# The Knights Tour

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(Improved ContT r (ST s) code) |
m (link to FBackTrack module) |

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## Latest revision as of 10:10, 2 December 2008

The Knight's Tour is a mathematical problem involving a knight on a chessboard. The knight is placed on the empty board and, moving according to the rules of chess, must visit each square exactly once.

Here are some Haskell implementations.

## Contents |

## [edit] 1 First Solution

-- -- Quick implementation by dmwit on #haskell -- Faster, shorter, uses less memory than the Python version. -- import Control.Arrow import Control.Monad import Data.List import Data.Maybe import Data.Ord import System.Environment import qualified Data.Map as M sortOn f = map snd . sortBy (comparing fst) . map (f &&& id) clip coord size = coord >= 0 && coord < size valid size solution xy@(x, y) = and [clip x size, clip y size, isNothing (M.lookup xy solution)] neighbors size solution xy = length . filter (valid size solution) $ sequence moves xy moves = do f <- [(+), subtract] g <- [(+), subtract] (x, y) <- [(1, 2), (2, 1)] [f x *** g y] solve size solution n xy = do guard (valid size solution xy) let solution' = M.insert xy n solution sortedMoves = sortOn (neighbors size solution) (sequence moves xy) if n == size * size then [solution'] else sortedMoves >>= solve size solution' (n+1) printBoard size solution = board [0..size-1] where sqSize = size * size elemSize = length (show sqSize) separator = intercalate (replicate elemSize '-') (replicate (size + 1) "+") pad n s = replicate (elemSize - length s) ' ' ++ s elem xy = pad elemSize . show $ solution M.! xy line y = concat . intersperseWrap "|" $ [elem (x, y) | x <- [0..size-1]] board = unlines . intersperseWrap separator . map line intersperseWrap s ss = s : intersperse s ss ++ [s] go size = case solve size M.empty 1 (0, 0) of [] -> "No solution found" (s:_) -> printBoard size s main = do args <- getArgs name <- getProgName putStrLn $ case map reads args of [] -> go 8 [[(size, "")]] -> go size _ -> "Usage: " ++ name ++ " <size>"

## [edit] 2 Using Continuations

An efficient version (some 10x faster than the example Python solution) using continuations.

This is about as direct a translation of the Python algorithm as you'll get without sticking the whole thing in IO. The Python version prints the board and exits immediately upon finding it, so it can roll back changes if that doesn't happen. Instead, this version sets up an exit continuation using callCC and calls that to immediately return the first solution found. The Logic version below takes around 50% more time.

import Control.Monad.Cont import Control.Monad.ST import Data.Array.ST import Data.List import Data.Ord import Data.Ix import System.Environment type Square = (Int, Int) type Board s = STUArray s (Int,Int) Int type ChessM r s = ContT r (ST s) type ChessK r s = String -> ChessM r s () successors :: Int -> Board s -> Square -> ChessM r s [Square] successors n b = sortWith (fmap length . succs) <=< succs where sortWith f l = map fst `fmap` sortBy (comparing snd) `fmap` mapM (\x -> (,) x `fmap` f x) l succs (i,j) = filterM (empty b) [ (i', j') | (dx,dy) <- [(1,2),(2,1)] , i' <- [i+dx,i-dx] , j' <- [j+dy, j-dy] , inRange ((1,1),(n,n)) (i',j') ] empty :: Board s -> Square -> ChessM r s Bool empty b s = fmap (<1) . lift $ readArray b s mark :: Square -> Int -> Board s -> ChessM r s () mark s k b = lift $ writeArray b s k tour :: Int -> Int -> ChessK r s -> Square -> Board s -> ChessM r s () tour n k exit s b | k > n*n = showBoard n b >>= exit | otherwise = successors n b s >>= mapM_ (\x -> do mark x k b tour n (k+1) exit x b -- failed mark x 0 b) showBoard :: Int -> Board s -> ChessM r s String showBoard n b = fmap unlines . forM [1..n] $ \i -> fmap unwords . forM [1..n] $ \j -> pad `fmap` lift (readArray b (i,j)) where k = ceiling . logBase 10 . fromIntegral $ n*n + 1 pad i = let s = show i in replicate (k-length s) ' ' ++ s main = do (n:_) <- map read `fmap` getArgs s <- stToIO . flip runContT return $ (do b <- lift $ newArray ((1,1),(n,n)) 0 mark (1,1) 1 b callCC $ \k -> tour n 2 k (1,1) b >> fail "No solution!") putStrLn s

