Sudoku

From Rosetta Code
Task
Sudoku
You are encouraged to solve this task according to the task description, using any language you may know.
Task

Solve a partially filled-in normal   9x9   Sudoku grid   and display the result in a human-readable format.


Algorithmics of Sudoku   may help implement this.

ALGOL 68[edit]

Translation of: D
Note: This specimen retains the original D coding style.
Works with: ALGOL 68 version Revision 1 - no extensions to language used.
Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny.
MODE AVAIL = [9]BOOL;
MODE BOX = [3, 3]CHAR;
 
FORMAT row fmt = $"|"3(" "3(g" ")"|")l$;
FORMAT line = $"+"3(7"-","+")l$;
FORMAT puzzle fmt = $f(line)3(3(f(row fmt))f(line))$;
 
AVAIL gen full = (TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE);
 
OP REPR = (AVAIL avail)STRING: (
STRING out := "";
FOR i FROM LWB avail TO UPB avail DO
IF avail[i] THEN out +:= REPR(ABS "0" + i) FI
OD;
out
);
 
CHAR empty = "_";
 
OP -:= = (REF AVAIL set, CHAR index)VOID: (
set[ABS index - ABS "0"]:=FALSE
);
 
# these two functions assume that the number has not already been found #
PROC avail slice = (REF[]CHAR slice, REF AVAIL available)REF AVAIL:(
FOR ele FROM LWB slice TO UPB slice DO
IF slice[ele] /= empty THEN available-:=slice[ele] FI
OD;
available
);
 
PROC avail box = (INT x, y, REF AVAIL available)REF AVAIL:(
# x designates row, y designates column #
# get a base index for the boxes #
INT bx := x - (x-1) MOD 3;
INT by := y - (y-1) MOD 3;
REF BOX box = puzzle[bx:bx+2, by:by+2];
FOR i FROM LWB box TO UPB box DO
FOR j FROM 2 LWB box TO 2 UPB box DO
IF box[i, j] /= empty THEN available-:=box[i, j] FI
OD
OD;
available
);
 
[9, 9]CHAR puzzle;
PROC solve = ([,]CHAR in puzzle)VOID:(
puzzle := in puzzle;
TO UPB puzzle UP 2 DO
BOOL done := TRUE;
FOR i FROM LWB puzzle TO UPB puzzle DO
FOR j FROM 2 LWB puzzle TO 2 UPB puzzle DO
CHAR ele := puzzle[i, j];
IF ele = empty THEN
# poke at the elements that are "_" #
AVAIL remaining := avail box(i, j,
avail slice(puzzle[i, ],
avail slice(puzzle[, j],
LOC AVAIL := gen full)));
STRING s = REPR remaining;
IF UPB s = 1 THEN puzzle[i, j] := s[LWB s]
ELSE done := FALSE
FI
FI
OD
OD;
IF done THEN break FI
OD;
break:
# write out completed puzzle #
printf(($gl$, "Completed puzzle:"));
printf((puzzle fmt, puzzle))
);
main:(
solve(("394__267_",
"___3__4__",
"5__69__2_",
"_45___9__",
"6_______7",
"__7___58_",
"_1__67__8",
"__9__8___",
"_264__735"))
CO # note: This codes/algorithm does not [yet] solve: #
solve(("9__2__5__",
"_4__6__3_",
"__3_____6",
"___9__2__",
"____5__8_",
"__7__4__3",
"7_____1__",
"_5__2__4_",
"__1__6__9"))
END CO
)
Output:
Completed puzzle:
+-------+-------+-------+
| 3 9 4 | 8 5 2 | 6 7 1 |
| 2 6 8 | 3 7 1 | 4 5 9 |
| 5 7 1 | 6 9 4 | 8 2 3 |
+-------+-------+-------+
| 1 4 5 | 7 8 3 | 9 6 2 |
| 6 8 2 | 9 4 5 | 3 1 7 |
| 9 3 7 | 1 2 6 | 5 8 4 |
+-------+-------+-------+
| 4 1 3 | 5 6 7 | 2 9 8 |
| 7 5 9 | 2 3 8 | 1 4 6 |
| 8 2 6 | 4 1 9 | 7 3 5 |
+-------+-------+-------+

AutoHotkey[edit]

#SingleInstance, Force
SetBatchLines, -1
SetTitleMatchMode, 3
 
Loop 9 {
r := A_Index, y := r*17-8 + (A_Index >= 7 ? 4 : A_Index >= 4 ? 2 : 0)
Loop 9 {
c := A_Index, x := c*17+5 + (A_Index >= 7 ? 4 : A_Index >= 4 ? 2 : 0)
Gui, Add, Edit, x%x% y%y% w17 h17 v%r%_%c% Center Number Limit1 gNext
}
}
Gui, Add, Button, vButton gSolve w175 x10 Center, Solve
Gui, Add, Text, vMsg r3, Enter Sudoku puzzle and click Solve
Gui, Show,, Sudoku Solver
Return
 
Solve:
Gui, Submit, NoHide
Loop 9
{
r := A_Index
Loop 9
If (%r%_%A_Index% = "")
puzzle .= "@"
Else
puzzle .= %r%_%A_Index%
}
s := A_TickCount
answer := Sudoku(puzzle)
iterations := ErrorLevel
e := A_TickCount
seconds := (e-s)/1000
StringSplit, a, answer, |
Loop 9
{
r := A_Index
Loop 9
{
b := (r*9)+A_Index-9
GuiControl,, %r%_%A_Index%, % a%b%
GuiControl, +ReadOnly, %r%_%A_Index%
}
}
if answer
GuiControl,, Msg, Solved!`nTime: %seconds%s`nIterations: %iterations%
else
GuiControl,, Msg, Failed! :(`nTime: %seconds%s`nIterations: %iterations%
GuiControl,, Button, Again!
GuiControl, +gAgain, Button
return
 
GuiClose:
ExitApp
 
Again:
Reload
 
#IfWinActive, Sudoku Solver
~*Enter::GoSub % GetKeyState( "Shift", "P" ) ? "~Up" : "~Down"
~Up::
GuiControlGet, f, focus
StringTrimLeft, f, f, 4
f := ((f >= 1 && f <= 9) ? f+72 : f-9)
GuiControl, Focus, Edit%f%
return
~Down::
GuiControlGet, f, focus
StringTrimLeft, f, f, 4
f := ((f >= 73 && f <= 81) ? f-72 : f + 9)
GuiControl, Focus, Edit%f%
return
~Left::
GuiControlGet, f, focus
StringTrimLeft, f, f, 4
f := Mod(f + 79, 81) + 1
GuiControl, Focus, Edit%f%
return
Next:
~Right::
GuiControlGet, f, focus
StringTrimLeft, f, f, 4
f := Mod(f, 81) + 1
GuiControl, Focus, Edit%f%
return
#IfWinActive
 
; Functions Start here
 
Sudoku( p ) { ;ErrorLevel contains the number of iterations
p := RegExReplace(p, "[^1-9@]"), ErrorLevel := 0 ;format puzzle as single line string
return Sudoku_Display(Sudoku_Solve(p))
}
 
Sudoku_Solve( p, d = 0 ) { ;d is 0-based
; http://www.autohotkey.com/forum/topic46679.html
; p: 81 character puzzle string
; (concat all 9 rows of 9 chars each)
; givens represented as chars 1-9
; fill-ins as any non-null, non 1-9 char
; d: used internally. omit on initial call
;
; returns: 81 char string with non-givens replaced with valid solution
;
If (d >= 81), ErrorLevel++
return p ;this is 82nd iteration, so it has successfully finished iteration 81
If InStr( "123456789", SubStr(p, d+1, 1) ) ;this depth is a given, skip through
return Sudoku_Solve(p, d+1)
m := Sudoku_Constraints(p,d) ;a string of this level's constraints.
; (these will not change for all 9 loops)
Loop 9
{
If InStr(m, A_Index)
Continue
NumPut(Asc(A_Index), p, d, "Char")
If r := Sudoku_Solve(p, d+1)
return r
}
return 0
}
 
Sudoku_Constraints( ByRef p, d ) {
; returns a string of the constraints for a particular position
c := Mod(d,9)
, r := (d - c) // 9
, b := r//3*27 + c//3*3 + 1
;convert to 1-based
, c++
return ""
; row:
. SubStr(p, r * 9 + 1, 9)
; column:
. SubStr(p,c ,1) SubStr(p,c+9 ,1) SubStr(p,c+18,1)
. SubStr(p,c+27,1) SubStr(p,c+36,1) SubStr(p,c+45,1)
. SubStr(p,c+54,1) SubStr(p,c+63,1) SubStr(p,c+72,1)
;box
. SubStr(p, b, 3) SubStr(p, b+9, 3) SubStr(p, b+18, 3)
}
 
Sudoku_Display( p ) {
If StrLen(p) = 81
loop 81
r .= SubStr(p, A_Index, 1) . "|"
return r
}

BBC BASIC[edit]

Sudoku bbc.gif
      VDU 23,22,453;453;8,20,16,128
*FONT Arial,28
 
DIM Board%(8,8)
Board%() = %111111111
 
FOR L% = 0 TO 9:P% = L%*100
LINE 2,P%+2,902,P%+2
IF (L% MOD 3)=0 LINE 2,P%,902,P% : LINE 2,P%+4,902,P%+4
LINE P%+2,2,P%+2,902
IF (L% MOD 3)=0 LINE P%,2,P%,902 : LINE P%+4,2,P%+4,902
NEXT
 
DATA " 4 5 6 "
DATA " 6 1 8 9"
DATA "3 7 "
DATA " 8 5 "
DATA " 4 3 "
DATA " 6 7 "
DATA " 2 6"
DATA "1 5 4 3 "
DATA " 2 7 1 "
 
FOR R% = 8 TO 0 STEP -1
READ A$
FOR C% = 0 TO 8
A% = ASCMID$(A$,C%+1) AND 15
IF A% Board%(R%,C%) = 1 << (A%-1)
NEXT
NEXT R%
 
GCOL 4
PROCshow
WAIT 200
dummy% = FNsolve(Board%(), TRUE)
GCOL 2
PROCshow
REPEAT WAIT 1 : UNTIL FALSE
END
 
DEF PROCshow
LOCAL C%,P%,R%
FOR C% = 0 TO 8
FOR R% = 0 TO 8
P% = Board%(R%,C%)
IF (P% AND (P%-1)) = 0 THEN
IF P% P% = LOGP%/LOG2+1.5
MOVE C%*100+30,R%*100+90
VDU 5,P%+48,4
ENDIF
NEXT
NEXT
ENDPROC
 
DEF FNsolve(P%(),F%)
LOCAL C%,D%,M%,N%,R%,X%,Y%,Q%()
DIM Q%(8,8)
REPEAT
Q%() = P%()
FOR R% = 0 TO 8
FOR C% = 0 TO 8
D% = P%(R%,C%)
IF (D% AND (D%-1))=0 THEN
M% = NOT D%
FOR X% = 0 TO 8
IF X%<>C% P%(R%,X%) AND= M%
IF X%<>R% P%(X%,C%) AND= M%
NEXT
FOR X% = C%DIV3*3 TO C%DIV3*3+2
FOR Y% = R%DIV3*3 TO R%DIV3*3+2
IF X%<>C% IF Y%<>R% P%(Y%,X%) AND= M%
NEXT
NEXT
ENDIF
NEXT
NEXT
Q%() -= P%()
UNTIL SUMQ%()=0
M% = 10
FOR R% = 0 TO 8
FOR C% = 0 TO 8
D% = P%(R%,C%)
IF D%=0 M% = 0
IF D% AND (D%-1) THEN
N% = 0
REPEAT N% += D% AND 1
D% DIV= 2
UNTIL D% = 0
IF N%<M% M% = N% : X% = C% : Y% = R%
ENDIF
NEXT
NEXT
IF M%=0 THEN = 0
IF M%=10 THEN = 1
D% = 0
FOR M% = 0 TO 8
IF P%(Y%,X%) AND (2^M%) THEN
Q%() = P%()
Q%(Y%,X%) = 2^M%
C% = FNsolve(Q%(),F%)
D% += C%
IF C% IF F% P%() = Q%() : = D%
ENDIF
NEXT
= D%

BCPL[edit]

// This can be run using Cintcode BCPL freely available from www.cl.cam.ac.uk/users/mr10.
// Implemented by Martin Richards.
 
// This is a really naive program to solve SuDoku problems. Even so it is usually quite fast.
 
// SuDoku consists of a 9x9 grid of cells. Each cell should contain
// a digit in the range 1..9. Every row, column and major 3x3
// square should contain all the digits 1..9. Some cells have
// given values. The problem is to find digits to place in
// the unspecified cells satisfying the constraints.
 
// A typical problem is:
 
// - - - 6 3 8 - - -
// 7 - 6 - - - 3 - 5
// - 1 - - - - - 4 -
 
// - - 8 7 1 2 4 - -
// - 9 - - - - - 5 -
// - - 2 5 6 9 1 - -
 
// - 3 - - - - - 1 -
// 1 - 5 - - - 6 - 8
// - - - 1 8 4 - - -
 
SECTION "sudoku"
 
GET "libhdr"
 
GLOBAL { count:ug
 
// The 9x9 board
 
a1; a2; a3; a4; a5; a6; a7; a8; a9
b1; b2; b3; b4; b5; b6; b7; b8; b9
c1; c2; c3; c4; c5; c6; c7; c8; c9
d1; d2; d3; d4; d5; d6; d7; d8; d9
e1; e2; e3; e4; e5; e6; e7; e8; e9
f1; f2; f3; f4; f5; f6; f7; f8; f9
g1; g2; g3; g4; g5; g6; g7; g8; g9
h1; h2; h3; h4; h5; h6; h7; h8; h9
i1; i2; i3; i4; i5; i6; i7; i8; i9
}
 
MANIFEST {
N1=1<<0; N2=1<<1; N3=1<<2;
N4=1<<3; N5=1<<4; N6=1<<5;
N7=1<<6; N8=1<<7; N9=1<<8
}
 
LET start() = VALOF
{ count := 0
initboard()
prboard()
ta1()
writef("*n*nTotal number of solutions: %n*n", count)
RESULTIS 0
}
 
AND initboard() BE {
a1, a2, a3, a4, a5, a6, a7, a8, a9 := 0, 0, 0, N6,N3,N8, 0, 0, 0
b1, b2, b3, b4, b5, b6, b7, b8, b9 := N7, 0,N6, 0, 0, 0, N3, 0,N5
c1, c2, c3, c4, c5, c6, c7, c8, c9 := 0,N1, 0, 0, 0, 0, 0,N4, 0
d1, d2, d3, d4, d5, d6, d7, d8, d9 := 0, 0,N8, N7,N1,N2, N4, 0, 0
e1, e2, e3, e4, e5, e6, e7, e8, e9 := 0,N9, 0, 0, 0, 0, 0,N5, 0
f1, f2, f3, f4, f5, f6, f7, f8, f9 := 0, 0,N2, N5,N6,N9, N1, 0, 0
g1, g2, g3, g4, g5, g6, g7, g8, g9 := 0,N3, 0, 0, 0, 0, 0,N1, 0
h1, h2, h3, h4, h5, h6, h7, h8, h9 := N1, 0,N5, 0, 0, 0, N6, 0,N8
i1, i2, i3, i4, i5, i6, i7, i8, i9 := 0, 0, 0, N1,N8,N4, 0, 0, 0
 
// Un-comment the following to test that the backtracking works
// giving 184 solutions.
//h1, h2, h3, h4, h5, h6, h7, h8, h9 := N1, 0,N5, 0, 0, 0, N6, 0, 0
//i1, i2, i3, i4, i5, i6, i7, i8, i9 := 0, 0, 0, 0, 0, 0, 0, 0, 0
}
 
AND c(n) = VALOF SWITCHON n INTO
{ DEFAULT: RESULTIS '?'
CASE 0: RESULTIS '-'
CASE N1: RESULTIS '1'
CASE N2: RESULTIS '2'
CASE N3: RESULTIS '3'
CASE N4: RESULTIS '4'
CASE N5: RESULTIS '5'
CASE N6: RESULTIS '6'
CASE N7: RESULTIS '7'
CASE N8: RESULTIS '8'
CASE N9: RESULTIS '9'
}
 
AND prboard() BE
{ LET form = "%c %c %c  %c %c %c  %c %c %c*n"
writef("*ncount = %n*n", count)
newline()
writef(form, c(a1),c(a2),c(a3),c(a4),c(a5),c(a6),c(a7),c(a8),c(a9))
writef(form, c(b1),c(b2),c(b3),c(b4),c(b5),c(b6),c(b7),c(b8),c(b9))
writef(form, c(c1),c(c2),c(c3),c(c4),c(c5),c(c6),c(c7),c(c8),c(c9))
newline()
writef(form, c(d1),c(d2),c(d3),c(d4),c(d5),c(d6),c(d7),c(d8),c(d9))
writef(form, c(e1),c(e2),c(e3),c(e4),c(e5),c(e6),c(e7),c(e8),c(e9))
writef(form, c(f1),c(f2),c(f3),c(f4),c(f5),c(f6),c(f7),c(f8),c(f9))
newline()
writef(form, c(g1),c(g2),c(g3),c(g4),c(g5),c(g6),c(g7),c(g8),c(g9))
writef(form, c(h1),c(h2),c(h3),c(h4),c(h5),c(h6),c(h7),c(h8),c(h9))
writef(form, c(i1),c(i2),c(i3),c(i4),c(i5),c(i6),c(i7),c(i8),c(i9))
 
newline()
 
//abort(1000)
}
 
AND try(p, f, row, col, sq) BE
{ LET x = !p
TEST x
THEN f()
ELSE { LET bits = row|col|sq
//prboard()
// writef("x=%n %b9*n", x, bits)
//abort(1000)
IF (N1&bits)=0 DO { !p:=N1; f() }
IF (N2&bits)=0 DO { !p:=N2; f() }
IF (N3&bits)=0 DO { !p:=N3; f() }
IF (N4&bits)=0 DO { !p:=N4; f() }
IF (N5&bits)=0 DO { !p:=N5; f() }
IF (N6&bits)=0 DO { !p:=N6; f() }
IF (N7&bits)=0 DO { !p:=N7; f() }
IF (N8&bits)=0 DO { !p:=N8; f() }
IF (N9&bits)=0 DO { !p:=N9; f() }
 !p := 0
}
}
 
AND ta1() BE try(@a1, ta2, a1+a2+a3+a4+a5+a6+a7+a8+a9,
a1+b1+c1+d1+e1+f1+g1+h1+i1,
a1+a2+a3+b1+b2+b3+c1+c2+c3)
AND ta2() BE try(@a2, ta3, a1+a2+a3+a4+a5+a6+a7+a8+a9,
a2+b2+c2+d2+e2+f2+g2+h2+i2,
a1+a2+a3+b1+b2+b3+c1+c2+c3)
AND ta3() BE try(@a3, ta4, a1+a2+a3+a4+a5+a6+a7+a8+a9,
a3+b3+c3+d3+e3+f3+g3+h3+i3,
a1+a2+a3+b1+b2+b3+c1+c2+c3)
AND ta4() BE try(@a4, ta5, a1+a2+a3+a4+a5+a6+a7+a8+a9,
a4+b4+c4+d4+e4+f4+g4+h4+i4,
a4+a5+a6+b4+b5+b6+c4+c5+c6)
AND ta5() BE try(@a5, ta6, a1+a2+a3+a4+a5+a6+a7+a8+a9,
a5+b5+c5+d5+e5+f5+g5+h5+i5,
a4+a5+a6+b4+b5+b6+c4+c5+c6)
AND ta6() BE try(@a6, ta7, a1+a2+a3+a4+a5+a6+a7+a8+a9,
a6+b6+c6+d6+e6+f6+g6+h6+i6,
a4+a5+a6+b4+b5+b6+c4+c5+c6)
AND ta7() BE try(@a7, ta8, a1+a2+a3+a4+a5+a6+a7+a8+a9,
a7+b7+c7+d7+e7+f7+g7+h7+i7,
a7+a8+a9+b7+b8+b9+c7+c8+c9)
AND ta8() BE try(@a8, ta9, a1+a2+a3+a4+a5+a6+a7+a8+a9,
a8+b8+c8+d8+e8+f8+g8+h8+i8,
a7+a8+a9+b7+b8+b9+c7+c8+c9)
AND ta9() BE try(@a9, tb1, a1+a2+a3+a4+a5+a6+a7+a8+a9,
a9+b9+c9+d9+e9+f9+g9+h9+i9,
a7+a8+a9+b7+b8+b9+c7+c8+c9)
 
AND tb1() BE try(@b1, tb2, b1+b2+b3+b4+b5+b6+b7+b8+b9,
a1+b1+c1+d1+e1+f1+g1+h1+i1,
a1+a2+a3+b1+b2+b3+c1+c2+c3)
AND tb2() BE try(@b2, tb3, b1+b2+b3+b4+b5+b6+b7+b8+b9,
a2+b2+c2+d2+e2+f2+g2+h2+i2,
a1+a2+a3+b1+b2+b3+c1+c2+c3)
AND tb3() BE try(@b3, tb4, b1+b2+b3+b4+b5+b6+b7+b8+b9,
a3+b3+c3+d3+e3+f3+g3+h3+i3,
a1+a2+a3+b1+b2+b3+c1+c2+c3)
AND tb4() BE try(@b4, tb5, b1+b2+b3+b4+b5+b6+b7+b8+b9,
a4+b4+c4+d4+e4+f4+g4+h4+i4,
a4+a5+a6+b4+b5+b6+c4+c5+c6)
AND tb5() BE try(@b5, tb6, b1+b2+b3+b4+b5+b6+b7+b8+b9,
a5+b5+c5+d5+e5+f5+g5+h5+i5,
a4+a5+a6+b4+b5+b6+c4+c5+c6)
AND tb6() BE try(@b6, tb7, b1+b2+b3+b4+b5+b6+b7+b8+b9,
a6+b6+c6+d6+e6+f6+g6+h6+i6,
a4+a5+a6+b4+b5+b6+c4+c5+c6)
AND tb7() BE try(@b7, tb8, b1+b2+b3+b4+b5+b6+b7+b8+b9,
a7+b7+c7+d7+e7+f7+g7+h7+i7,
a7+a8+a9+b7+b8+b9+c7+c8+c9)
AND tb8() BE try(@b8, tb9, b1+b2+b3+b4+b5+b6+b7+b8+b9,
a8+b8+c8+d8+e8+f8+g8+h8+i8,
a7+a8+a9+b7+b8+b9+c7+c8+c9)
AND tb9() BE try(@b9, tc1, b1+b2+b3+b4+b5+b6+b7+b8+b9,
a9+b9+c9+d9+e9+f9+g9+h9+i9,
a7+a8+a9+b7+b8+b9+c7+c8+c9)
 
AND tc1() BE try(@c1, tc2, c1+c2+c3+c4+c5+c6+c7+c8+c9,
a1+b1+c1+d1+e1+f1+g1+h1+i1,
a1+a2+a3+b1+b2+b3+c1+c2+c3)
AND tc2() BE try(@c2, tc3, c1+c2+c3+c4+c5+c6+c7+c8+c9,
a2+b2+c2+d2+e2+f2+g2+h2+i2,
a1+a2+a3+b1+b2+b3+c1+c2+c3)
AND tc3() BE try(@c3, tc4, c1+c2+c3+c4+c5+c6+c7+c8+c9,
a3+b3+c3+d3+e3+f3+g3+h3+i3,
a1+a2+a3+b1+b2+b3+c1+c2+c3)
AND tc4() BE try(@c4, tc5, c1+c2+c3+c4+c5+c6+c7+c8+c9,
a4+b4+c4+d4+e4+f4+g4+h4+i4,
a4+a5+a6+b4+b5+b6+c4+c5+c6)
AND tc5() BE try(@c5, tc6, c1+c2+c3+c4+c5+c6+c7+c8+c9,
a5+b5+c5+d5+e5+f5+g5+h5+i5,
a4+a5+a6+b4+b5+b6+c4+c5+c6)
AND tc6() BE try(@c6, tc7, c1+c2+c3+c4+c5+c6+c7+c8+c9,
a6+b6+c6+d6+e6+f6+g6+h6+i6,
a4+a5+a6+b4+b5+b6+c4+c5+c6)
AND tc7() BE try(@c7, tc8, c1+c2+c3+c4+c5+c6+c7+c8+c9,
a7+b7+c7+d7+e7+f7+g7+h7+i7,
a7+a8+a9+b7+b8+b9+c7+c8+c9)
AND tc8() BE try(@c8, tc9, c1+c2+c3+c4+c5+c6+c7+c8+c9,
a8+b8+c8+d8+e8+f8+g8+h8+i8,
a7+a8+a9+b7+b8+b9+c7+c8+c9)
AND tc9() BE try(@c9, td1, c1+c2+c3+c4+c5+c6+c7+c8+c9,
a9+b9+c9+d9+e9+f9+g9+h9+i9,
a7+a8+a9+b7+b8+b9+c7+c8+c9)
 
AND td1() BE try(@d1, td2, d1+d2+d3+d4+d5+d6+d7+d8+d9,
a1+b1+c1+d1+e1+f1+g1+h1+i1,
d1+d2+d3+e1+e2+e3+f1+f2+f3)
AND td2() BE try(@d2, td3, d1+d2+d3+d4+d5+d6+d7+d8+d9,
a2+b2+c2+d2+e2+f2+g2+h2+i2,
d1+d2+d3+e1+e2+e3+f1+f2+f3)
AND td3() BE try(@d3, td4, d1+d2+d3+d4+d5+d6+d7+d8+d9,
a3+b3+c3+d3+e3+f3+g3+h3+i3,
d1+d2+d3+e1+e2+e3+f1+f2+f3)
AND td4() BE try(@d4, td5, d1+d2+d3+d4+d5+d6+d7+d8+d9,
a4+b4+c4+d4+e4+f4+g4+h4+i4,
d4+d5+d6+e4+e5+e6+f4+f5+f6)
AND td5() BE try(@d5, td6, d1+d2+d3+d4+d5+d6+d7+d8+d9,
a5+b5+c5+d5+e5+f5+g5+h5+i5,
d4+d5+d6+e4+e5+e6+f4+f5+f6)
AND td6() BE try(@d6, td7, d1+d2+d3+d4+d5+d6+d7+d8+d9,
a6+b6+c6+d6+e6+f6+g6+h6+i6,
d4+d5+d6+e4+e5+e6+f4+f5+f6)
AND td7() BE try(@d7, td8, d1+d2+d3+d4+d5+d6+d7+d8+d9,
a7+b7+c7+d7+e7+f7+g7+h7+i7,
d7+d8+d9+e7+e8+e9+f7+f8+f9)
AND td8() BE try(@d8, td9, d1+d2+d3+d4+d5+d6+d7+d8+d9,
a8+b8+c8+d8+e8+f8+g8+h8+i8,
d7+d8+d9+e7+e8+e9+f7+f8+f9)
AND td9() BE try(@d9, te1, d1+d2+d3+d4+d5+d6+d7+d8+d9,
a9+b9+c9+d9+e9+f9+g9+h9+i9,
d7+d8+d9+e7+e8+e9+f7+f8+f9)
 
AND te1() BE try(@e1, te2, e1+e2+e3+e4+e5+e6+e7+e8+e9,
a1+b1+c1+d1+e1+f1+g1+h1+i1,
d1+d2+d3+e1+e2+e3+f1+f2+f3)
AND te2() BE try(@e2, te3, e1+e2+e3+e4+e5+e6+e7+e8+e9,
a2+b2+c2+d2+e2+f2+g2+h2+i2,
d1+d2+d3+e1+e2+e3+f1+f2+f3)
AND te3() BE try(@e3, te4, e1+e2+e3+e4+e5+e6+e7+e8+e9,
a3+b3+c3+d3+e3+f3+g3+h3+i3,
d1+d2+d3+e1+e2+e3+f1+f2+f3)
AND te4() BE try(@e4, te5, e1+e2+e3+e4+e5+e6+e7+e8+e9,
a4+b4+c4+d4+e4+f4+g4+h4+i4,
d4+d5+d6+e4+e5+e6+f4+f5+f6)
AND te5() BE try(@e5, te6, e1+e2+e3+e4+e5+e6+e7+e8+e9,
a5+b5+c5+d5+e5+f5+g5+h5+i5,
d4+d5+d6+e4+e5+e6+f4+f5+f6)
AND te6() BE try(@e6, te7, e1+e2+e3+e4+e5+e6+e7+e8+e9,
a6+b6+c6+d6+e6+f6+g6+h6+i6,
d4+d5+d6+e4+e5+e6+f4+f5+f6)
AND te7() BE try(@e7, te8, e1+e2+e3+e4+e5+e6+e7+e8+e9,
a7+b7+c7+d7+e7+f7+g7+h7+i7,
d7+d8+d9+e7+e8+e9+f7+f8+f9)
AND te8() BE try(@e8, te9, e1+e2+e3+e4+e5+e6+e7+e8+e9,
a8+b8+c8+d8+e8+f8+g8+h8+i8,
d7+d8+d9+e7+e8+e9+f7+f8+f9)
AND te9() BE try(@e9, tf1, e1+e2+e3+e4+e5+e6+e7+e8+e9,
a9+b9+c9+d9+e9+f9+g9+h9+i9,
d7+d8+d9+e7+e8+e9+f7+f8+f9)
 
AND tf1() BE try(@f1, tf2, f1+f2+f3+f4+f5+f6+f7+f8+f9,
a1+b1+c1+d1+e1+f1+g1+h1+i1,
d1+d2+d3+e1+e2+e3+f1+f2+f3)
AND tf2() BE try(@f2, tf3, f1+f2+f3+f4+f5+f6+f7+f8+f9,
a2+b2+c2+d2+e2+f2+g2+h2+i2,
d1+d2+d3+e1+e2+e3+f1+f2+f3)
AND tf3() BE try(@f3, tf4, f1+f2+f3+f4+f5+f6+f7+f8+f9,
a3+b3+c3+d3+e3+f3+g3+h3+i3,
d1+d2+d3+e1+e2+e3+f1+f2+f3)
AND tf4() BE try(@f4, tf5, f1+f2+f3+f4+f5+f6+f7+f8+f9,
a4+b4+c4+d4+e4+f4+g4+h4+i4,
d4+d5+d6+e4+e5+e6+f4+f5+f6)
AND tf5() BE try(@f5, tf6, f1+f2+f3+f4+f5+f6+f7+f8+f9,
a5+b5+c5+d5+e5+f5+g5+h5+i5,
d4+d5+d6+e4+e5+e6+f4+f5+f6)
AND tf6() BE try(@f6, tf7, f1+f2+f3+f4+f5+f6+f7+f8+f9,
a6+b6+c6+d6+e6+f6+g6+h6+i6,
d4+d5+d6+e4+e5+e6+f4+f5+f6)
AND tf7() BE try(@f7, tf8, f1+f2+f3+f4+f5+f6+f7+f8+f9,
a7+b7+c7+d7+e7+f7+g7+h7+i7,
d7+d8+d9+e7+e8+e9+f7+f8+f9)
AND tf8() BE try(@f8, tf9, f1+f2+f3+f4+f5+f6+f7+f8+f9,
a8+b8+c8+d8+e8+f8+g8+h8+i8,
d7+d8+d9+e7+e8+e9+f7+f8+f9)
AND tf9() BE try(@f9, tg1, f1+f2+f3+f4+f5+f6+f7+f8+f9,
a9+b9+c9+d9+e9+f9+g9+h9+i9,
d7+d8+d9+e7+e8+e9+f7+f8+f9)
 
AND tg1() BE try(@g1, tg2, g1+g2+g3+g4+g5+g6+g7+g8+g9,
a1+b1+c1+d1+e1+f1+g1+h1+i1,
g1+g2+g3+h1+h2+h3+i1+i2+i3)
AND tg2() BE try(@g2, tg3, g1+g2+g3+g4+g5+g6+g7+g8+g9,
a2+b2+c2+d2+e2+f2+g2+h2+i2,
g1+g2+g3+h1+h2+h3+i1+i2+i3)
AND tg3() BE try(@g3, tg4, g1+g2+g3+g4+g5+g6+g7+g8+g9,
a3+b3+c3+d3+e3+f3+g3+h3+i3,
g1+g2+g3+h1+h2+h3+i1+i2+i3)
AND tg4() BE try(@g4, tg5, g1+g2+g3+g4+g5+g6+g7+g8+g9,
a4+b4+c4+d4+e4+f4+g4+h4+i4,
g4+g5+g6+h4+h5+h6+i4+i5+i6)
AND tg5() BE try(@g5, tg6, g1+g2+g3+g4+g5+g6+g7+g8+g9,
a5+b5+c5+d5+e5+f5+g5+h5+i5,
g4+g5+g6+h4+h5+h6+i4+i5+i6)
AND tg6() BE try(@g6, tg7, g1+g2+g3+g4+g5+g6+g7+g8+g9,
a6+b6+c6+d6+e6+f6+g6+h6+i6,
g4+g5+g6+h4+h5+h6+i4+i5+i6)
AND tg7() BE try(@g7, tg8, g1+g2+g3+g4+g5+g6+g7+g8+g9,
a7+b7+c7+d7+e7+f7+g7+h7+i7,
g7+g8+g9+h7+h8+h9+i7+i8+i9)
AND tg8() BE try(@g8, tg9, g1+g2+g3+g4+g5+g6+g7+g8+g9,
a8+b8+c8+d8+e8+f8+g8+h8+i8,
g7+g8+g9+h7+h8+h9+i7+i8+i9)
AND tg9() BE try(@g9, th1, g1+g2+g3+g4+g5+g6+g7+g8+g9,
a9+b9+c9+d9+e9+f9+g9+h9+i9,
g7+g8+g9+h7+h8+h9+i7+i8+i9)
 
AND th1() BE try(@h1, th2, h1+h2+h3+h4+h5+h6+h7+h8+h9,
a1+b1+c1+d1+e1+f1+g1+h1+i1,
g1+g2+g3+h1+h2+h3+i1+i2+i3)
AND th2() BE try(@h2, th3, h1+h2+h3+h4+h5+h6+h7+h8+h9,
a2+b2+c2+d2+e2+f2+g2+h2+i2,
g1+g2+g3+h1+h2+h3+i1+i2+i3)
AND th3() BE try(@h3, th4, h1+h2+h3+h4+h5+h6+h7+h8+h9,
a3+b3+c3+d3+e3+f3+g3+h3+i3,
g1+g2+g3+h1+h2+h3+i1+i2+i3)
AND th4() BE try(@h4, th5, h1+h2+h3+h4+h5+h6+h7+h8+h9,
a4+b4+c4+d4+e4+f4+g4+h4+i4,
g4+g5+g6+h4+h5+h6+i4+i5+i6)
AND th5() BE try(@h5, th6, h1+h2+h3+h4+h5+h6+h7+h8+h9,
a5+b5+c5+d5+e5+f5+g5+h5+i5,
g4+g5+g6+h4+h5+h6+i4+i5+i6)
AND th6() BE try(@h6, th7, h1+h2+h3+h4+h5+h6+h7+h8+h9,
a6+b6+c6+d6+e6+f6+g6+h6+i6,
g4+g5+g6+h4+h5+h6+i4+i5+i6)
AND th7() BE try(@h7, th8, h1+h2+h3+h4+h5+h6+h7+h8+h9,
a7+b7+c7+d7+e7+f7+g7+h7+i7,
g7+g8+g9+h7+h8+h9+i7+i8+i9)
AND th8() BE try(@h8, th9, h1+h2+h3+h4+h5+h6+h7+h8+h9,
a8+b8+c8+d8+e8+f8+g8+h8+i8,
g7+g8+g9+h7+h8+h9+i7+i8+i9)
AND th9() BE try(@h9, ti1, h1+h2+h3+h4+h5+h6+h7+h8+h9,
a9+b9+c9+d9+e9+f9+g9+h9+i9,
g7+g8+g9+h7+h8+h9+i7+i8+i9)
 
AND ti1() BE try(@i1, ti2, i1+i2+i3+i4+i5+i6+i7+i8+i9,
a1+b1+c1+d1+e1+f1+g1+h1+i1,
g1+g2+g3+h1+h2+h3+i1+i2+i3)
AND ti2() BE try(@i2, ti3, i1+i2+i3+i4+i5+i6+i7+i8+i9,
a2+b2+c2+d2+e2+f2+g2+h2+i2,
g1+g2+g3+h1+h2+h3+i1+i2+i3)
AND ti3() BE try(@i3, ti4, i1+i2+i3+i4+i5+i6+i7+i8+i9,
a3+b3+c3+d3+e3+f3+g3+h3+i3,
g1+g2+g3+h1+h2+h3+i1+i2+i3)
AND ti4() BE try(@i4, ti5, i1+i2+i3+i4+i5+i6+i7+i8+i9,
a4+b4+c4+d4+e4+f4+g4+h4+i4,
g4+g5+g6+h4+h5+h6+i4+i5+i6)
AND ti5() BE try(@i5, ti6, i1+i2+i3+i4+i5+i6+i7+i8+i9,
a5+b5+c5+d5+e5+f5+g5+h5+i5,
g4+g5+g6+h4+h5+h6+i4+i5+i6)
AND ti6() BE try(@i6, ti7, i1+i2+i3+i4+i5+i6+i7+i8+i9,
a6+b6+c6+d6+e6+f6+g6+h6+i6,
g4+g5+g6+h4+h5+h6+i4+i5+i6)
AND ti7() BE try(@i7, ti8, i1+i2+i3+i4+i5+i6+i7+i8+i9,
a7+b7+c7+d7+e7+f7+g7+h7+i7,
g7+g8+g9+h7+h8+h9+i7+i8+i9)
AND ti8() BE try(@i8, ti9, i1+i2+i3+i4+i5+i6+i7+i8+i9,
a8+b8+c8+d8+e8+f8+g8+h8+i8,
g7+g8+g9+h7+h8+h9+i7+i8+i9)
AND ti9() BE try(@i9, suc, i1+i2+i3+i4+i5+i6+i7+i8+i9,
a9+b9+c9+d9+e9+f9+g9+h9+i9,
g7+g8+g9+h7+h8+h9+i7+i8+i9)
 
AND suc() BE
{ count := count + 1
prboard()
}

Bracmat[edit]

The program:

{sudokuSolver.bra
 
Solves any 9x9 sudoku, using backtracking.
Not a simple brute force algorithm!}
 
sudokuSolver=
( sudoku
= ( new
= create
. ( create
= a
.  !arg:%(<3:?a) ?arg
& ( !a
.  !arg:
& 1 2 3 4 5 6 7 8 9
| create$!arg
)
create$(!a+1 !arg)
|
)
& create$(0 0 0 0):?(its.Tree)
& ( init
= cell remainingCells remainingRows x y
.  !arg
 : ( ?y
. ?x
. (.%?cell ?remainingCells) ?remainingRows
)
& (  !cell:#
& ( !cell
. mod$(!x,3)
div$(!x,3)
mod$(!y,3)
div$(!y,3)
)
|
)
(  !remainingCells:
& init$(!y+1.0.!remainingRows)
| init
$ ( !y
. !x+1
. (.!remainingCells) !remainingRows
)
)
|
)
& out$!arg
& (its.Set)$(!(its.Tree).init$(0.0.!arg))
 : ?(its.Tree)
)
( Display
= val
. put$(str$("|~~~|~~~|~~~|" \n))
&  !(its.Tree)
 :  ?
( ?
.  ?
( ?&put$"|"
.  ?
( ?
.  ?
( ( ?
.  ?val
& !val:% %
& put$"-"
|  !val:
& put$" "
| put$!val
)
& ~
)
 ?
| ?&put$"|"&~
)
 ?
| ?&put$\n&~
)
 ?
|  ?
& put$(str$("|~~~|~~~|~~~|" \n))
& ~
)
 ?
|
)
( Set
= update certainValue a b c d
, tree branch todo DOING loop dcba minlen len minp
. ( update
= path rempath value tr
, k z x y trc p v branch s n
.  !arg:(?path.?value.?tr.?trc)
& (  !path:%?path ?rempath
& `(  !tr
 : ?k (!path:?p.?branch) ?z
& `( update$(!rempath.!value.!branch.!p !trc)
 : ?s
& update
$ (!path !rempath.!value.!z.!trc)
 : ?n
& !k (!p.!s) !n
)
| !tr
)
| !DOING:(?.!trc)&!value
|  !tr:?x !value ?y
& `( !x !y
 : ( ~:@
& (  !todo:? (?v.!trc) ?
& ( !v:!x !y
| out
$ (mismatch v !v "<>" x y !x !y)
& get'
)
| (!x !y.!trc) !todo:?todo
)
| % %
| &!DOING:(?.!trc)
)
)
| !tr
)
)
& !arg:(?tree.?todo)
& ( loop
=  !todo:
|  !todo
 : ((?certainValue.%?d %?c %?b %?a):?DOING) ?todo
& update$(!a ? !c ?.!certainValue.!tree.)
 : ?tree
& update$(!a !b <>!c ?.!certainValue.!tree.)
 : ?tree
& update$(<>!a ? !c !d.!certainValue.!tree.)
 : ?tree
& !loop
)
& !loop
& ( ~( !tree
 :  ?
(?.? (?.? (?.? (?.% %) ?) ?) ?)
 ?
)
| 9:?minlen
& :?minp
& ( len
=
.  !arg:% %?arg&1+len$!arg
| 1
)
& (  !tree
 :  ?
( ?a
.  ?
( ?b
.  ?
( ?c
.  ?
( ?d
.  % %:?p
& len$!p:<!minlen:?minlen
& !d !c !b !a:?dcba
& !p:?:?minp
& ~
)
 ?
)
 ?
)
 ?
)
 ?
|  !minp
 :  ?
( %@?n
& (its.Set)$(!tree.!n.!dcba):?tree
)
 ?
)
)
& !tree
)
(Tree=)
)
( new
= puzzle
. new$((its.sudoku),!arg):?puzzle
& (puzzle..Display)$
);

Solve a sudoku that is hard for a brute force solver:

new'( sudokuSolver
, (.- - - - - - - - -)
(.- - - - - 3 - 8 5)
(.- - 1 - 2 - - - -)
(.- - - 5 - 7 - - -)
(.- - 4 - - - 1 - -)
(.- 9 - - - - - - -)
(.5 - - - - - - 7 3)
(.- - 2 - 1 - - - -)
(.- - - - 4 - - - 9)
);

Solution:

|~~~|~~~|~~~|
|987|654|321|
|246|173|985|
|351|928|746|
|~~~|~~~|~~~|
|128|537|694|
|634|892|157|
|795|461|832|
|~~~|~~~|~~~|
|519|286|473|
|472|319|568|
|863|745|219|
|~~~|~~~|~~~|

C[edit]

See e.g. this GPLed solver written in C.

