Subset sum problem: Difference between revisions

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 None bruteforce: the list of numbers used here are different, and difficult for a bruteforce method.
-<lang haskell>subsum ls = sub_ [(0,[])] ls where
+<lang haskell>subsum = foldl sub_ [(0,[])] where
-	sub_ a [] = a
+	sub_ a x = merge a $ map f a where
-	sub_ a (x:xs) = sub_ (merge a $ map (f x) a) xs
+		f (a,l) = (a+x, l++[x])
-	f x (a,l) = (a+x, l++[x])
 	-- keep list of sums sorted. Don't use standard sort!