Sudoku
You are encouraged to solve this task according to the task description, using any language you may know.
- Task
Solve a partially filled-in normal 9x9 Sudoku grid and display the result in a human-readable format.
- references
- Algorithmics of Sudoku may help implement this.
- Python Sudoku Solver Computerphile video.
11l
<lang 11l>T Sudoku
solved = 0B grid = [0] * 81
F (rows)
assert(rows.len == 9 & all(rows.map(row -> row.len == 9)), ‘Grid must be 9 x 9’)
L(i) 9
L(j) 9
.grid[9 * i + j] = Int(rows[i][j])
F solve()
print("Starting grid:\n\n"(.))
.placeNumber(0)
print(I .solved {"Solution:\n\n"(.)} E ‘Unsolvable!’)
F placeNumber(pos)
I .solved
R
I pos == 81
.solved = 1B
R
I .grid[pos] > 0
.placeNumber(pos + 1)
R
L(n) 1..9
I .checkValidity(n, pos % 9, pos I/ 9)
.grid[pos] = n
.placeNumber(pos + 1)
I .solved
R
.grid[pos] = 0
F checkValidity(v, x, y)
L(i) 9
I .grid[y * 9 + i] == v |
.grid[i * 9 + x] == v
R 0B
V startX = (x I/ 3) * 3
V startY = (y I/ 3) * 3
L(i) startY .< startY + 3
L(j) startX .< startX + 3
I .grid[i * 9 + j] == v
R 0B
R 1B
F String()
V s = ‘’
L(i) 9
L(j) 9
s ‘’= .grid[i * 9 + j]‘ ’
I j C (2, 5)
s ‘’= ‘| ’
s ‘’= "\n"
I i C (2, 5)
s ‘’= "------+-------+------\n"
R s
V rows = [‘850002400’,
‘720000009’,
‘004000000’,
‘000107002’,
‘305000900’,
‘040000000’,
‘000080070’,
‘017000000’,
‘000036040’]
Sudoku(rows).solve()</lang>
- Output:
Starting grid: 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 Solution: 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
8th
<lang 8th> \ \ Simple iterative backtracking Sudoku solver for 8th \ needs array/each-slice
[ 00, 00, 00, 03, 03, 03, 06, 06, 06,
00, 00, 00, 03, 03, 03, 06, 06, 06, 00, 00, 00, 03, 03, 03, 06, 06, 06, 27, 27, 27, 30, 30, 30, 33, 33, 33, 27, 27, 27, 30, 30, 30, 33, 33, 33, 27, 27, 27, 30, 30, 30, 33, 33, 33, 54, 54, 54, 57, 57, 57, 60, 60, 60, 54, 54, 54, 57, 57, 57, 60, 60, 60, 54, 54, 54, 57, 57, 57, 60, 60, 60 ] constant top-left-cell
\ Bit number presentations a:new 2 b:new b:clear a:push ( 2 b:new b:clear swap 1 b:bit! a:push ) 0 8 loop constant posbit
- posbit? \ n -- s
posbit swap a:@ nip ;
- search \ b -- n
null swap ( dup -rot b:bit@ if rot drop break else nip then ) 0 8 loop swap ;
- b-or \ b b -- b
' n:bor b:op ;
- b-and \ b b -- b
' n:band b:op ;
- b-xor \ b b -- b
b:xor [ xff, x01 ] b:new b-and ;
- b-not \ b -- b
xff b:xor [ xff, x01 ] b:new b-and ;
- b-any \ a -- b
' b-or 0 posbit? a:reduce ;
- row \ a row -- a
9 n:* 9 a:slice ;
- col \ a col -- a
-1 9 a:slice+ ;
\ For testing sub boards
- sub \ a n -- a
top-left-cell swap a:@ nip over over 3 a:slice -rot 9 n:+ 2dup 3 a:slice -rot 9 n:+ 3 a:slice a:+ a:+ ;
a:new 0 args "Give Sudoku text file as param" thrownull f:slurp "Cannot read file" thrownull >s "" s:/ ' >n a:map ( posbit? a:push ) a:each! drop constant board
- display-board
board ( search nip -1 ?: n:1+ ) a:map "+-----+-----+-----+\n" "|%d %d %d|%d %d %d|%d %d %d|\n" s:+ "|%d %d %d|%d %d %d|%d %d %d|\n" s:+ "|%d %d %d|%d %d %d|%d %d %d|\n" s:+ "+-----+-----+-----+\n" s:+ "|%d %d %d|%d %d %d|%d %d %d|\n" s:+ "|%d %d %d|%d %d %d|%d %d %d|\n" s:+ "|%d %d %d|%d %d %d|%d %d %d|\n" s:+ "+-----+-----+-----+\n" s:+ "|%d %d %d|%d %d %d|%d %d %d|\n" s:+ "|%d %d %d|%d %d %d|%d %d %d|\n" s:+ "|%d %d %d|%d %d %d|%d %d %d|\n" s:+ "+-----+-----+-----+\n" s:+ s:strfmt . ;
\ Store move history a:new constant history
\ Possible numbers for a cell
- candidates? \ n -- s
dup dup 9 n:/ n:int swap 9 n:mod \ row col
board swap col b-any
board rot row b-any
b-or
board rot sub b-any
b-or
b-not ;
\ If found: -- n T \ If not found: -- T
- find-free-cell
false board ( 0 posbit? b:= if nip true break else drop then ) a:each drop ;
- validate
true
board
( dup -rot a:@ swap 2 pick 0 posbit? a:! 2 pick candidates? 2 pick b:= if
-rot a:!
else
2drop drop
false swap
break
then ) 0 80 loop drop ;
- solve
repeat
find-free-cell if
dup candidates?
repeat
search null? if
drop board -rot a:! drop
history a:len 0 n:= if
drop false ;;
then
a:pop nip
a:open
else
n:1+ posbit?
