# Sudoku

### From HaskellWiki

Here are a few sodoku solvers coded up in Haskell...

## Contents |

## 1 Serious, Non-Deterministic Solver

Here is a solver by CaleGibbard [1]. It possibly looks even more naïve than it actually is. This does a backtracking search, trying possibilities until it finds one which works, and backtracking when it can no longer make a legal move.

import MonadNondet (option) import Sudoku import System import Control.Monad forM = flip mapM solve = forM [(i,j) | i <- [1..9], j <- [1..9]] $ \(i,j) -> do v <- valAt (i,j) -- ^ for each board position when (v == 0) $ do -- if it's empty (we represent that with a 0) a <- option [1..9] -- pick a number place (i,j) a -- and try to put it there main = do [f] <- getArgs xs <- readFile f putStrLn $ evalSudoku $ do { readSudoku xs; solve; showSudoku }

Now, to the meat of the thing, the monad which makes the above look so nice. We construct a monad which is suitable for maintaining Sudoku grids and trying options nondeterministically. Note that outside of this module, it's impossible to create a state which has an invalid Sudoku grid, since the only way to update the state handles the check to ensure that the move is legal.

{-# OPTIONS_GHC -fglasgow-exts #-} module Sudoku (Sudoku, readSudoku, runSudoku, evalSudoku, execSudoku, showSudoku, valAt, rowAt, colAt, boxAt, place) where import Data.Array.Diff import MonadNondet import Control.Monad.State -- Nondet here is a drop-in replacement for [] (the list monad) which just runs a little faster. newtype Sudoku a = Sudoku (StateT (DiffUArray (Int,Int) Int) Nondet a) deriving (Functor, Monad, MonadPlus) {- -- That is, we could also use the following, which works exactly the same way. newtype Sudoku a = Sudoku (StateT (DiffUArray (Int,Int) Int) [] a) deriving (Functor, Monad, MonadPlus) -} initialSudokuArray = listArray ((1,1),(9,9)) [0,0..] runSudoku (Sudoku k) = runNondet (runStateT k initialSudokuArray) evalSudoku = fst . runSudoku execSudoku = snd . runSudoku showSudoku = Sudoku $ do a <- get return $ unlines [unwords [show (a ! (i,j)) | j <- [1..9]] | i <- [1..9]] readSudoku :: String -> Sudoku () readSudoku xs = sequence_ $ do (i,ys) <- zip [1..9] (lines xs) (j,n) <- zip [1..9] (words ys) return $ place (i,j) (read n) valAt' (i,j) = do a <- get return (a ! (i,j)) rowAt' (i,j) = mapM valAt' [(i, k) | k <- [1..9]] colAt' (i,j) = mapM valAt' [(k, j) | k <- [1..9]] boxAt' (i,j) = mapM valAt' [(i' + u, j' + v) | u <- [1..3], v <- [1..3]] where i' = ((i-1) `div` 3) * 3 j' = ((j-1) `div` 3) * 3 valAt = Sudoku . valAt' rowAt = Sudoku . rowAt' colAt = Sudoku . colAt' boxAt = Sudoku . boxAt' -- This is the least trivial part. -- It just guards to make sure that the move is legal, -- and updates the array in the state if it is. place :: (Int,Int) -> Int -> Sudoku () place (i,j) n = Sudoku $ do v <- valAt' (i,j) when (v == 0 && n /= 0) $ do rs <- rowAt' (i,j) cs <- colAt' (i,j) bs <- boxAt' (i,j) guard $ not . any (== n) $ rs ++ cs ++ bs a <- get put (a // [((i,j),n)])

This is a fast NonDeterminism monad. It's a drop-in replacement for the list monad in this case. It's twice as fast when compiled with optimisations but a little slower without. You can also find it on the wiki at NonDeterminism.

I've made a few small modifications to this one to hopefully make it more concretely readable.

