Sudoku
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(Added list comprehension solver.) 

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Revision as of 08:36, 11 April 2012
Here are a few Sudoku solvers coded up in Haskell...
1 Monadic nondeterministic solver
Here is a solver by CaleGibbard. 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 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 fglasgowexts #} 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 dropin 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' = ((i1) `div` 3) * 3 j' = ((j1) `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 $ (n `notElem`) $ rs ++ cs ++ bs a < get put (a // [((i,j),n)])
This is a fast NonDeterminism monad. It's a dropin 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 fglasgowexts #} 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 nondeterministic monad...
Note: this solver is exhaustive and will output all of the solutions, not just the first one. In order to make it nonexhaustive, 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 do 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 * ((x1) `quot` 3)) + 1 y' = (3 * ((y1) `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*(y1)) writeBoard a (x,y) e = writeArray a (x+9*(y1)) 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, isDigit, digitToInt, intToDigit) 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 sublevels. 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 (isDigit c) (digitToInt c) elemToChar :: Maybe Element > Char elemToChar = maybe ' ' intToDigit { * helper functions } nest :: Int > (a > a) > a > a nest 0 _ x = x nest n f x = f (nest (n1) 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 No guessing
By Simon Peyton Jones.
Since this page is here I thought I'd add a solver I wrote sometime last year. The main constraint I imposed is that it never guesses, and that it outputs a humancomprehensible explanation of every step of its reasoning. That means there are some puzzles it can't solve. I'd be interested to know if there are any puzzles that it gets stuck on where there is a noguessing way forward. I made no attempt to make it fast.
There are two files: Media:SudokuPJ.hs and Media:TestPJ.hs. The latter just contains a bunch of test cases; I was too lazy to write a proper parser.
The main entry point is:
run1 :: Verbosity > [String] > Doc data Verbosity = All  Terse  Final
The [String] the starting board configuration (see the tests file).
5 Just guessing
By ChrisKuklewicz
This solver is an implementation of Knuth's "Dancing Links" algorithm for solving binarycover problems. This algorithm represents the constraints as a sparse binary matrix, with 1's as linked nodes. The nodes are in a vertical and a horizontal doubly linked list, and each vertical list is headed by another node that represents one of the constraints. It is interesting as an example of the rare beast in Haskell: a mutable data structure. The code has been rewritten and cleaned up here Media:DancingSudoku.lhs. Its main routine is designed to handle the input from sudoku17 on stdin. Currently it only returns the first solution or calls an error, it can be modified (see comments in the file) to return all solutions in a list. An earlier version used ST.Lazy instead of ST.Strict which made operating on puzzles with many solutions more tractable.
Other trivia: It uses "mdo" and lazyness to initialize some of the doubly linked lists.
6 Very smart, with only a little guessing
by ChrisKuklewicz
This solver does its best to avoid the branch and guess approach. On the 36628 puzzles of length 17 it resorts to guessing on only 164. This extra strength comes from examining the constraints that can only be solved in exactly two ways, and how these constraints overlap and interact with each other and remaining possibilities.
The source code compiles to take a list of puzzles as input and produces a description of the number of (good and total) guesses required, as well as a shuffled version of the input. If there was guessing, then the shuffled version could be sent back into the solver to see how the difficulty depended on luck. The list of 164 hard puzzles is included with the source code. The Deduce.hs file contains comments.
The data is stored in a 9x9x9 boolean array, and the only operations are turning off possibilities and branching. For performance the array is thawed, mutated, and frozen. On the set of 36628 puzzles the speed averages 9.4 puzzles solved per second on a 1.33 GHz G4 (ghc6.4.1 on OS X). I liked the 9x9x9 array since it emphasized the symmetry of the problem.
7 Only guessing without dancing links
by AndrewBromage
This solver uses a different implementation of Knuth's algorithm, without using pointers. It instead relies on the fact that in Haskell, treelike data structure (in this case, a Priority Search Queue) "undo" operations are essentially free.
8 Generalized solver
By Thorkil Naur
This Su Doku solver is able to solve classes of Su Doku puzzles that extend the ordinary 9*9 puzzles. The documentation describes the solver and also some (to the present author at least) surprising properties of various reduction strategies used when solving Su Doku puzzles.
The following files comprise the Su Doku solver and related code:
Media:Format.hs Media:Merge.hs Media:SdkMSol2.hs Media:SortByF.hs Media:SuDoku.hs Media:t40.hs Media:t44.hs Media:Test.hs
For an example of use, the command
runhugs SdkMSol2 \ tn1 \ Traditional 3 \ #123456789 \ 1539 \ 6 \ 271 \ 82 \ 487 \ 53 \ 23 \ 759 \ 684
produces output that, among other things, contain
tn1: Solutions: 1 7 5 3 2 8 4 9 6 9 4 2 6 7 1 3 8 5 3 6 8 5 9 4 2 7 1 8 2 9 1 3 5 6 4 7 6 5 3 4 8 7 9 1 2 7 1 4 9 6 2 5 3 8 2 3 1 8 4 6 7 5 9 4 8 7 2 5 9 1 6 3 5 9 6 7 1 3 8 2 4
9 Simple small solver
I haven't looked at the other solvers in detail yet, so I'm not sure what is good or bad about mine, but here it is:
http://darcs.brianweb.net/sudoku/Sudoku.pdf http://darcs.brianweb.net/sudoku/src/Sudoku.lhs
Brian Alliet <brian@brianweb.net>
10 Backtrack monad solver
This is a simple but fast solver that uses standard monads from the MonadTemplateLibrary in the StandardLibraries.
