Sudoku
You are encouraged to solve this task according to the task description, using any language you may know.
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
Note: This specimen retains the original D coding style.
<lang algol68>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 )</lang>
- 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
<lang AutoHotkey>#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
}</lang>
BBC BASIC
<lang bbcbasic> 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%</lang>
BCPL
<lang BCPL>// This can be run using Cintcode BCPL freely available from www.cl.cam.ac.uk/users/mr10.
// This is a really naive program to solve Su Doku 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()
}</lang>
Bracmat
The program: <lang bracmat>{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$((its.sudoku),!arg):?puzzle
& (puzzle..Display)$
);</lang>
Solve a sudoku that is hard for a brute force solver: <lang bracmat>new'( sudokuSolver
, (.- - - - - - - - -)
(.- - - - - 3 - 8 5)
(.- - 1 - 2 - - - -)
(.- - - 5 - 7 - - -)
(.- - 4 - - - 1 - -)
(.- 9 - - - - - - -)
(.5 - - - - - - 7 3)
(.- - 2 - 1 - - - -)
(.- - - - 4 - - - 9)
);</lang>
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
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. <lang c>#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; }</lang>
C_sharp
“Manual” Solution
<lang csharp>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();
}
}</lang>
“Automatic” Solution
<lang csharp>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]);
}
}
}
}</lang> 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++
<lang cpp>#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();
}</lang>
Clojure
<lang clojure>(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)))))))</lang>
<lang clojure>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</lang>
Common Lisp
A simple solver without optimizations (except for pre-computing the possible entries of a cell). <lang lisp>(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))))))</lang>
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
Copied from Curry: Example Programs. <lang curry>----------------------------------------------------------------------------- --- 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"]</lang>
Alternative version
Minimal w/o read or show utilities. <lang curry>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</lang>
- 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
A little over-engineered solution, that shows some strong static typing useful in larger programs. <lang d>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;
}</lang>
- 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
Adapted from: http://code.activestate.com/recipes/576725-brute-force-sudoku-solver/ <lang d>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));
}</lang>
- 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
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. <lang d>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));
}</lang> Same output.
Delphi
Example taken from C++ <lang delphi>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;</lang> Usage: <lang delphi>var
SudokuSolver: TSudokuSolver;
begin
SudokuSolver := TSudokuSolver.Create('850002400' +
'720000009' +
'004000000' +
'000107002' +
'305000900' +
'040000000' +
'000080070' +
'017000000' +
'000036040');
try
SudokuSolver.Solve;
finally
FreeAndNil(SudokuSolver);
end;
end;</lang>
Elixir
<lang elixir>defmodule Sudoku do
def display( grid ), do: ( for y <- 1..9, do: display_row(y, grid) )
def start( knowns ), do: :dict.from_list( knowns )
def solve( grid ) do
sure = solve_all_sure( grid )
solve_unsure( potentials(sure), sure )
end
def task do
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 )
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( :dict.store(position, x, grid) ) )
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
for x <- [start, start+1, start+2], do: :io.fwrite(" ~c", [display_value(x, row, grid)])
IO.write( " " )
end
defp display_row_nl( n ) when n == 3 or n == 6 or n == 9, do: IO.puts "\n"
defp display_row_nl( _N ), do: IO.puts ""
defp display_value( x, y, grid ), do: display_value( :dict.find({x, y}, grid) )
defp display_value( :error ), do: ?.
defp display_value( {:ok, value} ), do: value + ?0
defp is_all_correct( grid ), do: :dict.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: Enum.any?( values_all_columns(grid), fn x-> is_not_allowed_values(x) end)
defp is_not_allowed_groups( grid ), do: Enum.any?( values_all_groups(grid), fn x-> is_not_allowed_values(x) end)
defp is_not_allowed_rows( grid ), do: Enum.any?( values_all_rows(grid), fn x-> is_not_allowed_values(x) end)
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} )
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 = :dict.fetch_keys( grid )
for x <- 1..9, y <- 1..9, not Enum.member?(keys, {x, y}), do: {x, y}
end
defp potentials_one_position( grid ) do
[{_shortest, position, values} | _t] = Enum.sort( for {position, values} <- potentials( grid ), do: {length(values), position, values} )
{position, values}
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, (for value <- :lists.seq(1, 9) -- useds, do: value) }
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 = List.delete( group_positions({x, y}), {x, y} ) |> potentials_values( grid )
row_values ++ column_values ++ group_values
end
defp potentials_values( keys, grid ) do
row_values_unfiltered = for x <- keys, do: :dict.find(x, grid)
for {:ok, value} <- row_values_unfiltered, do: value
end
defp values_all_columns( grid ), do: ( for x <- 1..9, do: values_all_columns(x, grid) )
defp values_all_columns( x, grid ) 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+1, x+2], do: values_all_groups(x, x_offset, grid) )
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: values_all_rows(y, grid) )
defp values_all_rows( y, grid ) 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( List.foldl(sures, grid, fn(x,acc)-> solve_all_sure_store(x,acc) 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: :dict.store( position, value, acc )
defp solve_unsure( [], grid ), do: grid
defp solve_unsure( _potentials, grid ) do
try do
bt( grid )
catch
{:ok, board} -> board
end
end
defp task( knowns ) do
IO.puts "start"
start = start( knowns )
display( start )
IO.puts "solved"
solved = solve( start )
display( solved )
IO.puts ""
end
end
Sudoku.task</lang>
- 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
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. <lang Erlang> -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(). </lang>
- 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
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.
<lang ERRE> !-------------------------------------------------------------------- ! 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
</lang>
Forth
<lang forth>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</lang>
Fortran
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.