{-# OPTIONS_GHC -fglasgow-exts #-} module MonadNondet where import Control.Monad import Control.Monad.Trans import Control.Monad.Identity newtype NondetT m a = NondetT { foldNondetT :: (forall b. (a -> m b -> m b) -> m b -> m b) } runNondetT :: (Monad m) => NondetT m a -> m a runNondetT m = foldNondetT m (\x xs -> return x) (error "No solution found.") instance (Functor m) => Functor (NondetT m) where fmap f (NondetT g) = NondetT (\cons nil -> g (cons . f) nil) instance (Monad m) => Monad (NondetT m) where return a = NondetT (\cons nil -> cons a nil) m >>= k = NondetT (\cons nil -> foldNondetT m (\x -> foldNondetT (k x) cons) nil) instance (Monad m) => MonadPlus (NondetT m) where mzero = NondetT (\cons nil -> nil) m1 `mplus` m2 = NondetT (\cons -> foldNondetT m1 cons . foldNondetT m2 cons) instance MonadTrans NondetT where lift m = NondetT (\cons nil -> m >>= \a -> cons a nil) newtype Nondet a = Nondet (NondetT Identity a) deriving (Functor, Monad, MonadPlus) runNondet (Nondet x) = runIdentity (runNondetT x) foldNondet :: Nondet a -> (a -> b -> b) -> b -> b foldNondet (Nondet nd) cons nil = runIdentity $ foldNondetT nd (\x xs -> return (cons x (runIdentity xs))) (return nil) option :: (MonadPlus m) => [a] -> m a option = msum . map return

## 2 Simple Solver

By AlsonKemp. This solver is probably similar to Cale's but I don't grok the non-deterministic monad...

Note: this solver is exhaustive and will output all of the solutions, not just the first one. In order to make it non-exchaustive, add a case statement to solve' in order to check "r" and branch on the result.

import System import Control.Monad import Data.List import Data.Array.IO type SodokuBoard = IOArray Int Int main = do [f] <- getArgs a <- newArray (1, 81) 0 readFile f >>= readSodokuBoard a putStrLn "Original:" printSodokuBoard a putStrLn "Solutions:" solve a (1,1) readSodokuBoard a xs = sequence_ $ do (i,ys) <- zip [1..9] (lines xs) (j,n) <- zip [1..9] (words ys) return $ writeBoard a (j,i) (read n) printSodokuBoard a = let printLine a y = mapM (\x -> readBoard a (x,y)) [1..9] >>= mapM_ (putStr . show) in putStrLn "-----------" >> mapM_ (\y -> putStr "|" >> printLine a y >> putStrLn "|") [1..9] >> putStrLn "-----------" -- the meat of the program. Checks the current square. -- If 0, then get the list of nums and try to "solve' " -- Otherwise, go to the next square. solve :: SodokuBoard -> (Int, Int) -> IO (Maybe SodokuBoard) solve a (10,y) = solve a (1,y+1) solve a (_, 10)= printSodokuBoard a >> return (Just a) solve a (x,y) = do v <- readBoard a (x,y) case v of 0 -> availableNums a (x,y) >>= solve' a (x,y) _ -> solve a (x+1,y) -- solve' handles the backtacking where solve' a (x,y) [] = return Nothing solve' a (x,y) (v:vs) = do writeBoard a (x,y) v -- put a guess onto the board r <- solve a (x+1,y) writeBoard a (x,y) 0 -- remove the guess from the board solve' a (x,y) vs -- recurse over the remainder of the list -- get the "taken" numbers from a row, col or box. getRowNums a y = sequence [readBoard a (x',y) | x' <- [1..9]] getColNums a x = sequence [readBoard a (x,y') | y' <- [1..9]] getBoxNums a (x,y) = sequence [readBoard a (x'+u, y'+v) | u <- [0..2], v <- [0..2]] where x' = (3 * ((x-1) `quot` 3)) + 1 y' = (3 * ((y-1) `quot` 3)) + 1 -- return the numbers that are available for a particular square availableNums a (x,y) = do r <- getRowNums a y c <- getColNums a x b <- getBoxNums a (x,y) return $ [0..9] \\ (r `union` c `union` b) -- aliases of read and write array that flatten the index readBoard a (x,y) = readArray a (x+9*(y-1)) writeBoard a (x,y) e = writeArray a (x+9*(y-1)) e

## 3 Complete decision tree

By Henning Thielemann.