dup
board 4 pick rot a:! drop
b-xor
2 a:close
history swap a:push drop
break
then
again
else
validate
break
then
again ;
- app:main
"Sudoku puzzle:\n" . display-board cr solve if "Sudoku solved:\n" . display-board else "No solution!\n" . then ;
</lang>
Ada
<lang ada> with Ada.Text_IO;
procedure Sudoku is
type sudoku_ar_t is array ( integer range 0..80 ) of integer range 0..9; FINISH_EXCEPTION : exception;
procedure prettyprint(sudoku_ar: sudoku_ar_t); function checkValidity( val : integer; x : integer; y : integer; sudoku_ar: in sudoku_ar_t) return Boolean; procedure placeNumber(pos: Integer; sudoku_ar: in out sudoku_ar_t); procedure solve(sudoku_ar: in out sudoku_ar_t);
function checkValidity( val : integer; x : integer; y : integer; sudoku_ar: in sudoku_ar_t) return Boolean
is
begin
for i in 0..8 loop
if ( sudoku_ar( y * 9 + i ) = val or sudoku_ar( i * 9 + x ) = val ) then
return False;
end if;
end loop;
declare
startX : constant integer := ( x / 3 ) * 3;
startY : constant integer := ( y / 3 ) * 3;
begin
for i in startY..startY+2 loop
for j in startX..startX+2 loop
if ( sudoku_ar( i * 9 +j ) = val ) then
return False;
end if;
end loop;
end loop;
return True;
end;
end checkValidity;
procedure placeNumber(pos: Integer; sudoku_ar: in out sudoku_ar_t)
is
begin
if ( pos = 81 ) then
raise FINISH_EXCEPTION;
end if;
if ( sudoku_ar(pos) > 0 ) then
placeNumber(pos+1, sudoku_ar);
return;
end if;
for n in 1..9 loop
if( checkValidity( n, pos mod 9, pos / 9 , sudoku_ar ) ) then
sudoku_ar(pos) := n;
placeNumber(pos + 1, sudoku_ar );
sudoku_ar(pos) := 0;
end if;
end loop;
end placeNumber;
procedure solve(sudoku_ar: in out sudoku_ar_t)
is
begin
placeNumber( 0, sudoku_ar );
Ada.Text_IO.Put_Line("Unresolvable !");
exception
when FINISH_EXCEPTION =>
Ada.Text_IO.Put_Line("Finished !");
prettyprint(sudoku_ar);
end solve;
procedure prettyprint(sudoku_ar: sudoku_ar_t)
is
line_sep : constant String := "------+------+------";
begin
for i in sudoku_ar'Range loop
Ada.Text_IO.Put(sudoku_ar(i)'Image);
if (i+1) mod 3 = 0 and not((i+1) mod 9 = 0) then
Ada.Text_IO.Put("|");
end if;
if (i+1) mod 9 = 0 then
Ada.Text_IO.Put_Line("");
end if;
if (i+1) mod 27 = 0 then
Ada.Text_IO.Put_Line(line_sep);
end if;
end loop;
end prettyprint;
sudoku_ar : sudoku_ar_t :=
(
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
);
begin
solve( sudoku_ar );
end Sudoku; </lang>
- Output:
Finished ! 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
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>
AWK
<lang AWK>
- syntax: GAWK -f SUDOKU_RC.AWK
BEGIN {
- row1 row2 row3 row4 row5 row6 row7 row8 row9
- puzzle = "111111111 111111111 111111111 111111111 111111111 111111111 111111111 111111111 111111111" # NG duplicate hints
- puzzle = "1........ ..274.... ...5....4 .3....... 75....... .....96.. .4...6... .......71 .....1.30" # NG can't use zero
- puzzle = "1........ ..274.... ...5....4 .3....... 75....... .....96.. .4...6... .......71 .....1.39" # no solution
- puzzle = "1........ ..274.... ...5....4 .3....... 75....... .....96.. .4...6... .......71 .....1.3." # OK
puzzle = "123456789 456789123 789123456 ......... ......... ......... ......... ......... ........." # OK
gsub(/ /,"",puzzle)
if (length(puzzle) != 81) { error("length of puzzle is not 81") }
if (puzzle !~ /^[1-9\.]+$/) { error("only 1-9 and . are valid") }
if (gsub(/[1-9]/,"&",puzzle) < 17) { error("too few hints") }
if (errors > 0) {
exit(1)
}
plot(puzzle,"unsolved")
if (dup_hints_check(puzzle) == 1) {
if (solve(puzzle) == 1) {
dup_hints_check(sos)
plot(sos,"solved")
printf("\nbef: %s\naft: %s\n",puzzle,sos)
exit(0)
}
else {
error("no solution")
}
}
exit(1)
} function dup_hints_check(ss, esf,msg,Rarr,Carr,Barr,i,r_row,r_col,r_pos,r_hint,c_row,c_col,c_pos,c_hint,box) {
esf = errors # errors so far
for (i=0; i<81; i++) {
# row
r_row = int(i/9) + 1 # determine row: 1..9
r_col = i%9 + 1 # determine column: 1..9
r_pos = i + 1 # determine hint position: 1..81
r_hint = substr(ss,r_pos,1) # extract 1 character; the hint
Rarr[r_row,r_hint]++ # save row
# column
c_row = i%9 + 1 # determine row: 1..9
c_col = int(i/9) + 1 # determine column: 1..9
c_pos = (c_row-1) * 9 + c_col # determine hint position: 1..81
c_hint = substr(ss,c_pos,1) # extract 1 character; the hint
Carr[c_col,c_hint]++ # save column
# box (there has to be a better way)
if ((r_row r_col) ~ /[123][123]/) { box = 1 }
else if ((r_row r_col) ~ /[123][456]/) { box = 2 }
else if ((r_row r_col) ~ /[123][789]/) { box = 3 }
else if ((r_row r_col) ~ /[456][123]/) { box = 4 }
else if ((r_row r_col) ~ /[456][456]/) { box = 5 }
else if ((r_row r_col) ~ /[456][789]/) { box = 6 }
else if ((r_row r_col) ~ /[789][123]/) { box = 7 }
else if ((r_row r_col) ~ /[789][456]/) { box = 8 }
else if ((r_row r_col) ~ /[789][789]/) { box = 9 }
else { box = 0 }
Barr[box,r_hint]++ # save box
}
dup_hints_print(Rarr,"row")
dup_hints_print(Carr,"column")
dup_hints_print(Barr,"box")
return((errors == esf) ? 1 : 0)
} function dup_hints_print(arr,rcb, hint,i) {
- rcb - Row Column Box
for (i=1; i<=9; i++) { # "i" is either the row, column, or box
for (hint=1; hint<=9; hint++) { # 1..9 only; don't care about "." place holder
if (arr[i,hint]+0 > 1) { # was a digit specified more than once
error(sprintf("duplicate hint in %s %d",rcb,i))
}
}
}
} function plot(ss,text1,text2, a,b,c,d,ou) {
- 1st call prints the unsolved puzzle.
- 2nd call prints the solved puzzle
printf("| - - - + - - - + - - - | %s\n",text1)
for (a=0; a<3; a++) {
for (b=0; b<3; b++) {
ou = "|"
for (c=0; c<3; c++) {
for (d=0; d<3; d++) {
ou = sprintf("%s %1s",ou,substr(ss,1+d+3*c+9*b+27*a,1))
}
ou = ou " |"
}
print(ou)
}
printf("| - - - + - - - + - - - | %s\n",(a==2)?text2:"")
}
} function solve(ss, a,b,c,d,e,r,co,ro,bi,bl,nss) {
i = 0
- first, use some simple logic to fill grid as much as possible
do {
i++
didit = 0
delete nr
delete nc
delete nb
delete ca
for (a=0; a<81; a++) {
b = substr(ss,a+1,1)
if (b == ".") { # construct row, column and block at cell
c = a % 9
r = int(a/9)
ro = substr(ss,r*9+1,9)
co = ""
for (d=0; d<9; d++) { co = co substr(ss,d*9+c+1,1) }
bi = int(c/3)*3+(int(r/3)*3)*9+1
bl = ""
for (d=0; d<3; d++) { bl = bl substr(ss,bi+d*9,3) }
e = 0
- count non-occurrences of digits 1-9 in combined row, column and block, per row, column and block, and flag cell/digit as candidate
for (d=1; d<10; d++) {
if (index(ro co bl, d) == 0) {
e++
nr[r,d]++
nc[c,d]++
nb[bi,d]++
ca[c,r,d] = bi
}
}
if (e == 0) { # in case no candidate is available, give up
return(0)
}
}
}
- go through all cell/digit candidates
- hidden singles
for (crd in ca) {
- a candidate may have been deleted after the loop started
if (ca[crd] != "") {
split(crd,spl,SUBSEP)
c = spl[1]
r = spl[2]
d = spl[3]
bi = ca[crd]
a = c + r * 9
- unique solution if at least one non-occurrence counter is exactly 1
if ((nr[r,d] == 1) || (nc[c,d] == 1) || (nb[bi,d] == 1)) {
ss = substr(ss,1,a) d substr(ss,a+2,length(ss))
didit = 1
- remove candidates from current row, column, block
for (e=0; e<9; e++) {
delete ca[c,e,d]
delete ca[e,r,d]
}
for (e=0; e<3; e++) {
for (b=0; b<3; b++) {
delete ca[int(c/3)*3+b,int(r/3)*3+e,d]
}
}
}
}
}
} while (didit == 1)
- second, pick a viable solution for the next empty cell and see if it leads to a solution
a = index(ss,".")-1
if (a == -1) { # no more empty cells, done
sos = ss
return(1)
}
else {
c = a % 9
r = int(a/9)
- concatenate current row, column and block
- row
co = substr(ss,r*9+1,9)
- column
for (d=0; d<9; d++) { co = co substr(ss,d*9+c+1,1) }
- block
c = int(c/3)*3+(int(r/3)*3)*9+1
for (d=0; d<3; d++) { co = co substr(ss,c+d*9,3) }
for (b=1; b<10; b++) { # get a viable digit
if (index(co,b) == 0) { # check if candidate digit already exists
- is viable, put in cell
nss = substr(ss,1,a) b substr(ss,a+2,length(ss))
d = solve(nss) # try to solve
if (d == 1) { # if successful, return
return(1)
}
}
}
}
return(0) # no digits viable, no solution
} function error(message) { printf("error: %s\n",message) ; errors++ } </lang>
- Output:
| - - - + - - - + - - - | unsolved | 1 2 3 | 4 5 6 | 7 8 9 | | 4 5 6 | 7 8 9 | 1 2 3 | | 7 8 9 | 1 2 3 | 4 5 6 | | - - - + - - - + - - - | | . . . | . . . | . . . | | . . . | . . . | . . . | | . . . | . . . | . . . | | - - - + - - - + - - - | | . . . | . . . | . . . | | . . . | . . . | . . . | | . . . | . . . | . . . | | - - - + - - - + - - - | | - - - + - - - + - - - | solved | 1 2 3 | 4 5 6 | 7 8 9 | | 4 5 6 | 7 8 9 | 1 2 3 | | 7 8 9 | 1 2 3 | 4 5 6 | | - - - + - - - + - - - | | 2 1 4 | 3 6 5 | 8 9 7 | | 3 6 5 | 8 9 7 | 2 1 4 | | 8 9 7 | 2 1 4 | 3 6 5 | | - - - + - - - + - - - | | 5 3 1 | 6 4 2 | 9 7 8 | | 6 4 2 | 9 7 8 | 5 3 1 | | 9 7 8 | 5 3 1 | 6 4 2 | | - - - + - - - + - - - | bef: 123456789456789123789123456...................................................... aft: 123456789456789123789123456214365897365897214897214365531642978642978531978531642
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. // Implemented by Martin Richards.