<lang fortran>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</lang>
- 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|+-----+-----+-----+
Go
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. <lang go>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
}</lang>
- 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
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. <lang groovy>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
}</lang> Test/Benchmark Cases
Mentions of "exceptionally difficult" example in Wikipedia refer to this page: Exceptionally difficult Sudokus <lang groovy>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 Ada: ~ 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"
}</lang>
- 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
Visit the Haskell wiki Sudoku
J
See Solving Sudoku in J.
Java
<lang java>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(" -----------------------");
}
}</lang>
Lua
<lang lua>--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</lang> 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
<lang mathematica>solve[sudoku_] :=
NestWhile[
Join @@ Table[
Table[ReplacePart[s, #1 -> n], {n, #2}] & @@
First@SortBy[{#,
Complement[Range@9, sFirst@#, s;; , Last@#,
Catenate@
Extract[Partition[s, {3, 3}], Quotient[#, 3, -2]]]} & /@
Position[s, 0, {2}],
Length@Last@# &], {s, #}] &, {sudoku}, ! FreeQ[#, 0] &]</lang>
Example: <lang>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}}]</lang>
- 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
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" <lang MATLAB>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) = Template:(1:3),(1:3);
subBoxes(4:6,:)= Template:(4:6),(1:3);
subBoxes(7:9,:)= Template:(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</lang> Test Input: All empty cells must have a value of NaN. <lang MATLAB>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]</lang>
Output: <lang MATLAB>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</lang>
OCaml
uses the library ocamlgraph <lang ocaml>(* 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 ()</lang>
Oz
Using built-in constraint propagation and search. <lang oz>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}</lang>
Perl
<lang Perl>#!/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();</lang>
- 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
<lang perl6>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;</lang>
- 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.
<lang perl6>#!/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 "
}
}
}</lang>
- 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
PicoLisp
<lang PicoLisp>(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) ) )</lang>
- 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
<lang PicoLisp>(go)</lang>
- 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
Working PL/I version, derived from the Rosetta Fortran version. <lang pli>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; </lang>
- 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. <lang pli>
- 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;
</lang>
Prolog
<lang Prolog>:- 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]]).</lang>
GNU Prolog version
<lang Prolog>:- 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], Row) :- transpose(Col,[Row]). transpose(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.</lang>
- 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
A brute force method is used, it seemed the fastest as well as the simplest. <lang PureBasic>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</lang>
- 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
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 <lang python> 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() </lang>
Racket
Rascal
A sudoku is represented as a matrix, see Rascal solutions to matrix related problems for examples.
<lang Rascal>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> };</lang>
Example
rascal>displaySudoku(sudoku(sudokuA)) See picture
REXX
The SUDOKU REXX programs (and output) are included here ──► Sudoku/REXX.
RPN (HP-15c)
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
Example of a back-tracking solver, from wp:Algorithmics of sudoku
<lang ruby>def read_matrix(data)
lines = data.each_line.to_a # ver 2.0 later 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]
# Same as another in the column isn't permissible...
9.times do |i2|
ok[matrix[i2][j]] = nil if matrix[i2][j].nonzero?
end
# Same as another in the row isn't permissible...
9.times do |j2|
ok[matrix[i][j2]] = nil if matrix[i][j2].nonzero?
end
# Same as another in the 3x3 block isn't permissible...
irange = (ig = (i / 3) * 3) .. ig + 2
jrange = (jg = (j / 3) * 3) .. jg + 2
irange.each do |i2|
jrange.each do |j2|
ok[matrix[i2][j2]] = nil if matrix[i2][j2].nonzero?
end
end
# 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))</lang>
- 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| +-----+-----+-----+
Scala
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.
<lang scala>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)})
}</lang>
- 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:
<lang scala>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))
}
}</lang>
- 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
<lang ruby>__USE_INTNUM__
func solve(board) {
board.range.each { |i|
!board[i] || next;
var t = board[
board.range.grep {|j|
(j.div(9) == i.div(9)) ||
(j.mod(9) == i.mod(9)) ||
(j.div(27) == i.div(27) &&
(j.mod(9).div(3) == i.mod(9).div(3)))
}
];
1..9 grep {!(.~~t)}
each {|k| board[i] = k; solve(board)};
board[i] = 0;
return;
};
board.range.each { |i|
print "#{board[i]} ";
{print " "} if (i+1 %% 3);
{print "\n"} if (i+1 %% 9);
{print "\n"} if (i+1 %% 27);
};
}
var board = %w(
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
).map{.to_i};
solve(board);</lang>
- 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
<lang Swift>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()</lang>
- 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
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.
or
<lang tcl>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
}
}</lang> Demonstration code: <lang tcl>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</lang>
- 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:
<lang tcl>oo::objdefine sudoku method Log msg {puts $msg}</lang>
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
<lang Ursala>#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;,|=≪+-,
~&rgg&& ~&irtPFXlrjrXPS; ~&lrK2tkZ2g&& ~&llrSL2rDrlPrrPljXSPTSL)+-,
//~&p ^|DlrDSLlrlPXrrPDSL(~&,num*+ rep2 block3)*= num block27 ~&iiK0 iota9,
* `0?=\~&iNC ! ~&t digits+-</lang>
test program: <lang Ursala>#show+
example =
sudoku
-[ 394002670 000300400 500690020 045000900 600000007 007000580 010067008 009008000 026400735]-</lang>
- 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
<lang VB>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</lang>
- 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
To run in console mode with cscript. <lang vb>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</lang>
- 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
This is a translation of the C example, but with a solution that can be verified by several other examples.
<lang XPL0>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 ");
]</lang>
- 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
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