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.

# 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]

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.

# 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)


# Search for sum through descending numbers (more efficient)
print(anySum(196, range(100, 0, -1)))
# -> [100, 96]

# Search for sum through ascending numbers (less efficient)
print(anySum(196, range(1, 101)))
# -> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 25]

print(anySum(7, [6, 3]))
# -> []

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

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]