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Hello!
I keep getting only 4/5 tests passed, i just can’t figure out what is missing on my code, for me it works 100%.
Can anyone please point out what i am missing?

def scale_of_truth_n(n):
c = 0
while n != 1:
n = n//2
c += 1
return c
scale_of_truth_n(3)

def scale_of_truth_n(n):
if n < 1:
return "Error"
else:
count = 0
while n != 1:
while n % 3 != 0:
n += 1
n /= 3
count += 1
return count
print(scale_of_truth_n(120))

if you put in 0 you would get an endless loop thats probably why. But i think your method is wrong, for 81 as an example you split it into 3 groups, 27 now, then again, 9 now, then again,3 now and finally once more to get the one. thats 4 times whereas your code would say 8.
read the prompt again to see why its 3 groups not 2

import math
def scale_of_truth_n(n):
if n < 2: return 0
return math.ceil(math.log(n, 3))
print(scale_of_truth_n(3))

Can’t be too creative here, just handing in my own solution which is basically identical to what @text9831554157 's most recently posted for the reasons @text9831554157 posted before.

Could someone please explain why I have to use log with a base of 3 instead of base 2? It seemed more logical to me to divide the number of N by 2 until I reached a setting in which I would have only 1 ball on either side of the scale. Consequently, the number of weighs would be math.ceil(math.log(n,2)).

def scale_of_truth_n(n):
from math import floor, log
if n==1: return 0
try:
x = floor(log(n-1,3))+1
except:
return 1
return x
print(scale_of_truth_n(3))

def scale_of_truth_n(n):
if n <= 1:
return 0
else:
c = 1
while n >= 3:
if n % 3 != 0:
n -= 1
if n % 3 != 0:
n-= 1
n /= 3
c += 1
return c
print(scale_of_truth_n(10))

This is basically the implementation of the log3 without actually using it…
If you keep finding yourself at 4/5 test passed, one possibility may be that you didn’t account for cases like n = 9 where it is actually optimal to form group of threes which allows for a maximum of two weight comparisons while the log2 algorithm needs three.