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?

Answer 1

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

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

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