finding nearest square number with a limit

Question

Hey so I want to write code for finding nearest square number less than or equal to a certain number(x). I've tried the following:

m = 0 
base = 0 
while m <= x:
   if m > x:
      break
   m = base**2
   base += 1

print(m)

This, however, gives the nearest square right after the number x despite putting the break since a m > x has already been assigned to m. how i do stop the loop before m > x?

Of course, you can do this without a loop: `math.floor(math.sqrt(x)) ** 2` — Mark, Sep 14 '20 at 16:20

score 3 · Accepted Answer · answered Sep 14 '20 at 16:16

The if m > x is redundant. The loop body will only be entered if m <= x is True, and if that's True, m > x never will be, so that break should never run.

To answer your question though, without knowing more about the math, I'd just introduce a second variable prev_m, then use that:

m = 0
prev_m = m
base = 0 
while m <= x:
    prev_m = m
    m = base**2
    base += 1

print(prev_m)  # Then use prev_m instead

score 0 · Answer 2 · answered Sep 14 '20 at 16:28

There's an easier way to do this using math module:

import math 
y = math.sqrt(x) #find the square root of x
if y.is_integer() == False: #check y is not a whole number
    print(math.pow(int(y),2)) # find the square root of the nearest whole number to y

score 0 · Answer 3 · answered Sep 14 '20 at 16:50

Most answers are based on calculating one square too many and then using the previous one.
But is there a possibility to know if the next square will be too big, without calculating it?

m = 0
base = -1
while base*2 + m < x:
    base += 1
    m = base**2

print(m)

Disclaimer: Just a fun answer to think about why this works.

finding nearest square number with a limit

3 Answers3