# Sudoku

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

## Contents

- 1 Monadic non-deterministic solver
- 2 Simple solver
- 3 Complete decision tree
- 4 No guessing
- 5 Just guessing
- 6 Very smart, with only a little guessing
- 7 Only guessing without dancing links
- 8 Generalized solver
- 9 Simple small solver
- 10 Backtrack monad solver
- 11 In-flight entertainment
- 12 Sudoku incrementally, à la Bird
- 13 607 bytes / 12 lines
- 14 A parallel solver
- 15 Another simple solver
- 16 Solving Every Sudoku Puzzle
- 17 Concurrent STM Solver
- 18 Chaining style Solver
- 19 Finite Domain Constraint Solver
- 20 Add your own
- 21 Test boards

## Monadic non-deterministic 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
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
```

## 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-exhaustive, 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
```

## 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))
```

## 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 human-comprehensible 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 no-guessing 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).

## Just guessing

By ChrisKuklewicz

This solver is an implementation of Knuth's "Dancing Links" algorithm for solving binary-cover 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.

## 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 (ghc-6.4.1 on OS X). I liked the 9x9x9 array since it emphasized the symmetry of the problem.

## 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, tree-like data structure (in this case, a Priority Search Queue) "undo" operations are essentially free.

## 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 \ 1-53---9- \ ---6----- \ ------271 \ 82------- \ ---487--- \ ------53- \ 23------- \ --7-59--- \ --6---8-4

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

## 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>

## 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 somewhat-nontrivial 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 re-initialize to the original state on each iteration".

I have several (Unix command line) front-ends to this module, available upon request. The one I use most creates and prints six new Sudoku puzzles on a page, with fine-grain control over the difficulty of the puzzle. This has made me quite popular among friends and extended family.

- YitzGale

```
{-# OPTIONS_GHC -fglasgow-exts #-}
-- 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) -- One-based 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..size-sqrtSize],
y <- [0,sqrtSize..size-sqrtSize]]
-- The one-based 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 re-initialized 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 1-based.
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 2-dimensional list.
-- Coordinates are 1-based.
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 _ _ _ _ = []
```

## In-flight entertainment

By Lennart Augustsson

When on a Lufthansa trans-atlantic flight in 2005 I picked up the in-flight 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 non-clues
zip [ (c, r) | r <- ['1'..'9'], c <- ['1'..'9'] ] . -- pair up the digits with their coordinates
filter (`elem` ['0'..'9']) -- get rid of non-digits
```

## 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 talk-through 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.

## 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 (a-1) 3 == div (b-1) 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)

## 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

## 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 (<x-3) [0,3,6]
m = head $ dropWhile (<y-3) [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 ( flip 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..num-1]]
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.

## Solving Every Sudoku Puzzle

By Manu

This is an attempt to implement in Haskell, Peter Norvig's sudoku solver : more explanations here : http://norvig.com/sudoku.html

This program greatly benefited from Daniel Fischer's refactoring (80 times faster !). It takes just above 1 second, on a MacBook Pro 2.33GHz Intel Core 2 Duo, to solve the 95 puzzles found here : http://norvig.com/top95.txt

```
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, elem s 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 (\c -> (elem c "0.-123456789")) g
then (Just g)
else Nothing
--------------------------------------------------
-- Propagating Constraints
assign :: Grid -> (Square, Digit) -> Maybe Grid
assign g (s,d) = if (elem d 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 not (elem d 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 (elem d . (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 = take n xs : sublist n (drop n xs)
line = hyphens ++ "+" ++ hyphens ++ "+" ++ hyphens
hyphens = take 9 (repeat '-')
--------------------------------------------------
main :: IO ()
main = do
grids <- fmap lines $ readFile "top95.txt"
mapM_ printGrid $ mapMaybe solve grids
```

## 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.

## 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 (n-1) 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(\(a,b) -> b == 0).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
```

## 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 = concat . map (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
```

## 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.

## 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."