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copied!<p>Looks like you're trying to solve some logic puzzle by unique selection from finite domain. Consult these:</p> <ul> <li><a href="https://stackoverflow.com/questions/9831374/euler-43-is-there-a-monad-to-help-write-this-list-comprehension/9889702#9889702">Euler 43 - is there a monad to help write this list comprehension?</a></li> <li><a href="https://stackoverflow.com/questions/12869097/splitting-list-into-a-list-of-possible-tuples/12872133#12872133">Splitting list into a list of possible tuples</a></li> </ul> <p>The way this helps us is, we carry our domain around while we're making picks from it; and the next pick is made from the narrowed domain containing what's left after the previous pick, so a chain is naturally formed. E.g.</p> <pre><code>p43 = sum [ fromDigits [v0,v1,v2,v3,v4,v5,v6,v7,v8,v9] | (dom5,v5) <- one_of [0,5] [0..9] -- [0..9] is the , (dom6,v6) <- pick_any dom5 -- initial domain , (dom7,v7) <- pick_any dom6 , rem (100*d5+10*d6+d7) 11 == 0 .... -- all possibilities of picking one elt from a domain pick_any :: [a] -> [([a], a)] pick_any [] = [] pick_any (x:xs) = (xs,x) : [ (x:dom,y) | (dom,y) <- pick_any xs] -- all possibilities of picking one of provided elts from a domain -- (assume unique domains, i.e. no repetitions) one_of :: (Eq a) => [a] -> [a] -> [([a], a)] one_of ns xs = [ (ys,y) | let choices = pick_any xs, n <- ns, (ys,y) <- take 1 $ filter ((==n).snd) choices ] </code></pre> <hr> <p>You can trivially check a number of elements in your answer as a part of your list comprehension:</p> <pre><code>s = [answer | a <- .... , let answer=[....] , length answer==4 ] </code></pre> <p>or just create different answers based on a condition,</p> <pre><code>s = [answer | a <- .... , let answer=if condition then [a,b,c] else [a]] </code></pre>

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