Talk:Weird numbers

My version.

I saw there's already a Python program listed here for doing this, but I (with help from some others) created this one. Using the "sum of factors" formula it tests if a number is abundant BEFORE generating the list of each factors. If the number is deficient, the program returns so and saves memory. The program is alnmost twice as fast as the original, taking 0.16 s for the first 50.

<lang python># anySum :: Int -> [Int] -> [Int]

""" Last updated 5/8/24 """

from itertools import combinations, takewhile from math import prod from time import time

start = time()

primitivesp_nos = {6} # No multiple of a semiperfect number is weird

def main(): # Range of nos [number1, number2]

   weird_nos = []
   n = 50 # Find n weird numbers 
   x = 1 # Number to be tested
   while n > 0:
       if isweird(x) == 1:
           weird_nos.append(x)
           n = n - 1
       x = x + 1        
   print("First", len(weird_nos), "weird nos:\n", weird_nos)

def get_prime_fctrs(n): # Wheel factorization

   """ Code from Jerome Richard """
   """ stackoverflow.com/questions/70635382/
   fastest-way-to-produce-a-list-of-all-divisors-of-a-number """
   fctrs = [] # Empty list
   if n % 6 == 0: # 6 is primitive semiperfect, equals 2 * 3 
       return "Semiperfect" 
   while n % 2 == 0: # Divides by 2 (adds 2, 2...) to prime fctrs 
       fctrs.append(2) # Append 2 
       n //= 2
   t = 2 ** (len(fctrs) + 1) - 1 # Test 
   while n % 3 == 0: # Divides by 3 (adds 3, 3...) to prime fctrs
       fctrs.append(3) # Append 3 
       n //= 3 
   i = 5
   while i*i <= n: # Repeats above process 
       for k in (i, i+2):
           while n % k == 0:
               while k <= t:
                   """ 2^k * p is never weird """ 
                   return "Semiperfect"
               fctrs.append(k) # Append k
               n //= k
       i += 6
   if n > 1:
       fctrs.append(n) # Append n  
   return fctrs # Passes prime fctrs to isweird 

def isweird(n): # Checks if n is weird

   global primitivesp_nos # Retrieves list of primitive semiperfect nos
   prime_fctrs = get_prime_fctrs(n)
   if prime_fctrs == "Semiperfect":
       return 0 
   sum_fctrs = 1 # Sum of all factors based on formula
   fctrs = set(prime_fctrs) # Set of all fctrs
   for i in fctrs:
       sum_fctrs = sum_fctrs * (i ** (prime_fctrs.count(i) + 1) - 1)//(i - 1)
   difference = sum_fctrs - n  - n # Difference between sum of fctrs and target n 
   if difference < 0: # If difference < 0, n is deficient 
       return 0
   if difference == 0: # If difference = 0, n is perfect 
       primitivesp_nos.add(n) # n is primitive semiperfect 
       return 0
   for i in range(2, len(prime_fctrs)): 
       for j in combinations(prime_fctrs, i): # All combinations of prime fctrs
           product = prod(j) # Product 
           if product not in primitivesp_nos: # Factor product added to set 
               fctrs.add(product)
           else: # If factor is semiperfect, n cannot be weird 
               return 0
   fctrs.add(1) # All numbers have 1 as a factor
   fctrs = sorted(fctrs) # Sorts fctrs in order 
   fctrs = set(takewhile(lambda x:x <= difference, fctrs)) # Remaining fctrs set 
   ns = n - (difference + n - sum(fctrs)) # Stores in variable to save space
   if ns < 0:
       return 1 # n is weird
   """ Code from Stefan2:
   https://discuss.python.org/t/a-python-program-for-
   finding-weird-numbers/48654/6 """ 
   prime_fctrs = 1 # Overwrites list, saves space
   for d in fctrs:
       prime_fctrs |= prime_fctrs << d
   if not prime_fctrs >> ns & 1: # Checks if combos set contains ns 
       return 1
   else:
       primitivesp_nos.add(n)
       return 0
   

main() # Start program

end = time() print("Execution time: ", round(end - start, 2), "s")

-- -> [100,96]</lang>

A faster and less ambitious algorithm ?

I noticed yesterday that entries here were sparse.

Perhaps some were abandoned after first-sketch exhaustive searches appeared interminably slow ? And possibly the references to number theory in the Perl 6 founding example could look a bit daunting to some ?

Here is a theoretically unambitious approach, which seems, for example, to compress the (functionally composed) Python version down to c. 300 ms for 50 weirds (about half that for 25 weirds), on a system which needs c. 24 seconds to find 25 weirds with the initial Perl 6 draft.

1. Choose a smaller target for the sum-search.
2. Use a recursive hasSum(target, divisorList) predicate which fails early.
3. Generate the properDivisors in descending order of magnitude.

A smaller target should, I think, involve a smaller number of possible sums. The obvious candidate in an abundant number is the difference between the sum of the proper divisors and the number considered. If a sum to that difference exists, then removing the participants in the smaller sum will leave a set which sums to the abundant number itself.

For possible early-failing implementations of hasSum, see the Python, Haskell, JavaScript and even AppleScript drafts.

If hasSum considers large divisors first, it can soon exclude all those those too big to sum to a smaller target. Hout (talk) 11:52, 24 March 2019 (UTC)

PS I think we may be able to see and test the operation of hasSum more clearly if we enrich its type from Bool to [Int] (with a return value of the empty list for False, and the integers of the first sum found for True). Let's call this richer-typed variant anySum, and sketch it in Python 3.

<lang python># anySum :: Int -> [Int] -> [Int]

def anySum(n, xs):

   First subset of xs found to sum to n.
      (Probably more efficient where xs is sorted in
      descending order of magnitude)
   def go(n, xs):
       if xs:
           # Assumes Python 3 for list deconstruction
           # Otherwise: h, t = xs[0], xs[1:]
           h, *t = xs
           if n < h:
               return go(n, t)
           else:
               if n == h:
                   return [h]
               else:
                   ys = go(n - h, t)
                   return [h] + ys if ys else go(n, t)
       else:
           return []
   return go(n, xs)

1. Search for sum through descending numbers (more efficient)

print(anySum(196, range(100, 0, -1)))

1. -> [100, 96]

1. Search for sum through ascending numbers (less efficient)

print(anySum(196, range(1, 101)))

1. -> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 25]

print(anySum(7, [6, 3]))

1. -> []</lang>

or similarly, rewriting hasSum to anySum (with comments) in a Haskell idiom:

<lang haskell>module AnySum where

hasSum :: Int -> [Int] -> Bool hasSum _ [] = False hasSum n (x:xs)

 | n < x = hasSum n xs
 | otherwise = (n == x) || hasSum (n - x) xs || hasSum n xs

-- Or, enriching the return type from Bool to [Int] -- with [] for False and a populated list (first sum found) for True

anySum :: Int -> [Int] -> [Int] anySum _ [] = [] anySum n (x:xs)

 | n < x = anySum n xs  -- x too large for a sum to n
 | n == x = [x] -- We have a sum
 | otherwise =
   let ys = anySum (n - x) xs -- Any sum for (n - x) in the tail ?
   in if null ys
        then anySum n xs -- Any sum (not involving x) in the tail ?
        else x : ys  -- x and the rest of the sum that was found.
        

main :: IO () main = do

 -- xs ascending - the test is less efficient
 print $ anySum 196 [1 .. 100]
 -- -> [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,25]
 
 -- xs descending - the test is more efficent
 print $ anySum 196 [100,99 ..]
 -- -> [100,96]</lang>