// If you have cintcode BCPL installed on a Linux system you can compile and run this program // execute the following sequence of commands.
// cd $BCPLROOT // cintsys // c bc sudoku // sudoku
// This is a really naive program to solve SuDoku problems. Even so it is usually quite fast.
// SuDoku consists of a 9x9 grid of cells. Each cell should contain // a digit in the range 1..9. Every row, column and major 3x3 // square should contain all the digits 1..9. Some cells have // given values. The problem is to find digits to place in // the unspecified cells satisfying the constraints.
// A typical problem is:
// - - - 6 3 8 - - - // 7 - 6 - - - 3 - 5 // - 1 - - - - - 4 -
// - - 8 7 1 2 4 - - // - 9 - - - - - 5 - // - - 2 5 6 9 1 - -
// - 3 - - - - - 1 - // 1 - 5 - - - 6 - 8 // - - - 1 8 4 - - -
SECTION "sudoku"
GET "libhdr"
GLOBAL { count:ug
// The 9x9 board
a1; a2; a3; a4; a5; a6; a7; a8; a9 b1; b2; b3; b4; b5; b6; b7; b8; b9 c1; c2; c3; c4; c5; c6; c7; c8; c9 d1; d2; d3; d4; d5; d6; d7; d8; d9 e1; e2; e3; e4; e5; e6; e7; e8; e9 f1; f2; f3; f4; f5; f6; f7; f8; f9 g1; g2; g3; g4; g5; g6; g7; g8; g9 h1; h2; h3; h4; h5; h6; h7; h8; h9 i1; i2; i3; i4; i5; i6; i7; i8; i9 }
MANIFEST { N1=1<<0; N2=1<<1; N3=1<<2; N4=1<<3; N5=1<<4; N6=1<<5; N7=1<<6; N8=1<<7; N9=1<<8 }
LET start() = VALOF { count := 0
initboard()
prboard()
ta1()
writef("*n*nTotal number of solutions: %n*n", count)
RESULTIS 0
}
AND initboard() BE { a1, a2, a3, a4, a5, a6, a7, a8, a9 := 0, 0, 0, N6,N3,N8, 0, 0, 0 b1, b2, b3, b4, b5, b6, b7, b8, b9 := N7, 0,N6, 0, 0, 0, N3, 0,N5 c1, c2, c3, c4, c5, c6, c7, c8, c9 := 0,N1, 0, 0, 0, 0, 0,N4, 0 d1, d2, d3, d4, d5, d6, d7, d8, d9 := 0, 0,N8, N7,N1,N2, N4, 0, 0 e1, e2, e3, e4, e5, e6, e7, e8, e9 := 0,N9, 0, 0, 0, 0, 0,N5, 0 f1, f2, f3, f4, f5, f6, f7, f8, f9 := 0, 0,N2, N5,N6,N9, N1, 0, 0 g1, g2, g3, g4, g5, g6, g7, g8, g9 := 0,N3, 0, 0, 0, 0, 0,N1, 0 h1, h2, h3, h4, h5, h6, h7, h8, h9 := N1, 0,N5, 0, 0, 0, N6, 0,N8 i1, i2, i3, i4, i5, i6, i7, i8, i9 := 0, 0, 0, N1,N8,N4, 0, 0, 0
// Un-comment the following to test that the backtracking works // giving 184 solutions. //h1, h2, h3, h4, h5, h6, h7, h8, h9 := N1, 0,N5, 0, 0, 0, N6, 0, 0 //i1, i2, i3, i4, i5, i6, i7, i8, i9 := 0, 0, 0, 0, 0, 0, 0, 0, 0 }
AND c(n) = VALOF SWITCHON n INTO { DEFAULT: RESULTIS '?'
CASE 0: RESULTIS '-' CASE N1: RESULTIS '1' CASE N2: RESULTIS '2' CASE N3: RESULTIS '3' CASE N4: RESULTIS '4' CASE N5: RESULTIS '5' CASE N6: RESULTIS '6' CASE N7: RESULTIS '7' CASE N8: RESULTIS '8' CASE N9: RESULTIS '9'
}
AND prboard() BE { LET form = "%c %c %c %c %c %c %c %c %c*n"
writef("*ncount = %n*n", count)
newline()
writef(form, c(a1),c(a2),c(a3),c(a4),c(a5),c(a6),c(a7),c(a8),c(a9))
writef(form, c(b1),c(b2),c(b3),c(b4),c(b5),c(b6),c(b7),c(b8),c(b9))
writef(form, c(c1),c(c2),c(c3),c(c4),c(c5),c(c6),c(c7),c(c8),c(c9))
newline()
writef(form, c(d1),c(d2),c(d3),c(d4),c(d5),c(d6),c(d7),c(d8),c(d9))
writef(form, c(e1),c(e2),c(e3),c(e4),c(e5),c(e6),c(e7),c(e8),c(e9))
writef(form, c(f1),c(f2),c(f3),c(f4),c(f5),c(f6),c(f7),c(f8),c(f9))
newline()
writef(form, c(g1),c(g2),c(g3),c(g4),c(g5),c(g6),c(g7),c(g8),c(g9))
writef(form, c(h1),c(h2),c(h3),c(h4),c(h5),c(h6),c(h7),c(h8),c(h9))
writef(form, c(i1),c(i2),c(i3),c(i4),c(i5),c(i6),c(i7),c(i8),c(i9))
newline()
//abort(1000) }
AND try(p, f, row, col, sq) BE { LET x = !p
TEST x
THEN f()
ELSE { LET bits = row|col|sq
//prboard() // writef("x=%n %b9*n", x, bits) //abort(1000)
IF (N1&bits)=0 DO { !p:=N1; f() }
IF (N2&bits)=0 DO { !p:=N2; f() }
IF (N3&bits)=0 DO { !p:=N3; f() }
IF (N4&bits)=0 DO { !p:=N4; f() }
IF (N5&bits)=0 DO { !p:=N5; f() }
IF (N6&bits)=0 DO { !p:=N6; f() }
IF (N7&bits)=0 DO { !p:=N7; f() }
IF (N8&bits)=0 DO { !p:=N8; f() }
IF (N9&bits)=0 DO { !p:=N9; f() }
!p := 0
}
}
AND ta1() BE try(@a1, ta2, a1+a2+a3+a4+a5+a6+a7+a8+a9,
a1+b1+c1+d1+e1+f1+g1+h1+i1,
a1+a2+a3+b1+b2+b3+c1+c2+c3)
AND ta2() BE try(@a2, ta3, a1+a2+a3+a4+a5+a6+a7+a8+a9,
a2+b2+c2+d2+e2+f2+g2+h2+i2,
a1+a2+a3+b1+b2+b3+c1+c2+c3)
AND ta3() BE try(@a3, ta4, a1+a2+a3+a4+a5+a6+a7+a8+a9,
a3+b3+c3+d3+e3+f3+g3+h3+i3,
a1+a2+a3+b1+b2+b3+c1+c2+c3)
AND ta4() BE try(@a4, ta5, a1+a2+a3+a4+a5+a6+a7+a8+a9,
a4+b4+c4+d4+e4+f4+g4+h4+i4,
a4+a5+a6+b4+b5+b6+c4+c5+c6)
AND ta5() BE try(@a5, ta6, a1+a2+a3+a4+a5+a6+a7+a8+a9,
a5+b5+c5+d5+e5+f5+g5+h5+i5,
a4+a5+a6+b4+b5+b6+c4+c5+c6)
AND ta6() BE try(@a6, ta7, a1+a2+a3+a4+a5+a6+a7+a8+a9,
a6+b6+c6+d6+e6+f6+g6+h6+i6,
a4+a5+a6+b4+b5+b6+c4+c5+c6)
AND ta7() BE try(@a7, ta8, a1+a2+a3+a4+a5+a6+a7+a8+a9,
a7+b7+c7+d7+e7+f7+g7+h7+i7,
a7+a8+a9+b7+b8+b9+c7+c8+c9)
AND ta8() BE try(@a8, ta9, a1+a2+a3+a4+a5+a6+a7+a8+a9,
a8+b8+c8+d8+e8+f8+g8+h8+i8,
a7+a8+a9+b7+b8+b9+c7+c8+c9)