The following code is really only good for size 3 puzzles. A longer, even less readable version here could handle size 4s.

#include <stdio.h>
 
void show(int *x)
{
int i, j;
for (i = 0; i < 9; i++) {
if (!(i % 3)) putchar('\n');
for (j = 0; j < 9; j++)
printf(j % 3 ? "%2d" : "%3d", *x++);
putchar('\n');
}
}
 
int trycell(int *x, int pos)
{
int row = pos / 9;
int col = pos % 9;
int i, j, used = 0;
 
if (pos == 81) return 1;
if (x[pos]) return trycell(x, pos + 1);
 
for (i = 0; i < 9; i++)
used |= 1 << (x[i * 9 + col] - 1);
 
for (j = 0; j < 9; j++)
used |= 1 << (x[row * 9 + j] - 1);
 
row = row / 3 * 3;
col = col / 3 * 3;
for (i = row; i < row + 3; i++)
for (j = col; j < col + 3; j++)
used |= 1 << (x[i * 9 + j] - 1);
 
for (x[pos] = 1; x[pos] <= 9; x[pos]++, used >>= 1)
if (!(used & 1) && trycell(x, pos + 1)) return 1;
 
x[pos] = 0;
return 0;
}
 
void solve(const char *s)
{
int i, x[81];
for (i = 0; i < 81; i++)
x[i] = s[i] >= '1' && s[i] <= '9' ? s[i] - '0' : 0;
 
if (trycell(x, 0))
show(x);
else
puts("no solution");
}
 
int main(void)
{
solve( "5x..7...."
"6..195..."
".98....6."
"8...6...3"
"4..8.3..1"
"7...2...6"
".6....28."
"...419..5"
"....8..79" );
 
return 0;
}

C_sharp[edit]

“Manual” Solution[edit]

Translation of: Java
using System;
 
class SudokuSolver
{
private int[] grid;
 
public SudokuSolver(String s)
{
grid = new int[81];
for (int i = 0; i < s.Length; i++)
{
grid[i] = int.Parse(s[i].ToString());
}
}
 
public void solve()
{
try
{
placeNumber(0);
Console.WriteLine("Unsolvable!");
}
catch (Exception ex)
{
Console.WriteLine(ex.Message);
Console.WriteLine(this);
}
}
 
public void placeNumber(int pos)
{
if (pos == 81)
{
throw new Exception("Finished!");
}
if (grid[pos] > 0)
{
placeNumber(pos + 1);
return;
}
for (int n = 1; n <= 9; n++)
{
if (checkValidity(n, pos % 9, pos / 9))
{
grid[pos] = n;
placeNumber(pos + 1);
grid[pos] = 0;
}
}
}
 
public bool checkValidity(int val, int x, int y)
{
for (int i = 0; i < 9; i++)
{
if (grid[y * 9 + i] == val || grid[i * 9 + x] == val)
return false;
}
int startX = (x / 3) * 3;
int startY = (y / 3) * 3;
for (int i = startY; i < startY + 3; i++)
{
for (int j = startX; j < startX + 3; j++)
{
if (grid[i * 9 + j] == val)
return false;
}
}
return true;
}
 
public override string ToString()
{
string sb = "";
for (int i = 0; i < 9; i++)
{
for (int j = 0; j < 9; j++)
{
sb += (grid[i * 9 + j] + " ");
if (j == 2 || j == 5)
sb += ("| ");
}
sb += ('\n');
if (i == 2 || i == 5)
sb += ("------+-------+------\n");
}
return sb;
}
 
public static void Main(String[] args)
{
new SudokuSolver("850002400" +
"720000009" +
"004000000" +
"000107002" +
"305000900" +
"040000000" +
"000080070" +
"017000000" +
"000036040").solve();
Console.Read();
}
}

“Automatic” Solution[edit]

using Microsoft.SolverFoundation.Solvers;
 
namespace Sudoku
{
class Program
{
private static int[,] B = new int[,] {{9,7,0, 3,0,0, 0,6,0},
{0,6,0, 7,5,0, 0,0,0},
{0,0,0, 0,0,8, 0,5,0},
 
{0,0,0, 0,0,0, 6,7,0},
{0,0,0, 0,3,0, 0,0,0},
{0,5,3, 9,0,0, 2,0,0},
 
{7,0,0, 0,2,5, 0,0,0},
{0,0,2, 0,1,0, 0,0,8},
{0,4,0, 0,0,7, 3,0,0}};
 
private static CspTerm[] GetSlice(CspTerm[][] sudoku, int Ra, int Rb, int Ca, int Cb)
{
CspTerm[] slice = new CspTerm[9];
int i = 0;
for (int row = Ra; row < Rb + 1; row++)
for (int col = Ca; col < Cb + 1; col++)
{
{
slice[i++] = sudoku[row][col];
}
}
return slice;
}
 
static void Main(string[] args)
{
ConstraintSystem S = ConstraintSystem.CreateSolver();
CspDomain Z = S.CreateIntegerInterval(1, 9);
CspTerm[][] sudoku = S.CreateVariableArray(Z, "cell", 9, 9);
for (int row = 0; row < 9; row++)
{
for (int col = 0; col < 9; col++)
{
if (B[row, col] > 0)
{
S.AddConstraints(S.Equal(B[row, col], sudoku[row][col]));
}
}
S.AddConstraints(S.Unequal(GetSlice(sudoku, row, row, 0, 8)));
}
for (int col = 0; col < 9; col++)
{
S.AddConstraints(S.Unequal(GetSlice(sudoku, 0, 8, col, col)));
}
for (int a = 0; a < 3; a++)
{
for (int b = 0; b < 3; b++)
{
S.AddConstraints(S.Unequal(GetSlice(sudoku, a * 3, a * 3 + 2, b * 3, b * 3 + 2)));
}
}
ConstraintSolverSolution soln = S.Solve();
object[] h = new object[9];
for (int row = 0; row < 9; row++)
{
if ((row % 3) == 0) System.Console.WriteLine();
for (int col = 0; col < 9; col++)
{
soln.TryGetValue(sudoku[row][col], out h [col]);
}
System.Console.WriteLine("{0}{1}{2} {3}{4}{5} {6}{7}{8}", h[0],h[1],h[2],h[3],h[4],h[5],h[6],h[7],h[8]);
}
}
}
}

Produces:

975 342 861
861 759 432
324 168 957

219 584 673
487 236 519
653 971 284

738 425 196
592 613 748
146 897 325

C++[edit]

Translation of: Java
#include <iostream>
using namespace std;
 
class SudokuSolver {
private:
int grid[81];
 
public:
 
SudokuSolver(string s) {
for (unsigned int i = 0; i < s.length(); i++) {
grid[i] = (int) (s[i] - '0');
}
}
 
void solve() {
try {
placeNumber(0);
cout << "Unsolvable!" << endl;
} catch (char* ex) {
cout << ex << endl;
cout << this->toString() << endl;
}
}
 
void placeNumber(int pos) {
if (pos == 81) {
throw (char*) "Finished!";
}
if (grid[pos] > 0) {
placeNumber(pos + 1);
return;
}
for (int n = 1; n <= 9; n++) {
if (checkValidity(n, pos % 9, pos / 9)) {
grid[pos] = n;
placeNumber(pos + 1);
grid[pos] = 0;
}
}
}
 
bool checkValidity(int val, int x, int y) {
for (int i = 0; i < 9; i++) {
if (grid[y * 9 + i] == val || grid[i * 9 + x] == val)
return false;
}
int startX = (x / 3) * 3;
int startY = (y / 3) * 3;
for (int i = startY; i < startY + 3; i++) {
for (int j = startX; j < startX + 3; j++) {
if (grid[i * 9 + j] == val)
return false;
}
}
return true;
}
 
string toString() {
string sb;
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
char c[2];
c[0] = grid[i * 9 + j] + '0';
c[1] = '\0';
sb.append(c);
sb.append(" ");
if (j == 2 || j == 5)
sb.append("| ");
}
sb.append("\n");
if (i == 2 || i == 5)
sb.append("------+-------+------\n");
}
return sb;
}
 
};
 
int main() {
SudokuSolver ss(
(string) "850002400" +
(string) "720000009" +
(string) "004000000" +
(string) "000107002" +
(string) "305000900" +
(string) "040000000" +
(string) "000080070" +
(string) "017000000" +
(string) "000036040"
);
ss.solve();
}

Clojure[edit]

(ns rosettacode.sudoku
(:use [clojure.pprint :only (cl-format)]))
 
(defn- compatible? [m x y n]
(let [n= #(= n (get-in m [%1 %2]))]
(or (n= y x)
(let [c (count m)]
(and (zero? (get-in m [y x]))
(not-any? #(or (n= y %) (n= % x)) (range c))
(let [zx (* c (quot x c)), zy (* c (quot y c))]
(every? false?
(map n= (range zy (+ zy c)) (range zx (+ zx c))))))))))
 
(defn solve [m]
(let [c (count m)]
(loop [m m, x 0, y 0]
(if (= y c) m
(let [ng (->> (range 1 c)
(filter #(compatible? m x y %))
first
(assoc-in m [y x]))]
(if (= x (dec c))
(recur ng 0 (inc y))
(recur ng (inc x) y)))))))
sudoku>(cl-format true "~{~{~a~^ ~}~%~}"
(solve [[3 9 4 0 0 2 6 7 0]
[0 0 0 3 0 0 4 0 0]
[5 0 0 6 9 0 0 2 0]
[0 4 5 0 0 0 9 0 0]
[6 0 0 0 0 0 0 0 7]
[0 0 7 0 0 0 5 8 0]
[0 1 0 0 6 7 0 0 8]
[0 0 9 0 0 8 0 0 0]
[0 2 6 4 0 0 7 3 5]])
3 9 4 8 5 2 6 7 1
2 6 8 3 7 1 4 5 9
5 7 1 6 9 4 8 2 3
1 4 5 7 8 3 9 6 2
6 8 2 9 4 5 3 1 7
9 3 7 1 2 6 5 8 4
4 1 3 5 6 7 2 9 8
7 5 9 2 3 8 1 4 6
8 2 6 4 1 9 7 3 5
 
nil

Common Lisp[edit]

A simple solver without optimizations (except for pre-computing the possible entries of a cell).

(defun row-neighbors (row column grid &aux (neighbors '()))
(dotimes (i 9 neighbors)
(let ((x (aref grid row i)))
(unless (or (eq '_ x) (= i column))
(push x neighbors)))))
 
(defun column-neighbors (row column grid &aux (neighbors '()))
(dotimes (i 9 neighbors)
(let ((x (aref grid i column)))
(unless (or (eq x '_) (= i row))
(push x neighbors)))))
 
(defun square-neighbors (row column grid &aux (neighbors '()))
(let* ((rmin (* 3 (floor row 3))) (rmax (+ rmin 3))
(cmin (* 3 (floor column 3))) (cmax (+ cmin 3)))
(do ((r rmin (1+ r))) ((= r rmax) neighbors)
(do ((c cmin (1+ c))) ((= c cmax))
(let ((x (aref grid r c)))
(unless (or (eq x '_) (= r row) (= c column))
(push x neighbors)))))))
 
(defun choices (row column grid)
(nset-difference
(list 1 2 3 4 5 6 7 8 9)
(nconc (row-neighbors row column grid)
(column-neighbors row column grid)
(square-neighbors row column grid))))
 
(defun solve (grid &optional (row 0) (column 0))
(cond
((= row 9)
grid)
((= column 9)
(solve grid (1+ row) 0))
((not (eq '_ (aref grid row column)))
(solve grid row (1+ column)))
(t (dolist (choice (choices row column grid) (setf (aref grid row column) '_))
(setf (aref grid row column) choice)
(when (eq grid (solve grid row (1+ column)))
(return grid))))))

Example:

> (defparameter *puzzle*
  #2A((3 9 4    _ _ 2    6 7 _)
      (_ _ _    3 _ _    4 _ _)
      (5 _ _    6 9 _    _ 2 _)
    
      (_ 4 5    _ _ _    9 _ _)
      (6 _ _    _ _ _    _ _ 7)
      (_ _ 7    _ _ _    5 8 _)
    
      (_ 1 _    _ 6 7    _ _ 8)
      (_ _ 9    _ _ 8    _ _ _)
      (_ 2 6    4 _ _    7 3 5)))
*PUZZLE*

> (pprint (solve *puzzle*))

#2A((3 9 4 8 5 2 6 7 1)
    (2 6 8 3 7 1 4 5 9)
    (5 7 1 6 9 4 8 2 3)
    (1 4 5 7 8 3 9 6 2)
    (6 8 2 9 4 5 3 1 7)
    (9 3 7 1 2 6 5 8 4)
    (4 1 3 5 6 7 2 9 8)
    (7 5 9 2 3 8 1 4 6)
    (8 2 6 4 1 9 7 3 5))

Curry[edit]

Copied from Curry: Example Programs.

-----------------------------------------------------------------------------
--- Solving Su Doku puzzles in Curry with FD constraints
---
--- @author Michael Hanus
--- @version December 2005
-----------------------------------------------------------------------------
 
import CLPFD
import List
 
-- Solving a Su Doku puzzle represented as a matrix of numbers (possibly free
-- variables):
sudoku :: [[Int]] -> Success
sudoku m =
domain (concat m) 1 9 & -- define domain of all digits
foldr1 (&) (map allDifferent m) & -- all rows contain different digits
foldr1 (&) (map allDifferent (transpose m)) & -- all columns have different digits
foldr1 (&) (map allDifferent (squaresOfNine m)) & -- all 3x3 squares are different
labeling [FirstFailConstrained] (concat m)
 
-- translate a matrix into a list of small 3x3 squares
squaresOfNine :: [[a]] -> [[a]]
squaresOfNine [] = []
squaresOfNine (l1:l2:l3:ls) = group3Rows [l1,l2,l3] ++ squaresOfNine ls
 
group3Rows l123 = if null (head l123) then [] else
concatMap (take 3) l123 : group3Rows (map (drop 3) l123)
 
-- read a Su Doku specification written as a list of strings containing digits
-- and spaces
readSudoku :: [String] -> [[Int]]
readSudoku s = map (map transDigit) s
where
transDigit c = if c==' ' then x else ord c - ord '0'
where x free
 
-- show a solved Su Doku matrix
showSudoku :: [[Int]] -> String
showSudoku = unlines . map (concatMap (\i->[chr (i + ord '0'),' ']))
 
-- the main function, e.g., evaluate (main s1):
main s | sudoku m = putStrLn (showSudoku m)
where m = readSudoku s
 
s1 = ["9 2 5 ",
" 4 6 3 ",
" 3 6",
" 9 2 ",
" 5 8 ",
" 7 4 3",
"7 1 ",
" 5 2 4 ",
" 1 6 9"]
 
s2 = ["819 5 ",
" 2 75 ",
" 371 4 6 ",
"4 59 1 ",
"7 3 8 2",
" 3 62 7",
" 5 7 921 ",
" 64 9 ",
" 2 438"]


Alternative version[edit]

Works with: PAKCS

Minimal w/o read or show utilities.

import CLPFD
import Constraint (allC)
import List (transpose)
 
 
sudoku :: [[Int]] -> Success
sudoku rows =
domain (concat rows) 1 9
& different rows
& different (transpose rows)
& different blocks
& labeling [] (concat rows)
where
different = allC allDifferent
 
blocks = [concat ys | xs <- each3 rows
, ys <- transpose $ map each3 xs
]
each3 xs = case xs of
(x:y:z:rest) -> [x,y,z] : each3 rest
rest -> [rest]
 
 
test = [ [_,_,3,_,_,_,_,_,_]
, [4,_,_,_,8,_,_,3,6]
, [_,_,8,_,_,_,1,_,_]
, [_,4,_,_,6,_,_,7,3]
, [_,_,_,9,_,_,_,_,_]
, [_,_,_,_,_,2,_,_,5]
, [_,_,4,_,7,_,_,6,8]
, [6,_,_,_,_,_,_,_,_]
, [7,_,_,6,_,_,5,_,_]
]
main | sudoku xs = xs where xs = test
Output:
Execution time: 0 msec. / elapsed: 10 msec.
[[1,2,3,4,5,6,7,8,9],[4,5,7,1,8,9,2,3,6],[9,6,8,3,2,7,1,5,4],[2,4,9,5,6,1,8,7,3],[5,7,6,9,3,8,4,1,2],[8,3,1,7,4,2,6,9,5],[3,1,4,2,7,5,9,6,8],[6,9,5,8,1,4,3,2,7],[7,8,2,6,9,3,5,4,1]]

D[edit]

Translation of: C++

A little over-engineered solution, that shows some strong static typing useful in larger programs.

import std.stdio, std.range, std.string, std.algorithm, std.array,
std.ascii, std.typecons;
 
struct Digit {
immutable char d;
 
this(in char d_) pure nothrow @safe @nogc
in { assert(d_ >= '0' && d_ <= '9'); }
body { this.d = d_; }
 
this(in int d_) pure nothrow @safe @nogc
in { assert(d_ >= '0' && d_ <= '9'); }
body { this.d = cast(char)d_; } // Required cast.
 
alias d this;
}
 
enum size_t sudokuUnitSide = 3;
enum size_t sudokuSide = sudokuUnitSide ^^ 2; // Sudoku grid side.
alias SudokuTable = Digit[sudokuSide ^^ 2];
 
 
Nullable!SudokuTable sudokuSolver(in ref SudokuTable problem)
pure nothrow {
alias Tgrid = uint;
Tgrid[SudokuTable.length] grid = void;
problem[].map!(c => c - '0').copy(grid[]);
 
// DMD doesn't inline this function. Performance loss.
Tgrid access(in size_t x, in size_t y) nothrow @safe @nogc {
return grid[y * sudokuSide + x];
}
 
// DMD doesn't inline this function. If you want to retain
// the same performance as the C++ entry and you use the DMD
// compiler then this function must be manually inlined.
bool checkValidity(in Tgrid val, in size_t x, in size_t y)
pure nothrow @safe @nogc {
/*static*/ foreach (immutable i; staticIota!(0, sudokuSide))
if (access(i, y) == val || access(x, i) == val)
return false;
 
immutable startX = (x / sudokuUnitSide) * sudokuUnitSide;
immutable startY = (y / sudokuUnitSide) * sudokuUnitSide;
 
/*static*/ foreach (immutable i; staticIota!(0, sudokuUnitSide))
/*static*/ foreach (immutable j; staticIota!(0, sudokuUnitSide))
if (access(startX + j, startY + i) == val)
return false;
 
return true;
}
 
bool canPlaceNumbers(in size_t pos=0) nothrow @safe @nogc {
if (pos == SudokuTable.length)
return true;
if (grid[pos] > 0)
return canPlaceNumbers(pos + 1);
 
foreach (immutable n; 1 .. sudokuSide + 1)
if (checkValidity(n, pos % sudokuSide, pos / sudokuSide)) {
grid[pos] = n;
if (canPlaceNumbers(pos + 1))
return true;
grid[pos] = 0;
}
 
return false;
}
 
if (canPlaceNumbers) {
//return typeof(return)(grid[]
// .map!(c => Digit(c + '0'))
// .array);
immutable SudokuTable result = grid[]
.map!(c => Digit(c + '0'))
.array;
return typeof(return)(result);
} else
return typeof(return)();
}
 
string representSudoku(in ref SudokuTable sudo)
pure nothrow @safe out(result) {
assert(result.countchars("1-9") == sudo[].count!q{a != '0'});
assert(result.countchars(".") == sudo[].count!q{a == '0'});
} body {
static assert(sudo.length == 81,
"representSudoku works only with a 9x9 Sudoku.");
string result;
 
foreach (immutable i; 0 .. sudokuSide) {
foreach (immutable j; 0 .. sudokuSide) {
result ~= sudo[i * sudokuSide + j];
result ~= ' ';
if (j == 2 || j == 5)
result ~= "| ";
}
result ~= "\n";
if (i == 2 || i == 5)
result ~= "------+-------+------\n";
}
 
return result.replace("0", ".");
}
 
void main() {
enum ValidateCells(string s) = s.map!Digit.array;
 
immutable SudokuTable problem = ValidateCells!("
850002400
720000009
004000000
000107002
305000900
040000000
000080070
017000000
000036040"
.removechars(whitespace));
problem.representSudoku.writeln;
 
immutable solution = problem.sudokuSolver;
if (solution.isNull)
writeln("Unsolvable!");
else
solution.get.representSudoku.writeln;
}
Output:
8 5 . | . . 2 | 4 . .
7 2 . | . . . | . . 9
. . 4 | . . . | . . .
------+-------+------
. . . | 1 . 7 | . . 2
3 . 5 | . . . | 9 . .
. 4 . | . . . | . . .
------+-------+------
. . . | . 8 . | . 7 .
. 1 7 | . . . | . . .
. . . | . 3 6 | . 4 . 

8 5 9 | 6 1 2 | 4 3 7 
7 2 3 | 8 5 4 | 1 6 9 
1 6 4 | 3 7 9 | 5 2 8 
------+-------+------
9 8 6 | 1 4 7 | 3 5 2 
3 7 5 | 2 6 8 | 9 1 4 
2 4 1 | 5 9 3 | 7 8 6 
------+-------+------
4 3 2 | 9 8 1 | 6 7 5 
6 1 7 | 4 2 5 | 8 9 3 
5 9 8 | 7 3 6 | 2 4 1 

Short Version[edit]

Adapted from: http://code.activestate.com/recipes/576725-brute-force-sudoku-solver/

import std.stdio, std.algorithm, std.range;
 
const(int)[] solve(immutable int[] s) pure nothrow @safe {
immutable i = s.countUntil(0);
if (i == -1)
return s;
 
enum B = (int i, int j) => i / 27 ^ j / 27 | (i%9 / 3 ^ j%9 / 3);
immutable c = iota(81)
.filter!(j => !((i - j) % 9 * (i/9 ^ j/9) * B(i, j)))
.map!(j => s[j]).array;
 
foreach (immutable v; 1 .. 10)
if (!c.canFind(v)) {
const r = solve(s[0 .. i] ~ v ~ s[i + 1 .. $]);
if (!r.empty)
return r;
}
return null;
}
 
void main() {
immutable problem = [
8, 5, 0, 0, 0, 2, 4, 0, 0,
7, 2, 0, 0, 0, 0, 0, 0, 9,
0, 0, 4, 0, 0, 0, 0, 0, 0,
0, 0, 0, 1, 0, 7, 0, 0, 2,
3, 0, 5, 0, 0, 0, 9, 0, 0,
0, 4, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 8, 0, 0, 7, 0,
0, 1, 7, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 3, 6, 0, 4, 0];
writefln("%(%s\n%)", problem.solve.chunks(9));
}
Output:
[8, 5, 9, 6, 1, 2, 4, 3, 7]
[7, 2, 3, 8, 5, 4, 1, 6, 9]
[1, 6, 4, 3, 7, 9, 5, 2, 8]
[9, 8, 6, 1, 4, 7, 3, 5, 2]
[3, 7, 5, 2, 6, 8, 9, 1, 4]
[2, 4, 1, 5, 9, 3, 7, 8, 6]
[4, 3, 2, 9, 8, 1, 6, 7, 5]
[6, 1, 7, 4, 2, 5, 8, 9, 3]
[5, 9, 8, 7, 3, 6, 2, 4, 1]

No-Heap Version[edit]

This version is similar to the precedent one, but it shows idioms to avoid memory allocations on the heap. This is enforced by the use of the @nogc attribute.

import std.stdio, std.algorithm, std.range, std.typecons;
 
Nullable!(const ubyte[81]) solve(in ubyte[81] s) pure nothrow @safe @nogc {
immutable i = s[].countUntil(0);
if (i == -1)
return typeof(return)(s);
 
static immutable B = (in int i, in int j) pure nothrow @safe @nogc =>
i / 27 ^ j / 27 | (i % 9 / 3 ^ j % 9 / 3);
 
ubyte[81] c = void;
size_t len = 0;
foreach (immutable int j; 0 .. c.length)
if (!((i - j) % 9 * (i/9 ^ j/9) * B(i, j)))
c[len++] = s[j];
 
foreach (immutable ubyte v; 1 .. 10)
if (!c[0 .. len].canFind(v)) {
ubyte[81] s2 = void;
s2[0 .. i] = s[0 .. i];
s2[i] = v;
s2[i + 1 .. $] = s[i + 1 .. $];
const r = solve(s2);
if (!r.isNull)
return typeof(return)(r);
}
return typeof(return)();
}
 
void main() {
immutable ubyte[81] problem = [
8, 5, 0, 0, 0, 2, 4, 0, 0,
7, 2, 0, 0, 0, 0, 0, 0, 9,
0, 0, 4, 0, 0, 0, 0, 0, 0,
0, 0, 0, 1, 0, 7, 0, 0, 2,
3, 0, 5, 0, 0, 0, 9, 0, 0,
0, 4, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 8, 0, 0, 7, 0,
0, 1, 7, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 3, 6, 0, 4, 0];
writefln("%(%s\n%)", problem.solve.get[].chunks(9));
}

Same output.

Delphi[edit]

Example taken from C++

type
TIntArray = array of Integer;
 
{ TSudokuSolver }
 
TSudokuSolver = class
private
FGrid: TIntArray;
 
function CheckValidity(val: Integer; x: Integer; y: Integer): Boolean;
function ToString: string; reintroduce;
function PlaceNumber(pos: Integer): Boolean;
public
constructor Create(s: string);
 
procedure Solve;
end;
 
implementation
 
uses
Dialogs;
 
{ TSudokuSolver }
 
function TSudokuSolver.CheckValidity(val: Integer; x: Integer; y: Integer
): Boolean;
var
i: Integer;
j: Integer;
StartX: Integer;
StartY: Integer;
begin
for i := 0 to 8 do
begin
if (FGrid[y * 9 + i] = val) or
(FGrid[i * 9 + x] = val) then
begin
Result := False;
Exit;
end;
end;
StartX := (x div 3) * 3;
StartY := (y div 3) * 3;
for i := StartY to Pred(StartY + 3) do
begin
for j := StartX to Pred(StartX + 3) do
begin
if FGrid[i * 9 + j] = val then
begin
Result := False;
Exit;
end;
end;
end;
Result := True;
end;
 
function TSudokuSolver.ToString: string;
var
sb: string;
i: Integer;
j: Integer;
c: char;
begin
sb := '';
for i := 0 to 8 do
begin
for j := 0 to 8 do
begin
c := (IntToStr(FGrid[i * 9 + j]) + '0')[1];
sb := sb + c + ' ';
if (j = 2) or (j = 5) then sb := sb + '| ';
end;
sb := sb + #13#10;
if (i = 2) or (i = 5) then
sb := sb + '-----+-----+-----' + #13#10;
end;
Result := sb;
end;
 
function TSudokuSolver.PlaceNumber(pos: Integer): Boolean;
var
n: Integer;
begin
Result := False;
if Pos = 81 then
begin
Result := True;
Exit;
end;
if FGrid[pos] > 0 then
begin
Result := PlaceNumber(Succ(pos));
Exit;
end;
for n := 1 to 9 do
begin
if CheckValidity(n, pos mod 9, pos div 9) then
begin
FGrid[pos] := n;
Result := PlaceNumber(Succ(pos));
if not Result then
FGrid[pos] := 0;
end;
end;
end;
 
constructor TSudokuSolver.Create(s: string);
var
lcv: Cardinal;
begin
SetLength(FGrid, 81);
for lcv := 0 to Pred(Length(s)) do
FGrid[lcv] := StrToInt(s[Succ(lcv)]);
end;
 
procedure TSudokuSolver.Solve;
begin
if not PlaceNumber(0) then
ShowMessage('Unsolvable')
else
ShowMessage('Solved!');
end;
end;

Usage:

var
SudokuSolver: TSudokuSolver;
begin
SudokuSolver := TSudokuSolver.Create('850002400' +
'720000009' +
'004000000' +
'000107002' +
'305000900' +
'040000000' +
'000080070' +
'017000000' +
'000036040');
try
SudokuSolver.Solve;
finally
FreeAndNil(SudokuSolver);
end;
end;

Elixir[edit]

Translation of: Erlang
defmodule Sudoku do
def display( grid ), do: ( for y <- 1..9, do: display_row(y, grid) )
 
def start( knowns ), do: Enum.into( knowns, Map.new )
 
def solve( grid ) do
sure = solve_all_sure( grid )
solve_unsure( potentials(sure), sure )
end
 
def task( knowns ) do
IO.puts "start"
start = start( knowns )
display( start )
IO.puts "solved"
solved = solve( start )
display( solved )
IO.puts ""
end
 
defp bt( grid ), do: bt_reject( is_not_allowed(grid), grid )
 
defp bt_accept( true, board ), do: throw( {:ok, board} )
defp bt_accept( false, grid ), do: bt_loop( potentials_one_position(grid), grid )
 
defp bt_loop( {position, values}, grid ), do: ( for x <- values, do: bt( Map.put(grid, position, x) ) )
 
defp bt_reject( true, _grid ), do: :backtrack
defp bt_reject( false, grid ), do: bt_accept( is_all_correct(grid), grid )
 
defp display_row( row, grid ) do
for x <- [1, 4, 7], do: display_row_group( x, row, grid )
display_row_nl( row )
end
 
defp display_row_group( start, row, grid ) do
Enum.each(start..start+2, &IO.write " #{Map.get( grid, {&1, row}, ".")}")
IO.write " "
end
 
defp display_row_nl( n ) when n in [3,6,9], do: IO.puts "\n"
defp display_row_nl( _n ), do: IO.puts ""
 
defp is_all_correct( grid ), do: map_size( grid ) == 81
 
defp is_not_allowed( grid ) do
is_not_allowed_rows( grid ) or is_not_allowed_columns( grid ) or is_not_allowed_groups( grid )
end
 
defp is_not_allowed_columns( grid ), do: values_all_columns(grid) |> Enum.any?(&is_not_allowed_values/1)
 
defp is_not_allowed_groups( grid ), do: values_all_groups(grid) |> Enum.any?(&is_not_allowed_values/1)
 
defp is_not_allowed_rows( grid ), do: values_all_rows(grid) |> Enum.any?(&is_not_allowed_values/1)
 
defp is_not_allowed_values( values ), do: length( values ) != length( Enum.uniq(values) )
 
defp group_positions( {x, y} ) do
for colum <- group_positions_close(x), row <- group_positions_close(y), do: {colum, row}
end
 
defp group_positions_close( n ) when n < 4, do: [1,2,3]
defp group_positions_close( n ) when n < 7, do: [4,5,6]
defp group_positions_close( _n ) , do: [7,8,9]
 
defp positions_not_in_grid( grid ) do
keys = Map.keys( grid )
for x <- 1..9, y <- 1..9, not {x, y} in keys, do: {x, y}
end
 
defp potentials_one_position( grid ) do
Enum.min_by( potentials( grid ), fn {_position, values} -> length(values) end )
end
 
defp potentials( grid ), do: List.flatten( for x <- positions_not_in_grid(grid), do: potentials(x, grid) )
 
defp potentials( position, grid ) do
useds = potentials_used_values( position, grid )
{position, Enum.to_list(1..9) -- useds }
end
 
defp potentials_used_values( {x, y}, grid ) do
row_values = (for row <- 1..9, row != x, do: {row, y}) |> potentials_values( grid )
column_values = (for column <- 1..9, column != y, do: {x, column}) |> potentials_values( grid )
group_values = group_positions({x, y}) -- [ {x, y} ] |> potentials_values( grid )
row_values ++ column_values ++ group_values
end
 
defp potentials_values( keys, grid ) do
for x <- keys, val = grid[x], do: val
end
 
defp values_all_columns( grid ) do
for x <- 1..9, do:
( for y <- 1..9, do: {x, y} ) |> potentials_values( grid )
end
 
defp values_all_groups( grid ) do
[[g1,g2,g3], [g4,g5,g6], [g7,g8,g9]] = for x <- [1,4,7], do: values_all_groups(x, grid)
[g1,g2,g3,g4,g5,g6,g7,g8,g9]
end
 
defp values_all_groups( x, grid ) do
for x_offset <- x..x+2, do: values_all_groups(x, x_offset, grid)
end
 
defp values_all_groups( _x, x_offset, grid ) do
( for y_offset <- group_positions_close(x_offset), do: {x_offset, y_offset} )
|> potentials_values( grid )
end
 
defp values_all_rows( grid ) do
for y <- 1..9, do:
( for x <- 1..9, do: {x, y} ) |> potentials_values( grid )
end
 
defp solve_all_sure( grid ), do: solve_all_sure( solve_all_sure_values(grid), grid )
 
defp solve_all_sure( [], grid ), do: grid
defp solve_all_sure( sures, grid ) do
solve_all_sure( Enum.reduce(sures, grid, &solve_all_sure_store/2) )
end
 
defp solve_all_sure_values( grid ), do: (for{position, [value]} <- potentials(grid), do: {position, value} )
 
defp solve_all_sure_store( {position, value}, acc ), do: Map.put( acc, position, value )
 
defp solve_unsure( [], grid ), do: grid
defp solve_unsure( _potentials, grid ) do
try do
bt( grid )
catch
{:ok, board} -> board
end
end
end
 
simple = [{{1, 1}, 3}, {{2, 1}, 9}, {{3, 1},4}, {{6, 1}, 2}, {{7, 1}, 6}, {{8, 1}, 7},
{{4, 2}, 3}, {{7, 2}, 4},
{{1, 3}, 5}, {{4, 3}, 6}, {{5, 3}, 9}, {{8, 3}, 2},
{{2, 4}, 4}, {{3, 4}, 5}, {{7, 4}, 9},
{{1, 5}, 6}, {{9, 5}, 7},
{{3, 6}, 7}, {{7, 6}, 5}, {{8, 6}, 8},
{{2, 7}, 1}, {{5, 7}, 6}, {{6, 7}, 7}, {{9, 7}, 8},
{{3, 8}, 9}, {{6, 8}, 8},
{{2, 9}, 2}, {{3, 9}, 6}, {{4, 9}, 4}, {{7, 9}, 7}, {{8, 9}, 3}, {{9, 9}, 5}]
Sudoku.task( simple )
 
difficult = [{{6, 2}, 3}, {{8, 2}, 8}, {{9, 2}, 5},
{{3, 3}, 1}, {{5, 3}, 2},
{{4, 4}, 5}, {{6, 4}, 7},
{{3, 5}, 4}, {{7, 5}, 1},
{{2, 6}, 9},
{{1, 7}, 5}, {{8, 7}, 7}, {{9, 7}, 3},
{{3, 8}, 2}, {{5, 8}, 1},
{{5, 9}, 4}, {{9, 9}, 9}]
Sudoku.task( difficult )
Output:
start
 3 9 4  . . 2  6 7 .
 . . .  3 . .  4 . .
 5 . .  6 9 .  . 2 .

 . 4 5  . . .  9 . .
 6 . .  . . .  . . 7
 . . 7  . . .  5 8 .

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

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

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

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


start
 . . .  . . .  . . .
 . . .  . . 3  . 8 5
 . . 1  . 2 .  . . .

 . . .  5 . 7  . . .
 . . 4  . . .  1 . .
 . 9 .  . . .  . . .

 5 . .  . . .  . 7 3
 . . 2  . 1 .  . . .
 . . .  . 4 .  . . 9

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

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

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

Erlang[edit]

I first try to solve the Sudoku grid without guessing. For the guessing part I eschew spawning a process for each guess, instead opting for backtracking. It is fun trying new things.

 
-module( sudoku ).
 
-export( [display/1, start/1, solve/1, task/0] ).
 
display( Grid ) -> [display_row(Y, Grid) || Y <- lists:seq(1, 9)].
%% A known value is {{Column, Row}, Value}
%% Top left corner is {1, 1}, Bottom right corner is {9,9}
start( Knowns ) -> dict:from_list( Knowns ).
 
solve( Grid ) ->
Sure = solve_all_sure( Grid ),
solve_unsure( potentials(Sure), Sure ).
 
task() ->
Simple = [{{1, 1}, 3}, {{2, 1}, 9}, {{3, 1},4}, {{6, 1}, 2}, {{7, 1}, 6}, {{8, 1}, 7},
{{4, 2}, 3}, {{7, 2}, 4},
{{1, 3}, 5}, {{4, 3}, 6}, {{5, 3}, 9}, {{8, 3}, 2},
{{2, 4}, 4}, {{3, 4}, 5}, {{7, 4}, 9},
{{1, 5}, 6}, {{9, 5}, 7},
{{3, 6}, 7}, {{7, 6}, 5}, {{8, 6}, 8},
{{2, 7}, 1}, {{5, 7}, 6}, {{6, 7}, 7}, {{9, 7}, 8},
{{3, 8}, 9}, {{6, 8}, 8},
{{2, 9}, 2}, {{3, 9}, 6}, {{4, 9}, 4}, {{7, 9}, 7}, {{8, 9}, 3}, {{9, 9}, 5}],
task( Simple ),
Difficult = [{{6, 2}, 3}, {{8, 2}, 8}, {{9, 2}, 5},
{{3, 3}, 1}, {{5, 3}, 2},
{{4, 4}, 5}, {{6, 4}, 7},
{{3, 5}, 4}, {{7, 5}, 1},
{{2, 6}, 9},
{{1, 7}, 5}, {{8, 7}, 7}, {{9, 7}, 3},
{{3, 8}, 2}, {{5, 8}, 1},
{{5, 9}, 4}, {{9, 9}, 9}],
task( Difficult ).
 
