Of course the recursive version is trivial:
hanoi n = solve n 1 2 3 solve 0 _ _ _ =  solve n from help to = (solve (n-1) from to help) ++ [(from,to)] ++ (solve (n-1) help from to)
However my iterative version looks terrible with a lot of code repetition:
hanoi n = map rep $ solve [1..n]   where rep (x,y) | odd n = ([1,3,2] !! (x-1), [1,3,2] !! (y-1)) | otherwise = (x,y) solve from help to = head $ mapMaybe ready $ iterate step (from,help,to,) where step (1:xs,ys,zs,sol) = let (xs',zs',sol') = try xs zs 1 3 ((1,2):sol) in (xs',1:ys,zs',sol') step (xs,1:ys,zs,sol) = let (xs',ys',sol') = try xs ys 1 2 ((2,3):sol) in (xs',ys',1:zs,sol') step (xs,ys,1:zs,sol) = let (ys',zs',sol') = try ys zs 2 3 ((3,1):sol) in (1:xs,ys',zs',sol') try   _ _ sol = (,, sol) try (x:xs)  a b sol = (xs,[x], (a,b):sol) try  (y:ys) a b sol = ([y],ys, (b,a):sol) try (x:xs) (y:ys) a b sol | x < y = (xs,x:y:ys, (a,b):sol) | y < x = (y:x:xs,ys, (b,a):sol) ready (,,_,sol) = Just $ reverse sol ready (,_,,sol) = Just $ reverse sol ready _ = Nothing
Any ideas? More general, how to deal with situations like this where you have a lot of different cases and args?
With "iterative solution" I mean the algorithm described here: http://en.wikipedia.org/wiki/Tower_of_Hanoi#Iterative_solution