Besides being Yet Another Example of a Sudoko solver, I think it is also a nice somewhatnontrivial example of monads in practice.
The idea is that the monad StateT s [] does backtracking. It means "iterate over a list while keeping state, but reinitialize to the original state on each iteration".
I have several (Unix command line) frontends to this module, available upon request. The one I use most creates and prints six new Sudoku puzzles on a page, with finegrain control over the difficulty of the puzzle. This has made me quite popular among friends and extended family.
 YitzGale
{# OPTIONS_GHC fglasgowexts #}  Solve a Sudoku puzzle module Sudoku where import Control.Monad.State import Data.Maybe (maybeToList) import Data.List (delete) type Value = Int type Cell = (Int, Int)  Onebased coordinates type Puzzle = [[Maybe Value]] type Solution = [[Value]]  The size of the puzzle. sqrtSize :: Int sqrtSize = 3 size = sqrtSize * sqrtSize  Besides the rows and columns, a Sudoku puzzle contains s blocks  of s cells each, where s = size. blocks :: [[Cell]] blocks = [[(x + i, y + j)  i < [1..sqrtSize], j < [1..sqrtSize]]  x < [0,sqrtSize..sizesqrtSize], y < [0,sqrtSize..sizesqrtSize]]  The onebased number of the block that a cell is contained in. blockNum :: Cell > Int blockNum (row, col) = row  (row  1) `mod` sqrtSize + (col  1) `div` sqrtSize  When a Sudoku puzzle has been partially filled in, the following  data structure represents the remaining options for how to proceed. data Options = Options { cellOpts :: [[[Value]]],  For each cell, a list of possible values rowOpts :: [[[Cell ]]],  For each row and value, a list of cells colOpts :: [[[Cell ]]],  For each column and value, a list of cells blkOpts :: [[[Cell ]]]  For each block and value, a list of cells } deriving Show modifyCellOpts f = do {opts < get; put $ opts {cellOpts = f $ cellOpts opts}} modifyRowOpts f = do {opts < get; put $ opts {rowOpts = f $ rowOpts opts}} modifyColOpts f = do {opts < get; put $ opts {colOpts = f $ colOpts opts}} modifyBlkOpts f = do {opts < get; put $ opts {blkOpts = f $ blkOpts opts}}  The full set of initial options, before any cells are constrained initOptions :: Options initOptions = Options { cellOpts = [[[1..size]  _ < [1..size]]  _ < [1..size]], rowOpts = [[[(r, c)  c < [1..size]]  _ < [1..size]]  r < [1..size]], colOpts = [[[(r, c)  r < [1..size]]  _ < [1..size]]  c < [1..size]], blkOpts = [[b  _ < [1..size]]  b < blocks]} solve :: Puzzle > [Solution] solve puz = evalStateT (initPuzzle >> solutions) initOptions where initPuzzle = sequence_ [fixCell v (r, c)  (row, r) < zip puz [1..], (val, c) < zip row [1..], v < maybeToList val]  Build a list of all possible solutions given the current options.  We use a list monad INSIDE a state monad. That way,  the state is reinitialized on each element of the list iteration,  allowing backtracking when an attempt fails (with mzero). solutions :: StateT Options [] Solution solutions = solveFromRow 1 where solveFromRow r  r > size = return []  otherwise = do row < solveRowFromCol r 1 rows < solveFromRow $ r + 1 return $ row : rows solveRowFromCol r c  c > size = return []  otherwise = do vals < gets $ (!! (c  1)) . (!! (r  1)) . cellOpts val < lift vals fixCell val (r, c) row < solveRowFromCol r (c + 1) return $ val : row  Fix the value of a cell.  More specifically  update Options to reflect the given value at  the given cell, or mzero if that is not possible. fixCell :: (MonadState Options m, MonadPlus m) => Value > Cell > m () fixCell val cell@(row, col) = do vals < gets $ (!! (col  1)) . (!! (row  1)) . cellOpts guard $ val `elem` vals modifyCellOpts $ replace2 row col [val] modifyRowOpts $ replace2 row val [cell] modifyColOpts $ replace2 col val [cell] modifyBlkOpts $ replace2 blk val [cell] sequence_ [constrainCell v cell  v < [1..size], v /= val] sequence_ [constrainCell val (row, c)  c < [1..size], c /= col] sequence_ [constrainCell val (r, col)  r < [1..size], r /= row] sequence_ [constrainCell val c  c < blocks !! (blk  1), c /= cell] where blk = blockNum cell  Assert that the given value cannot occur in the given cell.  Fail with mzero if that means that there are no options left. constrainCell :: (MonadState Options m, MonadPlus m) => Value > Cell > m () constrainCell val cell@(row, col) = do constrainOpts row col val cellOpts modifyCellOpts (flip fixCell cell) constrainOpts row val cell rowOpts modifyRowOpts (fixCell val) constrainOpts col val cell colOpts modifyColOpts (fixCell val) constrainOpts blk val cell blkOpts modifyBlkOpts (fixCell val) where blk = blockNum cell constrainOpts x y z getOpts modifyOpts fixOpts = do zs < gets $ (!! (y  1)) . (!! (x  1)) . getOpts case zs of [z'] > guard (z' /= z) [_,_] > when (z `elem` zs) $ fixOpts (head $ delete z zs) (_:_) > modifyOpts $ replace2 x y (delete z zs) _ > mzero  Replace one element of a list.  Coordinates are 1based. replace :: Int > a > [a] > [a] replace i x (y:ys)  i > 1 = y : replace (i  1) x ys  otherwise = x : ys replace _ _ _ = []  Replace one element of a 2dimensional list.  Coordinates are 1based. replace2 :: Int > Int > a > [[a]] > [[a]] replace2 i j x (y:ys)  i > 1 = y : replace2 (i  1) j x ys  otherwise = replace j x y : ys replace2 _ _ _ _ = []
11 Inflight entertainment
By Lennart Augustsson
When on a Lufthansa transatlantic flight in 2005 I picked up the inflight magazine and found a Sudoku puzzle. I decided to finally try one. After solving half of it by hand I got bored. Surely, this mechanical task is better performed by a machine? So I pulled out my laptop and wrote a Haskell program.
