Sudoku
February 19, 2009
Sudoku puzzles are a simple and popular amusement given as a ninebynine grid of cells, some of them containing digits:











The challenge is to fill the empty cells with the digits 1 through 9 in such a way that no row, column, or threebythree subgrid contains the same digit two or more times.
Write a program to solve sudoku puzzles; your program may assume the puzzle is wellformed. What is the solution of the above puzzle?
Pages: 1 2
Some time back I had read Peter Norvig’s beautiful post Solving Every Sudoku Puzzle, and tried to convert his program into Haskell. His idea, in short, was to use constraint propagation to fill up as many values as possible; and then ultimately search.
Though the resulting Haskell program was not as succinct as his original Python, I thought it wasn’t too bad either. Perhaps using some monad trick (that I yet have to learn!) all the checks against ‘Nothing’ can be eliminated.
One (only?) answer, got in under a second, is:
7 8 9 1 3 5 6 2 4
6 2 3 9 4 7 8 1 5
4 5 1 2 8 6 3 9 7
++
2 3 7 4 1 8 5 6 9
8 4 5 6 9 3 2 7 1
9 1 6 7 5 2 4 8 3
++
1 7 8 5 2 4 9 3 6
5 6 2 3 7 9 1 4 8
3 9 4 8 6 1 7 5 2
And my Haskell translation is (gist):
import List (elem, nub, filter, delete, intersperse, replicate)
import Data.Map (Map, fromList, (!), insert, keys, elems, toList)
import Data.List (intercalate)
import Data.Maybe
type Val = Char  Value of a cell
type Square = String  twocharacter square ID
type Board = Maybe (Map Square String)  The board state
cross xs ys = [[x,y]  x < xs, y < ys]
rows = "ABCDEFGHI"
cols = "123456789"
digits = "123456789"
squares = cross rows cols
unitlist = [cross rows [c]  c < cols]
++ [cross [r] cols  r < rows]
++ [cross rs cs  rs < ["ABC", "DEF", "GHI"], cs < ["123","456","789"]]
units sq = filter (elem sq) unitlist
peers sq = delete sq . nub . concat $ units sq
 Check for places where d appears in the units of sq
checkplaces :: Board > Square > Val > Board
checkplaces b0 sq d = foldl f b0 (units sq)
where f b u
 isNothing b  len == 0 = Nothing
 len == 1 = assign b (head dplaces) d
 otherwise = b
where dplaces = [s  s < u, elem d ((fromJust b)!s)]
 dplaces is all squares in the unit u possibly containing d
len = length dplaces
eliminate :: Board > Square > Val > Board
eliminate Nothing _ _ = Nothing
eliminate (Just b0) sq d
 notElem d v = Just b0  Already Eliminated
 length v' == 0 = Nothing  Contradiction: Removed last value
 length v' == 1 = checkplaces b'' sq d  Only 1 left: Remove from peers
 otherwise = checkplaces (Just b') sq d
where v = b0 ! sq
v' = delete d v
b' = insert sq v' b0
h = (head v')
b'' = foldl (\b p > eliminate b p h) (Just b') (peers sq)
assign :: Board > Square > Val > Board
assign Nothing _ _ = Nothing
assign b0 sq d0 = foldl f b0 ((fromJust b0) ! sq)
where f b d = if d0 == d then b else eliminate b sq d
parsegr :: String > Board
parsegr s = foldl f b0 (zip squares s')
where s' = filter (\x > elem x "0.123456789") s
b0 = Just (fromList [(s, digits)  s < squares])
f b (sq, d)
 isNothing b = Nothing
 notElem d digits = b
 otherwise = assign b sq d
search :: Board > Board
search Nothing = Nothing
search (Just b) =
if all ((==1) . length) $ elems b then Just b else search' (b!s)
where minl = minimum [l  v < elems b, let l = length v, l > 1]
s = head [sq  (sq, v) < toList b, length v == minl]
search' [] = Nothing
search' (d:ds) = if isJust b' then b' else search' ds
where b' = search $ assign (Just b) s d
printgr :: Board > String
printgr Nothing = "Unsolvable"
printgr (Just b) = concat [(fr r) ++ (if elem r "CF" then line else "") ++ "\n"  r<rows]
where w = 1 + maximum [length (b!sq)  sq < squares]
line = (++) "\n" $ intercalate "+" $ replicate 3 $ replicate (3 * w) ''
fr r = concat [(fmt (b![r,c])) ++ (if elem c "36" then "" else "")  c<cols]
fmt s = s ++ replicate (w  (length s)) ' '
solve :: String > IO ()
solve s = putStrLn $ printgr $ search $ parsegr s
 Instance from http://programmingpraxis.com/2009/02/19/sudoku/2/
grid = "7..1......2.....15.....639.2...18....4..9..7....75...3.785.....56.....4......1..2"
main = solve grid
Here’s my implementation, based on the Dlx algorithm:
http://codepad.org/a1lsh3ga
Implementation of DLX in PLT Scheme. This solves the sample problem in about 20ms on my machine.
