), and the second one is \(sq^{1/2}\)
Using the int() is also a good idea, like what you had.
As you said, 1/2 with and without brackets does make a difference.

And the int(..) that I tried earlier does not give the right answer. It wrongly declares 12 as a square because the prog is p[rogrammed to declare all Integers as Squares!!! - line 5
 
Last edited:

xox

Joined Sep 8, 2017
623
FWIW, creating a function which prints AND returns a value is probably not the best approach here. (It should really do one or the other.)

Code:
import math

while True:
    value = int(input("Enter an integer: "))
    root = int(math.sqrt(value))
    if root**2 != value:
        print(value, "is NOT perfect square")
    else:
        print(value, "IS perfect square and the next one is", (root + 1)**2)
 
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