 
 
bt( Grid ) -> bt_reject( is_not_allowed(Grid), Grid ).
 
bt_accept( true, Board ) -> erlang:throw( {ok, Board} );
bt_accept( false, Grid ) -> bt_loop( potentials_one_position(Grid), Grid ).
 
bt_loop( {Position, Values}, Grid ) -> [bt( dict:store(Position, X, Grid) ) || X <- Values].
 
bt_reject( true, _Grid ) -> backtrack;
bt_reject( false, Grid ) -> bt_accept( is_all_correct(Grid), Grid ).
 
display_row( Row, Grid ) ->
[display_row_group( X, Row, Grid ) || X <- [1, 4, 7]],
display_row_nl( Row ).
 
display_row_group( Start, Row, Grid ) ->
[io:fwrite(" ~c", [display_value(X, Row, Grid)]) || X <- [Start, Start+1, Start+2]],
io:fwrite( " " ).
 
display_row_nl( N ) when N =:= 3; N =:= 6; N =:= 9 -> io:nl(), io:nl();
display_row_nl( _N ) -> io:nl().
 
display_value( X, Y, Grid ) -> display_value( dict:find({X, Y}, Grid) ).
 
display_value( error ) -> $.;
display_value( {ok, Value} ) -> Value + $0.
 
is_all_correct( Grid ) -> dict:size( Grid ) =:= 81.
 
is_not_allowed( Grid ) ->
is_not_allowed_rows( Grid )
orelse is_not_allowed_columns( Grid )
orelse is_not_allowed_groups( Grid ).
 
is_not_allowed_columns( Grid ) -> lists:any( fun is_not_allowed_values/1, values_all_columns(Grid) ).
 
is_not_allowed_groups( Grid ) -> lists:any( fun is_not_allowed_values/1, values_all_groups(Grid) ).
 
is_not_allowed_rows( Grid ) -> lists:any( fun is_not_allowed_values/1, values_all_rows(Grid) ).
 
is_not_allowed_values( Values ) -> erlang:length( Values ) =/= erlang:length( lists:usort(Values) ).
 
group_positions( {X, Y} ) -> [{Colum, Row} || Colum <- group_positions_close(X), Row <- group_positions_close(Y)].
 
group_positions_close( N ) when N < 4 -> [1,2,3];
group_positions_close( N ) when N < 7 -> [4,5,6];
group_positions_close( _N ) -> [7,8,9].
 
positions_not_in_grid( Grid ) ->
Keys = dict:fetch_keys( Grid ),
[{X, Y} || X <- lists:seq(1, 9), Y <- lists:seq(1, 9), not lists:member({X, Y}, Keys)].
 
potentials_one_position( Grid ) ->
[{_Shortest, Position, Values} | _T] = lists:sort( [{erlang:length(Values), Position, Values} || {Position, Values} <- potentials( Grid )] ),
{Position, Values}.
 
potentials( Grid ) -> lists:flatten( [potentials(X, Grid) || X <- positions_not_in_grid(Grid)] ).
 
potentials( Position, Grid ) ->
Useds = potentials_used_values( Position, Grid ),
{Position, [Value || Value <- lists:seq(1, 9) -- Useds]}.
 
potentials_used_values( {X, Y}, Grid ) ->
Row_positions = [{Row, Y} || Row <- lists:seq(1, 9), Row =/= X],
Row_values = potentials_values( Row_positions, Grid ),
Column_positions = [{X, Column} || Column <- lists:seq(1, 9), Column =/= Y],
Column_values = potentials_values( Column_positions, Grid ),
Group_positions = lists:delete( {X, Y}, group_positions({X, Y}) ),
Group_values = potentials_values( Group_positions, Grid ),
Row_values ++ Column_values ++ Group_values.
 
potentials_values( Keys, Grid ) ->
Row_values_unfiltered = [dict:find(X, Grid) || X <- Keys],
[Value || {ok, Value} <- Row_values_unfiltered].
 
values_all_columns( Grid ) -> [values_all_columns(X, Grid) || X <- lists:seq(1, 9)].
 
values_all_columns( X, Grid ) ->
Positions = [{X, Y} || Y <- lists:seq(1, 9)],
potentials_values( Positions, Grid ).
 
values_all_groups( Grid ) ->
[G123, G456, G789] = [values_all_groups(X, Grid) || X <- [1, 4, 7]],
[G1,G2,G3] = G123,
[G4,G5,G6] = G456,
[G7,G8,G9] = G789,
[G1,G2,G3,G4,G5,G6,G7,G8,G9].
 
values_all_groups( X, Grid ) ->[values_all_groups(X, X_offset, Grid) || X_offset <- [X, X+1, X+2]].
 
values_all_groups( _X, X_offset, Grid ) ->
Positions = [{X_offset, Y_offset} || Y_offset <- group_positions_close(X_offset)],
potentials_values( Positions, Grid ).
 
values_all_rows( Grid ) ->[values_all_rows(Y, Grid) || Y <- lists:seq(1, 9)].
 
values_all_rows( Y, Grid ) ->
Positions = [{X, Y} || X <- lists:seq(1, 9)],
potentials_values( Positions, Grid ).
 
solve_all_sure( Grid ) -> solve_all_sure( solve_all_sure_values(Grid), Grid ).
 
solve_all_sure( [], Grid ) -> Grid;
solve_all_sure( Sures, Grid ) -> solve_all_sure( lists:foldl(fun solve_all_sure_store/2, Grid, Sures) ).
 
solve_all_sure_values( Grid ) -> [{Position, Value} || {Position, [Value]} <- potentials(Grid)].
 
solve_all_sure_store( {Position, Value}, Acc ) -> dict:store( Position, Value, Acc ).
 
solve_unsure( [], Grid ) -> Grid;
solve_unsure( _Potentials, Grid ) ->
try
bt( Grid )
 
catch
_:{ok, Board} -> Board
 
end.
 
task( Knowns ) ->
io:fwrite( "Start~n" ),
Start = start( Knowns ),
display( Start ),
io:fwrite( "Solved~n" ),
Solved = solve( Start ),
display( Solved ),
io:nl().
 
Output:
5> sudoku:task().
Start
 3 9 4  . . 2  6 7 . 
 . . .  3 . .  4 . . 
 5 . .  6 9 .  . 2 . 

 . 4 5  . . .  9 . . 
 6 . .  . . .  . . 7 
 . . 7  . . .  5 8 . 

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

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

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

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


Start
 . . .  . . .  . . . 
 . . .  . . 3  . 8 5 
 . . 1  . 2 .  . . . 

 . . .  5 . 7  . . . 
 . . 4  . . .  1 . . 
 . 9 .  . . .  . . . 

 5 . .  . . .  . 7 3 
 . . 2  . 1 .  . . . 
 . . .  . 4 .  . . 9 

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

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

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

ERRE[edit]

Sudoku solver. Program solves Sudoku grid with an iterative method: it's taken from ERRE distribution disk and so comments are in Italian. Grid data are contained in the file SUDOKU.TXT

Example of SUDOKU.TXT

503600009

010002600

900000080

000700005

006804100

200003000

030000008

004300050

800006702

0 is the empty cell.

 
!--------------------------------------------------------------------
! risolve Sudoku: in input il file SUDOKU.TXT
! Metodo seguito : cancellazioni successive e quando non possibile
! ricerca combinatoria sulle celle con due valori
! possibili - max. 30 livelli di ricorsione
! Non risolve se,dopo l'analisi per la cancellazione,
! restano solo celle a 4 valori
!--------------------------------------------------------------------
 
PROGRAM SUDOKU
 
LABEL 76,77,88,91,97,99
 
DIM TAV$[9,9]  ! 81 caselle in nove quadranti
 ! cella non definita --> 0/. nel file SUDOKU.TXT
 ! diventa 123456789 dopo LEGGI_SCHEMA
 
!---------------------------------------------------------------------------
! tabelle per gestire la ricerca combinatoria
! (primo indice--> livelli ricorsione)
!---------------------------------------------------------------------------
DIM TAV2$[30,9,9],INFO[30,4]
 
!$INCLUDE="PC.LIB"
 
PROCEDURE MESSAGGI(MEX%)
CASE MEX% OF
1-> LOCATE(21,1) PRINT("Cancellazione successiva - liv. 1") END ->
2-> LOCATE(21,1) PRINT("Cancellazione successiva - liv. 2") END ->
3-> LOCATE(22,1) PRINT("Ricerca combinatoria - liv.";LIVELLO;" ") END ->
END CASE
END PROCEDURE
 
PROCEDURE VISUALIZZA_SCHEMA
LOCATE(1,1)
PRINT("+---+---+---+---+---+---+---+---+----+")
FOR I=1 TO 9 DO
FOR J=1 TO 9 DO
PRINT("|";)
IF LEN(TAV$[I,J])=1 THEN
PRINT(" ";TAV$[I,J];" ";)
ELSE
PRINT(" ";)
END IF
END FOR
PRINT("³")
IF I<>9 THEN PRINT("+---+---+---+---+---+---+---+---+----+") END IF
END FOR
PRINT("+---+---+---+---+---+---+---+---+----+")
END PROCEDURE
 
!------------------------------------------------------------------------
! in input la cella (riga,colonna)
! in output se ha un valore definito
!------------------------------------------------------------------------
PROCEDURE VALORE_DEFINITO
FLAG%=FALSE
IF LEN(TAV$[RIGA,COLONNA])=1 THEN FLAG%=TRUE END IF
END PROCEDURE
 
 
PROCEDURE SALVA_CONFIG
LIVELLO=LIVELLO+1
FOR R=1 TO 9 DO
FOR S=1 TO 9 DO
TAV2$[LIVELLO,R,S]=TAV$[R,S]
END FOR
END FOR
INFO[LIVELLO,0]=1 INFO[LIVELLO,1]=RIGA INFO[LIVELLO,2]=COLONNA
INFO[LIVELLO,3]=SECOND INFO[LIVELLO,4]=THIRD
END PROCEDURE
 
PROCEDURE RIPRISTINA_CONFIG
91:
LIVELLO=LIVELLO-1
IF INFO[LIVELLO,0]=3 THEN GOTO 91 END IF
FOR R=1 TO 9 DO
FOR S=1 TO 9 DO
TAV$[R,S]=TAV2$[LIVELLO,R,S]
END FOR
END FOR
RIGA=INFO[LIVELLO,1] COLONNA=INFO[LIVELLO,2]
SECOND=INFO[LIVELLO,3] THIRD=INFO[LIVELLO,4]
IF INFO[LIVELLO,0]=1 THEN
TAV$[RIGA,COLONNA]=MID$(STR$(SECOND),2)
END IF
IF INFO[LIVELLO,0]=2 THEN
IF THIRD<>0 THEN
TAV$[RIGA,COLONNA]=MID$(STR$(THIRD),2)
ELSE
GOTO 91
END IF
END IF
INFO[LIVELLO,0]=INFO[LIVELLO,0]+1
VISUALIZZA_SCHEMA
END PROCEDURE
 
PROCEDURE VERIFICA_SE_FINITO
COMPLETO%=TRUE
FOR RIGA=1 TO 9 DO
PRD#=1
FOR COLONNA=1 TO 9 DO
PRD#=PRD#*VAL(TAV$[RIGA,COLONNA])
END FOR
IF PRD#<>362880 THEN COMPLETO%=FALSE EXIT END IF
END FOR
IF NOT COMPLETO% THEN EXIT PROCEDURE END IF
FOR COLONNA=1 TO 9 DO
PRD#=1
FOR RIGA=1 TO 9 DO
PRD#=PRD#*VAL(TAV$[RIGA,COLONNA])
END FOR
IF PRD#<>362880 THEN COMPLETO%=FALSE EXIT END IF
END FOR
END PROCEDURE
 
!-------------------------------------------------------------------
! toglie i valore certi dalle celle sulla
! stessa riga-stessa colonna-stesso quadrante
!-------------------------------------------------------------------
PROCEDURE TOGLI_VALORE
 
!iniziamo a togliere il valore dalla stessa riga ....
FOR J=1 TO 9 DO
CH$=TAV$[RIGA,J] CH=VAL(Z$)
IF LEN(CH$)<>1 THEN
CHANGE(CH$,CH,"-"->CH$)
TAV$[RIGA,J]=CH$
END IF
END FOR
!... iniziamo a togliere il valore dalla stessa colonna ...
FOR I=1 TO 9 DO
CH$=TAV$[I,COLONNA] CH=VAL(Z$)
IF LEN(CH$)<>1 THEN
CHANGE(CH$,CH,"-"->CH$)
TAV$[I,COLONNA]=CH$
END IF
END FOR
!... iniziamo a togliere il valore dallo stesso quadrante
R=INT(RIGA/3.1)*3+1
S=INT(COLONNA/3.1)*3+1
FOR I=R TO R+2 DO
FOR J=S TO S+2 DO
CH$=TAV$[I,J] CH=VAL(Z$)
IF LEN(CH$)<>1 THEN
CHANGE(CH$,CH,"-"->CH$)
TAV$[I,J]=CH$
END IF
END FOR
END FOR
MESSAGGI(1)
END PROCEDURE
 
PROCEDURE ESAMINA_SCHEMA
FOR RIGA=1 TO 9 DO
FOR COLONNA=1 TO 9 DO
VALORE_DEFINITO
IF FLAG% THEN
Z$=TAV$[RIGA,COLONNA]
TOGLI_VALORE
END IF
END FOR
END FOR
END PROCEDURE
 
PROCEDURE IDENTIFICA_UNICO
FOR KL=1 TO 9 DO
KL$=MID$(STR$(KL),2)
NN=0
FOR H=1 TO LEN(ZZ$) DO
IF MID$(ZZ$,H,1)=KL$ THEN NN=NN+1 END IF
END FOR
IF NN=1 THEN Q=INSTR(ZZ$,KL$) KL=9 END IF
END FOR
END PROCEDURE
 
!----------------------------------------------------------------------------
! intercetta i valori unici per le celle ancora non definite
!----------------------------------------------------------------------------
PROCEDURE TOGLI_VALORE2
 
MESSAGGI(2)
! iniziamo dalle righe ....
OK%=FALSE
FOR RIGA=1 TO 9 DO
ZZ$=""
FOR COLONNA=1 TO 9 DO
IF LEN(TAV$[RIGA,COLONNA])<>1 THEN
ZZ$=ZZ$+TAV$[RIGA,COLONNA]
ELSE
ZZ$=ZZ$+STRING$(9," ")
END IF
END FOR
Q=0 IDENTIFICA_UNICO
IF Q<>0 THEN
COLONNA=INT(Q/9.1)+1
TAV$[RIGA,COLONNA]=KL$
OK%=TRUE EXIT
END IF
END FOR
IF OK% THEN GOTO 76 END IF
 
! .... poi dalle colonne ....
FOR COLONNA=1 TO 9 DO
ZZ$=""
FOR RIGA=1 TO 9 DO
IF LEN(TAV$[RIGA,COLONNA])<>1 THEN
ZZ$=ZZ$+TAV$[RIGA,COLONNA]
ELSE
ZZ$=ZZ$+STRING$(9," ")
END IF
END FOR
Q=0 IDENTIFICA_UNICO
IF Q<>0 THEN
RIGA=INT(Q/9.1)+1
TAV$[RIGA,COLONNA]=KL$ OK%=TRUE EXIT
END IF
END FOR
IF OK% THEN GOTO 76 END IF
 
!.... e infine i quadranti
FOR QUADRANTE=1 TO 9 DO
ZZ$=""
CASE QUADRANTE OF
1-> R=1 S=1 END ->
2-> R=1 S=4 END ->
3-> R=1 S=7 END ->
4-> R=4 S=1 END ->
5-> R=4 S=4 END ->
6-> R=4 S=7 END ->
7-> R=7 S=1 END ->
8-> R=7 S=4 END ->
9-> R=7 S=7 END ->
END CASE
FOR RIGA=R TO R+2 DO
FOR COLONNA=S TO S+2 DO
IF LEN(TAV$[RIGA,COLONNA])<>1 THEN
ZZ$=ZZ$+TAV$[RIGA,COLONNA]
ELSE
ZZ$=ZZ$+STRING$(9," ")
END IF
END FOR
END FOR
Q=0 IDENTIFICA_UNICO
IF Q<>0 THEN
CASE Q OF
1..9-> ALFA=R BETA=S END ->
10..18-> ALFA=R BETA=S+1 END ->
19..27-> ALFA=R BETA=S+2 END ->
28..36-> ALFA=R+1 BETA=S END ->
37..45-> ALFA=R+1 BETA=S+1 END ->
46..54-> ALFA=R+1 BETA=S+2 END ->
55..63-> ALFA=R+2 BETA=S END ->
64..72-> ALFA=R+2 BETA=S+1 END ->
OTHERWISE
ALFA=R+2 BETA=S+2
END CASE
77:
TAV$[ALFA,BETA]=KL$ EXIT
END IF
END FOR
76:
MESSAGGI(2)
END PROCEDURE
 
PROCEDURE CONVERTI_VALORE
FINE%=TRUE NESSUNO%=TRUE
FOR RIGA=1 TO 9 DO
FOR COLONNA=1 TO 9 DO
CH$=TAV$[RIGA,COLONNA]
IF LEN(CH$)<>1 THEN
FINE%=FALSE ! flag per fine partita -- trovati tutti
Q=0  ! conta i '-' nella stringa se ce ne sono 8,
 ! trovato valore
FOR Z=1 TO LEN(CH$) DO
IF MID$(CH$,Z,1)="-" THEN Q=Q+1 ELSE LAST=Z END IF
END FOR
IF Q=8 THEN
CH$=MID$(STR$(LAST),2)
TAV$[RIGA,COLONNA]=CH$
NESSUNO%=FALSE
END IF
END IF
END FOR
END FOR
END PROCEDURE
 
PROCEDURE LEGGI_SCHEMA
OPEN("I",1,"sudoku.txt")
FOR I=1 TO 9 DO
INPUT(LINE,#1,RIGA$)
FOR J=1 TO 9 DO
CH$=MID$(RIGA$,J,1)
IF CH$="0" OR CH$="." THEN
TAV$[I,J]="123456789"
ELSE
TAV$[I,J]=CH$
END IF
END FOR
END FOR
CLOSE(1)
END PROCEDURE
 
!---------------------------------------------------------------------------
! Praticamente - visita di un albero binario (caso con cella a 2 valori
! possibili)
!---------------------------------------------------------------------------
PROCEDURE RICERCA_COMBINATORIA
TRE%=TRUE
FOR RIGA=1 TO 9 DO
FOR COLONNA=1 TO 9 DO
CH$=TAV$[RIGA,COLONNA]
IF LEN(CH$)<>1 THEN
Q=0 FIRST=0 SECOND=0 THIRD=0
FOR Z=1 TO LEN(CH$) DO
IF MID$(CH$,Z,1)="-" THEN
Q=Q+1
ELSE
IF FIRST=0 THEN
FIRST=Z
ELSE
SECOND=Z
END IF
END IF
END FOR
IF Q=7 THEN
SALVA_CONFIG
TAV$[RIGA,COLONNA]=MID$(STR$(FIRST),2)
TRE%=FALSE
GOTO 97
END IF
END IF
END FOR
END FOR
IF TRE% THEN GOTO 88 END IF
97:
MESSAGGI(3)
EXIT PROCEDURE
88:
QUATTRO%=TRUE
FOR RIGA=1 TO 9 DO
FOR COLONNA=1 TO 9 DO
CH$=TAV$[RIGA,COLONNA]
IF LEN(CH$)<>1 THEN
Q=0 FIRST=0 SECOND=0 THIRD=0
FOR Z=1 TO LEN(CH$) DO
IF MID$(CH$,Z,1)="-" THEN
Q=Q+1
ELSE
IF FIRST=0 THEN
FIRST=Z
ELSE
IF SECOND=0 THEN
SECOND=Z
ELSE
THIRD=Z
END IF
END IF
END IF
END FOR
IF Q=6 THEN
SALVA_CONFIG
TAV$[RIGA,COLONNA]=MID$(STR$(FIRST),2)
QUATTRO%=FALSE
GOTO 97
END IF
END IF
END FOR
END FOR
IF QUATTRO% THEN
LIVELLO=LIVELLO+1
RIPRISTINA_CONFIG
GOTO 97
END IF
 ! se restano solo celle con 4 valori,forza la chiusura del ramo dell'albero
 !$RCODE="STOP"
END PROCEDURE
 
BEGIN
CLS
LIVELLO=1 NZ%=0
LEGGI_SCHEMA
WHILE TRUE DO
VISUALIZZA_SCHEMA
99:
NZ%=NZ%+1
ESAMINA_SCHEMA
CONVERTI_VALORE
EXIT IF FINE%
IF NESSUNO% THEN
TOGLI_VALORE2
IF OK%=0 THEN
RICERCA_COMBINATORIA  ! cerca altri celle da assegnare
END IF
END IF
END WHILE
VISUALIZZA_SCHEMA
VERIFICA_SE_FINITO
IF NOT COMPLETO% THEN
LIVELLO=LIVELLO+1
RIPRISTINA_CONFIG
GOTO 99
END IF
END PROGRAM
 
 
 

Forth[edit]

Works with: 4tH version 3.60.0
include lib/interprt.4th
include lib/istype.4th
include lib/argopen.4th
 
\ ---------------------
\ Variables
\ ---------------------
 
81 string sudokugrid
9 array sudoku_row
9 array sudoku_col
9 array sudoku_box
 
\ -------------
\ 4tH interface
\ -------------
 
: >grid ( n2 a1 n1 -- n3)
rot dup >r 9 chars * sudokugrid + dup >r swap
0 do ( a1 a2)
over i chars + c@ dup is-digit ( a1 a2 c f)
if [char] 0 - over c! char+ else drop then
loop ( a1 a2)
nip r> - 9 / r> + ( n3)
;
 
0
s" 090004007" >grid
s" 000007900" >grid
s" 800000000" >grid
s" 405800000" >grid
s" 300000002" >grid
s" 000009706" >grid
s" 000000004" >grid
s" 003500000" >grid
s" 200600080" >grid
drop
 
\ ---------------------
\ Logic
\ ---------------------
\ Basically :
\ Grid is parsed. All numbers are put into sets, which are
\ implemented as bitmaps (sudoku_row, sudoku_col, sudoku_box)
\ which represent sets of numbers in each row, column, box.
\ only one specific instance of a number can exist in a
\ particular set.
 
\ SOLVER is recursively called
\ SOLVER looks for the next best guess using FINDNEXTSPACE
\ tries this trail down... if fails, backtracks... and tries
\ again.
 
 
\ Grid Related
 
: xy 9 * + ; \ x y -- offset ;
: getrow 9 / ;
: getcol 9 mod ;
: getbox dup getrow 3 / 3 * swap getcol 3 / + ;
 
\ Puts and gets numbers from/to grid only
: setnumber sudokugrid + c! ; \ n position --
: getnumber sudokugrid + c@ ;
 
: cleargrid sudokugrid 81 bounds do 0 i c! loop ;
 
\ --------------
\ Set related: sets are sudoku_row, sudoku_col, sudoku_box
 
\ ie x y --  ; adds x into bitmap y
: addbits_row cells sudoku_row + dup @ rot 1 swap lshift or swap ! ;
: addbits_col cells sudoku_col + dup @ rot 1 swap lshift or swap ! ;
: addbits_box cells sudoku_box + dup @ rot 1 swap lshift or swap ! ;
 
\ ie x y --  ; remove number x from bitmap y
: removebits_row cells sudoku_row + dup @ rot 1 swap lshift invert and swap ! ;
: removebits_col cells sudoku_col + dup @ rot 1 swap lshift invert and swap ! ;
: removebits_box cells sudoku_box + dup @ rot 1 swap lshift invert and swap ! ;
 
\ clears all bitsmaps to 0
: clearbitmaps 9 0 do i cells
0 over sudoku_row + !
0 over sudoku_col + !
0 swap sudoku_box + !
loop ;
 
\ Adds number to grid and sets
: addnumber \ number position --
2dup setnumber
2dup getrow addbits_row
2dup getcol addbits_col
getbox addbits_box
;
 
\ Remove number from grid, and sets
: removenumber \ position --
dup getnumber swap
2dup getrow removebits_row
2dup getcol removebits_col
2dup getbox removebits_box
nip 0 swap setnumber
;
 
\ gets bitmap at position, ie
\ position -- bitmap
 
: getrow_bits getrow cells sudoku_row + @ ;
: getcol_bits getcol cells sudoku_col + @ ;
: getbox_bits getbox cells sudoku_box + @ ;
 
\ position -- composite bitmap (or'ed)
: getbits
dup getrow_bits
over getcol_bits
rot getbox_bits or or
;
 
\ algorithm from c.l.f circa 1995 ? Will Baden
: countbits ( number -- bits )
[HEX] DUP 55555555 AND SWAP 1 RSHIFT 55555555 AND +
DUP 33333333 AND SWAP 2 RSHIFT 33333333 AND +
DUP 0F0F0F0F AND SWAP 4 RSHIFT 0F0F0F0F AND +
[DECIMAL] 255 MOD
;
 
\ Try tests a number in a said position of grid
\ Returns true if it's possible, else false.
: try \ number position -- true/false
getbits 1 rot lshift and 0=
;
 
\ --------------
: parsegrid \ Parses Grid to fill sets.. Run before solver.
sudokugrid \ to ensure all numbers are parsed into sets/bitmaps
81 0 do
dup i + c@
dup if
dup i try if
i addnumber
else
unloop drop drop FALSE exit
then
else
drop
then
loop
drop
TRUE
;
 
\ Morespaces? manually checks for spaces ...
\ Obviously this can be optimised to a count var, done initially
\ Any additions/subtractions made to the grid could decrement
\ a 'spaces' variable.
 
: morespaces?
0 sudokugrid 81 bounds do i c@ 0= if 1+ then loop ;
 
: findnextmove \ -- n ; n = index next item, if -1 finished.
 
-1 10 \ index prev_possibilities --
\ err... yeah... local variables, kind of...
 
81 0 do
i sudokugrid + c@ 0= IF
i getbits countbits 9 swap -
 
\ get bitmap and see how many possibilities
\ stack diagram:
\ index prev_possibilities new_possiblities --
 
2dup > if
\ if new_possibilities < prev_possibilities...
nip nip i swap
\ new_index new_possibilies --
 
else \ else prev_possibilities < new possibilities, so:
 
drop \ new_index new_possibilies --
 
then
THEN
loop
drop
;
 
\ findnextmove returns index of best next guess OR returns -1
\ if no more guesses. You then have to check to see if there are
\ spaces left on the board unoccupied. If this is the case, you
\ need to back up the recursion and try again.
 
: solver
findnextmove
dup 0< if
morespaces? if
drop false exit
else
drop true exit
then
then
 
10 1 do
i over try if
i over addnumber
recurse if
drop unloop TRUE EXIT
else
dup removenumber
then
then
loop
 
drop FALSE
;
 
\ SOLVER
 
: startsolving
clearbitmaps \ reparse bitmaps and reparse grid
parsegrid \ just in case..
solver
AND
;
 
\ ---------------------
\ Display Grid
\ ---------------------
 
\ Prints grid nicely
 
: .sudokugrid
CR CR
sudokugrid
81 0 do
dup i + c@ .
i 1+
dup 3 mod 0= if
dup 9 mod 0= if
CR
dup 27 mod 0= if
dup 81 < if ." ------+-------+------" CR then
then
else
." | "
then
then
drop
loop
drop
CR
;
 
\ ---------------------
\ Higher Level Words
\ ---------------------
 
: checkifoccupied ( offset -- t/f)
sudokugrid + c@
;
 
: add ( n x y --)
xy 2dup
dup checkifoccupied if
dup removenumber
then
try if
addnumber
.sudokugrid
else
CR ." Not a valid move. " CR
2drop
then
;
 
: rm
xy removenumber
.sudokugrid
;
 
: clearit
cleargrid
clearbitmaps
.sudokugrid
;
 
: solveit
CR
startsolving
if
." Solution found!" CR .sudokugrid
else
." No solution found!" CR CR
then
;
 
: showit .sudokugrid ;
 
\ Print help menu
: help
CR
." Type clearit  ; to clear grid " CR
." 1-9 x y add ; to add 1-9 to grid at x y (0 based) " CR
." x y rm  ; to remove number at x y " CR
." showit  ; redisplay grid " CR
." solveit  ; to solve " CR
." help  ; for help " CR
CR
;
 
\ ---------------------
\ Execution starts here
\ ---------------------
 
: godoit
clearbitmaps
parsegrid if
CR ." Grid valid!"
else
CR ." Warning: grid invalid!"
then
.sudokugrid
help
;
 
\ -------------
\ 4tH interface
\ -------------
 
: read-sudoku
input 1 arg-open 0
begin dup 9 < while refill while 0 parse >grid repeat
drop close
;
 
: bye quit ;
 
create wordlist \ dictionary
," clearit" ' clearit ,
," add" ' add ,
," rm" ' rm ,
," showit" ' showit ,
," solveit" ' solveit ,
," quit" ' bye ,
," exit" ' bye ,
," bye" ' bye ,
," q" ' bye ,
," help" ' help ,
NULL ,
 
wordlist to dictionary
:noname ." Unknown command '" type ." '" cr ; is NotFound
\ sudoku interpreter
: sudoku
argn 1 > if read-sudoku then
godoit
begin
." OK" cr
refill drop ['] interpret
catch if ." Error" cr then
again
;
 
sudoku

Fortran[edit]

Works with: Fortran version 90 and later

This implementation uses a brute force method. The subroutine solve recursively checks valid entries using the rules defined in the function is_safe. When solve is called beyond the end of the sudoku, we know that all the currently entered values are valid. Then the result is displayed.

program sudoku
 
implicit none
integer, dimension (9, 9) :: grid
integer, dimension (9, 9) :: grid_solved
grid = reshape ((/ &
& 0, 0, 3, 0, 2, 0, 6, 0, 0, &
& 9, 0, 0, 3, 0, 5, 0, 0, 1, &
& 0, 0, 1, 8, 0, 6, 4, 0, 0, &
& 0, 0, 8, 1, 0, 2, 9, 0, 0, &
& 7, 0, 0, 0, 0, 0, 0, 0, 8, &
& 0, 0, 6, 7, 0, 8, 2, 0, 0, &
& 0, 0, 2, 6, 0, 9, 5, 0, 0, &
& 8, 0, 0, 2, 0, 3, 0, 0, 9, &
& 0, 0, 5, 0, 1, 0, 3, 0, 0/), &
& shape = (/9, 9/), &
& order = (/2, 1/))
call pretty_print (grid)
call solve (1, 1)
write (*, *)
call pretty_print (grid_solved)
 
contains
 
recursive subroutine solve (i, j)
implicit none
integer, intent (in) :: i
integer, intent (in) :: j
integer :: n
integer :: n_tmp
if (i > 9) then
grid_solved = grid
else
do n = 1, 9
if (is_safe (i, j, n)) then
n_tmp = grid (i, j)
grid (i, j) = n
if (j == 9) then
call solve (i + 1, 1)
else
call solve (i, j + 1)
end if
grid (i, j) = n_tmp
end if
end do
end if
end subroutine solve
 
function is_safe (i, j, n) result (res)
implicit none
integer, intent (in) :: i
integer, intent (in) :: j
integer, intent (in) :: n
logical :: res
integer :: i_min
integer :: j_min
if (grid (i, j) == n) then
res = .true.
return
end if
if (grid (i, j) /= 0) then
res = .false.
return
end if
if (any (grid (i, :) == n)) then
res = .false.
return
end if
if (any (grid (:, j) == n)) then
res = .false.
return
end if
i_min = 1 + 3 * ((i - 1) / 3)
j_min = 1 + 3 * ((j - 1) / 3)
if (any (grid (i_min : i_min + 2, j_min : j_min + 2) == n)) then
res = .false.
return
end if
res = .true.
end function is_safe
 
subroutine pretty_print (grid)
implicit none
integer, dimension (9, 9), intent (in) :: grid
integer :: i
integer :: j
character (*), parameter :: bar = '+-----+-----+-----+'
character (*), parameter :: fmt = '(3 ("|", i0, 1x, i0, 1x, i0), "|")'
write (*, '(a)') bar
do j = 0, 6, 3
do i = j + 1, j + 3
write (*, fmt) grid (i, :)
end do
write (*, '(a)') bar
end do
end subroutine pretty_print
 
end program sudoku
Output:

+-----+-----+-----+
|0 0 3|0 2 0|6 0 0|
|9 0 0|3 0 5|0 0 1|
|0 0 1|8 0 6|4 0 0|
+-----+-----+-----+
|0 0 8|1 0 2|9 0 0|
|7 0 0|0 0 0|0 0 8|
|0 0 6|7 0 8|2 0 0|
+-----+-----+-----+
|0 0 2|6 0 9|5 0 0|
|8 0 0|2 0 3|0 0 9|
|0 0 5|0 1 0|3 0 0|
+-----+-----+-----+

+-----+-----+-----+
|4 8 3|9 2 1|6 5 7|
|9 6 7|3 4 5|8 2 1|
|2 5 1|8 7 6|4 9 3|
+-----+-----+-----+
|5 4 8|1 3 2|9 7 6|
|7 2 9|5 6 4|1 3 8|
|1 3 6|7 9 8|2 4 5|
+-----+-----+-----+
|3 7 2|6 8 9|5 1 4|
|8 1 4|2 5 3|7 6 9|
|6 9 5|4 1 7|3 8 2|
+-----+-----+-----+

FutureBasic[edit]

First is a short version:

 
include "ConsoleWindow"
include "NSLog.incl"
include "Util_Containers.incl"
 
begin globals
dim as container gC
end globals
 
BeginCDeclaration
short solve_sudoku(short i);
short check_sudoku(short r, short c);
CFMutableStringRef print_sudoku();
EndC
 
BeginCFunction
short sudoku[9][9] = {
{3,0,0,0,0,1,4,0,9},
{7,0,0,0,0,4,2,0,0},
{0,5,0,2,0,0,0,1,0},
{5,7,0,0,4,3,0,6,0},
{0,9,0,0,0,0,0,3,0},
{0,6,0,7,9,0,0,8,5},
{0,8,0,0,0,5,0,4,0},
{0,0,6,4,0,0,0,0,7},
{9,0,5,6,0,0,0,0,3},
};
 
 
short check_sudoku( short r, short c )
{
short i;
short rr, cc;
 
for (i = 0; i < 9; i++)
{
if (i != c && sudoku[r][i] == sudoku[r][c]) return 0;
if (i != r && sudoku[i][c] == sudoku[r][c]) return 0;
rr = r/3 * 3 + i/3;
cc = c/3 * 3 + i%3;
if ((rr != r || cc != c) && sudoku[rr][cc] == sudoku[r][c]) return 0;
}
return -1;
}
 
 
short solve_sudoku( short i )
{
short r, c;
 
if (i < 0) return 0;
else if (i >= 81) return -1;
 
r = i / 9;
c = i % 9;
 
if (sudoku[r][c])
return check_sudoku(r, c) && solve_sudoku(i + 1);
else
for (sudoku[r][c] = 9; sudoku[r][c] > 0; sudoku[r][c]--)
{
if ( solve_sudoku(i) ) return -1;
}
return 0;
}
 
 
CFMutableStringRef print_sudoku()
{
short i, j;
CFMutableStringRef mutStr;
 
mutStr = CFStringCreateMutable( kCFAllocatorDefault, 0 );
 
for (i = 0; i < 9; i++)
{
for (j = 0; j < 9; j++)
{
CFStringAppendFormat( mutStr, NULL, (CFStringRef)@" %d", sudoku[i][j] );
}
CFStringAppendFormat( mutStr, NULL, (CFStringRef)@"\r" );
}
return( mutStr );
}
EndC
 
toolbox fn solve_sudoku( short i ) = short
toolbox fn check_sudoku( short r, short c ) = short
toolbox fn print_sudoku() = CFMutableStringRef
 
dim as short solution
dim as CFMutableStringRef cfRef
 
gC = " "
cfRef = fn print_sudoku()
fn ContainerCreateWithCFString( cfRef, gC )
print : print "Sudoku challenge:" : print : print gC
 
solution = fn solve_sudoku(0)
 
print : print "Sudoku solved:" : print
if ( solution )
gC = " "
cfRef = fn print_sudoku()
fn ContainerCreateWithCFString( cfRef, gC )
print gC
else
print "No solution found"
end if
 

Output:

Sudoku challenge:

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


Sudoku solved:

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

More code in this one, but faster execution:

include "ConsoleWindow"
include "Tlbx Timer.incl"

begin globals
_digits = 9
_setH = 3
_setV = 3
_nSetH = 3
_nSetV = 3

begin record Board
dim as boolean f(_digits,_digits,_digits)
dim as char    match(_digits,_digits)
dim as pointer previousBoard // singly-linked list used to discover repetitions
dim &&
end record

dim quiz as board
dim as long t
dim as double       sProgStartTime

end globals

// 'ordinary' timer used for playing
local fn Milliseconds as long // time in ms since prog start
'~'1
dim as UnsignedWide us

long if ( sProgStartTime == 0.0 )
Microseconds( @us )
sProgStartTime = 4294967296.0*us.hi + us.lo
end if
Microseconds( @us )
end fn = (4294967296.0*us.hi + us.lo - sProgStartTime)'*1e-3

local fn InitMilliseconds
'~'1
sProgStartTime = 0.0
fn Milliseconds
end fn

local mode
local fn CopyBoard( source as ^Board, dest as ^Board )
'~'1
BlockMoveData( source, dest, sizeof( Board ) )
dest.previousBoard = source // linked list
end fn

local fn prepare( b as ^Board )
'~'1
dim as short i, j, n

for i = 1 to _digits
for j = 1 to _digits
for n = 1 to _digits
b.match[i, j] = 0
b.f[i, j, n] = _true
next n
next j
next i
end fn

local fn printBoard( b as ^Board )
'~'1
dim as short i, j

for i = 1 to _digits
for j = 1 to _digits
Print b.match[i, j];
next j
print
next i
end fn

local fn verifica( b as ^Board )
'~'1
dim as short i, j, n, first, x, y, ii
dim as boolean check

check = _true

for i = 1 to _digits
for j = 1 to _digits
long if ( b.match[i, j] == 0 )
check = _false
for n = 1 to _digits
long if ( b.f[i, j, n] != _false )
check = _true
end if
next n
if ( check == _false ) then exit fn
end if
next j
next i

check = _true
for j = 1 to _digits
for n = 1 to _digits
first = 0
for i = 1 to _digits
long if ( b.match[i, j] == n )
long if ( first == 0 )
first = i
xelse
check = _false
exit fn
end if
end if
next i
next n
next j

for i = 1 to _digits
for n = 1 to _digits
first = 0
for j = 1 to _digits
long if ( b.match[i, j] == n )
long if ( first == 0 )
first = j
xelse
check = _false
exit fn
end if
end if
next j
next n
next i

for x = 0 to ( _nSetH - 1 )
for y = 0 to ( _nSetV - 1 )
first = 0
for ii = 0 to ( _digits - 1 )
i = x * _setH + ii mod _setH + 1
j = y * _setV + ii / _setH + 1
long if ( b.match[i, j] == n )
long if ( first == 0 )
first = j
xelse
check = _false
exit fn
end if
end if
next ii
next y
next x

end fn = check


local fn setCell( b as ^Board, x as short, y as short, n as short) as boolean
dim as short   i, j, rx, ry
dim as boolean check

b.match[x, y] = n
for i = 1 to _digits
b.f[x, i, n] = _false
b.f[i, y, n] = _false
next i

rx = (x - 1) / _setH
ry = (y - 1) / _setV

for i = 1 to _setH
for j = 1 to _setV
b.f[ rx * _setH + i, ry * _setV + j, n ] = _false
next j
next i

check = fn verifica( #b )
if ( check == _false ) then exit fn

end fn = check


local fn solve( b as ^Board )
dim as short i, j, n, first, x, y, ii, ppi, ppj
dim as boolean check

check = _true

for i = 1 to _digits
for j = 1 to _digits
long if ( b.match[i, j] == 0 )
first = 0
for n = 1 to _digits
long if ( b.f[i, j, n] != _false )
long if ( first == 0 )
first = n
xelse
first = -1
exit for
end if
end if
next n

long if ( first > 0 )
check = fn setCell( #b, i, j, first )
if ( check == _false ) then exit fn
check = fn solve(#b)
if ( check == _false ) then exit fn
end if

end if
next j
next i

for i = 1 to _digits
for n = 1 to _digits
first = 0

for j = 1 to _digits
if ( b.match[i, j] == n ) then exit for

long if ( b.f[i, j, n] != _false ) and ( b.match[i, j] == 0 )
long if ( first == 0 )
first = j
xelse
first = -1
exit for
end if

end if

next j

long if ( first > 0 )
check = fn setCell( #b, i, first, n )
if ( check == _false ) then exit fn
check = fn solve(#b)
if ( check == _false ) then exit fn
end if

next n
next i


for j = 1 to _digits
for n = 1 to _digits
first = 0

for i = 1 to _digits
if ( b.match[i, j] == n ) then exit for

long if ( b.f[i, j, n] != _false ) and ( b.match[i, j] == 0 )
long if ( first == 0 )
first = i
xelse
first = -1
exit for
end if

end if

next i

long if ( first > 0 )
check = fn setCell( #b, first, j, n )
if ( check == _false ) then exit fn
check = fn solve(#b)
if ( check == _false ) then exit fn
end if

next n
next j


for x = 0 to ( _nSetH - 1 )
for y = 0 to ( _nSetV - 1 )

for n = 1 to _digits
first = 0

for ii = 0 to ( _digits - 1 )

i = x * _setH + ii mod _setH + 1
j = y * _setV + ii / _setH + 1

if ( b.match[i, j] == n ) then exit for

long if ( b.f[i, j, n] != _false ) and ( b.match[i, j] == 0 )
long if ( first == 0 )
first = n
ppi = i
ppj = j
xelse
first = -1
exit for
end if
end if


next ii

long if ( first > 0 )
check = fn setCell( #b, ppi, ppj, n )
if ( check == _false ) then exit fn
check = fn solve(#b)
if ( check == _false ) then exit fn
end if

next n

next y
next x

end fn = check


local fn resolve( b as ^Board )
dim as boolean check, daFinire
dim as long i, j, n
dim as board localBoard

check = fn solve(b)

long if ( check == _false )
exit fn
end if

daFinire = _false

for i = 1 to _digits
for j = 1 to _digits
long if ( b.match[i, j] == 0 )

daFinire = _true

for n = 1 to _digits
long if ( b.f[i, j, n] != _false )

fn CopyBoard( b, @localBoard )

check = fn setCell(@localBoard, i, j, n)

long if ( check != _false )
check = fn resolve( @localBoard )
long if ( check == -1 )
fn CopyBoard( @localBoard, b )

exit fn
end if
end if

end if

next n

end if
next j
next i

long if daFinire
xelse
check = -1
end if

end fn = check


fn InitMilliseconds

fn prepare( @quiz )