module Sudoku where {- This is inspired by John Hughes "Why Functional Programming Matters". We build a complete decision tree. That is, all alternatives in a certain depth have the same number of determined values. At the bottom of the tree all possible solutions can be found. Actually the algorithm is very stupid: In each depth we look for the field with the least admissible choices of numbers and prune the alternative branches for the other fields. -} import Data.Char (ord, chr) import Data.Array (Array, range, (!), (//)) import Data.Tree (Tree) import qualified Data.Tree as Tree import Data.List (sort, minimumBy) import Data.Maybe (catMaybes, isNothing, fromMaybe, fromJust) import qualified Data.Array as Array {- Example: ghci -Wall Sudoku.hs *Sudoku> mapM putCLn (solutions exampleHawiki0) -} {- [[ATree]] contains a list of possible alternatives for each position -} data ATree a = ANode T [[ATree a]] type Coord = Int type Address = (Int,Int,Int,Int) type Element = Int type T = Array Address (Maybe Element) type Complete = Array Address Element fieldBounds :: (Address, Address) fieldBounds = ((0,0,0,0), (2,2,2,2)) squareRange :: [(Coord, Coord)] squareRange = range ((0,0), (2,2)) alphabet :: [Element] alphabet = [1..9] {- * solution -} {- Given two sorted lists, remove the elements of the first list from the second one. -} deleteSorted :: Ord a => [a] -> [a] -> [a] deleteSorted [] ys = ys deleteSorted _ [] = [] deleteSorted (x:xs) (y:ys) = case compare x y of EQ -> deleteSorted xs ys LT -> deleteSorted xs (y:ys) GT -> y : deleteSorted (x:xs) ys admissibleNumbers :: [[Maybe Element]] -> [Element] admissibleNumbers = foldl (flip deleteSorted) alphabet . map (sort . catMaybes) admissibleAdditions :: T -> Address -> [Element] admissibleAdditions sudoku (i,j,k,l) = admissibleNumbers (map ($ sudoku) [selectRow (i,k), selectColumn (j,l), selectSquare (i,j)]) allAdmissibleAdditions :: T -> [(Address, [Element])] allAdmissibleAdditions sudoku = let adds addr = (addr, admissibleAdditions sudoku addr) in map adds (map fst (filter (isNothing . snd) (Array.assocs sudoku))) solutionTree :: T -> ATree T solutionTree sudoku = let new (addr,elms) = map (\elm -> solutionTree (sudoku // [(addr, Just elm)])) elms in ANode sudoku (map new (allAdmissibleAdditions sudoku)) treeAltToStandard :: ATree T -> Tree T treeAltToStandard (ANode sudoku subs) = Tree.Node sudoku (concatMap (map treeAltToStandard) subs) {- Convert a tree with alternatives for each position (ATree) into a normal tree by choosing one position and its alternative values. We need to consider only one position per level because the remaining positions are processed in the sub-levels. With other words: Choosing more than one position would lead to multiple reports of the same solution. For reasons of efficiency we choose the position with the least number of alternatives. If this number is zero, the numbers tried so far are wrong. If this number is one, then the choice is unique, but maybe still wrong. If the number of alternatives is larger, we have to check each alternative. -} treeAltToStandardOptimize :: ATree T -> Tree T treeAltToStandardOptimize (ANode sudoku subs) = let chooseMinLen [] = [] chooseMinLen xs = minimumBy compareLength xs in Tree.Node sudoku (chooseMinLen (map (map treeAltToStandardOptimize) subs)) maybeComplete :: T -> Maybe Complete maybeComplete sudoku = fmap (Array.array fieldBounds) (mapM (uncurry (fmap . (,))) (Array.assocs sudoku)) {- All leafs are at the same depth, namely the number of undetermined fields. That's why we can safely select all Sudokus at the lowest level. -} solutions :: T -> [Complete] solutions sudoku = let err = error "The lowest level should contain complete Sudokus