AND ta9() BE try(@a9, tb1, a1+a2+a3+a4+a5+a6+a7+a8+a9,
a9+b9+c9+d9+e9+f9+g9+h9+i9,
a7+a8+a9+b7+b8+b9+c7+c8+c9)
AND tb1() BE try(@b1, tb2, b1+b2+b3+b4+b5+b6+b7+b8+b9,
a1+b1+c1+d1+e1+f1+g1+h1+i1,
a1+a2+a3+b1+b2+b3+c1+c2+c3)
AND tb2() BE try(@b2, tb3, b1+b2+b3+b4+b5+b6+b7+b8+b9,
a2+b2+c2+d2+e2+f2+g2+h2+i2,
a1+a2+a3+b1+b2+b3+c1+c2+c3)
AND tb3() BE try(@b3, tb4, b1+b2+b3+b4+b5+b6+b7+b8+b9,
a3+b3+c3+d3+e3+f3+g3+h3+i3,
a1+a2+a3+b1+b2+b3+c1+c2+c3)
AND tb4() BE try(@b4, tb5, b1+b2+b3+b4+b5+b6+b7+b8+b9,
a4+b4+c4+d4+e4+f4+g4+h4+i4,
a4+a5+a6+b4+b5+b6+c4+c5+c6)
AND tb5() BE try(@b5, tb6, b1+b2+b3+b4+b5+b6+b7+b8+b9,
a5+b5+c5+d5+e5+f5+g5+h5+i5,
a4+a5+a6+b4+b5+b6+c4+c5+c6)
AND tb6() BE try(@b6, tb7, b1+b2+b3+b4+b5+b6+b7+b8+b9,
a6+b6+c6+d6+e6+f6+g6+h6+i6,
a4+a5+a6+b4+b5+b6+c4+c5+c6)
AND tb7() BE try(@b7, tb8, b1+b2+b3+b4+b5+b6+b7+b8+b9,
a7+b7+c7+d7+e7+f7+g7+h7+i7,
a7+a8+a9+b7+b8+b9+c7+c8+c9)
AND tb8() BE try(@b8, tb9, b1+b2+b3+b4+b5+b6+b7+b8+b9,
a8+b8+c8+d8+e8+f8+g8+h8+i8,
a7+a8+a9+b7+b8+b9+c7+c8+c9)
AND tb9() BE try(@b9, tc1, b1+b2+b3+b4+b5+b6+b7+b8+b9,
a9+b9+c9+d9+e9+f9+g9+h9+i9,
a7+a8+a9+b7+b8+b9+c7+c8+c9)
AND tc1() BE try(@c1, tc2, c1+c2+c3+c4+c5+c6+c7+c8+c9,
a1+b1+c1+d1+e1+f1+g1+h1+i1,
a1+a2+a3+b1+b2+b3+c1+c2+c3)
AND tc2() BE try(@c2, tc3, c1+c2+c3+c4+c5+c6+c7+c8+c9,
a2+b2+c2+d2+e2+f2+g2+h2+i2,
a1+a2+a3+b1+b2+b3+c1+c2+c3)
AND tc3() BE try(@c3, tc4, c1+c2+c3+c4+c5+c6+c7+c8+c9,
a3+b3+c3+d3+e3+f3+g3+h3+i3,
a1+a2+a3+b1+b2+b3+c1+c2+c3)
AND tc4() BE try(@c4, tc5, c1+c2+c3+c4+c5+c6+c7+c8+c9,
a4+b4+c4+d4+e4+f4+g4+h4+i4,
a4+a5+a6+b4+b5+b6+c4+c5+c6)
AND tc5() BE try(@c5, tc6, c1+c2+c3+c4+c5+c6+c7+c8+c9,
a5+b5+c5+d5+e5+f5+g5+h5+i5,
a4+a5+a6+b4+b5+b6+c4+c5+c6)
AND tc6() BE try(@c6, tc7, c1+c2+c3+c4+c5+c6+c7+c8+c9,
a6+b6+c6+d6+e6+f6+g6+h6+i6,
a4+a5+a6+b4+b5+b6+c4+c5+c6)
AND tc7() BE try(@c7, tc8, c1+c2+c3+c4+c5+c6+c7+c8+c9,
a7+b7+c7+d7+e7+f7+g7+h7+i7,
a7+a8+a9+b7+b8+b9+c7+c8+c9)
AND tc8() BE try(@c8, tc9, c1+c2+c3+c4+c5+c6+c7+c8+c9,
a8+b8+c8+d8+e8+f8+g8+h8+i8,
a7+a8+a9+b7+b8+b9+c7+c8+c9)
AND tc9() BE try(@c9, td1, c1+c2+c3+c4+c5+c6+c7+c8+c9,
a9+b9+c9+d9+e9+f9+g9+h9+i9,
a7+a8+a9+b7+b8+b9+c7+c8+c9)
AND td1() BE try(@d1, td2, d1+d2+d3+d4+d5+d6+d7+d8+d9,
a1+b1+c1+d1+e1+f1+g1+h1+i1,
d1+d2+d3+e1+e2+e3+f1+f2+f3)
AND td2() BE try(@d2, td3, d1+d2+d3+d4+d5+d6+d7+d8+d9,
a2+b2+c2+d2+e2+f2+g2+h2+i2,
d1+d2+d3+e1+e2+e3+f1+f2+f3)
AND td3() BE try(@d3, td4, d1+d2+d3+d4+d5+d6+d7+d8+d9,
a3+b3+c3+d3+e3+f3+g3+h3+i3,
d1+d2+d3+e1+e2+e3+f1+f2+f3)
AND td4() BE try(@d4, td5, d1+d2+d3+d4+d5+d6+d7+d8+d9,
a4+b4+c4+d4+e4+f4+g4+h4+i4,
d4+d5+d6+e4+e5+e6+f4+f5+f6)
AND td5() BE try(@d5, td6, d1+d2+d3+d4+d5+d6+d7+d8+d9,
a5+b5+c5+d5+e5+f5+g5+h5+i5,
d4+d5+d6+e4+e5+e6+f4+f5+f6)
AND td6() BE try(@d6, td7, d1+d2+d3+d4+d5+d6+d7+d8+d9,
a6+b6+c6+d6+e6+f6+g6+h6+i6,
d4+d5+d6+e4+e5+e6+f4+f5+f6)
AND td7() BE try(@d7, td8, d1+d2+d3+d4+d5+d6+d7+d8+d9,
a7+b7+c7+d7+e7+f7+g7+h7+i7,
d7+d8+d9+e7+e8+e9+f7+f8+f9)
AND td8() BE try(@d8, td9, d1+d2+d3+d4+d5+d6+d7+d8+d9,
a8+b8+c8+d8+e8+f8+g8+h8+i8,
d7+d8+d9+e7+e8+e9+f7+f8+f9)
AND td9() BE try(@d9, te1, d1+d2+d3+d4+d5+d6+d7+d8+d9,
a9+b9+c9+d9+e9+f9+g9+h9+i9,
d7+d8+d9+e7+e8+e9+f7+f8+f9)
AND te1() BE try(@e1, te2, e1+e2+e3+e4+e5+e6+e7+e8+e9,
a1+b1+c1+d1+e1+f1+g1+h1+i1,
d1+d2+d3+e1+e2+e3+f1+f2+f3)
AND te2() BE try(@e2, te3, e1+e2+e3+e4+e5+e6+e7+e8+e9,
a2+b2+c2+d2+e2+f2+g2+h2+i2,
d1+d2+d3+e1+e2+e3+f1+f2+f3)
AND te3() BE try(@e3, te4, e1+e2+e3+e4+e5+e6+e7+e8+e9,
a3+b3+c3+d3+e3+f3+g3+h3+i3,
d1+d2+d3+e1+e2+e3+f1+f2+f3)
AND te4() BE try(@e4, te5, e1+e2+e3+e4+e5+e6+e7+e8+e9,
a4+b4+c4+d4+e4+f4+g4+h4+i4,
d4+d5+d6+e4+e5+e6+f4+f5+f6)
AND te5() BE try(@e5, te6, e1+e2+e3+e4+e5+e6+e7+e8+e9,
a5+b5+c5+d5+e5+f5+g5+h5+i5,
d4+d5+d6+e4+e5+e6+f4+f5+f6)
AND te6() BE try(@e6, te7, e1+e2+e3+e4+e5+e6+e7+e8+e9,
a6+b6+c6+d6+e6+f6+g6+h6+i6,
d4+d5+d6+e4+e5+e6+f4+f5+f6)
AND te7() BE try(@e7, te8, e1+e2+e3+e4+e5+e6+e7+e8+e9,
a7+b7+c7+d7+e7+f7+g7+h7+i7,
d7+d8+d9+e7+e8+e9+f7+f8+f9)
AND te8() BE try(@e8, te9, e1+e2+e3+e4+e5+e6+e7+e8+e9,
a8+b8+c8+d8+e8+f8+g8+h8+i8,
d7+d8+d9+e7+e8+e9+f7+f8+f9)
AND te9() BE try(@e9, tf1, e1+e2+e3+e4+e5+e6+e7+e8+e9,
a9+b9+c9+d9+e9+f9+g9+h9+i9,
d7+d8+d9+e7+e8+e9+f7+f8+f9)
AND tf1() BE try(@f1, tf2, f1+f2+f3+f4+f5+f6+f7+f8+f9,
a1+b1+c1+d1+e1+f1+g1+h1+i1,
d1+d2+d3+e1+e2+e3+f1+f2+f3)
AND tf2() BE try(@f2, tf3, f1+f2+f3+f4+f5+f6+f7+f8+f9,
a2+b2+c2+d2+e2+f2+g2+h2+i2,
d1+d2+d3+e1+e2+e3+f1+f2+f3)
AND tf3() BE try(@f3, tf4, f1+f2+f3+f4+f5+f6+f7+f8+f9,
a3+b3+c3+d3+e3+f3+g3+h3+i3,
d1+d2+d3+e1+e2+e3+f1+f2+f3)
AND tf4() BE try(@f4, tf5, f1+f2+f3+f4+f5+f6+f7+f8+f9,
a4+b4+c4+d4+e4+f4+g4+h4+i4,
d4+d5+d6+e4+e5+e6+f4+f5+f6)
AND tf5() BE try(@f5, tf6, f1+f2+f3+f4+f5+f6+f7+f8+f9,
