it is my second post in StackOverflow I'm having trouble understanding and applying this exercise on my own with python please could help me! getting this TypeError: '<' not supported between instances of 'int' and 'NoneType'

This is the exercise Algorithm:

Definition: An integer is said to be perfect if it is equal to the sum of all its divisors. Examples: 6 and 28 are perfect since 6 = 1 2 3 (knowing that 1, 2 and 3 are the divisors of 6 less than 6) 28= 1 2 4 7 14 (knowing that 1, 2, 4, 7 and 14 are the divisors of 28 less than 28)

1. Create a function liste_divisors(), which accepts an integer N as a parameter and returns the list of its divisors less than N (1 included).

2. Make a function is perfect(), which accepts a (positive) integer N as a parameter and returns “True” if it is perfect and “False” otherwise (use the function from the 1st question)

3. Create a Perfect List() function, which accepts a Limit parameter, then returns a list containing the perfect numbers less than Limit

I'm having trouble with the last question

This is my attempt so far, but there are errors, please help me to correct these errors and make it better

def liste_diviseur(N):
   t = []
   for i in range(1,N):
      if(N%i == 0):
         t.append(i)
   return t

def est_parfait(M):
   s = 0
   for i in liste_diviseur(M):
      s  = i 
   if(s == M):
      return True
   else:
      return False

def liste_parfait(Limite):
   t = []
   lis = print(liste_diviseur(Limite))
   if(Limite<lis):
      t.append(Limite)
   return t
m = int(input('Giving an number :'))
print(liste_parfait(m))

CodePudding user response：

Your error appears to be when you set lis = print(liste_diviseur(Limite)). print() is a void function which will return null. Removing the print() should fix your problem

CodePudding user response：

You need to create a range and then check if each number in the range is perfect.

def liste_diviseur(n):
   return [i for i in range(1,n//2 1) if n%i==0]


def est_parfait(n):
   return n == sum(liste_diviseur(n))
         

def liste_parfait(limite): 
   return [i for i in range(2, limite) if est_parfait(i)]
   

test = int(input('Giving an number: '))
print(liste_parfait(test))
# output for 8129
[6, 28, 496, 8128]

CodePudding user response：

The main part is generating the divisors. You can find some ideas for efficiency e.g. here.

For large numbers, one would use factorization of n into its factors (see e.g. an answer I wrote for another problem), and then generate all divisors by combining those factors.

Here, we'll go for simplicity:

from math import isqrt

def divisors(n):
    m = isqrt(n)
    for i in range(1, m   1):
        if n % i == 0:
            yield i
            j = n // i
            if i < j < n:
                yield n // i

Example:

>>> list(divisors(36))
[1, 2, 18, 3, 12, 4, 9, 6]

Note that divisors() returns a generator, not a list, and the divisors are produced out of order (more precisely: as pairs i and n // i). But this doesn't matter to see if a number if perfect or not:

def is_perfect(n):
    return sum(divisors(n)) == n

>>> is_perfect(5)
False

>>> is_perfect(6)
True

Now for a list of perfect numbers:

>>> [n for n in range(1, 10_000) if is_perfect(n)]
[6, 28, 496, 8128]