only." {- "last'" is more efficient than "last" here because the program does not have to check whether deeper levels exist. We know that the tree is as deep as the number of undefined fields. This means that dropMatch returns a singleton list. We don't check that because then we would lose the efficiency again. -} last' = head . dropMatch (filter isNothing (Array.elems sudoku)) in map (fromMaybe err . maybeComplete) (last' (Tree.levels (treeAltToStandardOptimize (solutionTree sudoku)))) {- * transformations (can be used for construction, too) -} standard :: Complete standard = Array.listArray fieldBounds (map (\(i,j,k,l) -> mod (j+k) 3 * 3 + mod (i+l) 3 + 1) (range fieldBounds)) exampleHawiki0, exampleHawiki1 :: T exampleHawiki0 = fromString (unlines [ " 5 6 1", " 48 7 ", "8 52", "2 57 3 ", " ", " 3 69 5", "79 8", " 1 65 ", "5 3 6 " ]) exampleHawiki1 = fromString (unlines [ " 6 8 ", " 2 ", " 1 ", " 7 1 2", "5 3 ", " 4 ", " 42 1 ", "3 7 6 ", " 5 " ]) check :: Complete -> Bool check sudoku = let checkParts select = all (\addr -> sort (select addr sudoku) == alphabet) squareRange in all checkParts [selectRow, selectColumn, selectSquare] selectRow, selectColumn, selectSquare :: (Coord,Coord) -> Array Address element -> [element] selectRow (i,k) sudoku = map (sudoku!) (range ((i,0,k,0), (i,2,k,2))) -- map (sudoku!) (map (\(j,l) -> (i,j,k,l)) squareRange) selectColumn (j,l) sudoku = map (sudoku!) (range ((0,j,0,l), (2,j,2,l))) selectSquare (i,j) sudoku = map (sudoku!) (range ((i,j,0,0), (i,j,2,2))) {- * conversion from and to strings -} put, putLn :: T -> IO () put sudoku = putStr (toString sudoku) putLn sudoku = putStrLn (toString sudoku) putC, putCLn :: Complete -> IO () putC sudoku = putStr (toString (fmap Just sudoku)) putCLn sudoku = putStrLn (toString (fmap Just sudoku)) fromString :: String -> T fromString str = Array.array fieldBounds (concat (zipWith (\(i,k) -> map (\((j,l),x) -> ((i,j,k,l),x))) squareRange (map (zip squareRange . map charToElem) (lines str)))) toString :: T -> String toString sudoku = unlines (map (\(i,k) -> map (\(j,l) -> elemToChar (sudoku!(i,j,k,l))) squareRange) squareRange) charToElem :: Char -> Maybe Element charToElem c = toMaybe ('0'<=c && c<='9') (ord c - ord '0') elemToChar :: Maybe Element -> Char elemToChar = maybe ' ' (\c -> chr (ord '0' + c)) {- * helper functions -} nest :: Int -> (a -> a) -> a -> a nest 0 _ x = x nest n f x = f (nest (n-1) f x) toMaybe :: Bool -> a -> Maybe a toMaybe False _ = Nothing toMaybe True x = Just x compareLength :: [a] -> [b] -> Ordering compareLength (_:xs) (_:ys) = compareLength xs ys compareLength [] [] = EQ compareLength (_:_) [] = GT compareLength [] (_:_) = LT {- | Drop as many elements as the first list is long -} dropMatch :: [b] -> [a] -> [a] dropMatch xs ys = map fromJust (dropWhile isNothing (zipWith (toMaybe . null) (iterate (drop 1) xs) ys))

## 4 Add Your Own

If you have a Sudoku solver you're proud of, put it here. This ought to be a good way of helping people learn some fun, intermediate-advanced techniques in Haskell.

## 5 Test Boards

Here's an input file to test the solvers on. Zeroes represent blanks.

0 5 0 0 6 0 0 0 1 0 0 4 8 0 0 0 7 0 8 0 0 0 0 0 0 5 2 2 0 0 0 5 7 0 3 0 0 0 0 0 0 0 0 0 0 0 3 0 6 9 0 0 0 5 7 9 0 0 0 0 0 0 8 0 1 0 0 0 6 5 0 0 5 0 0 0 3 0 0 6 0

A nefarious one:

0 0 0 0 6 0 0 8 0 0 2 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 7 0 0 0 0 1 0 2 5 0 0 0 3 0 0 0 0 0 0 0 0 0 0 4 0 0 0 0 4 2 0 1 0 0 0 3 0 0 7 0 0 6 0 0 0 0 0 0 0 0 0 5 0

Chris Kuklewicz writes, "You can go get the 36,628 distict minimal puzzles from csse.uwa.edu that have only 17 clues. Then you can run all of them through your program to locate the most evil ones, and use them on your associates."