Python implementation of square root without using built in sqrt() function

sqrtimplementation.py

num = 17956

numlength = len(str(abs(num)))
ans = ''
rem = 0

def getlast(list):
	if list:
		return list[-1]

def perfectsquare(num):
	global rem
	global ans
	xlist = []
	xsquaredlist = []
	for x in range(1,10):
		xsquared = x * x
		if xsquared <= num:
			xlist.append(x)
			xsquaredlist.append(xsquared)
	rem = (int(num) % getlast(xsquaredlist))
	ans += str(getlast(xlist))

def getdigit(numbers):
	global num
	numstring = str(num)
	numberslist = numbers.split(',')
	return int(numstring[int(numberslist[0]) : int(numberslist[1])])

def tempcalc(tempans, tempnum):
	global ans
	global rem
	currprods = []
	xvalues = []
	for x in range(1,10):
		tempoperand = str(int(tempans)*2) + str(x)
		currprod = int(tempoperand) * x
		if currprod <= int(tempnum):
			currprods.append(currprod)
			xvalues.append(x)
	rem = int(tempnum) % getlast(currprods)
	ans += str(getlast(xvalues))
	
dig = ''
for x in range(0,numlength):
	key = numlength - x
	if numlength % 2 == 0:
		if key % 2 == 0:
			dig += '-' + str(key)
		else:
			if key == 1:
				dig += ',' + str(numlength)
			else:
				dig += ',-' + str(key-1)
		if len(dig) == 4 or len(dig) == 5:
			if '-' + str(numlength) in dig:
				perfectsquare(getdigit(dig))
			else:
				tempnum = str(rem) + str(getdigit(dig))
				tempcalc(ans, tempnum)
			dig = ''
	else:
		if key % 2 == 0:
			dig = ''
			dig += '-' + str(key)
		else:
			if key == numlength:
				dig += '-' + str(key)
				dig += ',-' + str(key - 1)
			else:
				if key == 1: 
					dig += ',' + str(numlength)
				else:
					dig += ',-' + str(key - 1)
		if len(dig) == 4 or len(dig) == 5:
			if '-' + str(numlength) in dig:
				perfectsquare(getdigit(dig))
			else:
				tempnum = str(rem) + str(getdigit(dig))
				tempcalc(ans, tempnum)
				dig = ''
print(ans)

Example of how it is done manually

There is no implementation for dealing with numbers that contain zeros.

More examples