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

dim as short i, j, d
for i = 1 to _digits
for j = 1 to _digits
read d
fn setCell(@quiz, j, i, d)
next j
next i

Print : print "quiz:"
fn printBoard( @quiz )
print : print "-------------------" : print
dim as boolean check

t = fn Milliseconds
check = fn resolve(@quiz)
t = fn Milliseconds - t

if ( check )
print "solution:"; str$( t/1000.0 ) + " ms"
else
print "No solution found"
end if
fn printBoard( @quiz )

Output:

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

-------------------

solution: 6.956 ms
 3 8 6 5 7 9 4 1 2
 1 9 5 2 4 6 3 7 8
 4 7 2 3 1 8 9 6 5
 6 3 8 9 5 7 2 4 1
 2 5 1 8 6 4 7 9 3
 9 4 7 1 2 3 8 5 6
 5 2 4 7 3 1 6 8 9
 8 6 3 4 9 5 1 2 7
 7 1 9 6 8 2 5 3 4

Go[edit]

Solution using Knuth's DLX. This code follows his paper fairly closely. Input to function solve is an 81 character string. This seems to be a conventional computer representation for Sudoku puzzles.

package main
 
import "fmt"
 
// sudoku puzzle representation is an 81 character string
var puzzle = "" +
"394 267 " +
" 3 4 " +
"5 69 2 " +
" 45 9 " +
"6 7" +
" 7 58 " +
" 1 67 8" +
" 9 8 " +
" 264 735"
 
func main() {
printGrid("puzzle:", puzzle)
if s := solve(puzzle); s == "" {
fmt.Println("no solution")
} else {
printGrid("solved:", s)
}
}
 
// print grid (with title) from 81 character string
func printGrid(title, s string) {
fmt.Println(title)
for r, i := 0, 0; r < 9; r, i = r+1, i+9 {
fmt.Printf("%c %c %c | %c %c %c | %c %c %c\n", s[i], s[i+1], s[i+2],
s[i+3], s[i+4], s[i+5], s[i+6], s[i+7], s[i+8])
if r == 2 || r == 5 {
fmt.Println("------+-------+------")
}
}
}
 
// solve puzzle in 81 character string format.
// if solved, result is 81 character string.
// if not solved, result is the empty string.
func solve(u string) string {
// construct an dlx object with 324 constraint columns.
// other than the number 324, this is not specific to sudoku.
d := newDlxObject(324)
// now add constraints that define sudoku rules.
for r, i := 0, 0; r < 9; r++ {
for c := 0; c < 9; c, i = c+1, i+1 {
b := r/3*3 + c/3
n := int(u[i] - '1')
if n >= 0 && n < 9 {
d.addRow([]int{i, 81 + r*9 + n, 162 + c*9 + n,
243 + b*9 + n})
} else {
for n = 0; n < 9; n++ {
d.addRow([]int{i, 81 + r*9 + n, 162 + c*9 + n,
243 + b*9 + n})
}
}
}
}
// run dlx. not sudoku specific.
d.search()
// extract the sudoku-specific 81 character result from the dlx solution.
return d.text()
}
 
// Knuth's data object
type x struct {
c *y
u, d, l, r *x
// except x0 is not Knuth's. it's pointer to first constraint in row,
// so that the sudoku string can be constructed from the dlx solution.
x0 *x
}
 
// Knuth's column object
type y struct {
x
s int // size
n int // name
}
 
// an object to hold the matrix and solution
type dlx struct {
ch []y // all column headers
h *y // ch[0], the root node
o []*x // solution
}
 
// constructor creates the column headers but no rows.
func newDlxObject(nCols int) *dlx {
ch := make([]y, nCols+1)
h := &ch[0]
d := &dlx{ch, h, nil}
h.c = h
h.l = &ch[nCols].x
ch[nCols].r = &h.x
nh := ch[1:]
for i := range ch[1:] {
hi := &nh[i]
ix := &hi.x
hi.n = i
hi.c = hi
hi.u = ix
hi.d = ix
hi.l = &h.x
h.r = ix
h = hi
}
return d
}
 
// rows define constraints
func (d *dlx) addRow(nr []int) {
if len(nr) == 0 {
return
}
r := make([]x, len(nr))
x0 := &r[0]
for x, j := range nr {
ch := &d.ch[j+1]
ch.s++
np := &r[x]
np.c = ch
np.u = ch.u
np.d = &ch.x
np.l = &r[(x+len(r)-1)%len(r)]
np.r = &r[(x+1)%len(r)]
np.u.d, np.d.u, np.l.r, np.r.l = np, np, np, np
np.x0 = x0
}
}
 
// extracts 81 character sudoku string
func (d *dlx) text() string {
b := make([]byte, len(d.o))
for _, r := range d.o {
x0 := r.x0
b[x0.c.n] = byte(x0.r.c.n%9) + '1'
}
return string(b)
}
 
// the dlx algorithm
func (d *dlx) search() bool {
h := d.h
j := h.r.c
if j == h {
return true
}
c := j
for minS := j.s; ; {
j = j.r.c
if j == h {
break
}
if j.s < minS {
c, minS = j, j.s
}
}
 
cover(c)
k := len(d.o)
d.o = append(d.o, nil)
for r := c.d; r != &c.x; r = r.d {
d.o[k] = r
for j := r.r; j != r; j = j.r {
cover(j.c)
}
if d.search() {
return true
}
r = d.o[k]
c = r.c
for j := r.l; j != r; j = j.l {
uncover(j.c)
}
}
d.o = d.o[:len(d.o)-1]
uncover(c)
return false
}
 
func cover(c *y) {
c.r.l, c.l.r = c.l, c.r
for i := c.d; i != &c.x; i = i.d {
for j := i.r; j != i; j = j.r {
j.d.u, j.u.d = j.u, j.d
j.c.s--
}
}
}
 
func uncover(c *y) {
for i := c.u; i != &c.x; i = i.u {
for j := i.l; j != i; j = j.l {
j.c.s++
j.d.u, j.u.d = j, j
}
}
c.r.l, c.l.r = &c.x, &c.x
}
Output:
puzzle:
3 9 4 |     2 | 6 7  
      | 3     | 4    
5     | 6 9   |   2  
------+-------+------
  4 5 |       | 9    
6     |       |     7
    7 |       | 5 8  
------+-------+------
  1   |   6 7 |     8
    9 |     8 |      
  2 6 | 4     | 7 3 5
solved:
3 9 4 | 8 5 2 | 6 7 1
2 6 8 | 3 7 1 | 4 5 9
5 7 1 | 6 9 4 | 8 2 3
------+-------+------
1 4 5 | 7 8 3 | 9 6 2
6 8 2 | 9 4 5 | 3 1 7
9 3 7 | 1 2 6 | 5 8 4
------+-------+------
4 1 3 | 5 6 7 | 2 9 8
7 5 9 | 2 3 8 | 1 4 6
8 2 6 | 4 1 9 | 7 3 5

Groovy[edit]

Adaptive "Non-guessing Then Guessing" Solution

Non-guessing part is iterative. Guessing part is recursive. Implementation uses exception handling to back out of bad guesses.

final CELL_VALUES = ('1'..'9')
 
class GridException extends Exception {
GridException(String message) { super(message) }
}
 
def string2grid = { string ->
assert string.size() == 81
(0..8).collect { i -> (0..8).collect { j -> string[9*i+j] } }
}
 
def gridRow = { grid, slot -> grid[slot.i] as Set }
 
def gridCol = { grid, slot -> grid.collect { it[slot.j] } as Set }
 
def gridBox = { grid, slot ->
def t, l; (t, l) = [slot.i.intdiv(3)*3, slot.j.intdiv(3)*3]
(0..2).collect { row -> (0..2).collect { col -> grid[t+row][l+col] } }.flatten() as Set
}
 
def slotList = { grid ->
def slots = (0..8).collect { i -> (0..8).findAll { j -> grid[i][j] == '.' } \
.collect {j -> [i: i, j: j] } }.flatten()
}
 
def assignCandidates = { grid, slots = slotList(grid) ->
slots.each { slot ->
def unavailable = [gridRow, gridCol, gridBox].collect { it(grid, slot) }.sum() as Set
slot.candidates = CELL_VALUES - unavailable
}
slots.sort { - it.candidates.size() }
if (slots && ! slots[-1].candidates) {
throw new GridException('Invalid Sudoku Grid, overdetermined slot: ' + slots[-1])
}
slots
}
 
def isSolved = { grid -> ! (grid.flatten().find { it == '.' }) }
 
def solve
solve = { grid ->
def slots = assignCandidates(grid)
if (! slots) { return grid }
while (slots[-1].candidates.size() == 1) {
def slot = slots.pop()
grid[slot.i][slot.j] = slot.candidates[0]
if (! slots) { return grid }
slots = assignCandidates(grid, slots)
}
if (! slots) { return grid }
def slot = slots.pop()
slot.candidates.each {
if (! isSolved(grid)) {
try {
def sGrid = grid.collect { row -> row.collect { cell -> cell } }
sGrid[slot.i][slot.j] = it
grid = solve(sGrid)
} catch (GridException ge) {
grid[slot.i][slot.j] = '.'
}
}
}
if (!isSolved(grid)) {
slots = assignCandidates(grid)
throw new GridException('Invalid Sudoku Grid, underdetermined slots: ' + slots)
}
grid
}

Test/Benchmark Cases

Mentions of "exceptionally difficult" example in Wikipedia refer to this (former) page: [Exceptionally difficult Sudokus]

def sudokus = [
//Used in Curry solution: ~ 0.1 seconds
'819..5.....2...75..371.4.6.4..59.1..7..3.8..2..3.62..7.5.7.921..64...9.....2..438',
 
//Used in Perl and PicoLisp solutions: ~ 0.1 seconds
'53..247....2...8..1..7.39.2..8.72.49.2.98..7.79.....8.....3.5.696..1.3...5.69..1.',
 
//Used in Fortran solution: ~ 0.1 seconds
'..3.2.6..9..3.5..1..18.64....81.29..7.......8..67.82....26.95..8..2.3..9..5.1.3..',
 
//Used in many other solutions, notably Algol 68: ~ 0.1 seconds
'394..267....3..4..5..69..2..45...9..6.......7..7...58..1..67..8..9..8....264..735',
 
//Used in C# solution: ~ 0.2 seconds
'97.3...6..6.75.........8.5.......67.....3.....539..2..7...25.....2.1...8.4...73..',
 
//Used in Oz solution: ~ 0.2 seconds
'4......6.5...8.9..3....1....2.7....1.9.....4.8....3.5....2....7..6.5...8.1......6',
 
//Used in many other solutions, notably C++: ~ 0.3 seconds
'85...24..72......9..4.........1.7..23.5...9...4...........8..7..17..........36.4.',
 
//Used in VBA solution: ~ 0.3 seconds
'..1..5.7.92.6.......8...6...9..2.4.1.........3.4.8..9...7...3.......7.69.1.8..7..',
 
//Used in Forth solution: ~ 0.8 seconds
'.9...4..7.....79..8........4.58.....3.......2.....97.6........4..35.....2..6...8.',
 
//3rd "exceptionally difficult" example in Wikipedia: ~ 2.3 seconds
'12.3....435....1....4........54..2..6...7.........8.9...31..5.......9.7.....6...8',
 
//Used in Curry solution: ~ 2.4 seconds
'9..2..5...4..6..3...3.....6...9..2......5..8...7..4..37.....1...5..2..4...1..6..9',
 
//"AL Escargot", so-called "hardest sudoku" (HA!): ~ 3.0 seconds
'1....7.9..3..2...8..96..5....53..9...1..8...26....4...3......1..4......7..7...3..',
 
//1st "exceptionally difficult" example in Wikipedia: ~ 6.5 seconds
'12.4..3..3...1..5...6...1..7...9.....4.6.3.....3..2...5...8.7....7.....5.......98',
 
//Used in Bracmat and Scala solutions: ~ 6.7 seconds
'..............3.85..1.2.......5.7.....4...1...9.......5......73..2.1........4...9',
 
//2nd "exceptionally difficult" example in Wikipedia: ~ 8.8 seconds
'.......39.....1..5..3.5.8....8.9...6.7...2...1..4.......9.8..5..2....6..4..7.....',
 
//Used in MATLAB solution: ~15 seconds
'....839..1......3...4....7..42.3....6.......4....7..1..2........8...92.....25...6',
 
//4th "exceptionally difficult" example in Wikipedia: ~29 seconds
'..3......4...8..36..8...1...4..6..73...9..........2..5..4.7..686........7..6..5..']
 
sudokus.each { sudoku ->
def grid = string2grid(sudoku)
println '\nPUZZLE'
grid.each { println it }
 
println '\nSOLUTION'
def start = System.currentTimeMillis()
def solution = solve(grid)
def elapsed = (System.currentTimeMillis() - start)/1000
solution.each { println it }
println "\nELAPSED: ${elapsed} seconds"
}
Output:
(last only):
PUZZLE
[., ., 3, ., ., ., ., ., .]
[4, ., ., ., 8, ., ., 3, 6]
[., ., 8, ., ., ., 1, ., .]
[., 4, ., ., 6, ., ., 7, 3]
[., ., ., 9, ., ., ., ., .]
[., ., ., ., ., 2, ., ., 5]
[., ., 4, ., 7, ., ., 6, 8]
[6, ., ., ., ., ., ., ., .]
[7, ., ., 6, ., ., 5, ., .]

SOLUTION
[1, 2, 3, 4, 5, 6, 7, 8, 9]
[4, 5, 7, 1, 8, 9, 2, 3, 6]
[9, 6, 8, 3, 2, 7, 1, 5, 4]
[2, 4, 9, 5, 6, 1, 8, 7, 3]
[5, 7, 6, 9, 3, 8, 4, 1, 2]
[8, 3, 1, 7, 4, 2, 6, 9, 5]
[3, 1, 4, 2, 7, 5, 9, 6, 8]
[6, 9, 5, 8, 1, 4, 3, 2, 7]
[7, 8, 2, 6, 9, 3, 5, 4, 1]

ELAPSED: 28.978 seconds

Haskell[edit]

Visit the Haskell wiki Sudoku

J[edit]

See Solving Sudoku in J.

Java[edit]

public class Sudoku
{
private int mBoard[][];
private int mBoardSize;
private int mBoxSize;
private boolean mRowSubset[][];
private boolean mColSubset[][];
private boolean mBoxSubset[][];
 
public Sudoku(int board[][]) {
mBoard = board;
mBoardSize = mBoard.length;
mBoxSize = (int)Math.sqrt(mBoardSize);
}
 
public void initSubsets() {
mRowSubset = new boolean[mBoardSize][mBoardSize];
mColSubset = new boolean[mBoardSize][mBoardSize];
mBoxSubset = new boolean[mBoardSize][mBoardSize];
for(int i = 0; i < mBoard.length; i++) {
for(int j = 0; j < mBoard.length; j++) {
int value = mBoard[i][j];
if(value != 0) {
setSubsetValue(i, j, value, true);
}
}
}
}
 
private void setSubsetValue(int i, int j, int value, boolean present) {
mRowSubset[i][value - 1] = present;
mColSubset[j][value - 1] = present;
mBoxSubset[computeBoxNo(i, j)][value - 1] = present;
}
 
public boolean solve() {
return solve(0, 0);
}
 
public boolean solve(int i, int j) {
if(i == mBoardSize) {
i = 0;
if(++j == mBoardSize) {
return true;
}
}
if(mBoard[i][j] != 0) {
return solve(i + 1, j);
}
for(int value = 1; value <= mBoardSize; value++) {
if(isValid(i, j, value)) {
mBoard[i][j] = value;
setSubsetValue(i, j, value, true);
if(solve(i + 1, j)) {
return true;
}
setSubsetValue(i, j, value, false);
}
}
 
mBoard[i][j] = 0;
return false;
}
 
private boolean isValid(int i, int j, int val) {
val--;
boolean isPresent = mRowSubset[i][val] || mColSubset[j][val] || mBoxSubset[computeBoxNo(i, j)][val];
return !isPresent;
}
 
private int computeBoxNo(int i, int j) {
int boxRow = i / mBoxSize;
int boxCol = j / mBoxSize;
return boxRow * mBoxSize + boxCol;
}
 
public void print() {
for(int i = 0; i < mBoardSize; i++) {
if(i % mBoxSize == 0) {
System.out.println(" -----------------------");
}
for(int j = 0; j < mBoardSize; j++) {
if(j % mBoxSize == 0) {
System.out.print("| ");
}
System.out.print(mBoard[i][j] != 0 ? ((Object) (Integer.valueOf(mBoard[i][j]))) : " ");
System.out.print(' ');
}
 
System.out.println("|");
}
 
System.out.println(" -----------------------");
}
}

Lua[edit]

--9x9 sudoku solver in lua
--based on a branch and bound solution
--fields are not tried in plain order
--but in a way to detect dead ends earlier
concat=table.concat
insert=table.insert
constraints = { } --contains a table with 3 constraints for every field
-- a contraint "cons" is a table containing all fields which must not have the same value
-- a field "f" is an integer from 1 to 81
columns = { } --contains all column-constraints variable "c"
rows = { } --contains all row-constraints variable "r"
blocks = { } --contains all block-constraints variable "b"
 
--initialize all constraints
for f = 1, 81 do
constraints[f] = { }
end
all_constraints = { } --union of colums, rows and blocks
for i = 1, 9 do
columns[i] = {
unknown = 9, --number of fields not yet solved
unknowns = { } --fields not yet solved
}
insert(all_constraints, columns[i])
rows[i] = {
unknown = 9, -- see l.15
unknowns = { } -- see l.16
}
insert(all_constraints, rows[i])
blocks[i] = {
unknown = 9, --see l.15
unknowns = { } --see l.16
}
insert(all_constraints, blocks[i])
end
constraints_by_unknown = { } --contraints sorted by their number of unknown fields
for i = 0, 9 do
constraints_by_unknown[i] = {
count = 0 --how many contraints are in here
}
end
for r = 1, 9 do
for c = 1, 9 do
local f = (r - 1) * 9 + c
insert(rows[r], f)
insert(constraints[f], rows[r])
insert(columns[c], f)
insert(constraints[f], columns[c])
end
end
for i = 1, 3 do
for j = 1, 3 do
local r = (i - 1) * 3 + j
for k = 1, 3 do
for l = 1, 3 do
local c = (k - 1) * 3 + l
local f = (r - 1) * 9 + c
local b = (i - 1) * 3 + k
insert(blocks[b], f)
insert(constraints[f], blocks[b])
end
end
end
end
working = { } --save the read values in here
function read() --read the values from stdin
local f = 1
local l = io.read("*a")
for d in l:gmatch("(%d)") do
local n = tonumber(d)
if n > 0 then
working[f] = n
for _,cons in pairs(constraints[f]) do
cons.unknown = cons.unknown - 1
end
else
for _,cons in pairs(constraints[f]) do
cons.unknowns[f] = f
end
end
f = f + 1
end
assert((f == 82), "Wrong number of digits")
end
read()
function printer(t) --helper function for printing a 1-81 table
local pattern = {1,2,3,false,4,5,6,false,7,8,9} --place seperators for better readability
for _,r in pairs(pattern) do
if r then
local function p(c)
return c and t[(r - 1) * 9 + c] or "|"
end
local line={}
for k,v in pairs(pattern) do
line[k]=p(v)
end
print(concat(line))
else
print("---+---+---")
end
end
end
order = { } --when to try a field
for _,cons in pairs(all_constraints) do --put all constraints in the corresponding constraints_by_unknown set
local level = constraints_by_unknown[cons.unknown]
level[cons] = cons
level.count = level.count + 1
end
function first(t) --helper function to get a value from a set
for k, v in pairs(t) do
if k == v then
return k
end
end
end
function establish_order() -- determine the sequence in which the fields are to be tried
local solved = constraints_by_unknown[0].count
while solved < 27 do --there 27 constraints
--contraints with no unknown fields are considered "solved"
--keep in mind the actual solving happens in function branch
local i = 1
while constraints_by_unknown[i].count == 0 do
i = i + 1
-- find a unsolved contraint with the least number of unsolved fields
end
local cons = first(constraints_by_unknown[i])
local f = first(cons.unknowns)
-- take one of its unknown fields and append it to "order"
insert(order, f)
for _,c in pairs(constraints[f]) do
--each constraint "c" of "f" is moved up one "level"
--delete "f" from the constraints unknown fields
--decrease unknown of "c"
c.unknowns[f] = nil
local level = constraints_by_unknown[c.unknown]
level[c] = nil
level.count = level.count - 1
c.unknown = c.unknown - 1
level = constraints_by_unknown[c.unknown]
level[c] = c
level.count = level.count + 1
constraints_by_unknown[c.unknown][c] = c
end
solved = constraints_by_unknown[0].count
end
end
establish_order()
max = #order --how many fields are to be solved
function bound(f,i)
for _,c in pairs(constraints[f]) do
for _,x in pairs(c) do
if i == working[x] then
return false --i is already used in fs column/row/block
end
end
end
return true
end
function branch(n)
local f = order[n] --recursively iterate over fields in order
if n > max then
return working --all fields solved without collision
else
for i = 1, 9 do --check all values
if bound(f, i) then --if there is no collision
working[f] = i
local res = branch(n + 1) --try next field
if res then
return res --all fields solved without collision
else
working[f] = nil --this lead to a dead end
end
else
working[f] = nil --reset field because of a collision
end
end
return false --this is a dead end
end
end
x = branch(1)
if x then
return printer(x)
end

Input:

003 000 000
400 080 036
008 000 100

040 060 073
000 900 000
000 002 005

004 070 068
600 000 000
700 600 500
Output:
123|456|789
457|189|236
968|327|154
---+---+---
249|561|873
576|938|412
831|742|695
---+---+---
314|275|968
695|814|327
782|693|541

Time with luajit: 9.245s

Mathematica[edit]

solve[sudoku_] := 
NestWhile[
Join @@ Table[
Table[ReplacePart[s, #1 -> n], {n, #2}] & @@
[email protected][{#,
Complement[[email protected], s[[First@#]], s[[;; , Last@#]],
Catenate@
Extract[Partition[s, {3, 3}], Quotient[#, 3, -2]]]} & /@
Position[s, 0, {2}],
[email protected]@# &], {s, #}] &, {sudoku}, ! FreeQ[#, 0] &]

Example:

solve[{{9, 7, 0, 3, 0, 0, 0, 6, 0},
{0, 6, 0, 7, 5, 0, 0, 0, 0},
{0, 0, 0, 0, 0, 8, 0, 5, 0},
{0, 0, 0, 0, 0, 0, 6, 7, 0},
{0, 0, 0, 0, 3, 0, 0, 0, 0},
{0, 5, 3, 9, 0, 0, 2, 0, 0},
{7, 0, 0, 0, 2, 5, 0, 0, 0},
{0, 0, 2, 0, 1, 0, 0, 0, 8},
{0, 4, 0, 0, 0, 7, 3, 0, 0}}]
Output:
{{{9, 7, 5, 3, 4, 2, 8, 6, 1}, {8, 6, 1, 7, 5, 9, 4, 3, 2}, {3, 2, 4, 
   1, 6, 8, 9, 5, 7}, {2, 1, 9, 5, 8, 4, 6, 7, 3}, {4, 8, 7, 2, 3, 6, 
   5, 1, 9}, {6, 5, 3, 9, 7, 1, 2, 8, 4}, {7, 3, 8, 4, 2, 5, 1, 9, 
   6}, {5, 9, 2, 6, 1, 3, 7, 4, 8}, {1, 4, 6, 8, 9, 7, 3, 2, 5}}}

MATLAB[edit]

This solution impliments a recursive, depth-first search of the possible values unfilled sudoku cells can take. The search tree is pruned using logical deduction rules and takes about a minute to solve some of the more difficult puzzles. This code can be cleaned by making the main code blocks, denoted by "%% [Block Title]," into their own separate functions. This can also be further improved by implementing a Sudoku class and making this solver a member function. There are also several lines of code that can be vectorized to improve efficiency, but at the expense of readability.

For this to work, this code must be placed in a file named "sudokuSolver.m"

function solution = sudokuSolver(sudokuGrid)
 
%Define what each of the sub-boxes of the sudoku grid are by defining
%the start and end coordinates of each sub-box. The indecies represent
%the column and row of a grid coordinate on the actual sudoku grid.
%The contents of each cell with the same grid coordinates contain the
%information to determine which sub-box that grid coordinate is
%contained in on the sudoku grid. The array in position 1, i.e.
%subBoxes{row,column}(1), represents the row indecies of the subbox.
%The array in position 2, i.e. subBoxes{row,column}(2),represents the
%column indecies of the subbox.
 
subBoxes(1:9,1:9) = {{(1:3),(1:3)}};
subBoxes(4:6,:)= {{(4:6),(1:3)}};
subBoxes(7:9,:)= {{(7:9),(1:3)}};
 
for column = (4:6)
for row = (1:9)
subBoxes{row,column}(2)= {4:6};
end
end
for column = (7:9)
for row = (1:9)
subBoxes{row,column}(2)= {7:9};
end
end
 
%Generate a cell of arrays which contain the possible values of the
%sudoku grid for each cell in the grid. The possible values a specific
%grid coordinate can take share the same indices as the sudoku grid
%coordinate they represent.
%For example sudokuGrid(m,n) can be possibly filled in by the
%values stored in the array at possibleValues(m,n).
possibleValues(1:9,1:9) = { (1:9) };
 
%Filter the possibleValues so that no entry exists for coordinates that
%have already been filled in. This will replace any array with an empty
%array in the possibleValues cell matrix at the coordinates of a grid
%already filled in the sudoku grid.
possibleValues( ~isnan(sudokuGrid) )={[]};
 
%Iterate through each grid coordinate and filter out the possible
%values for that grid point that aren't alowed by the rules given the
%current values that are filled in. Or, if there is only one possible
%value for the current coordinate, fill it in.
 
solution = sudokuGrid; %so the original sudoku input isn't modified
memory = 0; %contains the previous iterations possibleValues
dontStop = true; %stops the while loop when nothing else can be reasoned about the sudoku
 
while( dontStop )
 
%% Process of elimination deduction method
 
while( ~isequal(possibleValues,memory) ) %Stops using the process of elimination deduction method when this deduction rule stops working
 
memory = possibleValues; %Copies the current possibleValues into memory, for the above conditional on the next iteration.
 
%Iterate through everything
for row = (1:9)
for column = (1:9)
 
if isnan( solution(row,column) ) %If grid coordinate hasn't been filled in, try to determine it's value.
 
%Look at column to see what values have already
%been filled in and thus the current grid
%coordinate can't be
removableValues = solution( ~isnan(solution(:,column)),column );
 
%If there are any values that have been assigned to
%other cells in the same column, filter those out
%of the current cell's possiblValues
if ~isempty(removableValues)
for m = ( 1:numel(removableValues) )
possibleValues{row,column}( possibleValues{row,column}==removableValues(m) )=[];
end
end
 
%If the current grid coordinate can only atain one
%possible value, assign it that value
if numel( possibleValues{row,column} ) == 1
solution(row,column) = possibleValues{row,column};
possibleValues(row,column)={[]};
end
end %end if
 
if isnan( solution(row,column) ) %If grid coordinate hasn't been filled in, try to determine it's value.
 
%Look at row to see what values have already
%been filled in and thus the current grid
%coordinate can't be
removableValues = solution( row,~isnan(solution(row,:)) );
 
%If there are any values that have been assigned to
%other cells in the same row, filter those out
%of the current cell's possiblValues
if ~isempty(removableValues)
for m = ( 1:numel(removableValues) )
possibleValues{row,column}( possibleValues{row,column}==removableValues(m) )=[];
end
end
 
%If the current grid coordinate can only atain one
%possible value, assign it that value
if numel( possibleValues{row,column} ) == 1
solution(row,column) = possibleValues{row,column};
possibleValues(row,column)={[]};
end
end %end if
 
if isnan( solution(row,column) ) %If grid coordinate hasn't been filled in, try to determine it's value.
 
%Look at sub-box to see if any possible values can be
%filtered out. First pull the boundaries of the sub-box
%containing the current array coordinate
currentBoxBoundaries=subBoxes{row,column};
 
%Then pull the sub-boxes values out of the solution
box = solution(currentBoxBoundaries{:});
 
%Look at sub-box to see what values have already
%been filled in and thus the current grid
%coordinate can't be
removableValues = box( ~isnan(box) );
 
%If there are any values that have been assigned to
%other cells in the same sub-box, filter those out
%of the current cell's possiblValues
if ~isempty(removableValues)
for m = ( 1:numel(removableValues) )
possibleValues{row,column}( possibleValues{row,column}==removableValues(m) )=[];
end
end
 
%If the current grid coordinate can only atain one
%possible value, assign it that value
if numel( possibleValues{row,column} ) == 1
solution(row,column) = possibleValues{row,column};
possibleValues(row,column)={[]};
end
end %end if
 
end %end for column
end %end for row
end %stop process of elimination
 
%% Check that there are no contradictions in the solved grid coordinates.
 
%Check that each row at most contains one of each of the integers
%from 1 to 9
if ~isempty( find( histc( solution,(1:9),1 )>1 ) )
solution = false;
return
end
 
%Check that each column at most contains one of each of the integers
%from 1 to 9
if ~isempty( find( histc( solution,(1:9),2 )>1 ) )
solution = false;
return
end
 
%Check that each sub-box at most contains one of each of the integers
%from 1 to 9
subBoxBins = zeros(9,9);
counter = 0;
for row = [2 5 8]
for column = [2 5 8]
counter = counter +1;
 
%because the sub-boxes are extracted as square matricies,
%we need to reshape them into row vectors so all of the
%boxes can be input into histc simultaneously
subBoxBins(counter,:) = reshape( solution(subBoxes{row,column}{:}),1,9 );
end
end
if ~isempty( find( histc( subBoxBins,(1:9),2 )>1 ) )
solution = false;
return
end
 
%Check to make sure there are no grid coordinates that are not
%filled in and have no possible values.
 
[rowStack,columnStack] = find(isnan(solution)); %extracts the indicies of the unsolved grid coordinates
if (numel(rowStack) > 0)
 
for counter = (1:numel(rowStack))
if isempty(possibleValues{rowStack(counter),columnStack(counter)})
solution = false;
return
end
end
 
%if there are no more grid coordinates to be filed in then the
%sudoku is solved and we can return the solution without further
%computation
elseif (numel(rowStack) == 0)
return
end
 
%% Use the unique relative compliment of sets deduction method
 
%Because no more information can be determined by the process of
%ellimination we have to try a new method of reasoning. Now we will
%look at the possible values a cell can take. If there is a value that
%that grid coordinate can take but no other coordinates in the same row,
%column or sub-box can take that value then we assign that coordinate
%that value.
 
keepGoing = true; %signals to keep applying rules to the current grid-coordinate because it hasn't been solved using previous rules
dontStop = false; %if this method doesn't figure anything out, this will terminate the top level while loop
 
[rowStack,columnStack] = find(isnan(solution)); %This will also take care of the case where the sudoku is solved
counter = 0; %makes sure the loop terminates when there are no more cells to consider
 
while( keepGoing && (counter < numel(rowStack)) ) %stop this method of reasoning when the value of one of the cells has been determined and return to the process of elimination method
 
counter = counter + 1;
 
row = rowStack(counter);
column = columnStack(counter);
 
gridPossibles = [possibleValues{row,column}];
 
coords = (1:9);
coords(column) = [];
rowPossibles = [possibleValues{row,coords}]; %extract possible values for everything in the same row except the current grid coordinate
 
totalMatches = zeros( numel(gridPossibles),1 ); %preallocate for speed
 
%count how many times a possible value for the current cell
%appears as a possible value for the cells in the same row
for n = ( 1:numel(gridPossibles) )
totalMatches(n) = sum( (rowPossibles == gridPossibles(n)) );
end
 
%remove any possible values for the current cell that have
%matches in other cells
gridPossibles = gridPossibles(totalMatches==0);
 
%if there is only one possible value that the current cell can
%take that aren't shared by other cells, assign that value to
%the current cell.
if numel(gridPossibles) == 1
 
solution(row,column) = gridPossibles;
possibleValues(row,column)={[]};
keepGoing = false; %stop this method of deduction and return to the process of elimination
dontStop = true; %keep the top level loop going
 
end
 
if(keepGoing) %do the same as above but for the current cell's column
 
gridPossibles = [possibleValues{row,column}];
 
coords = (1:9);
coords(row) = [];
columnPossibles = [possibleValues{coords,column}];
 
totalMatches = zeros( numel(gridPossibles),1 );
for n = ( 1:numel(gridPossibles) )
totalMatches(n) = sum( (columnPossibles == gridPossibles(n)) );
end
 
gridPossibles = gridPossibles(totalMatches==0);
 
if numel(gridPossibles) == 1
 
solution(row,column) = gridPossibles;
possibleValues(row,column)={[]};
keepGoing = false;
dontStop = true;
 
end
end
 
if(keepGoing) %do the same as above but for the current cell's sub-box
 
gridPossibles = [possibleValues{row,column}];
 
currentBoxBoundaries = subBoxes{row,column};
subBoxPossibles = [];
for m = currentBoxBoundaries{1}
for n = currentBoxBoundaries{2}
if ~((m == row) && (n == column))
subBoxPossibles = [subBoxPossibles possibleValues{m,n}];
end
end
end
 
totalMatches = zeros( numel(gridPossibles),1 );
for n = ( 1:numel(gridPossibles) )
totalMatches(n) = sum( (subBoxPossibles == gridPossibles(n)) );
end
 
gridPossibles = gridPossibles(totalMatches==0);
 
if numel(gridPossibles) == 1
 
solution(row,column) = gridPossibles;
possibleValues(row,column)={[]};
keepGoing = false;
dontStop = true;
 
end
end %end
 
end %end set comliment rule while loop
end %end top-level while loop
 
%% Depth-first search of the solution tree
 
%There is no more reasoning that can solve the puzzle so now it is time
%for a depth-first search of the possible answers, basically
%guess-and-check. This is implimented recursively.
 
[rowStack,columnStack] = find(isnan(solution)); %Get all of the unsolved cells
 
if (numel(rowStack) > 0) %If all of the above stuff terminates then there will be at least one grid coordinate not filled in
 
%Treat the rowStack and columnStack like stacks, and pop the top
%value off the stack to act as the current node whose
%possibleValues to search through, then assign the possible values
%of that grid coordinate to a variable that holds that values to
%search through
searchTreeNodes = possibleValues{rowStack(1),columnStack(1)};
 
keepSearching = true; %used to continue the search
counter = 0; %counts the amount of possible values searched for the current node
tempSolution = solution; %used so that the solution is not overriden until a solution hase been found
 
while( keepSearching && (counter < numel(searchTreeNodes)) ) %stop recursing if we run out of possible values for the current node
 
counter = counter + 1;
tempSolution(rowStack(1),columnStack(1)) = searchTreeNodes(counter); %assign a possible value to the current node in the tree
tempSolution = sudokuSolver(tempSolution); %recursively call the solver with the current guess value for the current grid coordinate
 
if ~islogical(tempSolution) %if tempSolution is not a boolean but a valid sudoku stop recursing and set solution to tempSolution
keepSearching = false;
solution = tempSolution;
elseif counter == numel(searchTreeNodes) %if we have run out of guesses for the current node, stop recursing and return a value of "false" for the solution
solution = false;
else %reset tempSolution to the current state of the board and try the next guess for the possible value of the current cell
tempSolution = solution;
end
 
end %end recursion
end %end if
 
%% End of program
end %end sudokuSolver

Test Input: All empty cells must have a value of NaN.

sudoku = [NaN   NaN   NaN   NaN     8     3     9   NaN   NaN
1 NaN NaN NaN NaN NaN NaN 3 NaN
NaN NaN 4 NaN NaN NaN NaN 7 NaN
NaN 4 2 NaN 3 NaN NaN NaN NaN
6 NaN NaN NaN NaN NaN NaN NaN 4
NaN NaN NaN NaN 7 NaN NaN 1 NaN
NaN 2 NaN NaN NaN NaN NaN NaN NaN
NaN 8 NaN NaN NaN 9 2 NaN NaN
NaN NaN NaN 2 5 NaN NaN NaN 6]

Output:

solution =
 
7 6 5 4 8 3 9 2 1
1 9 8 7 2 6 4 3 5
2 3 4 9 1 5 6 7 8
8 4 2 5 3 1 7 6 9
6 1 7 8 9 2 3 5 4
3 5 9 6 7 4 8 1 2
9 2 6 1 4 7 5 8 3
5 8 1 3 6 9 2 4 7
4 7 3 2 5 8 1 9 6

OCaml[edit]

uses the library ocamlgraph

(* Ocamlgraph demo program: solving the Sudoku puzzle using graph coloring
Copyright 2004-2007 Sylvain Conchon, Jean-Christophe Filliatre, Julien Signoles
 
This software is free software; you can redistribute it and/or modify
it under the terms of the GNU Library General Public License version 2,
with the special exception on linking described in file LICENSE.
 
This software is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. *)

 
open Format
open Graph
 
(* We use undirected graphs with nodes containing a pair of integers
(the cell coordinates in 0..8 x 0..8).
The integer marks of the nodes will store the colors. *)

module G = Imperative.Graph.Abstract(struct type t = int * int end)
 
(* The Sudoku grid = a graph with 9x9 nodes *)
let g = G.create ()
 
(* We create the 9x9 nodes, add them to the graph and keep them in a matrix
for later access *)

let nodes =
let new_node i j = let v = G.V.create (i, j) in G.add_vertex g v; v in
Array.init 9 (fun i -> Array.init 9 (new_node i))
 
let node i j = nodes.(i).(j) (* shortcut for easier access *)
 
(* We add the edges:
two nodes are connected whenever they can't have the same value,
i.e. they belong to the same line, the same column or the same 3x3 group *)

let () =
for i = 0 to 8 do for j = 0 to 8 do
for k = 0 to 8 do
if k <> i then G.add_edge g (node i j) (node k j);
if k <> j then G.add_edge g (node i j) (node i k);
done;
let gi = 3 * (i / 3) and gj = 3 * (j / 3) in
for di = 0 to 2 do for dj = 0 to 2 do
let i' = gi + di and j' = gj + dj in
if i' <> i || j' <> j then G.add_edge g (node i j) (node i' j')
done done
done done
 
(* Displaying the current state of the graph *)
let display () =
for i = 0 to 8 do
for j = 0 to 8 do printf "%d" (G.Mark.get (node i j)) done;
printf "\n";
done;
printf "@?"
 