a5+b5+c5+d5+e5+f5+g5+h5+i5,
d4+d5+d6+e4+e5+e6+f4+f5+f6)
AND tf6() BE try(@f6, tf7, f1+f2+f3+f4+f5+f6+f7+f8+f9,
a6+b6+c6+d6+e6+f6+g6+h6+i6,
d4+d5+d6+e4+e5+e6+f4+f5+f6)
AND tf7() BE try(@f7, tf8, f1+f2+f3+f4+f5+f6+f7+f8+f9,
a7+b7+c7+d7+e7+f7+g7+h7+i7,
d7+d8+d9+e7+e8+e9+f7+f8+f9)
AND tf8() BE try(@f8, tf9, f1+f2+f3+f4+f5+f6+f7+f8+f9,
a8+b8+c8+d8+e8+f8+g8+h8+i8,
d7+d8+d9+e7+e8+e9+f7+f8+f9)
AND tf9() BE try(@f9, tg1, f1+f2+f3+f4+f5+f6+f7+f8+f9,
a9+b9+c9+d9+e9+f9+g9+h9+i9,
d7+d8+d9+e7+e8+e9+f7+f8+f9)
AND tg1() BE try(@g1, tg2, g1+g2+g3+g4+g5+g6+g7+g8+g9,
a1+b1+c1+d1+e1+f1+g1+h1+i1,
g1+g2+g3+h1+h2+h3+i1+i2+i3)
AND tg2() BE try(@g2, tg3, g1+g2+g3+g4+g5+g6+g7+g8+g9,
a2+b2+c2+d2+e2+f2+g2+h2+i2,
g1+g2+g3+h1+h2+h3+i1+i2+i3)
AND tg3() BE try(@g3, tg4, g1+g2+g3+g4+g5+g6+g7+g8+g9,
a3+b3+c3+d3+e3+f3+g3+h3+i3,
g1+g2+g3+h1+h2+h3+i1+i2+i3)
AND tg4() BE try(@g4, tg5, g1+g2+g3+g4+g5+g6+g7+g8+g9,
a4+b4+c4+d4+e4+f4+g4+h4+i4,
g4+g5+g6+h4+h5+h6+i4+i5+i6)
AND tg5() BE try(@g5, tg6, g1+g2+g3+g4+g5+g6+g7+g8+g9,
a5+b5+c5+d5+e5+f5+g5+h5+i5,
g4+g5+g6+h4+h5+h6+i4+i5+i6)
AND tg6() BE try(@g6, tg7, g1+g2+g3+g4+g5+g6+g7+g8+g9,
a6+b6+c6+d6+e6+f6+g6+h6+i6,
g4+g5+g6+h4+h5+h6+i4+i5+i6)
AND tg7() BE try(@g7, tg8, g1+g2+g3+g4+g5+g6+g7+g8+g9,
a7+b7+c7+d7+e7+f7+g7+h7+i7,
g7+g8+g9+h7+h8+h9+i7+i8+i9)
AND tg8() BE try(@g8, tg9, g1+g2+g3+g4+g5+g6+g7+g8+g9,
a8+b8+c8+d8+e8+f8+g8+h8+i8,
g7+g8+g9+h7+h8+h9+i7+i8+i9)
AND tg9() BE try(@g9, th1, g1+g2+g3+g4+g5+g6+g7+g8+g9,
a9+b9+c9+d9+e9+f9+g9+h9+i9,
g7+g8+g9+h7+h8+h9+i7+i8+i9)
AND th1() BE try(@h1, th2, h1+h2+h3+h4+h5+h6+h7+h8+h9,
a1+b1+c1+d1+e1+f1+g1+h1+i1,
g1+g2+g3+h1+h2+h3+i1+i2+i3)
AND th2() BE try(@h2, th3, h1+h2+h3+h4+h5+h6+h7+h8+h9,
a2+b2+c2+d2+e2+f2+g2+h2+i2,
g1+g2+g3+h1+h2+h3+i1+i2+i3)
AND th3() BE try(@h3, th4, h1+h2+h3+h4+h5+h6+h7+h8+h9,
a3+b3+c3+d3+e3+f3+g3+h3+i3,
g1+g2+g3+h1+h2+h3+i1+i2+i3)
AND th4() BE try(@h4, th5, h1+h2+h3+h4+h5+h6+h7+h8+h9,
a4+b4+c4+d4+e4+f4+g4+h4+i4,
g4+g5+g6+h4+h5+h6+i4+i5+i6)
AND th5() BE try(@h5, th6, h1+h2+h3+h4+h5+h6+h7+h8+h9,
a5+b5+c5+d5+e5+f5+g5+h5+i5,
g4+g5+g6+h4+h5+h6+i4+i5+i6)
AND th6() BE try(@h6, th7, h1+h2+h3+h4+h5+h6+h7+h8+h9,
a6+b6+c6+d6+e6+f6+g6+h6+i6,
g4+g5+g6+h4+h5+h6+i4+i5+i6)
AND th7() BE try(@h7, th8, h1+h2+h3+h4+h5+h6+h7+h8+h9,
a7+b7+c7+d7+e7+f7+g7+h7+i7,
g7+g8+g9+h7+h8+h9+i7+i8+i9)
AND th8() BE try(@h8, th9, h1+h2+h3+h4+h5+h6+h7+h8+h9,
a8+b8+c8+d8+e8+f8+g8+h8+i8,
g7+g8+g9+h7+h8+h9+i7+i8+i9)
AND th9() BE try(@h9, ti1, h1+h2+h3+h4+h5+h6+h7+h8+h9,
a9+b9+c9+d9+e9+f9+g9+h9+i9,
g7+g8+g9+h7+h8+h9+i7+i8+i9)
AND ti1() BE try(@i1, ti2, i1+i2+i3+i4+i5+i6+i7+i8+i9,
a1+b1+c1+d1+e1+f1+g1+h1+i1,
g1+g2+g3+h1+h2+h3+i1+i2+i3)
AND ti2() BE try(@i2, ti3, i1+i2+i3+i4+i5+i6+i7+i8+i9,
a2+b2+c2+d2+e2+f2+g2+h2+i2,
g1+g2+g3+h1+h2+h3+i1+i2+i3)
AND ti3() BE try(@i3, ti4, i1+i2+i3+i4+i5+i6+i7+i8+i9,
a3+b3+c3+d3+e3+f3+g3+h3+i3,
g1+g2+g3+h1+h2+h3+i1+i2+i3)
AND ti4() BE try(@i4, ti5, i1+i2+i3+i4+i5+i6+i7+i8+i9,
a4+b4+c4+d4+e4+f4+g4+h4+i4,
g4+g5+g6+h4+h5+h6+i4+i5+i6)
AND ti5() BE try(@i5, ti6, i1+i2+i3+i4+i5+i6+i7+i8+i9,
a5+b5+c5+d5+e5+f5+g5+h5+i5,
g4+g5+g6+h4+h5+h6+i4+i5+i6)
AND ti6() BE try(@i6, ti7, i1+i2+i3+i4+i5+i6+i7+i8+i9,
a6+b6+c6+d6+e6+f6+g6+h6+i6,
g4+g5+g6+h4+h5+h6+i4+i5+i6)
AND ti7() BE try(@i7, ti8, i1+i2+i3+i4+i5+i6+i7+i8+i9,
a7+b7+c7+d7+e7+f7+g7+h7+i7,
g7+g8+g9+h7+h8+h9+i7+i8+i9)
AND ti8() BE try(@i8, ti9, i1+i2+i3+i4+i5+i6+i7+i8+i9,
a8+b8+c8+d8+e8+f8+g8+h8+i8,
g7+g8+g9+h7+h8+h9+i7+i8+i9)
AND ti9() BE try(@i9, suc, i1+i2+i3+i4+i5+i6+i7+i8+i9,
a9+b9+c9+d9+e9+f9+g9+h9+i9,
g7+g8+g9+h7+h8+h9+i7+i8+i9)
AND suc() BE { count := count + 1
prboard()
}</lang>
Befunge
Input should be provided as a sequence of 81 digits (optionally separated by whitespace), with zero representing an unknown value.
<lang befunge>99*>1-:0>:#$"0"\# #~`#$_"0"-\::9%:9+00p3/\9/:99++10p3vv%2g\g01< 2%v|:p+9/9\%9:\p\g02\1p\g01\1:p\g00\1:+8:\p02+*93+*3/<>\20g\g#: v<+>:0\`>v >\::9%:9+00p3/\9/:99++10p3/3*+39*+20p\:8+::00g\g2%\^ v^+^pppp$_:|v<::<_>1-::9%\9/9+g.::9%!\3%+>>#v_>" "v..v,<<<+55<< 03!$v9:_>1v$>9%\v^|:<_v#<%<9<:<<_v#+%*93\!::<,,"|"<\/>:#^_>>>v^ p|<$0.0^!g+:#9/9<^@ ^,>#+5<5_>#!<>#$0"------+-------+-----":#<^ <>v$v1:::0<>"P"`!^>0g#0v#p+9/9\%9:p04:\pg03g021pg03g011pg03g001
- >^_:#<0#!:p#-\#1:#g0<>30g010g30g020g30g040g:9%\:9/9+\01-\1+0:</lang>
- Input:
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
- 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
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#
Backtracking
<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>
Best First Search
<lang csharp>using System.Linq; using static System.Linq.Enumerable; using System.Collections.Generic; using System; using System.Runtime.CompilerServices;