The program below is what I wrote on the plane, except for some comments that I've added. I have made no attempt as making it fast, so the nefarious test puzzle below takes a minute to solve.
First, the solver without user interface:
module Sudoku(Square, Board, ColDigit, RowDigit, BoxDigit, Digit, initialBoard, getBoard, mkSquare, setSquare, solveMany) where import Char(intToDigit, digitToInt) import List ((\\), sortBy)  A board is just a list of Squares. It always has all the squares. data Board = Board [Square] deriving (Show)  A Square contains its column (ColDigit), row (RowDigit), and  which 3x3 box it belongs to (BoxDigit). The box can be computed  from the row and column, but is kept for speed.  A Square also contains it's status: either a list of possible  digits that can be placed in the square OR a fixed digit (i.e.,  the square was given by a clue or has been solved). data Square = Square ColDigit RowDigit BoxDigit (Either [Digit] Digit) deriving (Show) type ColDigit = Digit type RowDigit = Digit type BoxDigit = Digit type Digit = Char  '1' .. '9'  The initial board, no clues given so all digits are possible in all squares. initialBoard :: Board initialBoard = Board [ Square col row (boxDigit col row) (Left allDigits)  row < allDigits, col < allDigits ]  Return a list of rows of a solved board.  If used on an unsolved board the return value is unspecified. getBoard :: Board > [[Char]] getBoard (Board sqs) = [ [ getDigit d  Square _ row' _ d < sqs, row' == row ]  row < allDigits ] where getDigit (Right d) = d getDigit _ = '0' allDigits :: [Char] allDigits = ['1' .. '9']  Compute the box from a column and row. boxDigit :: ColDigit > RowDigit > BoxDigit boxDigit c r = intToDigit $ (digitToInt c  1) `div` 3 + (digitToInt r  1) `div` 3 * 3 + 1  Given a column, row, and a digit make a (solved) square representing this. mkSquare :: ColDigit > RowDigit > Digit > Square mkSquare col row c  col `elem` allDigits && row `elem` allDigits && c `elem` allDigits = Square col row (boxDigit col row) (Right c) mkSquare _ _ _ = error "Bad mkSquare"  Place a given Square on a Board and return the new Board.  Illegal setSquare calls will just error out. The main work here  is to remove the placed digit from the other Squares on the board  that are in the same column, row, or box. setSquare :: Square > Board > Board setSquare sq@(Square scol srow sbox (Right d)) (Board sqs) = Board (map set sqs) where set osq@(Square col row box ds) = if col == scol && row == srow then sq else if col == scol  row == srow  box == sbox then (Square col row box (sub ds)) else osq sub (Left ds) = Left (ds \\ [d]) sub (Right d')  d == d' = error "Impossible setSquare" sub dd = dd setSquare _ _ = error "Bad setSquare"  Get the unsolved Squares from a Board. getLeftSquares :: Board > [Square] getLeftSquares (Board sqs) = [ sq  sq@(Square _ _ _ (Left _)) < sqs ]  Given an initial Board return all the possible solutions starting  from that Board.  Note, this all happens in the list monad and makes use of lazy evaluation  to avoid work. Using the list monad automatically handles all the backtracking  and enumeration of solutions. solveMany :: Board > [Board] solveMany brd = case getLeftSquares brd of [] > return brd  Nothing unsolved remains, we are done. sqs > do  Sort the unsolved Squares by the ascending length of the possible  digits. Pick the first of those so we always solve forced Squares  first. let Square c r b (Left ds) : _ = sortBy leftLen sqs leftLen (Square _ _ _ (Left ds1)) (Square _ _ _ (Left ds2)) = compare (length ds1) (length ds2) leftLen _ _ = error "bad leftLen" sq < [ Square c r b (Right d)  d < ds ]  Try all possible moves solveMany (setSquare sq brd)  And solve the extended Board.
