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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.
Choose a '''smaller target''' for the sum-search.
Use a recursive '''hasSum'''(''target'', ''divisorList'') predicate which fails early.
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.
If hasSum considers large divisors first, it can soon exclude all those those too big to sum to a smaller target. [[User:Hout|Hout]] ([[User talk: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, 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]