namespace SodukoFastMemoBFS {
internal readonly record struct Square (int Row, int Col);
internal record Constraints (IEnumerable<int> ConstrainedRange, Square Square);
internal class Cache : Dictionary<Square, Constraints> { };
internal record CacheGrid (int[][] Grid, Cache Cache);
internal static class SudokuFastMemoBFS {
internal static U Fwd<T, U>(this T data, Func<T, U> f) => f(data);
[MethodImpl(MethodImplOptions.AggressiveInlining)]
private static int RowCol(int rc) => rc <= 2 ? 0 : rc <= 5 ? 3 : 6;
private static bool Solve(this CacheGrid cg, Constraints constraints, int finished) {
var (row, col) = constraints.Square;
foreach (var i in constraints.ConstrainedRange) {
cg.Grid[row][col] = i;
if (cg.Cache.Count == finished || cg.Solve(cg.Next(constraints.Square), finished))
return true;
}
cg.Grid[row][col] = 0;
return false;
}
private static readonly int[] domain = Range(0, 9).ToArray();
private static readonly int[] range = Range(1, 9).ToArray();
private static bool Valid(this int[][] grid, int row, int col, int val) {
for (var i = 0; i < 9; i++)
if (grid[row][i] == val || grid[i][col] == val)
return false;
for (var r = RowCol(row); r < RowCol(row) + 3; r++)
for (var c = RowCol(col); c < RowCol(col) + 3; c++)
if (grid[r][c] == val)
return false;
return true;
}
private static IEnumerable<int> Constraints(this int[][] grid, int row, int col) =>
range.Where(val => grid.Valid(row, col, val));
private static Constraints Next(this CacheGrid cg, Square square) =>
cg.Cache.ContainsKey(square)
? cg.Cache[square]
: cg.Cache[square]=cg.Grid.SortedCells();
private static Constraints SortedCells(this int[][] grid) =>
(from row in domain
from col in domain
where grid[row][col] == 0
let cell = new Constraints(grid.Constraints(row, col), new Square(row, col))
orderby cell.ConstrainedRange.Count() ascending
select cell).First();
private static CacheGrid Parse(string input) =>
input
.Select((c, i) => (index: i, val: int.Parse(c.ToString())))
.GroupBy(id => id.index / 9)
.Select(grp => grp.Select(id => id.val).ToArray())
.ToArray()
.Fwd(grid => new CacheGrid(grid, new Cache()));
public static string AsString(this int[][] grid) =>
string.Join('\n', grid.Select(row => string.Concat(row)));
public static int[][] Run(string input) {
var cg = Parse(input);
var marked = cg.Grid.SelectMany(row => row.Where(c => c > 0)).Count();
return cg.Solve(cg.Grid.SortedCells(), 80 - marked) ? cg.Grid : new int[][] { Array.Empty<int>() };
}
}
}</lang> Usage <lang csharp>using System.Linq; using static System.Linq.Enumerable; using static System.Console; using System.IO;
namespace SodukoFastMemoBFS {
static class Program {
static void Main(string[] args) {
var num = int.Parse(args[0]);
var puzzles = File.ReadLines(@"sudoku17.txt").Take(num);
var single = puzzles.First();
var watch = new System.Diagnostics.Stopwatch();
watch.Start();
WriteLine(SudokuFastMemoBFS.Run(single).AsString());
watch.Stop();
WriteLine($"{single}: {watch.ElapsedMilliseconds} ms");
WriteLine($"Doing {num} puzzles");
var total = 0.0;
watch.Start();
foreach (var puzzle in puzzles) {
watch.Reset();
watch.Start();
SudokuFastMemoBFS.Run(puzzle);
watch.Stop();
total += watch.ElapsedMilliseconds;
Write(".");
}
watch.Stop();
WriteLine($"\nPuzzles:{num}, Total:{total} ms, Average:{total / num:0.00} ms");
ReadKey();
}
}
}</lang> Output
693784512 487512936 125963874 932651487 568247391 741398625 319475268 856129743 274836159 000000010400000000020000000000050407008000300001090000300400200050100000000806000: 336 ms Doing 100 puzzles .................................................................................................... Puzzles:100, Total:5316 ms, Average:53.16 ms
Solver
<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
"Dancing Links"/Algorithm X
<lang csharp>using System; using System.Collections.Generic; using System.Text; using static System.Linq.Enumerable;
public class Sudoku {
public static void Main2() {
string puzzle = "....7.94.....9...53....5.7...74..1..463...........7.8.8........7......28.5.26....";
string solution = new Sudoku().Solutions(puzzle).FirstOrDefault() ?? puzzle;
Print(puzzle, solution);
}
private DLX dlx;
public void Build() {
const int rows = 9 * 9 * 9, columns = 4 * 9 * 9;
dlx = new DLX(rows, columns);
for (int i = 0; i < columns; i++) dlx.AddHeader();
for (int cell = 0, row = 0; row < 9; row++) {
for (int column = 0; column < 9; column++) {
int box = row / 3 * 3 + column / 3;
for (int digit = 0; digit < 9; digit++) {
dlx.AddRow(cell, 81 + row * 9 + digit, 2 * 81 + column * 9 + digit, 3 * 81 + box * 9 + digit);
}
cell++;
}
}
}
public IEnumerable<string> Solutions(string puzzle) {