Second, a simple user interface (a different user interface that I have is an Excell addin):
module Main where import Sudoku  Col Row Digit solve :: [((Char, Char), Char)] > [[Char]] solve crds = let brd = foldr add initialBoard crds add ((c, r), d) = setSquare (mkSquare c r d) in case solveMany brd of [] > error "No solutions" b : _ > getBoard b  The parse assumes that squares without a clue  contain '0'. main = interact $ unlines .  turn it into lines map (concatMap (:" ")) .  add a space after each digit for readability solve .  solve the puzzle filter ((`elem` ['1'..'9']) . snd) .  get rid of nonclues zip [ (c, r)  r < ['1'..'9'], c < ['1'..'9'] ] .  pair up the digits with their coordinates filter (`elem` ['0'..'9'])  get rid of nondigits
12 Sudoku incrementally, à la Bird
 As part of a new Advanced Functional Programming course in Nottingham, Graham Hutton presented a Haskell approach to solving Sudoku puzzles, based upon notes from Richard Bird. The approach is classic Bird: start with a simple but impractical solver, whose efficiency is then improved in a series of steps. The end result is an elegant program that is able to solve any Sudoku puzzle in an instant. Its also an excellent example of what has been termed wholemeal programming focusing on entire data structures rather than their elements. (Transplanted from LtU.)
A full talkthrough of the evolution of the code may be found under the course page. Liyang 13:35, 27 July 2006 (UTC)
I've also written Media:sudokuWss.hs, a parallel version of this solver. It uses STM to prune the boxes, columns, and rows simultaneously, which is kind of cool. I'm pretty sure it can be optimized quite a bit... WouterSwierstra, August 2007.
13 607 bytes / 12 lines
A super quick attempt at a smallest solution, based on the 707 byte sudoku solver:
import List main = putStr . unlines . map disp . solve . return . input =<< getContents solve s = foldr (\p l > [mark (p,n) s  s < l, n < s p]) s idx mark (p@(i,j),n) s q@(x,y)  p == q = [n]  x == i  y == j  e x i && e y j = delete n (s q)  otherwise = s q where e a b = div (a1) 3 == div (b1) 3 disp s = unlines [unwords [show $ head $ s (i,j)  j < [1..9]]  i < [1..9]] input s = foldr mark (const [1..9]) $ [(p,n)  (p,n) < zip idx $ map read $ lines s >>= words, n>0] idx = [(i,j)  i < [1..9], j < [1..9]]
dons 07:54, 2 December 2006 (UTC)
14 A parallel solver
A parallel version of Richard Bird's function pearl solver by Wouter Swierstra:
http://www.haskell.org/sitewiki/images/1/12/SudokuWss.hs
15 Another simple solver
One day I wrote a completely naive sudoku solver which tried all possibilities to try arrays in Haskell. It works, however I doubt that I'll see it actually solve a puzzle during my remaining lifetime.
So I set out to improve it. The new version still tries all possibilities, but it starts with the cell that has a minimal number of possibilities.
import Array import List import System  ([Possible Entries], #Possible Entries) type Field = Array (Int,Int) ([Int], Int)  Fields are Strings of Numbers with 0 in empty cells readField ::String > Field readField f = listArray ((1,1),(9,9)) (map (\j > let n=read [j]::Int in if n==0 then ([0..9],9) else ([n],0)) f)  x y wrong way > reading wrong? no effect on solution though showField :: Field > String showField f = unlines [concat [show $ entry (f!(y,x))x<[1..9]]y<[1..9]] printField :: Maybe Field > String printField (Just f) = concat [concat [show $ entry f!(y,x))x<[1..9]]y<[1..9]] printField Nothing = "No solution"  true if cell is empty isEmpty :: ([Int],Int) > Bool isEmpty (xs,_) = xs == [0] entry :: ([Int],Int) > Int entry = head.fst  0 possibilties left, no emtpy fields done :: Field > Bool done a = let l=elems a in 0==foldr (\(_,x) y > x+y) 0 l && all (not.isEmpty) l return column/row/square containing coords (x,y), excluding (x,y) column::Field >(Int,Int) > [Int] column a ~(x,y)= [entry $ a!(i,y)i<[1..9],i/=x] row :: Field > (Int,Int) > [Int] row a ~(x,y)= [entry $ a!(x,j)j<[1..9],j/=y] square :: Field > (Int, Int)> [Int] square a ~(x,y) = block where n = head $ dropWhile (<x3) [0,3,6] m = head $ dropWhile (<y3) [0,3,6] block = [entry $ a!(i+n,j+m)i<[1..3],j<[1..3],x/=i+n  y/=j+m]  remove invalid possibilities remPoss :: Field > Field remPoss f =array ((1,1),(9,9)) $ map remPoss' (assocs f) where others xy= filter (/=0) $ row f xy ++ column f xy ++ square f xy remPoss' ~(i,(xs,n))  n/=0 = let nxs= filter ( `notElem` others i ) xs in (i,(nxs,length $ filter (/=0) nxs))  otherwise = (i,(xs,n))  remove invalid fields, i.e. contains empty cell without filling possibilities remInv :: [Field] > [Field] remInv = filter (all (\(_,(x,_)) > x/=[0]).assocs) genMoves :: (Int,Int) > Field > [Field] genMoves xy f = remInv $ map remPoss [f // [(xy,([poss!!i],0))]i<[0..num1]] where poss = tail $ fst (f!xy) num = snd (f!xy) always try the entry with least possibilties first moves :: Field > [Field] moves f = genMoves bestOne f where  remove all with 0 possibilities, select the one with minimum possibilities bestOne =fst $ minimumBy (\(_,(_,n)) (_,(_,m)) > compare n m) list list = ((filter (\(_,(_,x)) > x/=0).assocs) f) play :: [Field] > Maybe Field play (f:a)  done f= Just f  otherwise = play (moves f++a) play [] = Nothing  reads a file with puzzles, path as argument main :: IO () main = do path < getArgs inp<readFile (path!!0) let x=lines inp let erg=map (printField.play) (map ((\x>[x]).remPoss.readField) x) writeFile "./out.txt" (unlines erg)
I let it run on the 41747 minimal puzzles. On a 2.66 GHz Intel Xeon it took 15441m1.920s, which is about 22 seconds per puzzle. It could probably be further improved by making remPoss smarter. At the time of writing this the naive version from which I started is crunching for 20 days on a simple puzzle with 32 hints. I'd say that's quite a performance improvement.