http://github.com/d3zd3z/praxis/blob/master/2009/0219sudoku.ss
Solving sudoku is interseting.
How about designing sudoku puzzles that have exactly one solution?
Does anyone have an idea how to do that?
I can think of the following:
0: Initially all fields are unoccupied.
1: Randomly choose an unoccupied field.
2: Take all consistent digits in arbitrary order.
3: If there are no consistent digits backtrack from step 1.
4: Try the digits in arbitrary order.
5: For each trial solve the puzzle to a maximum of two solutions.
6: If there is no solution, try the next consistent digit or backtrack from step 1 if no digits are left.
7: If there is one solution we are ready.
9: If there is more than one solution, try to complete the puzzle from step 1.
Jos
https://github.com/gcapell/ProgrammingPraxis/blob/master/04_sudoku/sudoku.go
This is a solution in php that uses backtracking.
You start with a first empty cell and the least number that fits there (i.e. with no conflict from other numbers that are already supplied in the puzzle).
And apply the same logic to the next empty cell, till you’re not able to fill any number in an empty cell – at which time you backtrack to the previous cell, change the number there and try again.
So for the above Sudoku puzzle, since the first empty cell is in (1,2), the function below will first be invoked as:
solve(1,1,2)
$game_table is just an array of 9X9 that already contains supplied numbers in the puzzle
$complete will be set to FALSE before calling solve() for the first time
function solve($number,$row,$column)
{
global $game_table;
global $complete;
while($number<=9 && !$complete)
{
//check is this number can fit into this cell
if(check_horizontal($number,$row,$column) && check_vertical($number,$row,$column) && check_square($number,$row,$column))
{
$game_table[$row][$column]=$number; //no conflict, so fill the cell with this number
//find the next blank cell
$i=1;$j=1;$found=false;
for($i=1;$i<=9;$i++)
{
for($j=1;$j<=9;$j++)
{
if($game_table[$i][$j]==0)
{
$found=true;
break;
}
}
if ($found) break;
}
if($found)
solve(1,$i,$j); //recursive call
else
{
$complete= true;
return;
}
}
//recursive call will return here – so here either the puzzle is complete or there was no possibility to fill the next cell
if(!$complete)
$number++;
}
if(!$complete)
$game_table[$row][$column]=0; //reset this cell
}
function check_horizontal($n,$r,$c)
{
global $game_table;
for($j=1;$j<=9;$j++)
if($game_table[$r][$j]==$n) return false;
return true;
}
function check_vertical($n,$r,$c)
{
global $game_table;
for($i=1;$i<=9;$i++)
if($game_table[$i][$c]==$n) return false;
return true;
}
function check_square($n,$r,$c)
{
global $game_table;
$a = $r%3;
$b = floor($r/3);
if($a==0)
{
$r1 = (3*($b1))+1;
$r2 = $r;
}
else
{
$r1 = $r$a+1;
$r2 = 3*($b+1);
}
$a = $c%3;
$b = floor($c/3);
if($a==0)
{
$c1 = (3*($b1))+1;
$c2 = $c;
}
else
{
$c1 = $c$a+1;
$c2 = 3*($b+1);
}
for($i=$r1;$i<=$r2;$i++)
for($j=$c1;$j<=$c2;$j++)
if($game_table[$i][$j] == $n) return false;
return true;
}
golang
Python: https://github.com/ftt/programmingpraxis/blob/master/20090219sudoku/sudoku.py
This is roughly based on an article by P. Norvig I read once (Ashutosh Mehra mentions that article), but I’ve tried to implement it from memory, so the result is not as clean as the original.