(* We read the initial constraints from standard input and we display g *)
let () =
for i = 0 to 8 do
let s = read_line () in
for j = 0 to 8 do match s.[j] with
| '1'..'9' as ch -> G.Mark.set (node i j) (Char.code ch - Char.code '0')
| _ -> ()
done
done;
display ();
printf "---------@."
 
(* We solve the Sudoku by 9-coloring the graph g and we display the solution *)
module C = Coloring.Mark(G)
 
let () = C.coloring g 9; display ()

Oz[edit]

Using built-in constraint propagation and search.

declare
%% a puzzle is a function that returns an initial board configuration
fun {Puzzle1}
%% a board is a list of 9 rows
[[4 _ _ _ _ _ _ 6 _]
[5 _ _ _ 8 _ 9 _ _]
[3 _ _ _ _ 1 _ _ _]
 
[_ 2 _ 7 _ _ _ _ 1]
[_ 9 _ _ _ _ _ 4 _]
[8 _ _ _ _ 3 _ 5 _]
 
[_ _ _ 2 _ _ _ _ 7]
[_ _ 6 _ 5 _ _ _ 8]
[_ 1 _ _ _ _ _ _ 6]]
end
 
%% Returns a list of solutions for the given puzzle.
fun {Solve Puzzle}
{SearchAll {GetScript Puzzle}}
end
 
%% Creates a solver script for a puzzle.
fun {GetScript Puzzle}
proc {$ Board}
%% Every row is a list of nine finite domain vars
%% with the domain 1..9.
Board = {MapRange fun {$ _} {FD.list 9 1#9} end}
%% Post initial configuration.
Board = {Puzzle}
 
%% The core constraints:
{ForAll {Rows Board} FD.distinct}
{ForAll {Columns Board} FD.distinct}
{ForAll {Boxes Board} FD.distinct}
 
%% Search if necessary.
{FD.distribute ff {Flatten Board}}
end
end
 
%% Returns the board as a list of rows.
fun {Rows Board}
Board %% This is already the representation we have chosen.
end
 
%% Returns the board as a list of columns.
fun {Columns Board}
{MapRange fun {$ I} {Column Board I} end}
end
 
%% Returns the board as a list of boxes (sub-grids).
fun {Boxes Board}
{MapRange fun {$ I} {Box Board I} end}
end
 
%% Helper function: map the range 1..9 to something.
fun {MapRange F}
{Map [1 2 3 4 5 6 7 8 9] F}
end
 
%% Returns a column of the board as a list of fields.
fun {Column Board Index}
{Map Board
fun {$ Row}
{Nth Row Index}
end
}
end
 
%% Returns a box of the board as a list of fields.
fun {Box Board Index}
Index0 = Index-1
Fields = {Flatten Board}
Start = (Index0 div 3) * 27 + (Index0 mod 3)*3
in
{Flatten
for I in 0..2 collect:C do
{C {List.take {List.drop Fields Start+I*9} 3}}
end
}
end
in
{Inspect {Solve Puzzle1}.1}


PARI/GP[edit]

Build plugin for PARI's function interface from C code: sudoku.c

#include <pari/pari.h>
 
typedef int SUDOKU [9][9];
 
static inline int check_num(SUDOKU s, int row, int col, int num)
{
int i, r = (row/3)*3, c = (col/3)*3;
 
for (i = 0; i < 9; i++)
if (s[row][i] == num || s[i][col] == num || s[i%3 + r][i/3 + c] == num)
return 0;
 
return 1;
}
 
static int sudoku_solve(SUDOKU s, int row, int col)
{
int num;
 
if (row < 9 && col < 9) {
if (s[row][col]) {
if (col < 8)
return sudoku_solve(s, row, col+1);
if (row < 8)
return sudoku_solve(s, row+1, 0);
return 1;
}
else
for (num = 1; num < 10; num++)
if (check_num(s, row, col, num)) {
s[row][col] = num;
if (sudoku_solve(s, row, col))
return 1;
else
s[row][col] = 0;
}
return 0;
}
return 1;
}
 
GEN plug_sudoku(GEN M)
{
SUDOKU s;
GEN S;
int i, k;
 
if (typ(M) != t_MAT)
pari_err(e_MISC, "parameter not matrix");
 
S = matsize(M);
 
if (itos(gel(S, 1)) < 9 || itos(gel(S, 2)) < 9)
pari_err(e_MISC, "parameter not 9x9 matrix");
 
for (i = 0; i < 9; i++)
for (k = 0; k < 9; k++)
s[i][k] = itos(gcoeff(M, i+1, k+1)); /* get sudoku */
 
if (sudoku_solve(s, 0, 0)) { /* solve sudoku */
S = cgetg(10, t_MAT);
for (k = 0; k < 9; k++) { /* create 9x9 matrix */
gel(S, k+1) = cgetg(10, t_COL);
for (i = 0; i < 9; i++)
gcoeff(S, i+1, k+1) = stoi(s[i][k]); /* fill in elements */
}
return S;
}
return gen_0; /* no solution */
}
 

Compile plugin: gcc -O2 -Wall -fPIC -shared sudoku.c -o libsudoku.so -lpari

Install plugin from home directory and play:

install("plug_sudoku", "G", "sudoku", "~/libsudoku.so")
Output:
 gp > S=[5,3,0,0,7,0,0,0,0;6,0,0,1,9,5,0,0,0;0,9,8,0,0,0,0,6,0;8,0,0,0,6,0,0,0,3;4,0,0,8,0,3,0,0,1;7,0,0,0,2,0,0,0,6;0,6,0,0,0,0,2,8,0;0,0,0,4,1,9,0,0,5;0,0,0,0,8,0,0,7,9]
[5 3 0 0 7 0 0 0 0]
[6 0 0 1 9 5 0 0 0]
[0 9 8 0 0 0 0 6 0]
[8 0 0 0 6 0 0 0 3]
[4 0 0 8 0 3 0 0 1]
[7 0 0 0 2 0 0 0 6]
[0 6 0 0 0 0 2 8 0]
[0 0 0 4 1 9 0 0 5]
[0 0 0 0 8 0 0 7 9]

gp > sudoku(S)
[5 3 4 6 7 8 9 1 2]
[6 7 2 1 9 5 3 4 8]
[1 9 8 3 4 2 5 6 7]
[8 5 9 7 6 1 4 2 3]
[4 2 6 8 5 3 7 9 1]
[7 1 3 9 2 4 8 5 6]
[9 6 1 5 3 7 2 8 4]
[2 8 7 4 1 9 6 3 5]
[3 4 5 2 8 6 1 7 9]

Pascal[edit]

Works with: Free Pascal

Simple backtracking implimentation, therefor it must be fast to be competetive.With doReverse = true same sequence for trycell. nearly 5 times faster than C-Version.

Program soduko;
{$IFDEF FPC}
{$CODEALIGN proc=16,loop=8}
{$ENDIF}
uses
sysutils,crt;
const
carreeSize = 3;
maxCoor = carreeSize*carreeSize;
maxValue = maxCoor;
maxMask = 1 shl (maxCoor+1)-1;
type
tLimit = 0..maxCoor-1;
tValue = 0..maxCoor;
tSteps = 0..maxCoor*maxCoor;
tValField = array[tLimit,tLimit] of NativeInt;//tValue;
tBitrepr = 0..maxMask;
tcol = array[tLimit] of NativeInt;// tBitrepr;
trow = array[tLimit] of NativeInt;// tBitrepr;
tcar = array[tLimit] of NativeInt;// tBitrepr;
tpValue = ^NativeInt;//^tValue;
tpLimit = ^tLimit;
tpBitrepr= ^NativeInt;//^tBitrepr;
tchgVal = record
cvCol,
cvRow,
cvCar : tpBitrepr;
cvVal : tpValue;
end;
tpChgVal = ^tchgVal;
tchgList = array[tSteps] of tchgVal;
 
tField = record
fdChgList: tchgList;
fdCol : tcol;
fdRow : trow;
fdcar : tcar;
fdVal : tValField;
fdChgIdx : tSteps;
 
end;
const
Expl0:tValField = ((9,0,7,0,0,0,3,0,0),
(0,0,0,1,0,0,2,0,0),
(6,0,0,0,0,8,0,0,0),
(0,0,5,0,3,0,0,0,0),
(0,0,0,0,0,0,0,8,4),
(0,0,0,0,0,0,0,6,0),
(0,0,0,2,7,0,0,0,0),
(8,4,0,0,0,0,0,0,0),
(0,6,0,0,0,0,0,0,0));
Expl1:tValField=((0,0,0,1,0,0,0,3,8),
(2,0,0,0,0,5,0,0,0),
(0,0,0,0,0,0,0,0,0),
(0,5,0,0,0,0,4,0,0),
(4,0,0,0,3,0,0,0,0),
(0,0,0,7,0,0,0,0,6),
(0,0,1,0,0,0,0,5,0),
(0,0,0,0,6,0,2,0,0),
(0,6,0,0,0,4,0,0,0));
 
var
F,
solF : TField;
solCnt,
callCnt: NativeUint;
solFound : Boolean;
 
procedure OutField(const F:tField);
var
rw,cl : tLimit;
rowS: AnsiString;
Begin
GotoXy(1,1);
For rw := low(tLimit) to High(tLimit) do
Begin
rowS := ' ';
For cl := low(tLimit) to High(tLimit) do
RowS :=RowS+IntToStr(F.fdVal[rw,cl]);
writeln(RowS);
end;
end;
 
function CarIdx(rw,cl: NativeInt):NativeInt;
begin
CarIdx:= (rw DIV carreeSize)*carreeSize +cl DIV carreeSize;
end;
function InsertTest(const F:tField;rw,cl:tLimit;value:tValue):boolean;
var
msk: tBitrepr;
Begin
result := (Value = 0);
IF result then
EXIT;
msk := 1 shl (value-1);
with F do
Begin
result := fdRow[rw] AND msk = 0;
result := result AND (fdCol[cl] AND msk = 0);
rw :=CarIdx(rw,cl);
result := result AND (fdCar[rw] AND msk = 0);
end;
end;
 
function InitField(var F:tField;const InFd:tValField;DoReverse:boolean):boolean;
var
TmpchgVal:tchgVal;
rw,cl,
value,
msk : NativeInt;
leftSteps:tSteps;
Begin
Fillchar(F,SizeOf(F),#0);
leftSteps := High(tSteps)-1;
//unknown fields inserted from end
For rw := low(tLimit) to High(tLimit) do
For cl := low(tLimit) to High(tLimit) do
Begin
value := InFd[rw,cl];
IF InsertTest(F,rw,cl,value) then
Begin
with F do
Begin
if value > 0 then
Begin
msk := 1 shl (value-1);
//given state
//use pointer to the relevant places and mark as occupied
with fdChgList[fdChgIdx] do
begin
cvCol := @fdCol[cl];
cvCol^ +=Msk;
cvRow := @fdRow[rw];
cvRow^ +=Msk;
cvCar := @fdCar[CarIdx(rw,cl)];
cvCar^ +=Msk;
cvVal := @fdVal[rw,cl];
cvVal^ := value;
end;
inc(fdChgIdx);
end
else
Begin
//use pointer to the relevant places
with fdChgList[leftSteps] do
begin
cvCol := @fdCol[cl];
cvRow := @fdRow[rw];
cvCar := @fdCar[CarIdx(rw,cl)];
cvVal := @fdVal[rw,cl];
end;
dec(leftSteps);
end;
end
end
else
Begin
writeln(rw:10,cl:10,value:10);
Writeln(' not solvable SuDoKu ');
delay(2000);
result := false;
EXIT;
end;
end;
//reverse direction of left over
IF DoReverse then
Begin
leftSteps := High(tSteps)-1;
rw := F.fdChgIdx;
repeat
TmpchgVal:= F.fdChgList[leftSteps];
F.fdChgList[leftSteps]:= F.fdChgList[rw];
F.fdChgList[rw] :=TmpchgVal;
dec(leftSteps);
inc(rw);
until rw>=leftSteps;
end;
//OutField(F);
solFound := false;
result := true;
end;
procedure SolIsFound;
begin
solF := F;
inc(solCnt);
solFound := True;
end;
 
procedure TryCell(var ChgVal:tpchgVal);
var
value :NativeInt;
poss,msk: NativeInt;
Begin
IF solFound then EXIT;
with ChgVal^ do
poss:= (cvRow^ OR cvCol^ OR cvCar^) XOR maxMask;
IF Poss = 0 then
EXIT;
 
value := 1;
msk := 1;
 
repeat
IF Poss AND MSK <>0 then
Begin
inc(callCnt);
//insert test value
with ChgVal^ do
Begin
cvCol^ := cvCol^ OR msk;
cvRow^ := cvRow^ OR msk;
cvCar^ := cvCar^ OR msk;
cvVAl^ := value;
end;
//try next in list, if beyond last
inc(ChgVal);
 
IF ChgVal^.cvCol <> NIL then
TryCell(ChgVal)
else
SolIsFound;
//remove test value
dec(ChgVal);
with ChgVal^ do
Begin
cvCol^ := cvCol^ XOR msk;
cvRow^ := cvRow^ XOR msk;
cvCar^ := cvCar^ XOR msk;
cvVAl^ := 0;
end;
end;
inc(msk,msk);
inc(value);
until value> maxValue;
end;
 
var
ChangeBegin : tpChgVal;
k : NativeInt;
T1,T0: TDateTime;
begin
randomize;
ClrScr;
solCnt := 0;
callCnt:= 0;
T0 := time;
k := 0;
repeat
InitField(F,Expl1,FALSE);
ChangeBegin := @F.fdChgList[F.fdChgIdx];
TryCell(ChangeBegin);
inc(k);
until k >= 5;
T1 := time;
Outfield(solF);
writeln(86400*1000*(T1-T0)/k:10:3,' ms Test calls :',callCnt/k:8:0);
end.
Output:

Expl0
  927465318
  458193276
  613728459
  185634792
  376912584
  294587163
  539276841
  841359627
  762841935
// InitField  doReverse = true
    9.850 ms Test calls :  532466
// InitField  doReverse = false
  2609.000 ms Test calls :135196346
...
Expl1
  594126738
  237895164
  618473925
  859612473
  476539812
  123748596
  341287659
  985361247
  762954381
// InitField  doReverse = true
   857.600 ms Test calls :40980572
// InitField  doReverse = false    
    21.400 ms Test calls : 1089986

Perl[edit]

#!/usr/bin/perl
use integer;
use strict;
 
my @A = qw(
5 3 0 0 2 4 7 0 0
0 0 2 0 0 0 8 0 0
1 0 0 7 0 3 9 0 2
 
0 0 8 0 7 2 0 4 9
0 2 0 9 8 0 0 7 0
7 9 0 0 0 0 0 8 0
 
0 0 0 0 3 0 5 0 6
9 6 0 0 1 0 3 0 0
0 5 0 6 9 0 0 1 0
);
 
sub solve {
my $i;
foreach $i ( 0 .. 80 ) {
next if $A[$i];
my %t = map {
$_ / 9 == $i / 9 ||
$_ % 9 == $i % 9 ||
$_ / 27 == $i / 27 && $_ % 9 / 3 == $i % 9 / 3
? $A[$_] : 0,
1;
} 0 .. 80;
solve( $A[$i] = $_ ) for grep !$t{$_}, 1 .. 9;
return $A[$i] = 0;
}
$i = 0;
foreach (@A) {
print "-----+-----+-----\n" if !($i%27) && $i;
print !($i%9) ? '': $i%3 ? ' ' : '|', $_;
print "\n" unless ++$i%9;
}
}
solve();
Output:
5 3 9|8 2 4|7 6 1
6 7 2|1 5 9|8 3 4
1 8 4|7 6 3|9 5 2
-----+-----+-----
3 1 8|5 7 2|6 4 9
4 2 5|9 8 6|1 7 3
7 9 6|3 4 1|2 8 5
-----+-----+-----
8 4 1|2 3 7|5 9 6
9 6 7|4 1 5|3 2 8
2 5 3|6 9 8|4 1 7

Perl 6[edit]

Translation of: Perl
use v6;
my @A = <
5 3 0 0 2 4 7 0 0
0 0 2 0 0 0 8 0 0
1 0 0 7 0 3 9 0 2
 
0 0 8 0 7 2 0 4 9
0 2 0 9 8 0 0 7 0
7 9 0 0 0 0 0 8 0
 
0 0 0 0 3 0 5 0 6
9 6 0 0 1 0 3 0 0
0 5 0 6 9 0 0 1 0
>;
 
my &I = * div 9; # line number
my &J = * % 9; # column number
my &K = { ($_ div 27) * 3 + $_ % 9 div 3 }; # bloc number
 
sub solve {
for ^@A -> $i {
next if @A[$i];
my @taken-values = @A[
grep {
I($_) == I($i) || J($_) == J($i) || K($_) == K($i)
}, ^@A
];
for grep none(@taken-values), 1..9 {
@A[$i] = $_;
solve;
}
return @A[$i] = 0;
}
my $i = 1;
for ^@A {
print "@A[$_] ";
print " " if $i %% 3;
print "\n" if $i %% 9;
print "\n" if $i++ %% 27;
}
}
solve;
Output:
5 3 9  8 2 4  7 6 1  
6 7 2  1 5 9  8 3 4  
1 8 4  7 6 3  9 5 2  

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

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

This is an alternative solution that uses a more ellaborate set of choices instead of brute-forcing it.

#!/usr/bin/env perl6
use v6;
#
# In this code, a sudoku puzzle is represented as a two-dimentional
# array. The cells that are not yet solved are represented by yet
# another array of all the possible values.
#
# This implementation is not a simple brute force evaluation of all
# the options, but rather makes four extra attempts to guide the
# solution:
#
# 1) For every change in the grid, usually made by an attempt at a
# solution, we will reduce the search space of the possible values
# in all the other cells before going forward.
#
# 2) When a cell that is not yet resolved is the only one that can
# hold a specific value, resolve it immediately instead of
# performing the regular search.
#
# 3) Instead of trying from cell 1,1 and moving in sequence, this
# implementation will start trying on the cell that is the closest
# to being solved already.
#
# 4) Instead of trying all possible values in sequence, start with
# the value that is the most unique. I.e.: If the options for this
# cell are 1,4,6 and 6 is only a candidate for two of the
# competing cells, we start with that one.
#
 
# keep a list with all the cells, handy for traversal
my @cells = do for 0..8 X 0..8 -> $x, $y { [ $x, $y ] };
 
#
# Try to solve this puzzle and return the resolved puzzle if it is at
# all solvable in this configuration.
sub solve($sudoku, Int $level) {
# cleanup the impossible values first,
if (cleanup-impossible-values($sudoku, $level)) {
# try to find implicit answers
while (find-implicit-answers($sudoku, $level)) {
# and every time you find some, re-do the cleanup and try again
cleanup-impossible-values($sudoku, $level);
}
# Now let's actually try to solve a new value. But instead of
# going in sequence, we select the cell that is the closest to
# being solved already. This will reduce the overall number of
# guesses.
for sort { solution-complexity-factor($sudoku, $_[0], $_[1]) },
grep { $sudoku[$_[0]][$_[1]] ~~ Array },
@cells -> $cell
{
my Int ($x, $y) = @($cell);
# Now let's try the possible values in the order of
# uniqueness.
for sort { matches-in-competing-cells($sudoku, $x, $y, $_) }, @($sudoku[$x][$y]) -> $val {
trace $level, "Trying $val on "~($x+1)~","~($y+1)~" "~$sudoku[$x][$y].perl;
my $solution = clone-sudoku($sudoku);
$solution[$x][$y] = $val;
my $solved = solve($solution, $level+1);
if $solved {
trace $level, "Solved... ($val on "~($x+1)~","~($y+1)~")";
return $solved;
}
}
# if we fell through, it means that we found no valid
# value for this cell
trace $level, "Backtrack, path unsolvable... (on "~($x+1)~" "~($y+1)~")";
return 0;
}
# all cells are already solved.
return $sudoku;
} else {
# if the cleanup failed, it means this is an invalid grid.
return False;
}
}
 
# This function reduces the search space from values that are already
# assigned to competing cells.
sub cleanup-impossible-values($sudoku, Int $level = 1) {
my Bool $resolved;
repeat {
$resolved = False;
for grep { $sudoku[$_[0]][$_[1]] ~~ Array },
@cells -> $cell {
my Int ($x, $y) = @($cell);
# which block is this cell in
my Int $bx = Int($x / 3);
my Int $by = Int($y / 3);
 
# A unfilled cell is not resolved, so it shouldn't match
my multi match-resolved-cell(Array $other, Int $this) {
return 0;
}
my multi match-resolved-cell(Int $other, Int $this) {
return $other == $this;
}
 
# Reduce the possible values to the ones that are still
# valid
my @r =
grep { !match-resolved-cell($sudoku[any(0..2)+3*$bx][any(0..2)+3*$by], $_) }, # same block
grep { !match-resolved-cell($sudoku[any(0..8)][$y], $_) }, # same line
grep { !match-resolved-cell($sudoku[$x][any(0..8)], $_) }, # same column
@($sudoku[$x][$y]);
if (@r.elems == 1) {
# if only one element is left, then make it resolved
$sudoku[$x][$y] = @r[0];
$resolved = True;
} elsif (@r.elems == 0) {
# This is an invalid grid
return 0;
} else {
$sudoku[$x][$y] = @r;
}
}
} while $resolved; # repeat if there was any change
return 1;
}
 
sub solution-complexity-factor($sudoku, Int $x, Int $y) {
my Int $bx = Int($x / 3); # this block
my Int $by = Int($y / 3);
my multi count-values(Array $val) {
return $val.elems;
}
my multi count-values(Int $val) {
return 1;
}
# the number of possible values should take precedence
my Int $f = 1000 * count-values($sudoku[$x][$y]);
for 0..2 X 0..2 -> $lx, $ly {
$f += count-values($sudoku[$lx+$bx*3][$ly+$by*3])
}
for 0..^($by*3), (($by+1)*3)..8 -> $ly {
$f += count-values($sudoku[$x][$ly])
}
for 0..^($bx*3), (($bx+1)*3)..8 -> $lx {
$f += count-values($sudoku[$lx][$y])
}
return $f;
}
 
sub matches-in-competing-cells($sudoku, Int $x, Int $y, Int $val) {
my Int $bx = Int($x / 3); # this block
my Int $by = Int($y / 3);
# Function to decide which possible value to try first
my multi cell-matching(Int $cell) {
return $val == $cell ?? 1 !! 0;
}
my multi cell-matching(Array $cell) {
return $cell.grep({ $val == $_ }) ?? 1 !! 0;
}
my Int $c = 0;
for 0..2 X 0..2 -> $lx, $ly {
$c += cell-matching($sudoku[$lx+$bx*3][$ly+$by*3])
}
for 0..^($by*3), (($by+1)*3)..8 -> $ly {
$c += cell-matching($sudoku[$x][$ly])
}
for 0..^($bx*3), (($bx+1)*3)..8 -> $lx {
$c += cell-matching($sudoku[$lx][$y])
}
return $c;
}
 
sub find-implicit-answers($sudoku, Int $level) {
my Bool $resolved = False;
for grep { $sudoku[$_[0]][$_[1]] ~~ Array },
@cells -> $cell {
my Int ($x, $y) = @($cell);
for @($sudoku[$x][$y]) -> $val {
# If this is the only cell with this val as a possibility,
# just make it resolved already
if (matches-in-competing-cells($sudoku, $x, $y, $val) == 1) {
$sudoku[$x][$y] = $val;
$resolved = True;
}
}
}
return $resolved;
}
 
my $puzzle =
map { [ map { $_ == 0 ?? [1..9] !! $_+0 }, @($_) ] },
[ 0,0,0,0,3,7,6,0,0 ],
[ 0,0,0,6,0,0,0,9,0 ],
[ 0,0,8,0,0,0,0,0,4 ],
[ 0,9,0,0,0,0,0,0,1 ],
[ 6,0,0,0,0,0,0,0,9 ],
[ 3,0,0,0,0,0,0,4,0 ],
[ 7,0,0,0,0,0,8,0,0 ],
[ 0,1,0,0,0,9,0,0,0 ],
[ 0,0,2,5,4,0,0,0,0 ];
 
my $solved = solve($puzzle, 0);
if $solved {
print-sudoku($solved,0);
} else {
say "unsolvable.";
}
 
# Utility functions, not really part of the solution
 
sub trace(Int $level, Str $message) {
say '.' x $level, $message;
}
 
sub clone-sudoku($sudoku) {
my $clone;
for 0..8 X 0..8 -> $x, $y {
$clone[$x][$y] = $sudoku[$x][$y];
}
return $clone;
}
 
sub print-sudoku($sudoku, Int $level = 1) {
trace $level, '-' x 5*9;
for @($sudoku) -> $row {
trace $level, join " ", do for @($row) -> $cell {
$cell ~~ Array ?? "#{$cell.elems}#" !! " $cell "
}
}
}
Output:
Trying 8 on 9,1 [8, 9]
.Trying 6 on 9,2 [3, 6]
..Trying 7 on 9,9 [3, 7]
...Trying 1 on 9,8 [1, 3]
....Trying 4 on 8,1 [4, 5]
.....Trying 3 on 7,2 [3, 5]
......Trying 3 on 8,7 [2, 3]
.......Trying 6 on 8,9 [2, 6]
.......Trying 2 on 8,9 [2, 6]
.......Backtrack, path unsolvable... (on 8 9)
......Trying 2 on 8,7 [2, 3]
.......Trying 3 on 8,9 [3, 6]
.......Trying 6 on 8,9 [3, 6]
.......Backtrack, path unsolvable... (on 8 9)
......Backtrack, path unsolvable... (on 8 7)
.....Trying 5 on 7,2 [3, 5]
......Trying 5 on 8,7 [2, 5]
.......Trying 6 on 8,9 [2, 6]
.......Trying 2 on 8,9 [2, 6]
.......Backtrack, path unsolvable... (on 8 9)
......Trying 2 on 8,7 [2, 5]
.......Trying 5 on 8,9 [5, 6]
.......Trying 6 on 8,9 [5, 6]
.......Backtrack, path unsolvable... (on 8 9)
......Backtrack, path unsolvable... (on 8 7)
.....Backtrack, path unsolvable... (on 7 2)
....Trying 5 on 8,1 [4, 5]
.....Trying 3 on 8,3 [3, 4]
......Trying 6 on 8,9 [2, 6]
.......Trying 3 on 7,9 [3, 5]
........Trying 5 on 1,9 [2, 5]
.........Trying 3 on 3,8 [3, 7]
..........Trying 4 on 2,1 [1, 4]
..........Trying 1 on 2,1 [1, 4]
..........Backtrack, path unsolvable... (on 2 1)
.........Trying 7 on 3,8 [3, 7]
..........Trying 3 on 3,7 [1, 3]
..........Trying 1 on 3,7 [1, 3]
..........Backtrack, path unsolvable... (on 3 7)
.........Backtrack, path unsolvable... (on 3 8)
........Trying 2 on 1,9 [2, 5]
.........Trying 3 on 3,8 [3, 7]
..........Trying 7 on 2,7 [1, 7]
...........Trying 9 on 3,1 [2, 9]
...........Trying 2 on 3,1 [2, 9]
...........Backtrack, path unsolvable... (on 3 1)
..........Trying 1 on 2,7 [1, 7]
..........Backtrack, path unsolvable... (on 2 7)
.........Trying 7 on 3,8 [3, 7]
..........Trying 3 on 3,7 [1, 3]
..........Trying 1 on 3,7 [1, 3]
...........Trying 9 on 3,1 [2, 9]
...........Trying 2 on 3,1 [2, 9]
...........Backtrack, path unsolvable... (on 3 1)
..........Backtrack, path unsolvable... (on 3 7)
.........Backtrack, path unsolvable... (on 3 8)
........Backtrack, path unsolvable... (on 1 9)
.......Trying 5 on 7,9 [3, 5]
........Trying 8 on 1,9 [2, 8]
.........Trying 2 on 3,7 [1, 2]
.........Trying 1 on 3,7 [1, 2]
.........Backtrack, path unsolvable... (on 3 7)
........Trying 2 on 1,9 [2, 8]
.........Trying 7 on 3,8 [5, 7]
..........Trying 5 on 3,7 [1, 5]
...........Trying 2 on 2,1 [2, 4]
...........Trying 4 on 2,1 [2, 4]
...........Backtrack, path unsolvable... (on 2 1)
..........Trying 1 on 3,7 [1, 5]
...........Trying 9 on 3,1 [2, 9]
...........Trying 2 on 3,1 [2, 9]
...........Backtrack, path unsolvable... (on 3 1)
..........Backtrack, path unsolvable... (on 3 7)
.........Trying 5 on 3,8 [5, 7]
..........Trying 7 on 3,7 [1, 7]
...........Trying 2 on 2,1 [2, 4]
...........Trying 4 on 2,1 [2, 4]
...........Backtrack, path unsolvable... (on 2 1)
..........Trying 1 on 3,7 [1, 7]
..........Backtrack, path unsolvable... (on 3 7)
.........Backtrack, path unsolvable... (on 3 8)
........Backtrack, path unsolvable... (on 1 9)
.......Backtrack, path unsolvable... (on 7 9)
......Trying 2 on 8,9 [2, 6]
.......Trying 3 on 7,9 [3, 5]
........Trying 8 on 1,9 [5, 8]
.........Trying 3 on 3,8 [3, 7]
..........Trying 4 on 1,3 [1, 4]
...........Trying 9 on 1,1 [1, 9]
............Trying 1 on 3,1 [1, 2]
............Trying 2 on 3,1 [1, 2]
............Backtrack, path unsolvable... (on 3 1)
...........Trying 1 on 1,1 [1, 9]
...........Backtrack, path unsolvable... (on 1 1)
..........Trying 1 on 1,3 [1, 4]
...........Trying 9 on 1,1 [4, 9]
...........Trying 4 on 1,1 [4, 9]
...........Backtrack, path unsolvable... (on 1 1)
..........Backtrack, path unsolvable... (on 1 3)
.........Trying 7 on 3,8 [3, 7]
..........Trying 3 on 3,7 [1, 3]
..........Trying 1 on 3,7 [1, 3]
...........Trying 9 on 3,1 [2, 9]
...........Trying 2 on 3,1 [2, 9]
...........Backtrack, path unsolvable... (on 3 1)
..........Backtrack, path unsolvable... (on 3 7)
.........Backtrack, path unsolvable... (on 3 8)
........Trying 5 on 1,9 [5, 8]
........Backtrack, path unsolvable... (on 1 9)
.......Trying 5 on 7,9 [3, 5]
........Trying 5 on 1,8 [2, 5]
.........Trying 7 on 3,8 [2, 7]
..........Trying 4 on 1,3 [1, 4]
..........Trying 1 on 1,3 [1, 4]
..........Backtrack, path unsolvable... (on 1 3)
.........Trying 2 on 3,8 [2, 7]
..........Trying 9 on 3,1 [1, 9]
..........Trying 1 on 3,1 [1, 9]
..........Backtrack, path unsolvable... (on 3 1)
.........Backtrack, path unsolvable... (on 3 8)
........Trying 2 on 1,8 [2, 5]
.........Trying 7 on 3,8 [5, 7]
..........Trying 4 on 1,3 [1, 4]
...........Trying 9 on 1,1 [1, 9]
............Trying 1 on 3,1 [1, 2]
............Trying 2 on 3,1 [1, 2]
............Backtrack, path unsolvable... (on 3 1)
...........Trying 1 on 1,1 [1, 9]
...........Backtrack, path unsolvable... (on 1 1)
..........Trying 1 on 1,3 [1, 4]
...........Trying 9 on 1,1 [4, 9]
...........Trying 4 on 1,1 [4, 9]
...........Backtrack, path unsolvable... (on 1 1)
..........Backtrack, path unsolvable... (on 1 3)
.........Trying 5 on 3,8 [5, 7]
..........Trying 7 on 3,7 [1, 7]
...........Trying 2 on 2,1 [2, 4]
...........Trying 4 on 2,1 [2, 4]
...........Backtrack, path unsolvable... (on 2 1)
..........Trying 1 on 3,7 [1, 7]
..........Backtrack, path unsolvable... (on 3 7)
.........Backtrack, path unsolvable... (on 3 8)
........Backtrack, path unsolvable... (on 1 8)
.......Backtrack, path unsolvable... (on 7 9)
......Backtrack, path unsolvable... (on 8 9)
.....Trying 4 on 8,3 [3, 4]
......Trying 3 on 8,7 [2, 3]
.......Trying 6 on 8,9 [2, 6]
.......Trying 2 on 8,9 [2, 6]
.......Backtrack, path unsolvable... (on 8 9)
......Trying 2 on 8,7 [2, 3]
.......Trying 3 on 8,9 [3, 6]
.......Trying 6 on 8,9 [3, 6]
.......Backtrack, path unsolvable... (on 8 9)
......Backtrack, path unsolvable... (on 8 7)
.....Backtrack, path unsolvable... (on 8 3)
....Backtrack, path unsolvable... (on 8 1)
...Trying 3 on 9,8 [1, 3]
....Trying 4 on 8,1 [4, 5]
.....Trying 3 on 7,2 [3, 5]
.....Trying 5 on 7,2 [3, 5]
......Trying 6 on 7,9 [2, 6]
......Trying 2 on 7,9 [2, 6]
......Backtrack, path unsolvable... (on 7 9)
.....Backtrack, path unsolvable... (on 7 2)
....Trying 5 on 8,1 [4, 5]
.....Trying 6 on 8,9 [2, 6]
......Trying 8 on 1,9 [2, 8]
.......Trying 1 on 3,7 [1, 2]
.......Trying 2 on 3,7 [1, 2]
.......Backtrack, path unsolvable... (on 3 7)
......Trying 2 on 1,9 [2, 8]
.......Trying 7 on 3,8 [5, 7]
........Trying 5 on 3,7 [1, 5]
.........Trying 9 on 3,1 [1, 9]
.........Trying 1 on 3,1 [1, 9]
.........Backtrack, path unsolvable... (on 3 1)
........Trying 1 on 3,7 [1, 5]
.........Trying 9 on 3,1 [2, 9]
.........Trying 2 on 3,1 [2, 9]
.........Backtrack, path unsolvable... (on 3 1)
........Backtrack, path unsolvable... (on 3 7)
.......Trying 5 on 3,8 [5, 7]
........Trying 7 on 3,7 [1, 7]
.........Trying 9 on 3,1 [1, 9]
.........Trying 1 on 3,1 [1, 9]
.........Backtrack, path unsolvable... (on 3 1)
........Trying 1 on 3,7 [1, 7]
........Backtrack, path unsolvable... (on 3 7)
.......Backtrack, path unsolvable... (on 3 8)
......Backtrack, path unsolvable... (on 1 9)
.....Trying 2 on 8,9 [2, 6]
......Trying 5 on 1,8 [2, 5]
.......Trying 7 on 3,8 [2, 7]
........Trying 4 on 2,1 [1, 4]
........Trying 1 on 2,1 [1, 4]
........Backtrack, path unsolvable... (on 2 1)
.......Trying 2 on 3,8 [2, 7]
........Trying 9 on 3,1 [1, 9]
........Trying 1 on 3,1 [1, 9]
........Backtrack, path unsolvable... (on 3 1)
.......Backtrack, path unsolvable... (on 3 8)
......Trying 2 on 1,8 [2, 5]
.......Trying 7 on 3,8 [5, 7]
........Trying 1 on 3,7 [1, 5]
.........Trying 9 on 3,1 [2, 9]
.........Trying 2 on 3,1 [2, 9]
.........Backtrack, path unsolvable... (on 3 1)
........Trying 5 on 3,7 [1, 5]
.........Trying 9 on 3,1 [1, 9]
.........Trying 1 on 3,1 [1, 9]
.........Backtrack, path unsolvable... (on 3 1)
........Backtrack, path unsolvable... (on 3 7)
.......Trying 5 on 3,8 [5, 7]
........Trying 9 on 3,1 [1, 9]
........Trying 1 on 3,1 [1, 9]
........Backtrack, path unsolvable... (on 3 1)
.......Backtrack, path unsolvable... (on 3 8)
......Backtrack, path unsolvable... (on 1 8)
.....Backtrack, path unsolvable... (on 8 9)
....Backtrack, path unsolvable... (on 8 1)
...Backtrack, path unsolvable... (on 9 8)
..Trying 3 on 9,9 [3, 7]
...Trying 4 on 8,1 [4, 5]
....Trying 3 on 7,2 [3, 5]
....Trying 5 on 7,2 [3, 5]
.....Trying 6 on 7,9 [2, 6]
.....Trying 2 on 7,9 [2, 6]
.....Backtrack, path unsolvable... (on 7 9)
....Backtrack, path unsolvable... (on 7 2)
...Trying 5 on 8,1 [4, 5]
....Trying 6 on 8,9 [2, 6]
.....Trying 8 on 1,9 [2, 8]
......Trying 7 on 2,9 [2, 7]
.......Trying 1 on 3,7 [1, 2]
.......Trying 2 on 3,7 [1, 2]
.......Backtrack, path unsolvable... (on 3 7)
......Trying 2 on 2,9 [2, 7]
.......Trying 4 on 2,1 [1, 4]
.......Trying 1 on 2,1 [1, 4]
.......Backtrack, path unsolvable... (on 2 1)
......Backtrack, path unsolvable... (on 2 9)
.....Trying 2 on 1,9 [2, 8]
......Trying 3 on 3,8 [3, 5]
.......Trying 5 on 2,7 [1, 5]
........Trying 9 on 3,1 [2, 9]
........Trying 2 on 3,1 [2, 9]
........Backtrack, path unsolvable... (on 3 1)
.......Trying 1 on 2,7 [1, 5]
.......Backtrack, path unsolvable... (on 2 7)
......Trying 5 on 3,8 [3, 5]
.......Trying 3 on 3,7 [1, 3]
.......Trying 1 on 3,7 [1, 3]
.......Backtrack, path unsolvable... (on 3 7)
......Backtrack, path unsolvable... (on 3 8)
.....Backtrack, path unsolvable... (on 1 9)
....Trying 2 on 8,9 [2, 6]
.....Trying 5 on 1,8 [2, 5]
......Trying 3 on 3,8 [2, 3]
.......Trying 4 on 2,1 [1, 4]
.......Trying 1 on 2,1 [1, 4]
.......Backtrack, path unsolvable... (on 2 1)
......Trying 2 on 3,8 [2, 3]
.......Trying 9 on 3,1 [1, 9]
.......Trying 1 on 3,1 [1, 9]
.......Backtrack, path unsolvable... (on 3 1)
......Backtrack, path unsolvable... (on 3 8)
.....Trying 2 on 1,8 [2, 5]
......Trying 3 on 3,8 [3, 5]
.......Trying 4 on 1,3 [1, 4]
.......Trying 1 on 1,3 [1, 4]
.......Backtrack, path unsolvable... (on 1 3)
......Trying 5 on 3,8 [3, 5]
.......Trying 9 on 3,1 [1, 9]
.......Trying 1 on 3,1 [1, 9]
.......Backtrack, path unsolvable... (on 3 1)
......Backtrack, path unsolvable... (on 3 8)
.....Backtrack, path unsolvable... (on 1 8)
....Backtrack, path unsolvable... (on 8 9)
...Backtrack, path unsolvable... (on 8 1)
..Backtrack, path unsolvable... (on 9 9)
.Trying 3 on 9,2 [3, 6]
..Trying 7 on 9,9 [6, 7]
...Trying 1 on 9,8 [1, 6]
....Trying 4 on 8,1 [4, 5]
.....Trying 6 on 7,2 [5, 6]
......Trying 3 on 8,7 [2, 3]
.......Trying 6 on 8,9 [2, 6]
.......Trying 2 on 8,9 [2, 6]
.......Backtrack, path unsolvable... (on 8 9)
......Trying 2 on 8,7 [2, 3]
.......Trying 6 on 8,9 [3, 6]
.......Trying 3 on 8,9 [3, 6]
.......Backtrack, path unsolvable... (on 8 9)
......Backtrack, path unsolvable... (on 8 7)
.....Trying 5 on 7,2 [5, 6]
......Trying 4 on 1,2 [2, 4]
.......Trying 1 on 1,3 [1, 5]
........Trying 2 on 2,1 [2, 5]
........Trying 5 on 2,1 [2, 5]
........Backtrack, path unsolvable... (on 2 1)
.......Trying 5 on 1,3 [1, 5]
........Trying 1 on 2,1 [1, 2]
.........Trying 9 on 1,1 [2, 9]
..........Trying 8 on 1,9 [2, 8]
..........Trying 2 on 1,9 [2, 8]
..........Backtrack, path unsolvable... (on 1 9)
.........Trying 2 on 1,1 [2, 9]
.........Backtrack, path unsolvable... (on 1 1)
........Trying 2 on 2,1 [1, 2]
.........Trying 9 on 1,1 [1, 9]
..........Trying 8 on 1,9 [2, 8]
...........Trying 2 on 8,9 [2, 3]
...........Trying 3 on 8,9 [2, 3]
...........Backtrack, path unsolvable... (on 8 9)
..........Trying 2 on 1,9 [2, 8]
..........Backtrack, path unsolvable... (on 1 9)
.........Trying 1 on 1,1 [1, 9]
..........Trying 8 on 1,9 [2, 8]
...........Trying 2 on 8,9 [2, 3]
...........Trying 3 on 8,9 [2, 3]
...........Backtrack, path unsolvable... (on 8 9)
..........Trying 2 on 1,9 [2, 8]
..........Backtrack, path unsolvable... (on 1 9)
.........Backtrack, path unsolvable... (on 1 1)
........Backtrack, path unsolvable... (on 2 1)
.......Backtrack, path unsolvable... (on 1 3)
......Trying 2 on 1,2 [2, 4]
.......Trying 1 on 2,1 [1, 5]
........Trying 9 on 1,1 [5, 9]
.........Trying 8 on 1,9 [5, 8]
.........Trying 5 on 1,9 [5, 8]
.........Backtrack, path unsolvable... (on 1 9)
........Trying 5 on 1,1 [5, 9]
........Backtrack, path unsolvable... (on 1 1)
.......Trying 5 on 2,1 [1, 5]
........Trying 9 on 1,1 [1, 9]
.........Trying 8 on 1,9 [5, 8]
..........Trying 6 on 7,9 [3, 6]
..........Trying 3 on 7,9 [3, 6]
..........Backtrack, path unsolvable... (on 7 9)
.........Trying 5 on 1,9 [5, 8]
.........Backtrack, path unsolvable... (on 1 9)
........Trying 1 on 1,1 [1, 9]
.........Trying 8 on 1,9 [5, 8]
..........Trying 5 on 8,9 [3, 5]
..........Trying 3 on 8,9 [3, 5]
..........Backtrack, path unsolvable... (on 8 9)
.........Trying 5 on 1,9 [5, 8]
.........Backtrack, path unsolvable... (on 1 9)
........Backtrack, path unsolvable... (on 1 1)
.......Backtrack, path unsolvable... (on 2 1)
......Backtrack, path unsolvable... (on 1 2)
.....Backtrack, path unsolvable... (on 7 2)
....Trying 5 on 8,1 [4, 5]
.....Trying 6 on 8,3 [4, 6]
......Trying 3 on 8,9 [2, 3]
.......Trying 6 on 7,9 [5, 6]
........Trying 5 on 1,9 [2, 5]
.........Trying 3 on 3,8 [3, 7]
..........Trying 4 on 2,1 [1, 4]
..........Trying 1 on 2,1 [1, 4]
..........Backtrack, path unsolvable... (on 2 1)
.........Trying 7 on 3,8 [3, 7]
..........Trying 3 on 3,7 [1, 3]
..........Trying 1 on 3,7 [1, 3]
..........Backtrack, path unsolvable... (on 3 7)
.........Backtrack, path unsolvable... (on 3 8)
........Trying 2 on 1,9 [2, 5]
.........Trying 3 on 3,8 [3, 7]
..........Trying 7 on 2,7 [1, 7]
..........Trying 1 on 2,7 [1, 7]
...........Trying 4 on 2,1 [2, 4]
...........Trying 2 on 2,1 [2, 4]
...........Solved... (2 on 2,1)
..........Solved... (1 on 2,7)
.........Solved... (3 on 3,8)
........Solved... (2 on 1,9)
.......Solved... (6 on 7,9)
......Solved... (3 on 8,9)
.....Solved... (6 on 8,3)
....Solved... (5 on 8,1)
...Solved... (1 on 9,8)
..Solved... (7 on 9,9)
.Solved... (3 on 9,2)
Solved... (8 on 9,1)
---------------------------------------------
 9   5   4   1   3   7   6   8   2 
 2   7   3   6   8   4   1   9   5 
 1   6   8   2   9   5   7   3   4 
 4   9   5   7   2   8   3   6   1 
 6   8   1   4   5   3   2   7   9 
 3   2   7   9   6   1   5   4   8 
 7   4   9   3   1   2   8   5   6 
 5   1   6   8   7   9   4   2   3 
 8   3   2   5   4   6   9   1   7 