16 Constraint Propagation (a la Norvig)
By Manu
This is an Haskell implementation of Peter Norvig's sudoku solver (http://norvig.com/sudoku.html). It should solve, in a flash, the 95 puzzles found here : http://norvig.com/top95.txt Thanks to Daniel Fischer for helping and refactoring.
module Main where import Data.List hiding (lookup) import Data.Array import Control.Monad import Data.Maybe  Types type Digit = Char type Square = (Char,Char) type Unit = [Square]  We represent our grid as an array type Grid = Array Square [Digit]  Setting Up the Problem rows = "ABCDEFGHI" cols = "123456789" digits = "123456789" box = (('A','1'),('I','9')) cross :: String > String > [Square] cross rows cols = [ (r,c)  r < rows, c < cols ] squares :: [Square] squares = cross rows cols  [('A','1'),('A','2'),('A','3'),...] peers :: Array Square [Square] peers = array box [(s, set (units!s))  s < squares ] where set = nub . concat unitlist :: [Unit] unitlist = [ cross rows [c]  c < cols ] ++ [ cross [r] cols  r < rows ] ++ [ cross rs cs  rs < ["ABC","DEF","GHI"], cs < ["123","456","789"]]  this could still be done more efficiently, but what the heck... units :: Array Square [Unit] units = array box [(s, [filter (/= s) u  u < unitlist, s `elem` u ])  s < squares] allPossibilities :: Grid allPossibilities = array box [ (s,digits)  s < squares ]  Parsing a grid into an Array parsegrid :: String > Maybe Grid parsegrid g = do regularGrid g foldM assign allPossibilities (zip squares g) where regularGrid :: String > Maybe String regularGrid g = if all (`elem` "0.123456789") g then Just g else Nothing  Propagating Constraints assign :: Grid > (Square, Digit) > Maybe Grid assign g (s,d) = if d `elem` digits  check that we are assigning a digit and not a '.' then do let ds = g ! s toDump = delete d ds foldM eliminate g (zip (repeat s) toDump) else return g eliminate :: Grid > (Square, Digit) > Maybe Grid eliminate g (s,d) = let cell = g ! s in if d `notElem` cell then return g  already eliminated  else d is deleted from s' values else do let newCell = delete d cell newV = g // [(s,newCell)] newV2 < case newCell of  contradiction : Nothing terminates the computation [] > Nothing  if there is only one value left in s, remove it from peers [d'] > do let peersOfS = peers ! s foldM eliminate newV (zip peersOfS (repeat d'))  else : return the new grid _ > return newV  Now check the places where d appears in the peers of s foldM (locate d) newV2 (units ! s) locate :: Digit > Grid > Unit > Maybe Grid locate d g u = case filter ((d `elem`) . (g !)) u of [] > Nothing [s] > assign g (s,d) _ > return g  Search search :: Grid > Maybe Grid search g = case [(l,(s,xs))  (s,xs) < assocs g, let l = length xs, l /= 1] of [] > return g ls > do let (_,(s,ds)) = minimum ls msum [assign g (s,d) >>= search  d < ds] solve :: String > Maybe Grid solve str = do grd < parsegrid str search grd  Display solved grid printGrid :: Grid > IO () printGrid = putStrLn . gridToString gridToString :: Grid > String gridToString g = let l0 = elems g  [("1537"),("4"),...] l1 = (map (\s > " " ++ s ++ " ")) l0  ["1 "," 2 ",...] l2 = (map concat . sublist 3) l1  ["1 2 3 "," 4 5 6 ", ...] l3 = (sublist 3) l2  [["1 2 3 "," 4 5 6 "," 7 8 9 "],...] l4 = (map (concat . intersperse "")) l3  ["1 2 3  4 5 6  7 8 9 ",...] l5 = (concat . intersperse [line] . sublist 3) l4 in unlines l5 where sublist n [] = [] sublist n xs = ys : sublist n zs where (ys,zs) = splitAt n xs line = hyphens ++ "+" ++ hyphens ++ "+" ++ hyphens hyphens = replicate 9 '' main :: IO () main = do grids < fmap lines $ readFile "top95.txt" mapM_ printGrid $ mapMaybe solve grids
17 Concurrent STM Solver
Liyang wrote some applicative functor porn utilising STM. It's pretty but slow. Suggestions for speeding it up would be very welcome.
