G'day all.
Quoting Don Stewart
So, team, anyone want to implement a Knight's Tour solver in a list monad/list comprehension one liner? These little puzzles are made for fast languages with backtracking monads....
I conjecture that any one-liner won't be efficient. Anyway, here's my ~30 min attempt. The showBoard and main are both very quick and dirty, and I'm sure someone can do much better. I particularly like the fact that changing "Maybe" to "[]" will make knightsTour return all tours starting at the upper left-hand corner, rather than just one. Warm fuzzy things rule. Cheers, Andrew Bromage module Main where import qualified Data.Set as S import Data.List import Data.Function import Control.Arrow import Control.Monad import System knightsTour :: Int -> Maybe [(Int,Int)] knightsTour size = tour [(0,0)] (S.fromAscList [ (x,y) | x <- [0..size-1], y <- [0..size-1], x /= 0 || y /= 0 ]) where jumps = [(2,1),(1,2),(2,-1),(-1,2),(-2,1),(1,-2),(-2,-1),(-1,-2)] tour moves@(pos:_) blank | S.null blank = return (reverse moves) | otherwise = msum [ tour (npos:moves) (npos `S.delete` blank) | npos <- nextPositions pos ] where nextPositions = map snd . sortBy (compare `on` fst) . map (length . neighbours &&& id) . neighbours neighbours (x,y) = [ npos | (x',y') <- jumps, let { npos = (x+x',y+y') }, npos `S.member` blank ] showBoard :: Int -> [(Int,Int)] -> ShowS showBoard size = inter bdr . map (inter ('|':) . map (shows . fst)) . groupBy ((==) `on` fst.snd) . sortBy (compare `on` snd) . zip [1..] where bdr = ('\n':) . inter ('+':) (replicate size (replicate width '-' ++)) . ('\n':) width = length . show $ size*size pad s = \r -> replicate (width - length (s "")) ' ' ++ s r inter sep xs = sep . foldr (.) id [ pad x . sep | x <- xs ] main :: IO () main = do a <- getArgs size <- case a of [] -> return 8 (s:_) -> return (read s) putStrLn $ case knightsTour size of Nothing -> "No solution found." Just b -> showBoard size b ""