PHP[edit]

Translation of: C++
	class SudokuSolver {
protected $grid = [];
protected $emptySymbol;
public static function parseString($str, $emptySymbol = '0')
{
$grid = str_split($str);
foreach($grid as &$v)
{
if($v == $emptySymbol)
{
$v = 0;
}
else
{
$v = (int)$v;
}
}
return $grid;
}
 
public function __construct($str, $emptySymbol = '0') {
if(strlen($str) !== 81)
{
throw new \Exception('Error sudoku');
}
$this->grid = static::parseString($str, $emptySymbol);
$this->emptySymbol = $emptySymbol;
}
 
public function solve()
{
try
{
$this->placeNumber(0);
return false;
}
catch(\Exception $e)
{
return true;
}
}
 
protected function placeNumber($pos)
{
if($pos == 81)
{
throw new \Exception('Finish');
}
if($this->grid[$pos] > 0)
{
$this->placeNumber($pos+1);
return;
}
for($n = 1; $n <= 9; $n++)
{
if($this->checkValidity($n, $pos%9, floor($pos/9)))
{
$this->grid[$pos] = $n;
$this->placeNumber($pos+1);
$this->grid[$pos] = 0;
}
}
}
 
protected function checkValidity($val, $x, $y)
{
for($i = 0; $i < 9; $i++)
{
if(($this->grid[$y*9+$i] == $val) || ($this->grid[$i*9+$x] == $val))
{
return false;
}
}
$startX = (int) ((int)($x/3)*3);
$startY = (int) ((int)($y/3)*3);
 
for($i = $startY; $i<$startY+3;$i++)
{
for($j = $startX; $j<$startX+3;$j++)
{
if($this->grid[$i*9+$j] == $val)
{
return false;
}
}
}
return true;
}
 
public function display() {
$str = '';
for($i = 0; $i<9; $i++)
{
for($j = 0; $j<9;$j++)
{
$str .= $this->grid[$i*9+$j];
$str .= " ";
if($j == 2 || $j == 5)
{
$str .= "| ";
}
}
$str .= PHP_EOL;
if($i == 2 || $i == 5)
{
$str .= "------+-------+------".PHP_EOL;
}
}
echo $str;
}
 
public function __toString() {
foreach ($this->grid as &$item)
{
if($item == 0)
{
$item = $this->emptySymbol;
}
}
return implode('', $this->grid);
}
}
$solver = new SudokuSolver('009170000020600001800200000200006053000051009005040080040000700006000320700003900');
$solver->solve();
$solver->display();
Output:
3 6 9 | 1 7 5 | 8 4 2 
4 2 7 | 6 8 9 | 5 3 1 
8 5 1 | 2 3 4 | 6 9 7 
------+-------+------
2 1 8 | 7 9 6 | 4 5 3 
6 3 4 | 8 5 1 | 2 7 9 
9 7 5 | 3 4 2 | 1 8 6 
------+-------+------
1 4 3 | 9 2 8 | 7 6 5 
5 9 6 | 4 1 7 | 3 2 8 
7 8 2 | 5 6 3 | 9 1 4
(solved in 0.027s)

Phix[edit]

Simple brute force solution. Generally quite good but will struggle on some puzzles (eg see "the beast" below)

sequence board = split("""
.......39
.....1..5
..3.5.8..
..8.9...6
.7...2...
1..4.....
..9.8..5.
.2....6..
4..7.....""",'\n')
 
function valid_move(integer y, integer x, integer ch)
for i=1 to 9 do
if ch=board[i][x] then return 0 end if
if ch=board[y][i] then return 0 end if
end for
y -= mod(y-1,3)
x -= mod(x-1,3)
for ys=y to y+2 do
for xs=x to x+2 do
if ch=board[ys][xs] then return 0 end if
end for
end for
return 1
end function
 
sequence solution = {}
 
procedure brute_solve()
for y=1 to 9 do
for x=1 to 9 do
if board[y][x]<='0' then
for ch='1' to '9' do
if valid_move(y,x,ch) then
board[y][x] = ch
brute_solve()
board[y][x] = ' '
if length(solution) then return end if
end if
end for
return
end if
end for
end for
solution = board -- (already solved case)
end procedure
 
atom t0 = time()
brute_solve()
printf(1,"%s\n(solved in %3.2fs)\n",{join(solution,"\n"),time()-t0})
Output:
751846239
892371465
643259871
238197546
974562318
165438927
319684752
527913684
486725193
(solved in 0.95s)

OTT solution. Implements line/col and set exclusion, and x-wings. Blisteringly fast
The included program demo\rosetta\Sudoku.exw is an extended version of this that performs extended validation, contains 339 puzzles, can be run as a command-line or gui program, check for multiple solutions, and produce a more readable single-puzzle output (example below).

-- Working directly on 81-character strings ultimately proves easier: Originally I 
-- just wanted to simplify the final display, but later I realised that a 9x9 grid
-- encourages laborious indexing/looping everwhere whereas using a flat 81-element
-- approach encourages precomputation of index sets, and once you commit to that,
-- the rest of the code starts to get a whole lot cleaner. Below we create 27+18
-- sets and 5 tables of lookup indexes to locate them quickly.
 
sequence nines = {}, -- will be 27 in total
cols = repeat(0,9*9), -- remainder(i-1,9)+1
rows = repeat(0,9*9), -- floor((i-1)/9)+10
squares = repeat(0,9*9),
sixes = {}, -- will be 18 in total
dotcol = repeat(0,9*9), -- same col, diff square
dotrow = repeat(0,9*9) -- same row, diff square
 
procedure set_nines()
sequence nine, six
integer idx, ndx
for x=0 to 8 do -- columns
nine = {}
ndx = length(nines)+1
for y=1 to 81 by 9 do
idx = y+x
nine = append(nine,idx)
cols[idx] = ndx
end for
nines = append(nines,nine)
end for
for y=1 to 81 by 9 do -- rows
nine = {}
ndx = length(nines)+1
for x=0 to 8 do
idx = y+x
nine = append(nine,idx)
rows[idx] = ndx
end for
nines = append(nines,nine)
end for
if length(nines)!=18 then ?9/0 end if
for y=0 to 8 by 3 do -- small squares [19..27]
for x=0 to 8 by 3 do
nine = {}
ndx = length(nines)+1
for sy=y*9 to y*9+18 by 9 do
for sx=x to x+2 do
idx = sy+sx+1
nine = append(nine,idx)
squares[idx] = ndx
end for
end for
nines = append(nines,nine)
end for
end for
if length(nines)!=27 then ?9/0 end if
for i=1 to 9*9 do
six = {}
nine = nines[cols[i]] -- dotcol
for j=1 to length(nine) do
if squares[i]!=squares[nine[j]] then
six = append(six,nine[j])
end if
end for
ndx = find(six,sixes)
if ndx=0 then
sixes = append(sixes,six)
ndx = length(sixes)
end if
dotcol[i] = ndx
six = {}
nine = nines[rows[i]] -- dotrow
for j=1 to length(nine) do
if squares[i]!=squares[nine[j]] then
six = append(six,nine[j])
end if
end for
ndx = find(six,sixes)
if ndx=0 then
sixes = append(sixes,six)
ndx = length(sixes)
end if
dotrow[i] = ndx
end for
end procedure
set_nines()
 
integer improved = 0
 
function eliminate_in(sequence valid, sequence set, integer ch)
for i=1 to length(set) do
integer idx = set[i]
if string(valid[idx]) then
integer k = find(ch,valid[idx])
if k!=0 then
valid[idx][k..k] = ""
improved = 1
end if
end if
end for
return valid
end function
 
function test_comb(sequence chosen, sequence pool, sequence valid)
--
-- (see deep_logic()/set elimination)
-- chosen is a sequence of length 2..4 of integers 1..9: ordered elements of pool.
-- pool is a set of elements of the sequence valid, each of which is a sequence.
-- (note that elements of valid in pool not in chosen are not necessarily sequences)
--
sequence contains = repeat(0,9)
integer ccount = 0, ch
object set
 
for i=1 to length(chosen) do
set = valid[pool[chosen[i]]]
for j=1 to length(set) do
ch = set[j]-'0'
if contains[ch]=0 then
contains[ch] = 1
ccount += 1
end if
end for
end for
if ccount=length(chosen) then
for i=1 to length(pool) do
if find(i,chosen)=0 then
set = valid[pool[i]]
if sequence(set) then
-- (reverse order so deletions don't foul indexes)
for j=length(set) to 1 by -1 do
ch = set[j]-'0'
if contains[ch] then
valid[pool[i]][j..j] = ""
improved = 1
end if
end for
end if
end if
end for
end if
return valid
end function
 
-- from [[Combinations#Phix|Combinations]]
-- from http://rosettacode.org/wiki/Combinations#Phix
function comb(sequence pool, valid, integer needed, done=0, sequence chosen={})
-- (used by deep_logic()/set elimination)
if needed=0 then -- got a full set
return test_comb(chosen,pool,valid)
end if
if done+needed>length(pool) then return valid end if -- cannot fulfil
-- get all combinations with and without the next item:
done += 1
if sequence(valid[pool[done]]) then
valid = comb(pool,valid,needed-1,done,append(chosen,done))
end if
return comb(pool,valid,needed,done,chosen)
end function
 
function deep_logic(string board, sequence valid)
--
-- Create a grid of valid moves. Note this does not modify board, but instead creates
-- sets of permitted values for each cell, which can also be and are used for hints.
-- Apply standard eliminations of known cells, then try some more advanced tactics:
--
-- 1) row/col elimination
-- If in any of the 9 small squares a number can only occur in one row or column,
-- then that number cannot occur in that row or column in two other corresponding
-- small squares. Example (this one with significant practical benefit):
-- 000|000|036
-- 840|000|000
-- 000|000|020
-- ---+---+---
-- 000|203|000
-- 010|000|700
-- 000|600|400
-- ---+---+---
-- 000|410|050
-- 003|000|200
-- 600|000|000 <-- 3
-- ^-- 3
-- Naively, the br can contain a 3 in the four corners, but looking at mid-right and
-- mid-bottom leads us to eliminating 3s in column 9 and row 9, leaving 7,7 as the
-- only square in the br that can be a 3. Uses dotcol and dotrow.
-- Without this, brute force on the above takes ~8s, but with it ~0s
--
-- 2) set elimination
-- If in any 9-set there is a set of n blank squares that can only contain n digits,
-- then no other squares can contain those digits. Example (with some benefit):
-- 75.|.9.|.46
-- 961|...|352
-- 4..|...|79.
-- ---+---+---
-- 2..|6.1|..7
-- .8.|...|.2.
-- 1..|328|.65
-- ---+---+---
-- ...|...|... <-- [7,8] is {1,3,8}, [7,9] is {1,3,8}
-- 3.9|...|2.4 <-- [8,8] is {1,8}
-- 84.|.3.|.79
-- The three cells above the br 479 can only contain {1,3,8}, so the .. of the .2.
-- in column 7 of that square are {5,6} (not 1) and hence [9,4] must be a 1.
-- (Relies on plain_logic to spot that "must be a 1", and serves as a clear example
-- of why this routine should not bother to attempt updating the board itself - as
-- it spends almost all of its time looking in a completely different place.)
-- (One could argue that [7,7] and [9,7] are the only places that can hold {5,6} and
-- therefore we should eliminate all non-{5,6} from those squares, as an alternative
-- strategy. However I think that would be harder to code and cannot imagine a case
-- said complementary logic covers, that the above does not, cmiiw.)
--
-- 3) x-wings
-- If a pair of rows or columns can only contain a given number in two matching places,
-- then once filled they will occupy opposite diagonal corners, hence that said number
-- cannot occur elsewhere in those two columns/rows. Example (with a benefit):
-- .43|98.|25. <-- 6 in [1,{6,9}]
-- 6..|425|...
-- 2..|..1|.94
-- ---+---+---
-- 9..|..4|.7. <-- hence 6 not in [4,9]
-- 3..|6.8|...
-- 41.|2.9|..3
-- ---+---+---
-- 82.|5..|... <-- hence 6 not in [7,6],[7,9]
-- ...|.4.|..5 <-- hence 6 not in [8,6]
-- 534|89.|71. <-- 6 in [9,{6,9}]
-- A 6 must be in [1,6] or [1,9] and [9,6] or [9,9], hence [7,9] is not 6 and must be 9.
-- (we also eliminate 6 from [4,9], [7,6] and [8,6] to no great use)
-- In practice this offers little benefit over a single trial-and-error step, as
-- obviously trying either 6 in row 1 or 9 immediately pinpoints that 9 anyway.
--
-- 4) swordfish (not attempted)
-- There is an extension to x-wings known as swordfish: three (or more) pairs form
-- a staggered pair (or more) of rectangles that exhibit similar properties, eg:
-- 8-1|-5-|-3-
-- 953|-68|---
-- -4-|-*3|5*8
-- ---+---+---
-- 6--|9-2|---
-- -8-|-3-|-4-
-- 3*-|5-1|-*7 <-- hence [6,3] is not 9, must be 4
-- ---+---+---
-- 5*2|-*-|-8-
-- --8|37-|--9
-- -3-|82-|1--
-- ^---^---^-- 3 pairs of 9s (marked with *) on 3 rows (only)
-- It is not a swordfish if the 3 pairs are on >3 rows, I trust that is obvious.
-- Logically you can extend this to N pairs on N rows, however I cannot imagine a
-- case where this is not immediately solved by a single trial-step being invalid.
-- (eg above if you try [3,5]:=9 it is quickly proved to be invalid, and the same
-- goes for [6,8]:=9 and [7,2]:=9, since they are all entirely inter-dependent.)
-- Obviously where I have said rows, the same concept can be applied to columns.
-- Likewise there are "Alternate Pairs" and "Hook or X-Y wing" strategies, which
-- are easily solved with a single trial-and-error step, and of course the brute
-- force algorithm is going to select pairs first anyway. [Erm, no it doesn't,
-- it selects shortest - I've noted the possible improvement below.]
--
integer col, row
sequence c, r
sequence nine, prevsets, set
object vj
integer ch, k, idx, sx, sy, count
 
if length(valid)=0 then
-- initialise/start again from scratch
valid = repeat("123456789",9*9)
end if
--
-- First perform standard eliminations of any known cells:
-- (repeated every time so plain_logic() does not have to worry about it)
--
for i=1 to 9*9 do
ch = board[i]
if ch>'0'
and string(valid[i]) then
valid[i] = ch
valid = eliminate_in(valid,nines[cols[i]],ch)
valid = eliminate_in(valid,nines[rows[i]],ch)
valid = eliminate_in(valid,nines[squares[i]],ch)
end if
end for
--
-- 1) row/col elimination
--
for s=19 to 27 do
c = repeat(0,9) -- 0 = none seen, 1..9 this col only, -1: >1 col
r = repeat(0,9) -- "" row row
nine = nines[s]
for n=1 to 9 do
k = nine[n]
vj = valid[k]
if string(vj) then
for i=1 to length(vj) do
ch = vj[i]-'0'
col = dotcol[k]
row = dotrow[k]
c[ch] = iff(find(c[ch],{0,col})!=0?col:-1)
r[ch] = iff(find(r[ch],{0,row})!=0?row:-1)
end for
end if
end for
for i=1 to 9 do
ch = i+'0'
col = c[i]
if col>0 then
valid = eliminate_in(valid,sixes[col],ch)
end if
row = r[i]
if row>0 then
valid = eliminate_in(valid,sixes[row],ch)
end if
end for
end for
--
-- 2) set elimination
--
for i=1 to length(nines) do
--
-- Practical note: Meticulously counting empties to eliminate larger set sizes
-- would at best reduce 6642 tests to 972, not deemed worth it.
--
for set_size=2 to 4 do
--if floor(count_empties(nines[i])/2)>=set_size then -- (untested)
valid = comb(nines[i],valid,set_size)
--end if
end for
end for
--
-- 3) x-wings
--
for ch='1' to '9' do
prevsets = repeat(0,9)
for x=1 to 9 do
count = 0
set = repeat(0,9)
for y=0 to 8 do
idx = y*9+x
if sequence(valid[idx]) and find(ch,valid[idx]) then
set[y+1] = 1
count += 1
end if
end for
if count=2 then
k = find(set,prevsets)
if k!=0 then
for y=0 to 8 do
if set[y+1]=1 then
for sx=1 to 9 do
if sx!=k and sx!=x then
valid = eliminate_in(valid,{y*9+sx},ch)
end if
end for
end if
end for
else
prevsets[x] = set
end if
end if
end for
prevsets = repeat(0,9)
for y=0 to 8 do
count = 0
set = repeat(0,9)
for x=1 to 9 do
idx = y*9+x
if sequence(valid[idx]) and find(ch,valid[idx]) then
set[x] = 1
count += 1
end if
end for
if count=2 then
k = find(set,prevsets)
if k!=0 then
for x=1 to 9 do
if set[x]=1 then
for sy=0 to 8 do
if sy+1!=k and sy!=y then
valid = eliminate_in(valid,{sy*9+x},ch)
end if
end for
end if
end for
else
prevsets[y+1] = set
end if
end if
end for
end for
return valid
end function
 
function permitted_in(string board, sequence sets, sequence valid, integer ch)
sequence set
integer pos, idx, bch
for i=1 to 9 do
set = nines[sets[i]]
pos = 0
for j=1 to 9 do
idx = set[j]
bch = board[idx]
if bch>'0' then
if bch=ch then pos = -1 exit end if
elsif find(ch,valid[idx]) then
if pos!=0 then pos = -1 exit end if
pos = idx
end if
end for
if pos>0 then
board[pos] = ch
improved = 1
end if
end for
return board
end function
 
enum INVALID = -1, INCOMPLETE = 0, SOLVED = 1, MULTIPLE = 2, BRUTE = 3
 
function plain_logic(string board)
--
-- Responsible for:
-- 1) cells with only one option
-- 2) numbers with only one home
--
integer solved
sequence valid = {}
object vi
 
while 1 do
solved = SOLVED
improved = 0
valid = deep_logic(board,valid)
 
-- 1) cells with only one option:
for i=1 to length(valid) do
vi = valid[i]
if string(vi) then
if length(vi)=0 then return {board,{},INVALID} end if
if length(vi)=1 then
board[i] = vi[1]
improved = 1
end if
end if
if board[i]<='0' then
solved = INCOMPLETE
end if
end for
if solved=SOLVED then return {board,{},SOLVED} end if
 
-- 2) numbers with only one home
for ch='1' to '9' do
board = permitted_in(board,cols,valid,ch)
board = permitted_in(board,rows,valid,ch)
board = permitted_in(board,squares,valid,ch)
end for
if not improved then exit end if
end while
return {board,valid,solved}
end function
 
function validate(string board)
-- (sum9 should be sufficient - if you want, get rid of nine/nines)
integer ch, sum9
sequence nine, nines = tagset(9)
 
for x=0 to 8 do -- columns
sum9 = 0
nine = repeat(0,9)
for y=1 to 81 by 9 do
ch = board[y+x]-'0'
if ch<1 or ch>9 then return 0 end if
sum9 += ch
nine[ch] = ch
end for
if sum9!=45 then return 0 end if
if nine!=nines then return 0 end if
end for
for y=1 to 81 by 9 do -- rows
sum9 = 0
nine = repeat(0,9)
for x=0 to 8 do
ch = board[y+x]-'0'
sum9 += ch
nine[ch] = ch
end for
if sum9!=45 then return 0 end if
if nine!=nines then return 0 end if
end for
for y=0 to 8 by 3 do -- small squares
for x=0 to 8 by 3 do
sum9 = 0
nine = repeat(0,9)
for sy=y*9 to y*9+18 by 9 do
for sx=x to x+2 do
ch = board[sy+sx+1]-'0'
sum9 += ch
nine[ch] = ch
end for
end for
if sum9!=45 then return 0 end if
if nine!=nines then return 0 end if
end for
end for
return 1
end function
 
function solve(string board, sequence valid={})
sequence solution, solutions
integer solved
integer minopt, mindx
object vi
{solution,valid,solved} = plain_logic(board)
if solved=INVALID then return {{},INVALID} end if
if solved=SOLVED then return {{solution},SOLVED} end if
if solved=BRUTE then return {{solution},BRUTE} end if
if solved!=INCOMPLETE then ?9/0 end if
-- find the cell with the fewest options:
-- (a possible improvement here would be to select the shortest
-- with the "most pairs" set, see swordfish etc above.)
minopt = 10
for i=1 to 9*9 do
vi = valid[i]
if string(vi) then
if length(vi)<=1 then ?9/0 end if -- should be caught above
if length(vi)<minopt then
minopt = length(vi)
mindx = i
end if
end if
end for
solutions = {}
for i=1 to minopt do
board[mindx] = valid[mindx][i]
{solution,solved} = solve(board,valid)
if solved=MULTIPLE then
return {solution,MULTIPLE}
elsif solved=SOLVED
or solved=BRUTE then
if not find(solution[1],solutions)
and validate(solution[1]) then
solutions = append(solutions,solution[1])
end if
if length(solutions)>1 then
return {solutions,MULTIPLE}
elsif length(solutions) then
return {solutions,BRUTE}
end if
end if
end for
if length(solutions)=1 then
return {solutions,BRUTE}
end if
return {{},INVALID}
end function
 
function test_one(string board)
sequence solutions
string solution, desc
integer solved
{solutions,solved} = solve(board)
if solved=SOLVED then
desc = "(logic)"
elsif solved=BRUTE then
desc = "(brute force)"
else
desc = "???" -- INVALID/INCOMPLETE/MULTIPLE
end if
if length(solutions)=0 then
solution = board
desc = "*** NO SOLUTIONS ***"
elsif length(solutions)=1 then
solution = solutions[1]
if not validate(solution) then
desc = "*** ERROR ***" -- (should never happen)
end if
else
solution = board
desc = "*** MULTIPLE SOLUTIONS ***"
end if
return {solution,desc}
end function
 
--NB Blank cells can be represented by any character <'1'. Spaces are not recommended since
-- they can all too easily be converted to tabs by copy/paste/save. In particular, ? and
-- _ are NOT valid characters for representing a blank square. Use any of .0-* instead.
 
constant tests = {
"..............3.85..1.2.......5.7.....4...1...9.......5......73..2.1........4...9", -- (0.01s, (logic))
-- row/col elimination (was 8s w/o logic first)
"000000036840000000000000020000203000010000700000600400000410050003000200600000000", -- (0.04s, (brute force))
".......39.....1..5..3.5.8....8.9...6.7...2...1..4.......9.8..5..2....6..4..7.....", -- (1.12s, (brute force))
"000037600000600090008000004090000001600000009300000040700000800010009000002540000", -- (0.00s, (logic))
"....839..1......3...4....7..42.3....6.......4....7..1..2........8...92.....25...6", -- (0.04s, (brute force))
"..1..5.7.92.6.......8...6...9..2.4.1.........3.4.8..9...7...3.......7.69.1.8..7..", -- (0.00s, (logic))
-- (the following takes ~8s when checking for multiple solutions)
"--3------4---8--36--8---1---4--6--73---9----------2--5--4-7--686--------7--6--5--", -- (0.01s, (brute force))
"..3.2.6..9..3.5..1..18.64....81.29..7.......8..67.82....26.95..8..2.3..9..5.1.3..", -- (0.00s, (logic))
"--4-5--6--6-1--8-93----7----8----5-----4-3-----6----7----2----61-5--4-3--2--7-1--", -- (0.00s, (logic))
-- x-wings
".4398.25.6..425...2....1.949....4.7.3..6.8...41.2.9..382.5.........4...553489.71.", -- (0.00s, (logic))
".9...4..7.....79..8........4.58.....3.......2.....97.6........4..35.....2..6...8.", -- (0.00s, (logic))
-- "AL Escargot", so-called "hardest sudoku"
"1....7.9..3..2...8..96..5....53..9...1..8...26....4...3......1..4......7..7...3..", -- (0.26s, (brute force))
"12.3....435....1....4........54..2..6...7.........8.9...31..5.......9.7.....6...8", -- (0.48s, (brute force))
"12.4..3..3...1..5...6...1..7...9.....4.6.3.....3..2...5...8.7....7.....5.......98", -- (1.07s, (brute force))
"394..267....3..4..5..69..2..45...9..6.......7..7...58..1..67..8..9..8....264..735", -- (0.00s, (logic))
"4......6.5...8.9..3....1....2.7....1.9.....4.8....3.5....2....7..6.5...8.1......6", -- (0.01s, (brute force))
"5...7....6..195....98....6.8...6...34..8.3..17...2...6.6....28....419..5....8..79", -- (0.00s, (logic))
"503600009010002600900000080000700005006804100200003000030000008004300050800006702", -- (0.00s, (logic))
"53..247....2...8..1..7.39.2..8.72.49.2.98..7.79.....8.....3.5.696..1.3...5.69..1.", -- (0.00s, (logic))
"530070000600195000098000060800060003400803001700020006060000280000419005000080079", -- (0.00s, (logic))
-- set exclusion
"75..9..46961...3524.....79.2..6.1..7.8.....2.1..328.65.........3.9...2.484..3..79", -- (0.00s, (logic))
-- Worlds hardest sudoku:
"800000000003600000070090200050007000000045700000100030001000068008500010090000400", -- (0.21s, (brute force))
"819--5-----2---75--371-4-6-4--59-1--7--3-8--2--3-62--7-5-7-921--64---9-----2--438", -- (0.00s, (logic))
"85...24..72......9..4.........1.7..23.5...9...4...........8..7..17..........36.4.", -- (0.01s, (logic))
"9..2..5...4..6..3...3.....6...9..2......5..8...7..4..37.....1...5..2..4...1..6..9", -- (0.17s, (brute force))
"97.3...6..6.75.........8.5.......67.....3.....539..2..7...25.....2.1...8.4...73..", -- (0.00s, (logic))
-- "the beast" (an earlier algorithm took 318s (5min 18s) on this):
"000060080020000000001000000070000102500030000000000400004201000300700600000000050", -- (0.03s, (brute force))
$},
 
lt = length(tests),
run_one_test = 0
 
constant l = " x x x | x x x | x x x ",
s = "-------+-------+-------",
l3 = join({l,l,l},"\n"),
fmt = substitute(join({l3,s,l3,s,l3},"\n"),"x","%c")&"\n"
 
procedure print_board(string board)
printf(1,fmt,board)
end procedure
 
procedure test()
string board -- (81 characters)
string solution, desc
atom t0 = time()
if run_one_test then
board = tests[run_one_test]
print_board(board)
{solution,desc} = test_one(board)
if length(solution)!=0 then
printf(1,"solution:\n")
print_board(solution)
end if
printf(1,"%s, %3.2fs\n",{desc,time()-t0})
else
for i=1 to lt do
atom t1 = time()
board = tests[i]
{solution,desc} = test_one(board)
printf(1," \"%s\", -- (%3.2fs, %s)\n",{board,time()-t1,desc})
-- printf(1," \"%s\", -- (%3.2fs, %s)\n",{solution,time()-t1,desc})
end for
t0 = time()-t0
printf(1,"%d puzzles solved in %3.2fs (av %3.2fs)\n",{lt,t0,t0/lt})
end if
end procedure
test()
Output:
    "..............3.85..1.2.......5.7.....4...1...9.......5......73..2.1........4...9",    -- (0.02s, (logic))
    "000000036840000000000000020000203000010000700000600400000410050003000200600000000",    -- (0.03s, (brute force))
    ".......39.....1..5..3.5.8....8.9...6.7...2...1..4.......9.8..5..2....6..4..7.....",    -- (1.31s, (brute force))
    "000037600000600090008000004090000001600000009300000040700000800010009000002540000",    -- (0.00s, (logic))
    "....839..1......3...4....7..42.3....6.......4....7..1..2........8...92.....25...6",    -- (0.03s, (brute force))
    "..1..5.7.92.6.......8...6...9..2.4.1.........3.4.8..9...7...3.......7.69.1.8..7..",    -- (0.00s, (logic))
    "--3------4---8--36--8---1---4--6--73---9----------2--5--4-7--686--------7--6--5--",    -- (0.03s, (brute force))
    "..3.2.6..9..3.5..1..18.64....81.29..7.......8..67.82....26.95..8..2.3..9..5.1.3..",    -- (0.00s, (logic))
    "--4-5--6--6-1--8-93----7----8----5-----4-3-----6----7----2----61-5--4-3--2--7-1--",    -- (0.01s, (logic))
    ".4398.25.6..425...2....1.949....4.7.3..6.8...41.2.9..382.5.........4...553489.71.",    -- (0.00s, (logic))
    ".9...4..7.....79..8........4.58.....3.......2.....97.6........4..35.....2..6...8.",    -- (0.00s, (logic))
    "1....7.9..3..2...8..96..5....53..9...1..8...26....4...3......1..4......7..7...3..",    -- (0.26s, (brute force))
    "12.3....435....1....4........54..2..6...7.........8.9...31..5.......9.7.....6...8",    -- (0.40s, (brute force))
    "12.4..3..3...1..5...6...1..7...9.....4.6.3.....3..2...5...8.7....7.....5.......98",    -- (1.12s, (brute force))
    "394..267....3..4..5..69..2..45...9..6.......7..7...58..1..67..8..9..8....264..735",    -- (0.00s, (logic))
    "4......6.5...8.9..3....1....2.7....1.9.....4.8....3.5....2....7..6.5...8.1......6",    -- (0.03s, (brute force))
    "5...7....6..195....98....6.8...6...34..8.3..17...2...6.6....28....419..5....8..79",    -- (0.00s, (logic))
    "503600009010002600900000080000700005006804100200003000030000008004300050800006702",    -- (0.01s, (logic))
    "53..247....2...8..1..7.39.2..8.72.49.2.98..7.79.....8.....3.5.696..1.3...5.69..1.",    -- (0.00s, (logic))
    "530070000600195000098000060800060003400803001700020006060000280000419005000080079",    -- (0.00s, (logic))
    "75..9..46961...3524.....79.2..6.1..7.8.....2.1..328.65.........3.9...2.484..3..79",    -- (0.01s, (logic))
    "800000000003600000070090200050007000000045700000100030001000068008500010090000400",    -- (0.21s, (brute force))
    "819--5-----2---75--371-4-6-4--59-1--7--3-8--2--3-62--7-5-7-921--64---9-----2--438",    -- (0.00s, (logic))
    "85...24..72......9..4.........1.7..23.5...9...4...........8..7..17..........36.4.",    -- (0.01s, (logic))
    "9..2..5...4..6..3...3.....6...9..2......5..8...7..4..37.....1...5..2..4...1..6..9",    -- (0.18s, (brute force))
    "97.3...6..6.75.........8.5.......67.....3.....539..2..7...25.....2.1...8.4...73..",    -- (0.00s, (logic))
    "000060080020000000001000000070000102500030000000000400004201000300700600000000050",    -- (0.01s, (brute force))
27 puzzles solved in 3.74s (av 0.14s)

Running the fuller version mentioned above:

    "008002000000600040064000092017005004200000008800100730470000910080001000000900200",    -- (0.05s, *** MULTIPLE SOLUTIONS ***)
338 puzzles solved in 36.20s (av 0.11s)
--or
338 puzzles solved in 16.46s (av 0.05s) (w/o check for multiple solutions)

Running a single puzzle (run_one_test set non-zero) produces:

 . . . | . . . | . . .
 . . . | . . 3 | . 8 5
 . . 1 | . 2 . | . . .
-------+-------+-------
 . . . | 5 . 7 | . . .
 . . 4 | . . . | 1 . .
 . 9 . | . . . | . . .
-------+-------+-------
 5 . . | . . . | . 7 3
 . . 2 | . 1 . | . . .
 . . . | . 4 . | . . 9
solution:
 9 8 7 | 6 5 4 | 3 2 1
 2 4 6 | 1 7 3 | 9 8 5
 3 5 1 | 9 2 8 | 7 4 6
-------+-------+-------
 1 2 8 | 5 3 7 | 6 9 4
 6 3 4 | 8 9 2 | 1 5 7
 7 9 5 | 4 6 1 | 8 3 2
-------+-------+-------
 5 1 9 | 2 8 6 | 4 7 3
 4 7 2 | 3 1 9 | 5 6 8
 8 6 3 | 7 4 5 | 2 1 9
(logic), 0.02s

PicoLisp[edit]

(load "lib/simul.l")
 
### Fields/Board ###
# val lst
 
(setq
*Board (grid 9 9)
*Fields (apply append *Board) )
 
# Init values to zero (empty)
(for L *Board
(for This L
(=: val 0) ) )
 
# Build lookup lists
(for (X . L) *Board
(for (Y . This) L
(=: lst
(make
(let A (* 3 (/ (dec X) 3))
(do 3
(inc 'A)
(let B (* 3 (/ (dec Y) 3))
(do 3
(inc 'B)
(unless (and (= A X) (= B Y))
(link
(prop (get *Board A B) 'val) ) ) ) ) ) )
(for Dir '(`west `east `south `north)
(for (This (Dir This) This (Dir This))
(unless (memq (:: val) (made))
(link (:: val)) ) ) ) ) ) ) )
 
# Cut connections (for display only)
(for (X . L) *Board
(for (Y . This) L
(when (member X (3 6))
(con (car (val This))) )
(when (member Y (4 7))
(set (cdr (val This))) ) ) )
 
# Display board
(de display ()
(disp *Board 0
'((This)
(if (=0 (: val))
" "
(pack " " (: val) " ") ) ) ) )
 
# Initialize board
(de main (Lst)
(for (Y . L) Lst
(for (X . N) L
(put *Board X (- 10 Y) 'val N) ) )
(display) )
 
# Find solution
(de go ()
(unless
(recur (*Fields)
(with (car *Fields)
(if (=0 (: val))
(loop
(NIL
(or
(assoc (inc (:: val)) (: lst))
(recurse (cdr *Fields)) ) )
(T (= 9 (: val)) (=: val 0)) )
(recurse (cdr *Fields)) ) ) )
(display) ) )
 
(main
(quote
(5 3 0 0 7 0 0 0 0)
(6 0 0 1 9 5 0 0 0)
(0 9 8 0 0 0 0 6 0)
(8 0 0 0 6 0 0 0 3)
(4 0 0 8 0 3 0 0 1)
(7 0 0 0 2 0 0 0 6)
(0 6 0 0 0 0 2 8 0)
(0 0 0 4 1 9 0 0 5)
(0 0 0 0 8 0 0 7 9) ) )
Output:
   +---+---+---+---+---+---+---+---+---+
 9 | 5   3     |     7     |           |
   +   +   +   +   +   +   +   +   +   +
 8 | 6         | 1   9   5 |           |
   +   +   +   +   +   +   +   +   +   +
 7 |     9   8 |           |     6     |
   +---+---+---+---+---+---+---+---+---+
 6 | 8         |     6     |         3 |
   +   +   +   +   +   +   +   +   +   +
 5 | 4         | 8       3 |         1 |
   +   +   +   +   +   +   +   +   +   +
 4 | 7         |     2     |         6 |
   +---+---+---+---+---+---+---+---+---+
 3 |     6     |           | 2   8     |
   +   +   +   +   +   +   +   +   +   +
 2 |           | 4   1   9 |         5 |
   +   +   +   +   +   +   +   +   +   +
 1 |           |     8     |     7   9 |
   +---+---+---+---+---+---+---+---+---+
     a   b   c   d   e   f   g   h   i
(go)
Output:
   +---+---+---+---+---+---+---+---+---+
 9 | 5   3   4 | 6   7   8 | 9   1   2 |
   +   +   +   +   +   +   +   +   +   +
 8 | 6   7   2 | 1   9   5 | 3   4   8 |
   +   +   +   +   +   +   +   +   +   +
 7 | 1   9   8 | 3   4   2 | 5   6   7 |
   +---+---+---+---+---+---+---+---+---+
 6 | 8   5   9 | 7   6   1 | 4   2   3 |
   +   +   +   +   +   +   +   +   +   +
 5 | 4   2   6 | 8   5   3 | 7   9   1 |
   +   +   +   +   +   +   +   +   +   +
 4 | 7   1   3 | 9   2   4 | 8   5   6 |
   +---+---+---+---+---+---+---+---+---+
 3 | 9   6   1 | 5   3   7 | 2   8   4 |
   +   +   +   +   +   +   +   +   +   +
 2 | 2   8   7 | 4   1   9 | 6   3   5 |
   +   +   +   +   +   +   +   +   +   +
 1 | 3   4   5 | 2   8   6 | 1   7   9 |
   +---+---+---+---+---+---+---+---+---+
     a   b   c   d   e   f   g   h   i

PL/I[edit]

Working PL/I version, derived from the Rosetta Fortran version.

sudoku: procedure options (main); /* 27 July 2014 */
 
declare grid (9,9) fixed (1) static initial (
0, 0, 3, 0, 2, 0, 6, 0, 0,
9, 0, 0, 3, 0, 5, 0, 0, 1,
0, 0, 1, 8, 0, 6, 4, 0, 0,
0, 0, 8, 1, 0, 2, 9, 0, 0,
7, 0, 0, 0, 0, 0, 0, 0, 8,
0, 0, 6, 7, 0, 8, 2, 0, 0,
0, 0, 2, 6, 0, 9, 5, 0, 0,
8, 0, 0, 2, 0, 3, 0, 0, 9,
0, 0, 5, 0, 1, 0, 3, 0, 0 );
 
declare grid_solved (9,9) fixed (1);
 
call print_sudoku (grid);
call solve (1, 1);
put skip (2);
call print_sudoku (grid_solved);
 
solve: procedure (i, j) recursive options (reorder);
declare (i, j) fixed binary;
declare (n, n_tmp) fixed binary;
 
if i > 9 then
grid_solved = grid;
else
do n = 1 to 9;
if is_safe (i, j, n) then
do;
n_tmp = grid (i, j);
grid (i, j) = n;
if j = 9 then
call solve (i + 1, 1);
else
call solve (i, j + 1);
grid (i, j) = n_tmp;
end;
end;
 
end solve;
 
is_safe: procedure (i, j, n) returns (bit(1) aligned) options (reorder);
declare (i, j, n) fixed binary;
declare (true value ('1'b), false value ('0'b) ) bit (1);
declare (i_min, j_min, ii, jj) fixed binary;
declare kk bit(1) aligned;
 
if grid (i, j) = n then return (true);
if grid (i, j) ^= 0 then return (false);
if any (grid (i, *) = n) then return (false);
if any (grid (*, j) = n) then return (false);
 
/* i_min and j_min are the co-ordinates of the top left-hand corner */
/* of 3 x 3 grid in which element (i,j) exists. */
i_min = 1 + 3 * trunc((i - 1) / 3);
j_min = 1 + 3 * trunc((j - 1) / 3);
 
begin;
declare sub_grid(3,3) fixed (1) defined grid(1sub+i_min-1,2sub+j_min-1);
 
kk = true;
if any(sub_grid = n) then kk = false;
end;
return (kk);
end is_safe;
 
print_sudoku: procedure (grid);
declare grid (*,*) fixed (1);
declare ( i, j, ii) fixed binary;
declare bar character (19) initial ( '+-----+-----+-----+' );
declare frame (9) character (1) initial (' ', ' ', '|', ' ', ' ', '|', ' ', ' ', '|' );
 
put skip list (bar);
do i = 1 to 7 by 3;
do ii = i to i + 2;
put skip edit ( '|', (grid (ii, j), frame(j) do j = 1 to 9) ) (a, f(1));
end;
put skip list (bar);
end;
end print_sudoku;
 
end sudoku;
 
Output:
+-----+-----+-----+ 
|0 0 3|0 2 0|6 0 0|
|9 0 0|3 0 5|0 0 1|
|0 0 1|8 0 6|4 0 0|
+-----+-----+-----+ 
|0 0 8|1 0 2|9 0 0|
|7 0 0|0 0 0|0 0 8|
|0 0 6|7 0 8|2 0 0|
+-----+-----+-----+ 
|0 0 2|6 0 9|5 0 0|
|8 0 0|2 0 3|0 0 9|
|0 0 5|0 1 0|3 0 0|
+-----+-----+-----+ 


+-----+-----+-----+ 
|4 8 3|9 2 1|6 5 7|
|9 6 7|3 4 5|8 2 1|
|2 5 1|8 7 6|4 9 3|
+-----+-----+-----+ 
|5 4 8|1 3 2|9 7 6|
|7 2 9|5 6 4|1 3 8|
|1 3 6|7 9 8|2 4 5|
+-----+-----+-----+ 
|3 7 2|6 8 9|5 1 4|
|8 1 4|2 5 3|7 6 9|
|6 9 5|4 1 7|3 8 2|
+-----+-----+-----+ 

Another PL/I version, reads sudoku from the text data file as 81 character record.