18 Chaining style Solver
by jinjing
It uses some snippets and the dot hack
import Prelude hiding ((.)) import T.T import List import Data.Maybe import Data.Char import Data.Map(keys, elems) import qualified Data.Map as Map row i = i `div` 9 col i = i `mod` 9 row_list i positions = positions.select(on_i_row) where on_i_row pos = pos.row == i.row col_list i positions = positions.select(on_i_col) where on_i_col pos = pos.col == i.col grid_list i positions = positions.select(on_same_grid i) on_same_grid i j = on_same_row_grid i j && on_same_col_grid i j on_same_row_grid i j = ( i.row.mod.send_to(3)  j.row.mod.send_to(3) ) == i.row  j.row on_same_col_grid i j = ( i.col.mod.send_to(3)  j.col.mod.send_to(3) ) == i.col  j.col board = 0.upto 80 choices = 1.upto 9 related i positions = positions.row_list(i) ++ positions.col_list(i) ++ positions.grid_list(i) values moves positions = positions.mapMaybe (moves.let_receive Map.lookup) possible_moves i moves = let positions = moves.keys in choices \\ positions.related(i).values(moves) sudoku_move moves = let i = moves.next_pos in moves.possible_moves(i).map(Map.insert i).map_send_to(moves) next_pos moves = (board \\ moves.keys) .label_by(choice_size).sort.first.snd where choice_size i = moves.possible_moves(i).length solve solutions 0 = solutions solve solutions n = solve next_solutions (n1) where next_solutions = solutions.map(sudoku_move).concat parse_input line = line.words.join("") .map(\c > if '1' <= c && c <= '9' then c else '0') .map(digitToInt).zip([0..]).reject((==0).snd).Map.fromList pretty_output solution = solution.elems.map(show).in_group_of(9) .map(unwords).unlines sudoku line = solve [given] (81  given.Map.size).first.pretty_output where given = parse_input line
19 Finite Domain Constraint Solver
by David Overton
This solver uses a finite domain constraint solver monad described here. The core functions are shown below. A full explanation is here.
type Puzzle = [Int] sudoku :: Puzzle > [Puzzle] sudoku puzzle = runFD $ do vars < newVars 81 [1..9] zipWithM_ (\x n > when (n > 0) (x `hasValue` n)) vars puzzle mapM_ allDifferent (rows vars) mapM_ allDifferent (columns vars) mapM_ allDifferent (boxes vars) labelling vars rows, columns, boxes :: [a] > [[a]] rows = chunk 9 columns = transpose . rows boxes = concatMap (map concat . transpose) . chunk 3 . chunk 3 . chunk 3 chunk :: Int > [a] > [[a]] chunk _ [] = [] chunk n xs = ys : chunk n zs where (ys, zs) = splitAt n xs
20 Very fast Solver
by Frank Kuehnel
This solver implements constraint propagation with higher level logic and search. Solves the 49151 puzzles with 17 hints in less than 50 seconds! More detail and less optimized versions are here.
module Main where import qualified Data.Vector.Unboxed as V import qualified Data.Vector as BV (generate,(!)) import Data.List (foldl',sort,group) import Data.Char (chr, ord) import Data.Word import Data.Bits import Control.Monad import Data.Maybe import System (getArgs)  Types type Alphabet = Word8 type Hypothesis = Word32  Hypotheses space is a matrix of independed hypoteses type HypothesesSpace = V.Vector Hypothesis  Set up spatial transformers / discriminators to reflect the spatial  properties of a Sudoku puzzle ncells = 81  vector rearrangement functions rows :: HypothesesSpace > HypothesesSpace rows = id columns :: HypothesesSpace > HypothesesSpace columns vec = V.map (\cidx > vec `V.unsafeIndex` cidx) cIndices where cIndices = V.fromList [r*9 + c  c <[0..8], r<[0..8]] subGrids :: HypothesesSpace > HypothesesSpace subGrids vec= V.map (\idx > vec `V.unsafeIndex` idx) sgIndices where sgIndices = V.fromList [i + bc + br  br < [0,27,54], bc < [0,3,6], i<[0,1,2,9,10,11,18,19,20]]  needs to be boxed, because vector elements are not primitives peersDiscriminators = BV.generate ncells discriminator where discriminator idx1 = V.zipWith3 (\r c s > (r  c  s)) rDscr cDscr sDscr where rDscr = V.generate ncells (\idx2 > idx1 `div` 9 == idx2 `div` 9) cDscr = V.generate ncells (\idx2 > idx1 `mod` 9 == idx2 `mod` 9) sDscr = V.generate ncells (\idx2 > subGridOfidx1 == subGrid idx2) where subGridOfidx1 = subGrid idx1 subGrid idx = (idx `div` 27, (idx `div` 3) `mod` 3)  Let's implement the logic  Level 0 logic (enforce consistency):  We can't have multiple same solutions in a peer unit,  eliminate solutions from other hypotheses enforceConsistency :: HypothesesSpace > Maybe HypothesesSpace enforceConsistency hypS0 = do V.foldM solutionReduce hypS0 $ V.findIndices newSingle hypS0 solutionReduce :: HypothesesSpace > Int > Maybe HypothesesSpace solutionReduce hypS0 idx = let sol = hypS0 V.! idx peers = peersDiscriminators