if (puzzle == null) throw new ArgumentNullException(nameof(puzzle));
if (puzzle.Length != 81) throw new ArgumentException("The input is not of the correct length.");
if (dlx == null) Build();
for (int i = 0; i < puzzle.Length; i++) {
if (puzzle[i] == '0' || puzzle[i] == '.') continue;
if (puzzle[i] < '1' && puzzle[i] > '9') throw new ArgumentException($"Input contains an invalid character: ({puzzle[i]})");
int digit = puzzle[i] - '0' - 1;
dlx.Give(i * 9 + digit);
}
return Iterator();
IEnumerable<string> Iterator() {
var sb = new StringBuilder(new string('.', 81));
foreach (int[] rows in dlx.Solutions()) {
foreach (int r in rows) {
sb[r / 81 * 9 + r / 9 % 9] = (char)(r % 9 + '1');
}
yield return sb.ToString();
}
}
}
static void Print(string left, string right) {
foreach (string line in GetPrintLines(left).Zip(GetPrintLines(right), (l, r) => l + "\t" + r)) {
Console.WriteLine(line);
}
IEnumerable<string> GetPrintLines(string s) {
int r = 0;
foreach (string row in s.Cut(9)) {
yield return r == 0
? "╔═══╤═══╤═══╦═══╤═══╤═══╦═══╤═══╤═══╗"
: r % 3 == 0
? "╠═══╪═══╪═══╬═══╪═══╪═══╬═══╪═══╪═══╣"
: "╟───┼───┼───╫───┼───┼───╫───┼───┼───╢";
yield return "║ " + row.Cut(3).Select(segment => segment.DelimitWith(" │ ")).DelimitWith(" ║ ") + " ║";
r++;
}
yield return "╚═══╧═══╧═══╩═══╧═══╧═══╩═══╧═══╧═══╝";
}
}
}
public class DLX //Some functionality elided {
private readonly Header root = new Header(null, null) { Size = int.MaxValue };
private readonly List<Header> columns;
private readonly List<Node> rows;
private readonly Stack<Node> solutionNodes = new Stack<Node>();
private int initial = 0;
public DLX(int rowCapacity, int columnCapacity) {
columns = new List<Header>(columnCapacity);
rows = new List<Node>(rowCapacity);
}
public void AddHeader() {
Header h = new Header(root.Left, root);
h.AttachLeftRight();
columns.Add(h);
}
public void AddRow(params int[] newRow) {
Node first = null;
if (newRow != null) {
for (int i = 0; i < newRow.Length; i++) {
if (newRow[i] < 0) continue;
if (first == null) first = AddNode(rows.Count, newRow[i]);
else AddNode(first, newRow[i]);
}
}
rows.Add(first);
}
private Node AddNode(int row, int column) {
Node n = new Node(null, null, columns[column].Up, columns[column], columns[column], row);
n.AttachUpDown();
n.Head.Size++;
return n;
}
private void AddNode(Node firstNode, int column) {
Node n = new Node(firstNode.Left, firstNode, columns[column].Up, columns[column], columns[column], firstNode.Row);
n.AttachLeftRight();
n.AttachUpDown();
n.Head.Size++;
}
public void Give(int row) {
solutionNodes.Push(rows[row]);
CoverMatrix(rows[row]);
initial++;
}
public IEnumerable<int[]> Solutions() {
try {
Node node = ChooseSmallestColumn().Down;
do {
if (node == node.Head) {
if (node == root) {
yield return solutionNodes.Select(n => n.Row).ToArray();
}
if (solutionNodes.Count > initial) {
node = solutionNodes.Pop();
UncoverMatrix(node);
node = node.Down;
}
} else {
solutionNodes.Push(node);
CoverMatrix(node);
node = ChooseSmallestColumn().Down;
}
} while(solutionNodes.Count > initial || node != node.Head);
} finally {
Restore();
}
}
private void Restore() {
while (solutionNodes.Count > 0) UncoverMatrix(solutionNodes.Pop());
initial = 0;
}
private Header ChooseSmallestColumn() {
Header traveller = root, choice = root;
do {
traveller = (Header)traveller.Right;
if (traveller.Size < choice.Size) choice = traveller;
} while (traveller != root && choice.Size > 0);
return choice;
}
private void CoverRow(Node row) {
Node traveller = row.Right;
while (traveller != row) {
traveller.DetachUpDown();
traveller.Head.Size--;
traveller = traveller.Right;
}
}
private void UncoverRow(Node row) {
Node traveller = row.Left;
while (traveller != row) {
traveller.AttachUpDown();
traveller.Head.Size++;
traveller = traveller.Left;
}
}
private void CoverColumn(Header column) {
column.DetachLeftRight();
Node traveller = column.Down;
while (traveller != column) {
CoverRow(traveller);
traveller = traveller.Down;
}
}
private void UncoverColumn(Header column) {
Node traveller = column.Up;
while (traveller != column) {
UncoverRow(traveller);
traveller = traveller.Up;
}
column.AttachLeftRight();
}
private void CoverMatrix(Node node) {
Node traveller = node;
do {
CoverColumn(traveller.Head);
traveller = traveller.Right;
} while (traveller != node);
}
private void UncoverMatrix(Node node) {
Node traveller = node;
do {
traveller = traveller.Left;
UncoverColumn(traveller.Head);
} while (traveller != node);
}
private class Node
{
public Node(Node left, Node right, Node up, Node down, Header head, int row) {
Left = left ?? this;
Right = right ?? this;