 
*PROCESS MARGINS(1,120) LIBS(SINGLE,STATIC);
*PROCESS OPTIMIZE(2) DFT(REORDER);
 
 
sudoku: proc(parms) options(main);
dcl parms char (100) var;
 
define alias bits bit (9) aligned;
dcl total (81) type bits;
dcl matrix (9, 9) type bits based(addr(total));
dcl box (9, 3, 3) type bits defined (total(trunc((1sub-1) /3) * 27 + mod(1sub-1, 3) * 3 + (2sub-1) * 9 + 3sub));
 
dcl posbit (0:9) type bits init('000000000'b, '100000000'b, '010000000'b, '001000000'b,
'000100000'b, '000010000'b, '000001000'b, '000000100'b,
'000000010'b, '000000001'b);
 
dcl (i, j, k) fixed bin(31);
dcl (start, finish) float(18);
dcl result fixed dec(5,3);
 
dcl buffer char(81);
dcl in file;
 
/* ON UNIT for the Sudoku data conversion */
on conversion
begin;
put skip
list('Sudoku data not valid.');
stop;
end;
 
/* ON UNIT to display info about the usage */
on undefinedfile(in)
begin;
put skip
list('Usage: ' || procedurename() || ' /filename');
stop;
end;
 
open file(in)
title ('/'||parms||',type(fixed), recsize(81)') record input;
 
/* Ignore the endfile condition */
on endfile(in);
 
/* Read the Sudoku data into buffer as one record */
read file(in) into(buffer);
close file(in);
 
/* Convert numbers -> position bit presentation and assign into the Sudoku board */
do k = 1 to 81;
total(k) = posbit(substr(buffer, k, 1));
end;
 
/* Start solving the Sudoku */
start = secs();
if solve() then
do;
finish = secs();
result = finish - start + 0.0005;
put skip list('Sudoku solved! Time: ' || trim(result) || ' seconds');
put skip(2);
 
/* display the solved Sudoku if solution exist */
do i = 1 to 9;
do j = 1 to 9;
put edit(trim(index(matrix(i, j), '1'b))) (a(3));
end;
put skip(2);
end;
end;
else put skip list('Impossible!');
 
 
/*************************************/
/* Simple backtracking sudoku solver */
/*************************************/
solve: proc recursive returns(bit(1));
dcl (i, j, k) fixed bin(31);
dcl result type bits;
 
/* find free cell */
do i = 1 to 9;
do j = 1 to 9;
if matrix(i, j) = posbit(0) then goto skip;
end;
end;
 
/* No more free cells. Check if the completed Sudoku is valid. */
/* Number in the cell is valid if the matching position bit is set. */
do i = 1 to 9;
do j = 1 to 9;
k = index(matrix(i, j), '1'b);
matrix(i, j) = posbit(0);
result = ^(any(matrix(i, *)) | any(matrix(*, j)) | any(box(numbox(i, j), *, *)));
if substr(result, k, 1) = '0'b then return('0'b);
matrix(i, j) = posbit(k);
end;
end;
 
return('1'b);
skip:
 
/* Go through and test possible values for the free cell untill the Sudoku is completed */
result = ^(any(matrix(i, *)) | any(matrix(*, j)) | any(box(numbox(i, j), *, *)));
k = 0;
do forever;
k = search(result, '1'b, k+1);
if k = 0 then leave;
matrix(i, j) = posbit(k);
if solve() then return('1'b);
else matrix(i, j) = posbit(0);
end;
 
return('0'b);
end solve;
 
 
/********************************************/
/* Returns box number for the sudoku coords */
/********************************************/
numbox: proc(i, j) returns(fixed bin(31));
dcl (i, j) fixed bin(31);
 
dcl lookup (9, 9) fixed bin(31) static init( (3)1, (3)2, (3)3,
(3)1, (3)2, (3)3,
(3)1, (3)2, (3)3,
(3)4, (3)5, (3)6,
(3)4, (3)5, (3)6,
(3)4, (3)5, (3)6,
(3)7, (3)8, (3)9,
(3)7, (3)8, (3)9,
(3)7, (3)8, (3)9 );
 
return(lookup(i, j));
end numbox;
 
end sudoku;
 

Prolog[edit]

:- use_module(library(clpfd)).
 
sudoku(Rows) :-
length(Rows, 9), maplist(length_(9), Rows),
append(Rows, Vs), Vs ins 1..9,
maplist(all_distinct, Rows),
transpose(Rows, Columns), maplist(all_distinct, Columns),
Rows = [A,B,C,D,E,F,G,H,I],
blocks(A, B, C), blocks(D, E, F), blocks(G, H, I).
 
length_(L, Ls) :- length(Ls, L).
 
blocks([], [], []).
blocks([A,B,C|Bs1], [D,E,F|Bs2], [G,H,I|Bs3]) :-
all_distinct([A,B,C,D,E,F,G,H,I]),
blocks(Bs1, Bs2, Bs3).
 
problem(1, [[_,_,_,_,_,_,_,_,_],
[_,_,_,_,_,3,_,8,5],
[_,_,1,_,2,_,_,_,_],
[_,_,_,5,_,7,_,_,_],
[_,_,4,_,_,_,1,_,_],
[_,9,_,_,_,_,_,_,_],
[5,_,_,_,_,_,_,7,3],
[_,_,2,_,1,_,_,_,_],
[_,_,_,_,4,_,_,_,9]]).

GNU Prolog version[edit]

Works with: GNU Prolog version 1.4.4
:- initialization(main).
 
 
solve(Rows) :-
maplist(domain_1_9, Rows)
, different(Rows)
, transpose(Rows,Cols), different(Cols)
, blocks(Rows,Blocks) , different(Blocks)
, maplist(fd_labeling, Rows)
.
 
domain_1_9(Rows) :- fd_domain(Rows,1,9).
different(Rows) :- maplist(fd_all_different, Rows).
 
blocks(Rows,Blocks) :-
maplist(split3,Rows,Xs), transpose(Xs,Ys)
, concat(Ys,Zs), concat_map(split3,Zs,Blocks)
. % where
split3([X,Y,Z|L],[[X,Y,Z]|R]) :- split3(L,R).
split3([],[]).
 
 
% utils/list
concat_map(F,Xs,Ys) :- call(F,Xs,Zs), maplist(concat,Zs,Ys).
 
concat([],[]).
concat([X|Xs],Ys) :- append(X,Zs,Ys), concat(Xs,Zs).
 
transpose([],[]).
transpose([[X]|Col], [[X|Row]]) :- transpose(Col,[Row]).
transpose([[X|Row]], [[X]|Col]) :- transpose([Row],Col).
transpose([[X|Row]|Xs], [[X|Col]|Ys]) :-
maplist(bind_head, Row, Ys, YX)
, maplist(bind_head, Col, Xs, XY)
, transpose(XY,YX)
. % where
bind_head(H,[H|T],T).
bind_head([],[],[]).
 
 
% tests
test([ [_,_,3,_,_,_,_,_,_]
, [4,_,_,_,8,_,_,3,6]
, [_,_,8,_,_,_,1,_,_]
, [_,4,_,_,6,_,_,7,3]
, [_,_,_,9,_,_,_,_,_]
, [_,_,_,_,_,2,_,_,5]
, [_,_,4,_,7,_,_,6,8]
, [6,_,_,_,_,_,_,_,_]
, [7,_,_,6,_,_,5,_,_]
]).
 
main :- test(T), solve(T), maplist(show,T), halt.
show(X) :- write(X), nl.
Output:
[1,2,3,4,5,6,7,8,9]
[4,5,7,1,8,9,2,3,6]
[9,6,8,3,2,7,1,5,4]
[2,4,9,5,6,1,8,7,3]
[5,7,6,9,3,8,4,1,2]
[8,3,1,7,4,2,6,9,5]
[3,1,4,2,7,5,9,6,8]
[6,9,5,8,1,4,3,2,7]
[7,8,2,6,9,3,5,4,1]

Runs in: time: 0.02 memory: 68352 (adapted for gprolog 1.3.1)

PureBasic[edit]

A brute force method is used, it seemed the fastest as well as the simplest.

DataSection
puzzle:
Data.s "394002670"
Data.s "000300400"
Data.s "500690020"
Data.s "045000900"
Data.s "600000007"
Data.s "007000580"
Data.s "010067008"
Data.s "009008000"
Data.s "026400735"
EndDataSection
 
#IsPossible = 0
#IsNotPossible = 1
#Unknown = 0
Global Dim sudoku(8, 8)
;-declarations
Declare readSudoku()
Declare displaySudoku()
Declare.s buildpossible(x, y, Array possible.b(1))
Declare solvePuzzle(x = 0, y = 0)
 
;-procedures
Procedure readSudoku()
Protected a$, row, column
 
Restore puzzle
For row = 0 To 8
Read.s a$
For column = 0 To 8
sudoku(column, row) = Val(Mid(a$, column + 1, 1))
Next
Next
EndProcedure
 
Procedure displaySudoku()
Protected row, column
Static border.s = "+-----+-----+-----+"
For row = 0 To 8
If row % 3 = 0: PrintN(border): EndIf
For column = 0 To 8
If column % 3 = 0: Print("|"): Else: Print(" "): EndIf
If sudoku(column, row): Print(Str(sudoku(column, row))): Else: Print("."): EndIf
Next
PrintN("|")
Next
PrintN(border)
EndProcedure
 
Procedure.s buildpossible(x, y, Array possible.b(1))
Protected index, column, row, boxColumn = (x / 3) * 3, boxRow = (y / 3) * 3
Dim possible.b(9)
 
For index = 0 To 8
possible(sudoku(index, y)) = #IsNotPossible ;record possibles in column
possible(sudoku(x, index)) = #IsNotPossible ;record possibles in row
Next
 
;record possibles in box
For row = boxRow To boxRow + 2
For column = boxColumn To boxColumn + 2
possible(sudoku(column, row)) = #IsNotPossible
Next
Next
EndProcedure
 
Procedure solvePuzzle(x = 0, y = 0)
Protected row, column, spot, digit
Dim possible.b(9)
 
For row = y To 8
For column = x To 8
If sudoku(column, row) = #Unknown
buildpossible(column, row, possible())
 
For digit = 1 To 9
If possible(digit) = #IsPossible
sudoku(column, row) = digit
spot = row * 9 + column + 1
If solvePuzzle(spot % 9, spot / 9)
Break 3
EndIf
EndIf
Next
 
If digit = 10
sudoku(column, row) = #Unknown
ProcedureReturn #False
EndIf
EndIf
Next
x = 0 ;reset column start point
Next
ProcedureReturn #True
EndProcedure
 
If OpenConsole()
readSudoku()
displaySudoku()
If solvePuzzle()
PrintN("Solved.")
displaySudoku()
Else
PrintN("Unable to solve puzzle") ;due to bad starting data
EndIf
 
Print(#CRLF$ + #CRLF$ + "Press ENTER to exit")
Input()
CloseConsole()
EndIf
Output:
+-----+-----+-----+
|3 9 4|. . 2|6 7 .|
|. . .|3 . .|4 . .|
|5 . .|6 9 .|. 2 .|
+-----+-----+-----+
|. 4 5|. . .|9 . .|
|6 . .|. . .|. . 7|
|. . 7|. . .|5 8 .|
+-----+-----+-----+
|. 1 .|. 6 7|. . 8|
|. . 9|. . 8|. . .|
|. 2 6|4 . .|7 3 5|
+-----+-----+-----+
Solved.
+-----+-----+-----+
|3 9 4|8 5 2|6 7 1|
|2 6 8|3 7 1|4 5 9|
|5 7 1|6 9 4|8 2 3|
+-----+-----+-----+
|1 4 5|7 8 3|9 6 2|
|6 8 2|9 4 5|3 1 7|
|9 3 7|1 2 6|5 8 4|
+-----+-----+-----+
|4 1 3|5 6 7|2 9 8|
|7 5 9|2 3 8|1 4 6|
|8 2 6|4 1 9|7 3 5|
+-----+-----+-----+

Python[edit]

See Solving Sudoku puzzles with Python for GPL'd solvers of increasing complexity of algorithm.

A simple backtrack algorithm -- Quick but may take longer if the grid had been more than 9 x 9

 
def initiate():
box.append([0, 1, 2, 9, 10, 11, 18, 19, 20])
box.append([3, 4, 5, 12, 13, 14, 21, 22, 23])
box.append([6, 7, 8, 15, 16, 17, 24, 25, 26])
box.append([27, 28, 29, 36, 37, 38, 45, 46, 47])
box.append([30, 31, 32, 39, 40, 41, 48, 49, 50])
box.append([33, 34, 35, 42, 43, 44, 51, 52, 53])
box.append([54, 55, 56, 63, 64, 65, 72, 73, 74])
box.append([57, 58, 59, 66, 67, 68, 75, 76, 77])
box.append([60, 61, 62, 69, 70, 71, 78, 79, 80])
for i in range(0, 81, 9):
row.append(range(i, i+9))
for i in range(9):
column.append(range(i, 80+i, 9))
 
def valid(n, pos):
current_row = pos/9
current_col = pos%9
current_box = (current_row/3)*3 + (current_col/3)
for i in row[current_row]:
if (grid[i] == n):
return False
for i in column[current_col]:
if (grid[i] == n):
return False
for i in box[current_box]:
if (grid[i] == n):
return False
return True
 
def solve():
i = 0
proceed = 1
while(i < 81):
if given[i]:
if proceed:
i += 1
else:
i -= 1
else:
n = grid[i]
prev = grid[i]
while(n < 9):
if (n < 9):
n += 1
if valid(n, i):
grid[i] = n
proceed = 1
break
if (grid[i] == prev):
grid[i] = 0
proceed = 0
if proceed:
i += 1
else:
i -=1
 
def inputs():
nextt = 'T'
number = 0
pos = 0
while(not(nextt == 'N' or nextt == 'n')):
print "Enter the position:",
pos = int(raw_input())
given[pos - 1] = True
print "Enter the numerical:",
number = int(raw_input())
grid[pos - 1] = number
print "Do you want to enter another given?(Y, for yes: N, for no)"
nextt = raw_input()
 
 
grid = [0]*81
given = [False]*81
box = []
row = []
column = []
initiate()
inputs()
solve()
for i in range(9):
print grid[i*9:i*9+9]
raw_input()
 

Racket[edit]

A Sudoku Solver in Racket.

Rascal[edit]

Sudoku.jpeg

A sudoku is represented as a matrix, see Rascal solutions to matrix related problems for examples.

import Prelude;
import vis::Figure;
import vis::Render;
 
public rel[int,int,int] sudoku(rel[int x, int y, int v] sudoku){
annotated= annotateGrid(sudoku);
solved = {<0,0,0,0,{0}>};
 
while(!isEmpty(solved)){
for (n <- [0 ..8]){
column = domainR(annotated, {n});
annotated -= column;
annotated += reduceOptions(column);
 
row = {<x,y,v,g,p> | <x,y,v,g,p> <- annotated, y==n};
annotated -= row;
annotated += reduceOptions(row);
 
grid1 = {<x,y,v,g,p> | <x,y,v,g,p> <- annotated, g==n};
annotated -= grid1;
annotated += reduceOptions(grid1);
}
 
solved = {<x,y,v,g,p> | <x,y,v,g,p> <- annotated, size(p)==1};
annotated -= solved;
annotated += {<x,y,getOneFrom(p),g,{*[1 .. 9]}> | <x,y,v,g,p> <- solved};
}
 
result = {<x,y,v> | <x,y,v,g,p> <- annotated};
return result;
}
 
 
//adds gridnumber and default set of options
public rel[int,int,int,int,set[int]] annotateGrid(rel[int x, int y, int v] sudoku){
result = {};
for (<x, y, v> <- sudoku){
g = 0;
if (x<3 && y<3) g = 0;
if (2<x && x<6 && y<3) g = 1;
if (x>5 && y<3) g = 2;
 
if (x<3 && 2<y && y<6) g = 3;
if (2<x && x<6 && 2<y && y<6) g = 4;
if (x>5 && 2<y && y<6) g = 5;
 
if (x<3 && y>5) g=6;
if (2<x && x<6 && y>5) g=7;
if (x>5 && y>5) g=8;
 
result += <x,y,v,g,{*[1 .. 9]}>;
}
return result;
}
 
//reduces set of options
public rel[int,int,int,int,set[int]] reduceOptions(rel[int x, int y, int v, int g, set[int] p] subSudoku){
solved = {<x,y,v,g,p> | <x,y,v,g,p> <- subSudoku, v!=0};
numbers = {*[1 .. 9]} - {v | <x,y,v,g,p> <- solved};
remaining = {<x,y,v,g,numbers&p> | <x,y,v,g,p> <- subSudoku-solved};
result = remaining + solved;
return result;
}
 
//a function to visualize the result
public void displaySudoku(rel[int x, int y, int v] sudoku){
points = [box(text("<v>"), align(0.111111*(x+1),0.111111*(y+1)),shrink(0.1)) | <x,y,v> <- sudoku];
print(points);
render(overlay([*points], aspectRatio(1.0)));
}
 
//a sudoku
public rel[int, int, int] sudokuA =
{
<0,0,3>, <1,0,9>, <2,0,4>, <3,0,0>, <4,0,0>, <5,0,2>, <6,0,6>, <7,0,7>, <8,0,0>,
<0,1,0>, <1,1,0>, <2,1,0>, <3,1,3>, <4,1,0>, <5,1,0>, <6,1,4>, <7,1,0>, <8,1,0>,
<0,2,5>, <1,2,0>, <2,2,0>, <3,2,6>, <4,2,9>, <5,2,0>, <6,2,0>, <7,2,2>, <8,2,0>,
<0,3,0>, <1,3,4>, <2,3,5>, <3,3,0>, <4,3,0>, <5,3,0>, <6,3,9>, <7,3,0>, <8,3,0>,
<0,4,6>, <1,4,0>, <2,4,0>, <3,4,0>, <4,4,0>, <5,4,0>, <6,4,0>, <7,4,0>, <8,4,7>,
<0,5,0>, <1,5,0>, <2,5,7>, <3,5,0>, <4,5,0>, <5,5,0>, <6,5,5>, <7,5,8>, <8,5,0>,
<0,6,0>, <1,6,1>, <2,6,0>, <3,6,0>, <4,6,6>, <5,6,7>, <6,6,0>, <7,6,0>, <8,6,8>,
<0,7,0>, <1,7,0>, <2,7,9>, <3,7,0>, <4,7,0>, <5,7,8>, <6,7,0>, <7,7,0>, <8,7,0>,
<0,8,0>, <1,8,2>, <2,8,6>, <3,8,4>, <4,8,0>, <5,8,0>, <6,8,7>, <7,8,3>, <8,8,5>
};

Example

rascal>displaySudoku(sudoku(sudokuA))

See picture

REXX[edit]

The   SUDOKU   REXX programs (and output) are included here ──► Sudoku/REXX.

RPN (HP-15c)[edit]

This is a back-tracking solver written in RPN for the HP-15C calculator. It is highly optimized for size, rather than speed, as the target platform only has 448 bytes of memory for code and data combined.

Latest version and usage notes kept at: [Sudoku Solver for the HP 15-C]

;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;       
; Register And Flag Usage        
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;       
;        0        General purpose variable used for miscelaneous purposes
;        1        Current index (0-80) in the pseudo-recursion 
;        2        Row (0-8) of current index
;        3        Column (0-8) of current index
;        4        Block # (0-8) of current index
;        5        Power of 10 of current column index
;        6        Value in the test solution at current index
;        7        Value of start clue at current index (0 if not set)
;        8 – 16   Starting row data
;        17 – 25  Current test solution
;        26 – 34  Flag matrix (bit set if digit used in a row/column/block)
;
;        Flag 2   Indicates that a digit has been used in cur row/column/block
;        Flag 3   Input to Subroutine B (whether to set or clear flags)
        
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;       
; setU(x)        
; Set/clear flag matrix values (show that x is used in a row/column/block)
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;       
LBL D        
        GSB 5       ; calc bit value we need to set/clear in existing row
        RCL 2       ; Get the current row index into x
        GSB B       ; set flag matrix value and calc new bit value for the column
        RCL 3       ; Get the current column index into x
        GSB B       ; set flag matrix values and calc new bit value for the block
        RCL 4       ; Get the current block index into x
        
; MUST IMMEDIATELY FOLLOW PRECEEDING SUBROUTINE        
; utility subroutine for setting flag matrix values         

LBL B        
        GSB 1       ; get the current flag matrix row at index x
        
        RCL 0       ; get temp register (holds the bit value we will be setting)
        F? 3        ; flag 3 indicates if we are setting or clearing the flag
        CHS         ; if we are clearing, we will do a subtraction instead
        +           ; set/clear the flag
        
        X<>Y        ; bring the row index back into x
        2           ; 26 is the starting register for the flag matrix
        6        
        GSB 3       ; set I so that we are ready to store the new value
        STO (i)     ; store the new value into the flag matrix
        RDN         ; get rid of the new value to restore the stack
        9           ; the next bit value will be 9 bits to the left
        +           ; set the next bit index
        GTO 5       ; calculate the value with that bit set
                    ; we GTO instead of GSB and it will do the RTN
        
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;       
; putA(x)        
; Set the value x into the current row/column in the trial solution.         
; Does it by subtracting the previous value and adding the new one.        
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;       
LBL 7        
        X<>6        ; swap new value with register that holds current value
        STO 0       ; store the old value in the temp register
        RCL 2       ; Get the current row index into x
        1           ; 17 is the starting register for the current trial solution
        7        
        GSB 3       ; Set the indirect register
        RCL (i)     ; Get the current value for the entire row
        RCL 6       ; Get the new value
        RCL- 0      ; subtract the old value from the new value
        RCL* 5      ; shift the power of 10 to the appropriate column
        +           ; add to the old value
        STO (i)     ; store the new row value from where we got it
        RTN        
        
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;       
; change(x)        
; Increments or decrements the current position in the trial solution.        
; Updates the registers containing the current row, column and block index,
; and the one with the power of 10 factor for the current column and others
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;       
LBL 6        
        STO+ 1      ; x holds +1 or -1; Register 1 is the current index
        
        RCL 1       ; get the current index (0 to 80)
        RCL 1       ; get the current index (0 to 80)
        9           ; integer divide by 9 to get the row index (0 to 8)
        /           ; no integer divide on 15c so do a floating point divide
        INT         ; use the INT operator to finish of the integer divide
        STO 2       ; register 2 contains the current row index
        
        9        
        *        
        -           ; col = index - 9 * row
        STO 3       ; register 3 contains the current column index
        
        3           ; calculate the block index from the row & column indexes
        /           ; TODO: save a couple of bytes in this section of code
        RCL 2        
        3        
        /        
        INT          
        3        
        *        
        +        
        STO 4       ; register 4 holds the block index
        
        8           ; now calculate the power of 10 of the current column
        RCL- 3      ; Get the digit (from right) based on the column
        10^X        ; calculate the exponent
        STO 5       ; save in register 5 which is used throughout the code
        
        RCL 2       ; get the current row
        1           ; 17 is the start register of the current trial solution
        7        
        GSB 4       ; extract the value at the current column
        STO 6       ; reg 6: the current trial value at the current row/column
        
        RCL 2       ; get the current row
        8           ; 8 is the start register of the input data from the user
        GSB 4       ; extract the value at the current column
        STO 7       ; reg 7: starting value at the current row/column (0 if none)
        RTN        
        
; Extract value at the current column from the matrix indirectly specified by x&y
LBL 4        
        GSB 3       ; set the indirect register based on x & y
        RCL (i)     ; get the row from the matrix passed in
        RCL / 5     ; shift the row to the right
        INT         ; trim off the digits shifted to the right of the decimal
        1           ; we will do a modulus 10 to extract the last digit
        0        
        /           ; do the equivalent of a mod 10
        FRAC                
        1
        0
        *
        RTN
        
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;       
;  main()        
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;       
LBL A        
        CF 2        ; make sure flag 2 is unset - CLR REG does not do this
        CF 3        ; make sure flag 3 is unset - CLR REG does not do this
        1           ; start with a index in register 1 of -1 (0 to 80)
        CHS         ; that way we can start with an increment operation
        STO 1       ; and actually start at 0 where we want.
        
LBL 2               ; set the flags to show the input values are set
        1           ; go forward one position at a time
        GSB 6       ; go to the next position in the trial solution
        
        RCL 7       ; get the starting input value at this row/col 
        GSB 7       ; set the value in the trial solution
        RCL 7       ; get starting input value because the last call destroyed it
        TEST 1      ; if > 0 then the user input a value for this row/col
        GSB D       ; set the flags to indicate this value is set
        
        8           ; 80 is the upper bound of the indexes (9x9 = 80 = 0:80)
        0        
        RCL 1       ; get the current index
        TEST 6      ; if the current index hasn't reached 80
        GTO 2       ; do the next value
        1           ; reset the starting value
        CHS         ; to -1 as we did at the beginning of the program
        STO 1       ; register 1 holds the current index
        
LBL E               ; main solution loop
        8           ; when we reach the last index (80) we are done
        0        
        RCL 1       ; register 1 holds the current index
        TEST 5      ; see if we are at the end
        RTN         ;  finished        ; woohoo - we are done!
        1           ; Go forward one spot
        GSB 6       ; Do the position increment
        RCL 7       ; get the starting input value at this row/col 
        TEST 1      ; if it's > 0, the user specified a value here
        GTO E       ; go forward, since this value was specified by the user
        GSB 7       ; Set the value in the trial solution
        
LBL 8          
        9           ; check the possible digits in order 1-9.
        RCL 6       ; Get the current trial solution value
        TEST 5      ; Check to see if it is 9
        GTO C       ; If it is, backup one step
        1           ; We weren't at 9 yet, so increment the value by 1
        +        
        GSB 7       ; Set the value in the trial solution
        
        RCL 6       ; Get the current trial solution value
        GSB 5       ; Calc 2^x-1 to get the bit mask
        CF 2        ; Clear the flag thats used as a return value
        RCL 2       ; Get the current row index into x
        GSB 9       ; see if the current value has already been used in the row
        F? 2        ; If number has been used in the block, try the next value
        GTO 8        
        RCL 3       ; Get the current column index into x
        GSB 9       ; see if current value has already been used in the column
        F? 2        ; If number has been used in the block, try the next value
        GTO 8        
        RCL 4       ; Get the current block index into x
        GSB 9       ; see if the current value has already been used in the block
        F? 2        ; If number has been used in the block, try the next value
        GTO 8        
        RCL 6       ; Get the current trial solution value
        GSB D       ; set the flags to indicate this value is set
        GTO E       ; move on to the next position in the puzzle
        
LBL C               ; Come here to back up to the previous position
        1           ; We will go one spot backwards
        CHS        
        GSB 6       ; Set the new current position and all temp values
        TEST 1      ; previous call leaves the starting value in X
        GTO C       ; if value is > 0, it was set, backup one more spot
        RCL 6       ; Get the current trial solution value
        
        SF 3        ; flag 3: clear the flag matrix bits, instead of setting them
        GSB D       ; Set/Clear the flag matrix bits
        CF 3        ; unset the 3 flag
        GTO 8       ; check the next digit
        
LBL 9        
        GSB 1       ; get the appropriate row (x) from the flag matrix
        RCL /  0    ; divide by the temp register - right shifts value
        INT        
        2           ; if bit is set, fractional part will be non 0 when / 2
        /        
        FRAC        
        TEST 1      ; if bit is set, set flag 2 which is used as a return value
        SF 2        
        RDN         ; move the stack down to prepare the caller for the next call
        RDN         ; move the stack down to prepare the caller for the next call
        9           ; bit flags for row/col/block are << by 9 from each other
        +           ; calculates the appropriate bit offset for the next call
        GTO 5       ; calc 2^x-1 to get the bit mask
                    ; do a GTO instead of GSB and it will return for us

;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;       
; setPow2(x)        
; Sets the utility temp register to 2^(x-1). Leaves x in place.        
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;       
LBL 5        
        STO 0       ; store the input X in the temp register
        1           ; we want to subtract 1 from the exponent
        -           ; calculate x-1
        2           ; set the base as 2
        X<>Y        ; the y^x function wants x and y reversed
        y^x         ; calculate the value
        X<>0        ; stuff result in temp register and restore the input x
        RTN        
        
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;       
; getPart(x)        
; Returns the integer representing the entire Xth row of the flag matrix        
; Row numbers start at 0.        
; returns value in x - input parameter x ends up in y        
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;       
LBL 1        
        ENTER        
        ENTER       ; duplicate the parameter so we can leave it for the caller
        2           ; 26 is the starting register for the flag matrix
        6        
        GSB 3       ; set the indirect register to the row specified by x
        RCL (i)     ; retrieve the entire row from the flag matrix
        RTN        
        
; Set the indirect register and remove the parameters from the stack
LBL 3        
        +           ; x+y is the memory offset we want
        STO I       ; put it in the indirect register
        RDN         ; get rid of the sum from the stack
        RTN        

Ruby[edit]

Example of a back-tracking solver, from wp:Algorithmics of sudoku

Works with: Ruby version 2.0+
def read_matrix(data)
lines = data.lines
9.times.collect { |i| 9.times.collect { |j| lines[i][j].to_i } }
end
 
def permissible(matrix, i, j)
ok = [nil, *1..9]
check = ->(x,y) { ok[matrix[x][y]] = nil if matrix[x][y].nonzero? }
# Same as another in the column isn't permissible...
9.times { |x| check[x, j] }
# Same as another in the row isn't permissible...
9.times { |y| check[i, y] }
# Same as another in the 3x3 block isn't permissible...
xary = [ *(x = (i / 3) * 3) .. x + 2 ] #=> [0,1,2], [3,4,5] or [6,7,8]
yary = [ *(y = (j / 3) * 3) .. y + 2 ]
xary.product(yary).each { |x, y| check[x, y] }
# Gathering only permitted one
ok.compact
end
 
def deep_copy_sudoku(matrix)
matrix.collect { |row| row.dup }
end
 
def solve_sudoku(matrix)
loop do
options = []
9.times do |i|
9.times do |j|
next if matrix[i][j].nonzero?
p = permissible(matrix, i, j)
# If nothing is permissible, there is no solution at this level.
return if p.empty? # return nil
options << [i, j, p]
end
end
# If the matrix is complete, we have a solution...
return matrix if options.empty?
 
i, j, permissible = options.min_by { |x| x.last.length }
 
# If there is an option with only one solution, set it and re-check permissibility
if permissible.length == 1
matrix[i][j] = permissible[0]
next
end
 
# We have two or more choices. We need to search both...
permissible.each do |v|
mtmp = deep_copy_sudoku(matrix)
mtmp[i][j] = v
ret = solve_sudoku(mtmp)
return ret if ret
end
 
# We did an exhaustive search on this branch and nothing worked out.
return
end
end
 
def print_matrix(matrix)
puts "Impossible" or return unless matrix
 
border = "+-----+-----+-----+"
9.times do |i|
puts border if i%3 == 0
9.times do |j|
print j%3 == 0 ? "|" : " "
print matrix[i][j] == 0 ? "." : matrix[i][j]
end
puts "|"
end
puts border
end
 
data = <<EOS
394__267_
___3__4__
5__69__2_
_45___9__
6_______7
__7___58_
_1__67__8
__9__8___
_264__735
EOS

 
matrix = read_matrix(data)
print_matrix(matrix)
puts
print_matrix(solve_sudoku(matrix))
Output:
+-----+-----+-----+
|3 9 4|. . 2|6 7 .|
|. . .|3 . .|4 . .|
|5 . .|6 9 .|. 2 .|
+-----+-----+-----+
|. 4 5|. . .|9 . .|
|6 . .|. . .|. . 7|
|. . 7|. . .|5 8 .|
+-----+-----+-----+
|. 1 .|. 6 7|. . 8|
|. . 9|. . 8|. . .|
|. 2 6|4 . .|7 3 5|
+-----+-----+-----+

+-----+-----+-----+
|3 9 4|8 5 2|6 7 1|
|2 6 8|3 7 1|4 5 9|
|5 7 1|6 9 4|8 2 3|
+-----+-----+-----+
|1 4 5|7 8 3|9 6 2|
|6 8 2|9 4 5|3 1 7|
|9 3 7|1 2 6|5 8 4|
+-----+-----+-----+
|4 1 3|5 6 7|2 9 8|
|7 5 9|2 3 8|1 4 6|
|8 2 6|4 1 9|7 3 5|
+-----+-----+-----+

SAS[edit]

Use CLP solver in SAS/OR:

/* define SAS data set */
data Indata;
input C1-C9;
datalines;
. . 5 . . 7 . . 1
. 7 . . 9 . . 3 .
. . . 6 . . . . .
. . 3 . . 1 . . 5
. 9 . . 8 . . 2 .
1 . . 2 . . 4 . .
. . 2 . . 6 . . 9
. . . . 4 . . 8 .
8 . . 1 . . 5 . .
;
 
/* call OPTMODEL procedure in SAS/OR */
proc optmodel;
/* declare variables */
set ROWS = 1..9;
set COLS = ROWS;
var X {ROWS, COLS} >= 1 <= 9 integer;
 
/* declare nine row constraints */
con RowCon {i in ROWS}:
alldiff({j in COLS} X[i,j]);
 
/* declare nine column constraints */
con ColCon {j in COLS}:
alldiff({i in ROWS} X[i,j]);
 
/* declare nine 3x3 block constraints */
con BlockCon {s in 0..2, t in 0..2}:
alldiff({i in 3*s+1..3*s+3, j in 3*t+1..3*t+3} X[i,j]);
 
/* fix variables to cell values */
/* X[i,j] = c[i,j] if c[i,j] is not missing */
num c {ROWS, COLS};
read data indata into [_N_] {j in COLS} <c[_N_,j]=col('C'||j)>;
for {i in ROWS, j in COLS: c[i,j] ne .}
fix X[i,j] = c[i,j];
 
/* call CLP solver */
solve;
 
/* print solution */
print X;
quit;

Output:

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

Scala[edit]

I use the following slightly modified code for creating new sudokus and it seems to me usable for solving given sudokus. It doesn't look like elegant and functional programming - so what! it works! This solver works with normally 9x9 sudokus as well as with sudokus of jigsaw type or sudokus with additional condition like diagonal constraint.