BV.! idx hypS1 = V.zipWith reduceInUnit peers hypS0 where reduceInUnit p h  p && (h == sol) = setSolution sol  p = h `minus` sol  otherwise = h in if V.any empty hypS1 then return hypS1 else if V.any newSingle hypS1 then enforceConsistency hypS1  constraint propagation else return hypS1  Level 1 logic is rather simple:  We tally up all unknown values in a given unit,  if a value occurs only once, then it must be the solution! localizeSingles :: HypothesesSpace > Maybe HypothesesSpace localizeSingles unit = let known = maskChoices $ accumTally $ V.filter single unit in if dups known then Nothing else case (filterSingles $ accumTally $ V.filter (not . single) unit) `minus` known of 0 > return unit sl > return $ replaceWith unit sl where replaceWith :: V.Vector Hypothesis > Hypothesis > V.Vector Hypothesis replaceWith unit s = V.map (\u > if 0 /= maskChoices (s .&. u) then s `Main.intersect` u else u) unit  Level 2 logic is a bit more complicated:  Say in a given unit, we find exactly two places with the hypothesis {1,9}.  Then obviously, the value 1 and 9 can only occur in those two places.  All other ocurrances of the value 1 and 9 can eliminated. localizePairs :: HypothesesSpace > Maybe HypothesesSpace localizePairs unit = let pairs = V.toList $ V.filter pair unit in if nodups pairs then return unit else case map head $ filter lpair $ tally pairs of [] > return unit pl@(p:ps) > return $ foldl' eliminateFrom unit pl where  "subtract" pair out of a hypothesis eliminateFrom :: V.Vector Hypothesis > Hypothesis > V.Vector Hypothesis eliminateFrom unit p = V.map (\u > if u /= p then u `minus` p else u) unit  Level 3 logic resembles the level 2 logic:  If we find exactly three places with the hypothesis {1,7,8} in a given unit, then all other ...  you'll get the gist! localizeTriples :: HypothesesSpace > Maybe HypothesesSpace localizeTriples unit = let triples = V.toList $ V.filter triple unit in if nodups triples then return unit else case map head $ filter ltriple $ tally triples of [] > return unit tl@(t:ts) > return $ foldl' eliminateFrom unit tl where  "subtract" triple out of a hypothesis eliminateFrom :: V.Vector Hypothesis > Hypothesis > V.Vector Hypothesis eliminateFrom unit t = V.map (\u > if u /= t then u `minus` t else u) unit  Even higher order logic is easy to implement, but becomes rather useless in the general case!  Implement the whole nine yard: constraint propagation and search applySameDimensionLogic :: HypothesesSpace > Maybe HypothesesSpace applySameDimensionLogic hyp0 = do res1 < logicInDimensionBy rows chainedLogic hyp0 res2 < logicInDimensionBy columns chainedLogic res1 logicInDimensionBy subGrids chainedLogic res2 where chainedLogic = localizeSingles >=> localizePairs >=> localizeTriples logicInDimensionBy :: (HypothesesSpace > HypothesesSpace) > (HypothesesSpace > Maybe HypothesesSpace) > HypothesesSpace > Maybe HypothesesSpace logicInDimensionBy trafo logic hyp = liftM (trafo . V.concat) $ mapM (\ridx > do logic $ V.unsafeSlice ridx 9 hyp') [r*9  r< [0..8]] where hyp' :: HypothesesSpace hyp' = trafo hyp prune :: HypothesesSpace > Maybe HypothesesSpace prune hypS0 = do hypS1 < applySameDimensionLogic =<< enforceConsistency hypS0 if V.any newSingle hypS1 then prune hypS1  effectively implemented constraint propagation else do hypS2 < applySameDimensionLogic hypS1 if hypS1 /= hypS2 then prune hypS2  effectively implemented a fix point method else return hypS2 search :: HypothesesSpace > Maybe HypothesesSpace search hypS0  complete hypS0 = return hypS0  otherwise = do msum [prune hypS1 >>= search  hypS1 < expandFirst hypS0]  guessing order makes a big difference!! expandFirst :: HypothesesSpace > [HypothesesSpace] expandFirst hypS  suitable == [] = []  otherwise = let (_, idx) = minimum suitable  minimum is the preferred strategy! in map (\choice > hypS V.// [(idx, choice)]) (split $ hypS V.! idx) where suitable = filter ((>1) . fst) $ V.toList $ V.imap (\idx e > (numChoices e, idx)) hypS  Some very useful tools:  partition a list into sublists chop :: Int > [a] > [[a]] chop n [] = [] chop n xs = take n xs : chop n (drop n xs)  when does a list have no duplicates nodups :: Eq a => [a] > Bool nodups [] = True nodups (x:xs) = not (elem x xs) && nodups xs dups :: Hypothesis > Bool dups t = (filterDups t) /= 0 tally :: Ord a => [a] > [[a]] tally = group . sort empty :: Hypothesis > Bool empty n = (maskChoices n) == 0 single :: Hypothesis > Bool single n = (numChoices n) == 1 lsingle :: [a] > Bool lsingle [n] = True lsingle _ = False pair :: Hypothesis > Bool pair n = numChoices n == 2 lpair :: [a] > Bool lpair (x:xs) = lsingle xs lpair _ = False triple :: Hypothesis > Bool triple n = (numChoices n) == 3 ltriple :: [a] > Bool ltriple (x:xs) = lpair xs ltriple _ = False complete :: HypothesesSpace > Bool complete = V.all single  The bit gymnastics (wish some were implemented in Data.Bits)  bits 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 .. 