## [edit] 3 LogicT monad

A very short implementation using the LogicT monad

16 lines of code. 7 imports.

import Control.Monad.Logic import Data.List import Data.Maybe import Data.Ord import Data.Ix import qualified Data.Map as Map import System.Environment successors n b = sortWith (length . succs) . succs where sortWith f = map fst . sortBy (comparing snd) . map (\x -> (x, f x)) succs (i,j) = [ (i', j') | (dx,dy) <- [(1,2),(2,1)] , i' <- [i+dx,i-dx] , j' <- [j+dy, j-dy] , isNothing (Map.lookup (i',j') b) , inRange ((1,1),(n,n)) (i',j') ] tour n k s b | k > n*n = return b | otherwise = do next <- msum . map return $ successors n b s tour n (k+1) next $ Map.insert next k b showBoard n b = unlines . map (\i -> unwords . map (\j -> pad . fromJust $ Map.lookup (i,j) b) $ [1..n]) $ [1..n] where k = ceiling . logBase 10 . fromIntegral $ n*n + 1 pad i = let s = show i in replicate (k-length s) ' ' ++ s main = do (n:_) <- map read `fmap` getArgs let b = observe . tour n 2 (1,1) $ Map.singleton (1,1) 1 putStrLn $ showBoard n b

## [edit] 4 Oleg Kiselyov's Solution

Oleg provided a solution on haskell-cafe:

It seems the following pure functional (except for the final printout) version of the search has almost the same performance as the Dan Doel's latest version with the unboxed arrays and callCC. For the board of size 40, Dan Doel's version takes 0.047s on my computer; the version below takes 0.048s. For smaller boards, the difference is imperceptible. Interestingly, the file sizes of the compiled executables (ghc -O2, ghc 6.8.2) are similar too: 606093 bytes for Dan Doel's version, and 605938 bytes for the version below. The version below is essentially Dan Doel's earlier version. Since the problem involves only pure search (rather than committed choice), I took the liberty of substituting FBackTrack (efficient MonadPlus) for LogicT. FBackTrack can too be made the instance of LogicT; there has not been any demand for that though.

import Data.List import Data.Ord import qualified Data.IntMap as Map import System.Environment import FBackTrack import Control.Monad -- Emulate the 2-dimensional map as a nested 1-dimensional map initmap n = Map.fromList $ (1,Map.singleton 1 1):[ (k,Map.empty) | k <- [2..n] ] notMember (i,j) m = Map.notMember j $ Map.findWithDefault undefined i m insrt (i,j) v m = Map.update (Just . Map.insert j v) i m lkup (i,j) m = Map.findWithDefault undefined j $ Map.findWithDefault undefined i m successors n b = sortWith (length . succs) . succs where sortWith f = map fst . sortBy (comparing snd) . map (\x -> (x, f x)) succs (i,j) = [ (i', j') | (dx,dy) <- [(1,2),(2,1)] , i' <- [i+dx,i-dx] , j' <- [j+dy, j-dy] , i' >= 1, j' >= 1, i' <= n, j' <= n , notMember (i',j') b ] tour n k s b | k > n*n = return b | otherwise = do next <- foldl1 mplus.map return $ successors n b s tour n (k+1) next $ insrt next k b showBoard n b = unlines . map (\i -> unwords . map (\j -> pad $ lkup (i,j) b) $ [1..n]) $ [1..n] where k = length . show $ n*n + 1 pad i = let s = show i in replicate (k-length s) ' ' ++ s main = do (n:_) <- map read `fmap` getArgs let (b:_) = runM Nothing . tour n 2 (1,1) $ initmap n putStrLn $ showBoard n b