Works with: Scala version 2.9.1
object SudokuSolver extends App {
 
class Solver {
 
var solution = new Array[Int](81) //listOfFields toArray
 
val fp2m: Int => Tuple2[Int,Int] = pos => Pair(pos/9+1,pos%9+1) //get row, col from array position
val setAll = (1 to 9) toSet //all possibilities
 
val arrayGroups = new Array[List[List[Int]]](81)
val sv: Int => Int = (row: Int) => (row-1)*9 //start value group row
val ev: Int => Int = (row: Int) => sv(row)+8 //end value group row
val fgc: (Int,Int) => Int = (i,col) => i*9+col-1 //get group col
val fgs: Int => (Int,Int) = p => Pair(p, p/(27)*3+p%9/3) //get group square box
for (pos <- 0 to 80) {
val (row,col) = fp2m(pos)
val gRow = (sv(row) to ev(row)).toList
val gCol = ((0 to 8) toList) map (fgc(_,col))
val gSquare = (0 to 80 toList) map fgs filter (_._2==(fgs(pos))._2) map (_._1)
arrayGroups(pos) = List(gRow,gCol,gSquare)
}
val listGroups = arrayGroups toList
 
val fpv4s: (Int) => List[Int] = pos => { //get possible values for solving
val setRow = (listGroups(pos)(0) map (solution(_))).toSet
val setCol = listGroups(pos)(1).map(solution(_)).toSet
val setSquare = listGroups(pos)(2).map(solution(_)).toSet
val setG = setRow++setCol++setSquare--Set(0)
val setPossible = setAll--setG
setPossible.toList.sortWith(_<_)
}
 
 
//solve the riddle: Nil ==> solution does not exist
def solve(listOfFields: List[Int]): List[Int] = {
solution = listOfFields toArray
 
def checkSol(uncheckedSol: List[Int]): List[Int] = {
if (uncheckedSol == Nil) return Nil
solution = uncheckedSol toArray
val check = (0 to 80).map(fpv4s(_)).filter(_.size>0)
if (check == Nil) return uncheckedSol
return Nil
}
 
val f1: Int => Pair[Int,Int] = p => Pair(p,listOfFields(p))
val numFields = (0 to 80 toList) map f1 filter (_._2==0)
val iter = numFields map ((_: (Int,Int))._1)
var p_iter = 0
 
val first: () => Int = () => {
val ret = numFields match {
case Nil => -1
case _ => numFields(0)._1
}
ret
}
 
val last: () => Int = () => {
val ret = numFields match {
case Nil => -1
case _ => numFields(numFields.size-1)._1
}
ret
}
 
val hasPrev: () => Boolean = () => p_iter > 0
val prev: () => Int = () => {p_iter -= 1; iter(p_iter)}
val hasNext: () => Boolean = () => p_iter < iter.size-1
val next: () => Int = () => {p_iter += 1; iter(p_iter)}
val fixed: Int => Boolean = pos => listOfFields(pos) != 0
val possiArray = new Array[List[Int]](numFields.size)
val firstUF = first() //first unfixed
if (firstUF < 0) return checkSol(solution.toList) //that is it!
var pif = iter(p_iter) //pos in fields
val lastUF = last() //last unfixed
val (row,col) = fp2m(pif)
possiArray(p_iter) = fpv4s(pif).toList.sortWith(_<_)
 
while(pif <= lastUF) {
val (row,col) = fp2m(pif)
if (possiArray(p_iter) == null) possiArray(p_iter) = fpv4s(pif).toList.sortWith(_<_)
val possis = possiArray(p_iter)
if (possis.isEmpty) {
if (hasPrev()) {
possiArray(p_iter) = null
solution(pif) = 0
pif = prev()
} else {
return Nil
}
} else {
solution(pif) = possis(0)
possiArray(p_iter) = (possis.toSet - possis(0)).toList.sortWith(_<_)
if (hasNext()) {
pif = next()
} else {
return checkSol(solution.toList)
}
}
}
checkSol(solution.toList)
}
}
 
val f2Str: List[Int] => String = fields => {
val sepLine = "+---+---+---+"
val sepPoints = Set(2,5,8)
val fs: (Int, Int) => String = (i, v) => v.toString.replace("0"," ")+(if (sepPoints.contains(i%9)) "|" else "")
sepLine+"\n"+(0 to fields.size-1).map(i => (if (i%9==0) "|" else "")+fs(i,fields(i))+(if (i%9==8) if (sepPoints.contains(i/9)) "\n"+sepLine+"\n" else "\n" else "")).foldRight("")(_+_)
}
 
val solver = new Solver()
 
val riddle = List(3,9,4,0,0,2,6,7,0,
0,0,0,3,0,0,4,0,0,
5,0,0,6,9,0,0,2,0,
0,4,5,0,0,0,9,0,0,
6,0,0,0,0,0,0,0,7,
0,0,7,0,0,0,5,8,0,
0,1,0,0,6,7,0,0,8,
0,0,9,0,0,8,0,0,0,
0,2,6,4,0,0,7,3,5)
 
println("riddle:")
println(f2Str(riddle))
var solution = solver.solve(riddle)
 
println("solution:")
println(solution match {case Nil => "no solution!!!" case _ => f2Str(solution)})
 
}
Output:
riddle:
+---+---+---+
|394|  2|67 |
|   |3  |4  |
|5  |69 | 2 |
+---+---+---+
| 45|   |9  |
|6  |   |  7|
|  7|   |58 |
+---+---+---+
| 1 | 67|  8|
|  9|  8|   |
| 26|4  |735|
+---+---+---+

solution:
+---+---+---+
|394|852|671|
|268|371|459|
|571|694|823|
+---+---+---+
|145|783|962|
|682|945|317|
|937|126|584|
+---+---+---+
|413|567|298|
|759|238|146|
|826|419|735|
+---+---+---+

The implementation above doesn't work so effective for sudokus like Bracmat version, therefore I implemented a second version inspired by Java section:

Works with: Scala version 2.9.1
object SudokuSolver extends App {
 
object Solver {
var solution = new Array[Int](81)
 
val fap: (Int, Int) => Int = (row, col) => (row)*9+col //function array position
 
def solve(listOfFields: List[Int]): List[Int] = {
solution = listOfFields toArray
 
val mRowSubset = new Array[Boolean](81)
val mColSubset = new Array[Boolean](81)
val mBoxSubset = new Array[Boolean](81)
 
def initSubsets: Unit = {
for (row <- 0 to 8) {
for (col <- 0 to 8) {
val value = solution(fap(row, col))
if (value != 0)
setSubsetValue(row, col, value, true)
}
}
}
 
def setSubsetValue(r: Int, c: Int, value: Int, present: Boolean): Unit = {
mRowSubset(fap(r, value - 1)) = present
mColSubset(fap(c, value - 1)) = present
mBoxSubset(fap(computeBoxNo(r, c), value - 1)) = present
}
 
def computeBoxNo(r: Int, c: Int): Int = {
val boxRow = r / 3
val boxCol = c / 3
return boxRow * 3 + boxCol
}
 
def isValid(r: Int, c: Int, value: Int): Boolean = {
val vVal = value - 1
val isPresent = mRowSubset(fap(r, vVal)) || mColSubset(fap(c, vVal)) || mBoxSubset(fap(computeBoxNo(r, c), vVal))
return !isPresent
}
 
def solve(row: Int, col: Int): Boolean = {
var r = row
var c = col
 
if (r == 9) {
r = 0
c += 1
if (c == 9)
return true
}
 
if(solution(fap(r,c)) != 0)
return solve(r+1,c)
for(value <- 1 to 9)
if(isValid(r, c, value)) {
solution(fap(r,c)) = value
setSubsetValue(r, c, value, true)
if(solve(r+1,c))
return true
setSubsetValue(r, c, value, false)
}
solution(fap(r,c)) = 0
return false
}
 
def checkSol: Boolean = {
initSubsets
if ((mRowSubset.exists(_==false)) || (mColSubset.exists(_==false)) || (mBoxSubset.exists(_==false))) return false
true
}
 
initSubsets
val ret = solve(0,0)
if (ret)
if (checkSol) return solution.toList else Nil
else
return Nil
}
}
 
val f2Str: List[Int] => String = fields => {
val f2Stri: List[Int] => String = fields => {
val sepLine = "+---+---+---+"
val sepPoints = Set(2,5,8)
val fs: (Int, Int) => String = (i, v) => v.toString.replace("0"," ")+(if (sepPoints.contains(i%9)) "|" else "")
val s = sepLine+"\n"+(0 to fields.size-1).map(i => (if (i%9==0) "|" else "")+fs(i,fields(i))+(if (i%9==8) if (sepPoints.contains(i/9)) "\n"+sepLine+"\n" else "\n" else "")).foldRight("")(_+_)
s
}
val s = fields match {case Nil => "no solution!!!" case _ => f2Stri(fields)}
s
}
 
val elapsedtime: (=> Unit) => Long = f => {val s = System.currentTimeMillis; f; (System.currentTimeMillis - s)/1000}
 
var sol = List[Int]()
 
val sudokus = List(
("riddle used in Ada section:",
"394..267....3..4..5..69..2..45...9..6.......7..7...58..1..67..8..9..8....264..735"),
("riddle used in Bracmat section:",
"..............3.85..1.2.......5.7.....4...1...9.......5......73..2.1........4...9"),
("riddle from Groovy section: 4th exceptionally difficult example in Wikipedia: ~80 seconds",
"..3......4...8..36..8...1...4..6..73...9..........2..5..4.7..686........7..6..5.."),
("riddle used in Ada section with incorrect modifactions - it should fail:",
"3943.267....3..4..5..69..2..45...9..6.......7..7...58..1..67..8..9..8....264..735"),
("riddle constructed with mess - it should fail too:",
"123456789456789123789123456.45..89..6.......72.7...58.31..67..8..9..8....264..735"))
 
for (sudoku <- sudokus) {
val desc = sudoku._1
val riddle = sudoku._2.replace(".","0").toList.map(_.toString.toInt)
println(desc+"\n"+f2Str(riddle)+"\n"
+"elapsed time: "+elapsedtime(sol = Solver.solve(riddle))+" sec"+"\n"+"solution:"+"\n"+f2Str(sol)
+("\n"*2))
}
}
Output:
riddle used in Ada section:
+---+---+---+
|394|  2|67 |
|   |3  |4  |
|5  |69 | 2 |
+---+---+---+
| 45|   |9  |
|6  |   |  7|
|  7|   |58 |
+---+---+---+
| 1 | 67|  8|
|  9|  8|   |
| 26|4  |735|
+---+---+---+
elapsed time: 0 sec
solution:
+---+---+---+
|394|852|671|
|268|371|459|
|571|694|823|
+---+---+---+
|145|783|962|
|682|945|317|
|937|126|584|
+---+---+---+
|413|567|298|
|759|238|146|
|826|419|735|
+---+---+---+

riddle used in Bracmat section:
+---+---+---+
|   |   |   |
|   |  3| 85|
|  1| 2 |   |
+---+---+---+
|   |5 7|   |
|  4|   |1  |
| 9 |   |   |
+---+---+---+
|5  |   | 73|
|  2| 1 |   |
|   | 4 |  9|
+---+---+---+
elapsed time: 43 sec
solution:
+---+---+---+
|987|654|321|
|246|173|985|
|351|928|746|
+---+---+---+
|128|537|694|
|634|892|157|
|795|461|832|
+---+---+---+
|519|286|473|
|472|319|568|
|863|745|219|
+---+---+---+

riddle from Groovy section: 4th exceptionally difficult example in Wikipedia: ~80 seconds
+---+---+---+
|  3|   |   |
|4  | 8 | 36|
|  8|   |1  |
+---+---+---+
| 4 | 6 | 73|
|   |9  |   |
|   |  2|  5|
+---+---+---+
|  4| 7 | 68|
|6  |   |   |
|7  |6  |5  |
+---+---+---+
elapsed time: 3 sec
solution:
+---+---+---+
|123|456|789|
|457|189|236|
|968|327|154|
+---+---+---+
|249|561|873|
|576|938|412|
|831|742|695|
+---+---+---+
|314|275|968|
|695|814|327|
|782|693|541|
+---+---+---+

riddle used in Ada section with incorrect modifactions - it should fail:
+---+---+---+
|394|3 2|67 |
|   |3  |4  |
|5  |69 | 2 |
+---+---+---+
| 45|   |9  |
|6  |   |  7|
|  7|   |58 |
+---+---+---+
| 1 | 67|  8|
|  9|  8|   |
| 26|4  |735|
+---+---+---+
elapsed time: 0 sec
solution:
no solution!!!

riddle constructed with mess - it should fail too:
+---+---+---+
|123|456|789|
|456|789|123|
|789|123|456|
+---+---+---+
| 45|  8|9  |
|6  |   |  7|
|2 7|   |58 |
+---+---+---+
|31 | 67|  8|
|  9|  8|   |
| 26|4  |735|
+---+---+---+
elapsed time: 0 sec
solution:
no solution!!!

Sidef[edit]

Translation of: Perl 6
func _check(i, j) is cached {
var (id, im) = i.divmod(9);
var (jd, jm) = j.divmod(9);
 
jd == id && return true;
jm == im && return true;
 
var id2 = id//3;
var jd2 = jd//3;
 
jd2 == id2 || return false;
 
jm//3 == im//3
}
 
func solve(board) {
board.range.each { |i|
board[i] && next;
var t = board.items(
board.range.grep {|j|
_check(i, j)
}...
);
 
{ |k|
t.contains(k) && next;
board[i] = k;
solve(board);
} * 9;
 
board[i] = 0;
return;
}
 
for i in ^board {
print "#{board[i]} ";
print " " if (3 -> divides(i+1));
print "\n" if (9 -> divides(i+1));
print "\n" if (27 -> divides(i+1));
}
}
 
var board = %i(
5 3 0 0 2 4 7 0 0
0 0 2 0 0 0 8 0 0
1 0 0 7 0 3 9 0 2
 
0 0 8 0 7 2 0 4 9
0 2 0 9 8 0 0 7 0
7 9 0 0 0 0 0 8 0
 
0 0 0 0 3 0 5 0 6
9 6 0 0 1 0 3 0 0
0 5 0 6 9 0 0 1 0
);
 
solve(board);
Output:
5 3 9  8 2 4  7 6 1  
6 7 2  1 5 9  8 3 4  
1 8 4  7 6 3  9 5 2  

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

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

Swift[edit]

Translation of: Java
import Foundation
 
typealias SodukuPuzzle = [[Int]]
 
class Soduku {
let mBoardSize:Int!
let mBoxSize:Int!
var mBoard:SodukuPuzzle!
var mRowSubset:[[Bool]]!
var mColSubset:[[Bool]]!
var mBoxSubset:[[Bool]]!
 
init(board:SodukuPuzzle) {
mBoard = board
mBoardSize = board.count
mBoxSize = Int(sqrt(Double(mBoardSize)))
mRowSubset = [[Bool]](count: mBoardSize, repeatedValue: [Bool](count: mBoardSize, repeatedValue: false))
mColSubset = [[Bool]](count: mBoardSize, repeatedValue: [Bool](count: mBoardSize, repeatedValue: false))
mBoxSubset = [[Bool]](count: mBoardSize, repeatedValue: [Bool](count: mBoardSize, repeatedValue: false))
initSubsets()
}
 
func computeBoxNo(i:Int, _ j:Int) -> Int {
let boxRow = i / mBoxSize
let boxCol = j / mBoxSize
 
return boxRow * mBoxSize + boxCol
}
 
func initSubsets() {
for i in 0..<mBoard.count {
for j in 0..<mBoard.count {
let value = mBoard[i][j]
 
if value != 0 {
setSubsetValue(i, j, value, true);
}
}
}
}
 
func isValid(i:Int, _ j:Int, var _ val:Int) -> Bool {
val--
let isPresent = mRowSubset[i][val] || mColSubset[j][val] || mBoxSubset[computeBoxNo(i, j)][val]
return !isPresent
}
 
func printBoard() {
for i in 0..<mBoardSize {
if i % mBoxSize == 0 {
println(" -----------------------")
}
 
for j in 0..<mBoardSize {
if j % mBoxSize == 0 {
print("| ")
}
 
print(mBoard[i][j] != 0 ? String(mBoard[i][j]) : " ")
print(" ")
}
 
println("|")
}
 
println(" -----------------------")
}
 
func setSubsetValue(i:Int, _ j:Int, _ value:Int, _ present:Bool) {
mRowSubset[i][value - 1] = present
mColSubset[j][value - 1] = present
mBoxSubset[computeBoxNo(i, j)][value - 1] = present
}
 
func solve() {
solve(0, 0)
}
 
func solve(var i:Int, var _ j:Int) -> Bool {
if i == mBoardSize {
i = 0
j++
if j == mBoardSize {
return true
}
}
 
if mBoard[i][j] != 0 {
return solve(i + 1, j)
}
 
for value in 1...mBoardSize {
if isValid(i, j, value) {
mBoard[i][j] = value
setSubsetValue(i, j, value, true)
 
if solve(i + 1, j) {
return true
}
 
setSubsetValue(i, j, value, false)
}
}
 
mBoard[i][j] = 0
return false
}
}
 
let board = [
[4, 0, 0, 0, 0, 0, 0, 6, 0],
[5, 0, 0, 0, 8, 0, 9, 0, 0],
[3, 0, 0, 0, 0, 1, 0, 0, 0],
 
[0, 2, 0, 7, 0, 0, 0, 0, 1],
[0, 9, 0, 0, 0, 0, 0, 4, 0],
[8, 0, 0, 0, 0, 3, 0, 5, 0],
[0, 0, 0, 2, 0, 0, 0, 0, 7],
[0, 0, 6, 0, 5, 0, 0, 0, 8],
[0, 1, 0, 0, 0, 0, 0, 0, 6]
]
 
let puzzle = Soduku(board: board)
puzzle.solve()
puzzle.printBoard()
Output:
 -----------------------
| 4 8 2 | 9 7 5 | 1 6 3 |
| 5 6 1 | 3 8 2 | 9 7 4 |
| 3 7 9 | 6 4 1 | 8 2 5 |
 -----------------------
| 6 2 5 | 7 9 4 | 3 8 1 |
| 1 9 3 | 5 6 8 | 7 4 2 |
| 8 4 7 | 1 2 3 | 6 5 9 |
 -----------------------
| 9 5 8 | 2 1 6 | 4 3 7 |
| 7 3 6 | 4 5 9 | 2 1 8 |
| 2 1 4 | 8 3 7 | 5 9 6 |
 -----------------------

Tcl[edit]

Adapted from a page on the Tcler's Wiki to use a standard object system.

Note that you can implement more rules if you want. Just make another subclass of Rule and the solver will pick it up and use it automatically.

Works with: Tcl version 8.6
or
Library: TclOO
package require Tcl 8.6
oo::class create Sudoku {
variable idata
 
method clear {} {
for {set y 0} {$y < 9} {incr y} {
for {set x 0} {$x < 9} {incr x} {
my set $x $y {}
}
}
}
method load {data} {
set error "data must be a 9-element list, each element also being a\
list of 9 numbers from 1 to 9 or blank or an @ symbol."

if {[llength $data] != 9} {
error $error
}
for {set y 0} {$y<9} {incr y} {
set row [lindex $data $y]
if {[llength $row] != 9} {
error $error
}
for {set x 0} {$x<9} {incr x} {
set d [lindex $row $x]
if {![regexp {^[@1-9]?$} $d]} {
error $d-$error
}
if {$d eq "@"} {set d ""}
my set $x $y $d
}
}
}
method dump {} {
set rows {}
for {set y 0} {$y < 9} {incr y} {
lappend rows [my getRow 0 $y]
}
return $rows
}
 
method Log msg {
# Chance to print message
}
 
method set {x y value} {
if {[catch {set value [format %d $value]}]} {set value 0}
if {$value<1 || $value>9} {
set idata(sq$x$y) {}
} else {
set idata(sq$x$y) $value
}
}
method get {x y} {
if {![info exists idata(sq$x$y)]} {
return {}
}
return $idata(sq$x$y)
}
 
method getRow {x y} {
set row {}
for {set x 0} {$x<9} {incr x} {
lappend row [my get $x $y]
}
return $row
}
method getCol {x y} {
set col {}
for {set y 0} {$y<9} {incr y} {
lappend col [my get $x $y]
}
return $col
}
method getRegion {x y} {
set xR [expr {($x/3)*3}]
set yR [expr {($y/3)*3}]
set regn {}
for {set x $xR} {$x < $xR+3} {incr x} {
for {set y $yR} {$y < $yR+3} {incr y} {
lappend regn [my get $x $y]
}
}
return $regn
}
}
 
# SudokuSolver inherits from Sudoku, and adds the ability to filter
# possibilities for a square by looking at all the squares in the row, column,
# and region that the square is a part of. The method 'solve' contains a list
# of rule-objects to use, and iterates over each square on the board, applying
# each rule sequentially until the square is allocated.
 
oo::class create SudokuSolver {
superclass Sudoku
method validchoices {x y} {
if {[my get $x $y] ne {}} {
return [my get $x $y]
}
 
set row [my getRow $x $y]
set col [my getCol $x $y]
set regn [my getRegion $x $y]
set eliminate [list {*}$row {*}$col {*}$regn]
set eliminate [lsearch -all -inline -not $eliminate {}]
set eliminate [lsort -unique $eliminate]
 
set choices {}
for {set c 1} {$c < 10} {incr c} {
if {$c ni $eliminate} {
lappend choices $c
}
}
if {[llength $choices]==0} {
error "No choices left for square $x,$y"
}
return $choices
}
method completion {} {
return [expr {
81-[llength [lsearch -all -inline [join [my dump]] {}]]
}]
}
method solve {} {
foreach ruleClass [info class subclass Rule] {
lappend rules [$ruleClass new]
}
 
while {1} {
set begin [my completion]
for {set y 0} {$y < 9} {incr y} {
for {set x 0} {$x < 9} {incr x} {
if {[my get $x $y] eq ""} {
foreach rule $rules {
set c [$rule solve [self] $x $y]
if {$c} {
my set $x $y $c
my Log "[info object class $rule] solved [self] at $x,$y for $c"
break
}
}
}
}
}
set end [my completion]
if {$end==81} {
my Log "Finished solving!"
break
} elseif {$begin==$end} {
my Log "A round finished without solving any squares, giving up."
break
}
}
foreach rule $rules {
$rule destroy
}
}
}
 
# Rule is the template for the rules used in Solver. The other rule-objects
# apply their logic to the values passed in and return either '0' or a number
# to allocate to the requested square.
oo::class create Rule {
method solve {hSudoku x y} {
if {![info object isa typeof $hSudoku SudokuSolver]} {
error "hSudoku must be an instance of class SudokuSolver."
}
 
tailcall my Solve $hSudoku $x $y [$hSudoku validchoices $x $y]
}
}
 
# Get all the allocated numbers for each square in the the row, column, and
# region containing $x,$y. If there is only one unallocated number among all
# three groups, it must be allocated at $x,$y
oo::class create RuleOnlyChoice {
superclass Rule
method Solve {hSudoku x y choices} {
if {[llength $choices]==1} {
return $choices
} else {
return 0
}
}
}
 
# Test each column to determine if $choice is an invalid choice for all other
# columns in row $X. If it is, it must only go in square $x,$y.
oo::class create RuleColumnChoice {
superclass Rule
method Solve {hSudoku x y choices} {
foreach choice $choices {
set failed 0
for {set x2 0} {$x2<9} {incr x2} {
if {$x2 != $x && $choice in [$hSudoku validchoices $x2 $y]} {
set failed 1
break
}
}
if {!$failed} {return $choice}
}
return 0
}
}
 
# Test each row to determine if $choice is an invalid choice for all other
# rows in column $y. If it is, it must only go in square $x,$y.
oo::class create RuleRowChoice {
superclass Rule
method Solve {hSudoku x y choices} {
foreach choice $choices {
set failed 0
for {set y2 0} {$y2<9} {incr y2} {
if {$y2 != $y && $choice in [$hSudoku validchoices $x $y2]} {
set failed 1
break
}
}
if {!$failed} {return $choice}
}
return 0
}
}
 
# Test each square in the region occupied by $x,$y to determine if $choice is
# an invalid choice for all other squares in that region. If it is, it must
# only go in square $x,$y.
oo::class create RuleRegionChoice {
superclass Rule
method Solve {hSudoku x y choices} {
foreach choice $choices {
set failed 0
set regnX [expr {($x/3)*3}]
set regnY [expr {($y/3)*3}]
for {set y2 $regnY} {$y2 < $regnY+3} {incr y2} {
for {set x2 $regnX} {$x2 < $regnX+3} {incr x2} {
if {
($x2!=$x || $y2!=$y)
&& $choice in [$hSudoku validchoices $x2 $y2]
} then {
set failed 1
break
}
}
}
if {!$failed} {return $choice}
}
return 0
}
}

Demonstration code:

SudokuSolver create sudoku
sudoku load {
{3 9 4 @ @ 2 6 7 @}
{@ @ @ 3 @ @ 4 @ @}
{5 @ @ 6 9 @ @ 2 @}
 
{@ 4 5 @ @ @ 9 @ @}
{6 @ @ @ @ @ @ @ 7}
{@ @ 7 @ @ @ 5 8 @}
 
{@ 1 @ @ 6 7 @ @ 8}
{@ @ 9 @ @ 8 @ @ @}
{@ 2 6 4 @ @ 7 3 5}
}
sudoku solve
# Simple pretty-printer for completed sudokus
puts +-----+-----+-----+
foreach line [sudoku dump] postline {0 0 1 0 0 1 0 0 1} {
puts |[lrange $line 0 2]|[lrange $line 3 5]|[lrange $line 6 8]|
if {$postline} {
puts +-----+-----+-----+
}
}
sudoku destroy
Output:
+-----+-----+-----+
|3 9 4|8 5 2|6 7 1|
|2 6 8|3 7 1|4 5 9|
|5 7 1|6 9 4|8 2 3|
+-----+-----+-----+
|1 4 5|7 8 3|9 6 2|
|6 8 2|9 4 5|3 1 7|
|9 3 7|1 2 6|5 8 4|
+-----+-----+-----+
|4 1 3|5 6 7|2 9 8|
|7 5 9|2 3 8|1 4 6|
|8 2 6|4 1 9|7 3 5|
+-----+-----+-----+

If we'd added a logger method (after creating the sudoku object but before running the solver) like this:

oo::objdefine sudoku method Log msg {puts $msg}

Then this additional logging output would have been produced prior to the result being printed:

::RuleOnlyChoice solved ::sudoku at 8,0 for 1
::RuleColumnChoice solved ::sudoku at 1,1 for 6
::RuleRegionChoice solved ::sudoku at 4,1 for 7
::RuleRowChoice solved ::sudoku at 7,1 for 5
::RuleOnlyChoice solved ::sudoku at 8,1 for 9
::RuleColumnChoice solved ::sudoku at 1,2 for 7
::RuleColumnChoice solved ::sudoku at 5,2 for 4
::RuleRowChoice solved ::sudoku at 6,2 for 8
::RuleOnlyChoice solved ::sudoku at 8,2 for 3
::RuleColumnChoice solved ::sudoku at 3,3 for 7
::RuleRowChoice solved ::sudoku at 1,4 for 8
::RuleRowChoice solved ::sudoku at 5,4 for 5
::RuleRowChoice solved ::sudoku at 6,4 for 3
::RuleRowChoice solved ::sudoku at 0,5 for 9
::RuleOnlyChoice solved ::sudoku at 1,5 for 3
::RuleOnlyChoice solved ::sudoku at 0,6 for 4
::RuleOnlyChoice solved ::sudoku at 2,6 for 3
::RuleColumnChoice solved ::sudoku at 3,6 for 5
::RuleOnlyChoice solved ::sudoku at 6,6 for 2
::RuleOnlyChoice solved ::sudoku at 7,6 for 9
::RuleOnlyChoice solved ::sudoku at 0,7 for 7
::RuleOnlyChoice solved ::sudoku at 1,7 for 5
::RuleColumnChoice solved ::sudoku at 4,7 for 3
::RuleOnlyChoice solved ::sudoku at 6,7 for 1
::RuleOnlyChoice solved ::sudoku at 0,8 for 8
::RuleOnlyChoice solved ::sudoku at 4,8 for 1
::RuleOnlyChoice solved ::sudoku at 5,8 for 9
::RuleOnlyChoice solved ::sudoku at 3,0 for 8
::RuleOnlyChoice solved ::sudoku at 4,0 for 5
::RuleColumnChoice solved ::sudoku at 2,1 for 8
::RuleOnlyChoice solved ::sudoku at 5,1 for 1
::RuleOnlyChoice solved ::sudoku at 2,2 for 1
::RuleRowChoice solved ::sudoku at 0,3 for 1
::RuleColumnChoice solved ::sudoku at 4,3 for 8
::RuleColumnChoice solved ::sudoku at 5,3 for 3
::RuleOnlyChoice solved ::sudoku at 7,3 for 6
::RuleOnlyChoice solved ::sudoku at 8,3 for 2
::RuleOnlyChoice solved ::sudoku at 2,4 for 2
::RuleColumnChoice solved ::sudoku at 3,4 for 9
::RuleOnlyChoice solved ::sudoku at 4,4 for 4
::RuleOnlyChoice solved ::sudoku at 7,4 for 1
::RuleColumnChoice solved ::sudoku at 3,5 for 1
::RuleOnlyChoice solved ::sudoku at 4,5 for 2
::RuleOnlyChoice solved ::sudoku at 5,5 for 6
::RuleOnlyChoice solved ::sudoku at 8,5 for 4
::RuleOnlyChoice solved ::sudoku at 3,7 for 2
::RuleOnlyChoice solved ::sudoku at 7,7 for 4
::RuleOnlyChoice solved ::sudoku at 8,7 for 6
::RuleOnlyChoice solved ::sudoku at 0,1 for 2
Finished solving!

Ursala[edit]

#import std
#import nat
 
sudoku =
 
@FL mat0+ block3+ mat` *+ block3*+ block9+ -+
~&rSL+ (psort (nleq+)* <~&blrl,~&blrr>)+ ~&arg^& -+
~&al?\~&ar ~&aa^&~&afahPRPfafatPJPRY+ ~&farlthlriNCSPDPDrlCS2DlrTS2J,
^|J/~& ~&rt!=+ ^= ~&s+ ~&H(
-+.|=&lrr;,|=&lrl;,|=&ll;+-,
~&rgg&& ~&irtPFXlrjrXPS; ~&lrK2tkZ2g&& ~&llrSL2rDrlPrrPljXSPTSL)+-,
//~&p ^|DlrDSLlrlPXrrPDSL(~&,num*+ rep2 block3)*= num block27 ~&iiK0 iota9,
* `0?=\~&iNC ! ~&t digits+-

test program:

#show+
 
example =
 
sudoku
 
-[
394002670
000300400
500690020
045000900
600000007
007000580
010067008
009008000
026400735]-
Output:
394 852 671
268 371 459
571 694 823

145 783 962
682 945 317
937 126 584

413 567 298
759 238 146
826 419 735

VBA[edit]

Translation of: Fortran
Dim grid(9, 9)
Dim gridSolved(9, 9)
 
Public Sub Solve(i, j)
If i > 9 Then
'exit with gridSolved = Grid
For r = 1 To 9
For c = 1 To 9
gridSolved(r, c) = grid(r, c)
Next c
Next r
Exit Sub
End If
For n = 1 To 9
If isSafe(i, j, n) Then
nTmp = grid(i, j)
grid(i, j) = n
If j = 9 Then
Solve i + 1, 1
Else
Solve i, j + 1
End If
grid(i, j) = nTmp
End If
Next n
End Sub
 
Public Function isSafe(i, j, n) As Boolean
Dim iMin As Integer
Dim jMin As Integer
 
If grid(i, j) <> 0 Then
isSafe = (grid(i, j) = n)
Exit Function
End If
 
'grid(i,j) is an empty cell. Check if n is OK
'first check the row i
For c = 1 To 9
If grid(i, c) = n Then
isSafe = False
Exit Function
End If
Next c
 
'now check the column j
For r = 1 To 9
If grid(r, j) = n Then
isSafe = False
Exit Function
End If
Next r
 
'finally, check the 3x3 subsquare containing grid(i,j)
iMin = 1 + 3 * Int((i - 1) / 3)
jMin = 1 + 3 * Int((j - 1) / 3)
For r = iMin To iMin + 2
For c = jMin To jMin + 2
If grid(r, c) = n Then
isSafe = False
Exit Function
End If
Next c
Next r
 
'all tests were OK
isSafe = True
End Function
 
Public Sub Sudoku()
'main routine
'to use, fill in the grid and
'type "Sudoku" in the Immediate panel of the Visual Basic for Applications window

Dim s(9) As String
 
'initialise grid using 9 strings,one per row
s(1) = "001005070"
s(2) = "920600000"
s(3) = "008000600"
s(4) = "090020401"
s(5) = "000000000"
s(6) = "304080090"
s(7) = "007000300"
s(8) = "000007069"
s(9) = "010800700"
For i = 1 To 9
For j = 1 To 9
grid(i, j) = Int(Val(Mid$(s(i), j, 1)))
Next j
Next i
'solve it!
Solve 1, 1
'print solution
Debug.Print "Solution:"
For i = 1 To 9
For j = 1 To 9
Debug.Print Format$(gridSolved(i, j)); " ";
Next j
Debug.Print
Next i
End Sub
Output:
Sudoku
Solution:
6 3 1 2 4 5 9 7 8 
9 2 5 6 7 8 1 4 3 
4 7 8 3 1 9 6 5 2 
7 9 6 5 2 3 4 8 1 
1 8 2 9 6 4 5 3 7 
3 5 4 7 8 1 2 9 6 
8 6 7 4 9 2 3 1 5 
2 4 3 1 5 7 8 6 9 
5 1 9 8 3 6 7 2 4 

VBScript[edit]

Translation of: VBA

To run in console mode with cscript.

Dim grid(9, 9)
Dim gridSolved(9, 9)
 
Public Sub Solve(i, j)
If i > 9 Then
'exit with gridSolved = Grid
For r = 1 To 9
For c = 1 To 9
gridSolved(r, c) = grid(r, c)
Next 'c
Next 'r
Exit Sub
End If
For n = 1 To 9
If isSafe(i, j, n) Then
nTmp = grid(i, j)
grid(i, j) = n
If j = 9 Then
Solve i + 1, 1
Else
Solve i, j + 1
End If
grid(i, j) = nTmp
End If
Next 'n
End Sub 'Solve

Public Function isSafe(i, j, n)
If grid(i, j) <> 0 Then
isSafe = (grid(i, j) = n)
Exit Function
End If
'grid(i,j) is an empty cell. Check if n is OK
'first check the row i
For c = 1 To 9
If grid(i, c) = n Then
isSafe = False
Exit Function
End If
Next 'c
'now check the column j
For r = 1 To 9
If grid(r, j) = n Then
isSafe = False
Exit Function
End If
Next 'r
'finally, check the 3x3 subsquare containing grid(i,j)
iMin = 1 + 3 * Int((i - 1) / 3)
jMin = 1 + 3 * Int((j - 1) / 3)
For r = iMin To iMin + 2
For c = jMin To jMin + 2
If grid(r, c) = n Then
isSafe = False
Exit Function
End If
Next 'c
Next 'r
'all tests were OK
isSafe = True
End Function 'isSafe

Public Sub Sudoku()
'main routine
Dim s(9)
s(1) = "001005070"
s(2) = "920600000"
s(3) = "008000600"
s(4) = "090020401"
s(5) = "000000000"
s(6) = "304080090"
s(7) = "007000300"
s(8) = "000007069"
s(9) = "010800700"
For i = 1 To 9
For j = 1 To 9
grid(i, j) = Int(Mid(s(i), j, 1))
Next 'j
Next 'j
'print problem
Wscript.echo "Problem:"
For i = 1 To 9
c=""
For j = 1 To 9
c=c & grid(i, j) & " "
Next 'j
Wscript.echo c
Next 'i
'solve it!
Solve 1, 1
'print solution
Wscript.echo "Solution:"
For i = 1 To 9
c=""
For j = 1 To 9
c=c & gridSolved(i, j) & " "
Next 'j
Wscript.echo c
Next 'i
End Sub 'Sudoku

Call sudoku
Output:
Problem:
0 0 1 0 0 5 0 7 0
9 2 0 6 0 0 0 0 0
0 0 8 0 0 0 6 0 0
0 9 0 0 2 0 4 0 1
0 0 0 0 0 0 0 0 0
3 0 4 0 8 0 0 9 0
0 0 7 0 0 0 3 0 0
0 0 0 0 0 7 0 6 9
0 1 0 8 0 0 7 0 0
Solution:
6 3 1 2 4 5 9 7 8
9 2 5 6 7 8 1 4 3
4 7 8 3 1 9 6 5 2
7 9 6 5 2 3 4 8 1
1 8 2 9 6 4 5 3 7
3 5 4 7 8 1 2 9 6
8 6 7 4 9 2 3 1 5
2 4 3 1 5 7 8 6 9
5 1 9 8 3 6 7 2 4

XPL0[edit]

This is a translation of the C example, but with a solution that can be verified by several other examples.

Translation of: C
code    ChOut=8, CrLf=9, IntOut=11, Text=12;
 
proc Show(X);
char X;
int I, J;
[for I:= 0 to 8 do
[if rem(I/3) = 0 then CrLf(0);
for J:= 0 to 8 do
[if rem(J/3) = 0 then ChOut(0, ^ );
ChOut(0, ^ ); IntOut(0, X(0));
X:= X+1;
];
CrLf(0);
];
];
 
func TryCell(X, Pos);
char X;
int Pos;
int Row, Col, I, J, Used;
[Row:= Pos/9;
Col:= rem(0);
Used:= 0;
 
if Pos = 81 then return true;
if X(Pos) then return TryCell(X, Pos+1);
 
for I:= 0 to 8 do Used:= Used ! 1 << (X(I*9+Col)-1);
for J:= 0 to 8 do Used:= Used ! 1 << (X(Row*9+J)-1);
 
Row:= Row/3*3;
Col:= Col/3*3;
for I:= Row to Row+2 do
for J:= Col to Col+2 do
Used:= Used ! 1 << (X(I*9+J)-1);
 
for I:= 1 to 9 do
[X(Pos):= I;
if (Used&1)=0 & TryCell(X, Pos+1) then return true;
Used:= Used>>1;
];
X(Pos):= 0;
return false;
];
 
proc Solve(S);
char S;
int I, J, C;
char X(81);
[J:= 0;
for I:= 0 to 80 do
[repeat C:= S(J);
J:= J+1;
until C>=^1 & C<=^9 ! C=^.;
X(I):= if C=^. then 0 else C-^0;
];
 
if TryCell(X, 0) then Show(X)
else Text(0, "No solution");
];
 
[Solve("394 ..2 67.
... 3.. 4..
5.. 69. .2.
.45 ... 9..
6.. ... ..7
..7 ... 58.
.1. .67 ..8
..9 ..8 ...
.26 4.. 735 ");
]
Output:
  3 9 4  8 5 2  6 7 1
  2 6 8  3 7 1  4 5 9
  5 7 1  6 9 4  8 2 3

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

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

zkl[edit]

Translation of: C
Note: Unlike in the C solution, 1<<-1 is defined (as 0).
fcn trycell(sdku,pos=0){
row,col:=pos/9, pos%9;
 
if(pos==81) return(True);
if(sdku[pos]) return(trycell(sdku, pos + 1));
 
used:=0;
foreach r in (9){ used=used.bitOr((1).shiftLeft(sdku[r*9 + col] - 1)) }
foreach c in (9){ used=used.bitOr((1).shiftLeft(sdku[row*9 + c] - 1)) }
 
row,col = row/3*3, col/3*3;
foreach r,c in ([row..row+2], [col..col+2])
{ used=used.bitOr((1).shiftLeft(sdku[r*9 + c] - 1)) }
 
sdku[pos]=1; while(sdku[pos]<=9){
if(used.isEven and trycell(sdku, pos + 1)) return(True);
sdku[pos]+=1; used/=2;
}
 
sdku[pos]=0;
return(False);
}
problem:=
#<<<
" 5 3 0 0 7 0 0 0 0
6 0 0 1 9 5 0 0 0
0 9 8 0 0 0 0 6 0
8 0 0 0 6 0 0 0 3
4 0 0 8 0 3 0 0 1
7 0 0 0 2 0 0 0 6
0 6 0 0 0 0 2 8 0
0 0 0 4 1 9 0 0 5
0 0 0 0 8 0 0 7 9";
#<<<
s:=problem.split().apply("toInt").copy(); // writable list of 81 ints
trycell(s).println();
println("+-----+-----+-----+");
foreach n in (3){
s[n*27,27].pump(Console.println,T(Void.Read,8),("| " + "%s%s%s | "*3).fmt); // 3 lines
println("+-----+-----+-----+");
}
Output:
True
+-----+-----+-----+
| 534 | 678 | 912 | 
| 672 | 195 | 348 | 
| 198 | 342 | 567 | 
+-----+-----+-----+
| 859 | 761 | 423 | 
| 426 | 853 | 791 | 
| 713 | 924 | 856 | 
+-----+-----+-----+
| 961 | 537 | 284 | 
| 287 | 419 | 635 | 
| 345 | 286 | 179 | 
+-----+-----+-----+