27 28 29 30 31 represents  h  h  h  h  h  h  h  h  h  .. s l l l l  with  h : 1 iff element is part of the hypothesis set  l : 4 bits for the cached number of h bits set  s : 1 iff a single solution for the cell is found  experiment with different strategies split :: Hypothesis > [Hypothesis] split 0 = [] split n = [n `minus` bit1, (bit 28) .. bit1] where bit1 = (bit $ firstBit n) minus :: Hypothesis > Hypothesis > Hypothesis xs `minus` ys  maskChoices (xs .&. ys) == 0 = xs  otherwise = zs .. ((countBits zs) `shiftL` 28) where zs = maskChoices $ xs .&. (complement ys) numChoices :: Hypothesis > Word32 numChoices n = (n `shiftR` 28) newSingle :: Hypothesis > Bool newSingle n = (n `shiftR` 27) == 2 isSolution :: Hypothesis > Bool isSolution n = n `testBit` 27 setSolution :: Hypothesis > Hypothesis setSolution n = n `setBit` 27 maskChoices :: Hypothesis > Hypothesis maskChoices n = n .&. 0x07FFFFFF intersect :: Hypothesis > Hypothesis > Hypothesis intersect x y = z .. ((countBits z) `shiftL` 28) where z = maskChoices $ x .&. y countBits :: Word32 > Word32  would be wonderful if Data.Bits had such a function countBits 0 = 0 countBits n = (cBLH 16 0xFFFF . cBLH 8 0xFF00FF . cBLH 4 0x0F0F0F0F . cBLH 2 0x33333333 . cBLH 1 0x55555555) n cBLH :: Int > Word32 > Word32 > Word32 cBLH s mask n = (n .&. mask) + (n `shiftR` s) .&. mask firstBit :: Hypothesis > Int  should also be in Data.Bits firstBit 0 = 0  stop recursion !! firstBit n  n .&. 1 > 0 = 0  otherwise = (+) 1 $ firstBit $ n `shiftR` 1 accumTally :: V.Vector Hypothesis > Hypothesis accumTally nl = V.foldl' accumTally2 0 nl accumTally2 :: Word32 > Word32 > Word32 accumTally2 t n = (+) t $ n .&. (((complement t) .&. 0x02AAAAAA) `shiftR` 1) filterSingles :: Hypothesis > Hypothesis filterSingles t = t .&. (((complement t) .&. 0x02AAAAAA) `shiftR` 1) filterDups :: Hypothesis > Hypothesis filterDups t = (t .&. 0x02AAAAAA) `shiftR` 1 defaultHypothesis :: Hypothesis defaultHypothesis = 0x90015555  all nine alphabet elements are set mapAlphabet :: V.Vector Hypothesis mapAlphabet = V.replicate 256 defaultHypothesis V.// validDigits where validDigits :: [(Int, Hypothesis)] validDigits = [(ord i, (bit 28) .. (bit $ 2*(ord i  49)))  i < "123456789"] toChar :: Hypothesis > [Char] toChar s  single s = [normalize s]  otherwise = "." where normalize s = chr $ (+) 49 $ (firstBit s) `shiftR` 1 toCharDebug :: Hypothesis > [Char] toCharDebug s  isSolution s = ['!', normalize s]  single s = [normalize s]  otherwise = "{" ++ digits ++ "}" where normalize s = chr $ (+) 49 $ (firstBit s) `shiftR` 1 digits = zipWith test "123456789" $ iterate (\e > e `shiftR` 2) s test c e  e.&.1 == 1 = c  otherwise = '.'  Initial hypothesis space initialize :: String > Maybe HypothesesSpace initialize g = if all (`elem` "0.123456789") g then let hints = zip [0..] translated translated = map (\c > mapAlphabet V.! ord c) $ take ncells g in Just $ (V.replicate ncells defaultHypothesis) V.// hints else Nothing  Display (partial) solution printResultD :: HypothesesSpace > IO () printResultD = putStrLn . toString where toString :: HypothesesSpace > String toString hyp = unlines $ map translate . chop 9 $ V.toList hyp where translate = concatMap (\s > toCharDebug s ++ " ") printResult :: HypothesesSpace > IO () printResult = putStrLn . toString where toString :: HypothesesSpace > String toString hyp = translate (V.toList hyp) where translate = concatMap (\s > toChar s ++ "")  The entire solution process! solve :: String > Maybe HypothesesSpace solve str = do initialize str >>= prune >>= search main :: IO () main = do [f] < getArgs sudoku < fmap lines $ readFile f  "test.txt" mapM_ printResult $ mapMaybe solve sudoku
21 List comprehensions
by Ben Lynn.
Translated from my brute force solver in C:
module Main where f x s@(h:y)=let(r,c)=divMod(length x)9;m#n=m`div`3==n`div`3;e=[0..8]in [az<['1'..'9'],h==zh=='.'&¬Elem z(map((x++s)!!)[i*9+ji<e, j<e,i==rj==ci#r&&j#c]),a<f(x++[z])y] f x[]=[x] main=print$f[] "53..7....6..195....98....6.8...6...34..8.3..17...2...6.6....28....419..5....8..79"
22 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, intermediateadvanced techniques in Haskell.
23 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 get over 47,000 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."