Up = up ?? this;
Down = down ?? this;
Head = head ?? this as Header;
Row = row;
}
public Node Left { get; set; }
public Node Right { get; set; }
public Node Up { get; set; }
public Node Down { get; set; }
public Header Head { get; }
public int Row { get; }
public void AttachLeftRight() {
this.Left.Right = this;
this.Right.Left = this;
}
public void AttachUpDown() {
this.Up.Down = this;
this.Down.Up = this;
}
public void DetachLeftRight() {
this.Left.Right = this.Right;
this.Right.Left = this.Left;
}
public void DetachUpDown() {
this.Up.Down = this.Down;
this.Down.Up = this.Up;
}
}
private class Header : Node
{
public Header(Node left, Node right) : base(left, right, null, null, null, -1) { }
public int Size { get; set; }
}
}
static class Extensions {
public static IEnumerable<string> Cut(this string input, int length)
{
for (int cursor = 0; cursor < input.Length; cursor += length) {
if (cursor + length > input.Length) yield return input.Substring(cursor);
else yield return input.Substring(cursor, length);
}
}
public static string DelimitWith<T>(this IEnumerable<T> source, string separator) => string.Join(separator, source);
}</lang>
- Output:
╔═══╤═══╤═══╦═══╤═══╤═══╦═══╤═══╤═══╗ ╔═══╤═══╤═══╦═══╤═══╤═══╦═══╤═══╤═══╗ ║ . │ . │ . ║ . │ 7 │ . ║ 9 │ 4 │ . ║ ║ 2 │ 1 │ 5 ║ 8 │ 7 │ 6 ║ 9 │ 4 │ 3 ║ ╟───┼───┼───╫───┼───┼───╫───┼───┼───╢ ╟───┼───┼───╫───┼───┼───╫───┼───┼───╢ ║ . │ . │ . ║ . │ 9 │ . ║ . │ . │ 5 ║ ║ 6 │ 7 │ 8 ║ 3 │ 9 │ 4 ║ 2 │ 1 │ 5 ║ ╟───┼───┼───╫───┼───┼───╫───┼───┼───╢ ╟───┼───┼───╫───┼───┼───╫───┼───┼───╢ ║ 3 │ . │ . ║ . │ . │ 5 ║ . │ 7 │ . ║ ║ 3 │ 4 │ 9 ║ 1 │ 2 │ 5 ║ 8 │ 7 │ 6 ║ ╠═══╪═══╪═══╬═══╪═══╪═══╬═══╪═══╪═══╣ ╠═══╪═══╪═══╬═══╪═══╪═══╬═══╪═══╪═══╣ ║ . │ . │ 7 ║ 4 │ . │ . ║ 1 │ . │ . ║ ║ 5 │ 8 │ 7 ║ 4 │ 3 │ 2 ║ 1 │ 6 │ 9 ║ ╟───┼───┼───╫───┼───┼───╫───┼───┼───╢ ╟───┼───┼───╫───┼───┼───╫───┼───┼───╢ ║ 4 │ 6 │ 3 ║ . │ . │ . ║ . │ . │ . ║ ║ 4 │ 6 │ 3 ║ 9 │ 8 │ 1 ║ 7 │ 5 │ 2 ║ ╟───┼───┼───╫───┼───┼───╫───┼───┼───╢ ╟───┼───┼───╫───┼───┼───╫───┼───┼───╢ ║ . │ . │ . ║ . │ . │ 7 ║ . │ 8 │ . ║ ║ 1 │ 9 │ 2 ║ 6 │ 5 │ 7 ║ 3 │ 8 │ 4 ║ ╠═══╪═══╪═══╬═══╪═══╪═══╬═══╪═══╪═══╣ ╠═══╪═══╪═══╬═══╪═══╪═══╬═══╪═══╪═══╣ ║ 8 │ . │ . ║ . │ . │ . ║ . │ . │ . ║ ║ 8 │ 2 │ 6 ║ 7 │ 4 │ 3 ║ 5 │ 9 │ 1 ║ ╟───┼───┼───╫───┼───┼───╫───┼───┼───╢ ╟───┼───┼───╫───┼───┼───╫───┼───┼───╢ ║ 7 │ . │ . ║ . │ . │ . ║ . │ 2 │ 8 ║ ║ 7 │ 3 │ 4 ║ 5 │ 1 │ 9 ║ 6 │ 2 │ 8 ║ ╟───┼───┼───╫───┼───┼───╫───┼───┼───╢ ╟───┼───┼───╫───┼───┼───╫───┼───┼───╢ ║ . │ 5 │ . ║ 2 │ 6 │ . ║ . │ . │ . ║ ║ 9 │ 5 │ 1 ║ 2 │ 6 │ 8 ║ 4 │ 3 │ 7 ║ ╚═══╧═══╧═══╩═══╧═══╧═══╩═══╧═══╧═══╝ ╚═══╧═══╧═══╩═══╧═══╧═══╩═══╧═══╧═══╝
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("850002400"
"720000009"
"004000000"
"000107002"
"305000900"
"040000000"
"000080070"
"017000000"
"000036040");
ss.solve();
return EXIT_SUCCESS;
}</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>
EasyLang
<lang>len row[] 810 len col[] 810 len box[] 810 len grid[] 82
func init . .
for pos range 81
if pos mod 9 = 0
s$[] = str_split input " "
.
dig = number s$[pos mod 9]
grid[pos] = dig
r = pos div 9
c = pos mod 9
b = r div 3 * 3 + c div 3
row[r * 10 + dig] = 1
col[c * 10 + dig] = 1
box[b * 10 + dig] = 1
.
. call init
func display . .
for i range 81
write grid[i] & " "
if i mod 3 = 2
write " "
.
if i mod 9 = 8
print ""
.
if i mod 27 = 26
print ""
.
.
.
func solve pos . .
while grid[pos] <> 0
pos += 1
.
if pos = 81
# solved
call display
break 1
.
r = pos div 9
c = pos mod 9
b = r div 3 * 3 + c div 3
r *= 10
c *= 10
b *= 10
for d = 1 to 9
if row[r + d] = 0 and col[c + d] = 0 and box[b + d] = 0
grid[pos] = d
row[r + d] = 1
col[c + d] = 1
box[b + d] = 1
call solve pos + 1
row[r + d] = 0
col[c + d] = 0
box[b + d] = 0
.
.
grid[pos] = 0
. call solve 0
input_data 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</lang>
Elixir
<lang elixir>defmodule Sudoku do
def display( grid ), do: ( for y <- 1..9, do: display_row(y, grid) )
def start( knowns ), do: Enum.into( knowns, Map.new )
def solve( grid ) do
sure = solve_all_sure( grid )
solve_unsure( potentials(sure), sure )
end
def task( knowns ) do
IO.puts "start"
start = start( knowns )
display( start )
IO.puts "solved"
solved = solve( start )
display( solved )
IO.puts ""
end
defp bt( grid ), do: bt_reject( is_not_allowed(grid), grid )
defp bt_accept( true, board ), do: throw( {:ok, board} )
defp bt_accept( false, grid ), do: bt_loop( potentials_one_position(grid), grid )
defp bt_loop( {position, values}, grid ), do: ( for x <- values, do: bt( Map.put(grid, position, x) ) )
defp bt_reject( true, _grid ), do: :backtrack
defp bt_reject( false, grid ), do: bt_accept( is_all_correct(grid), grid )
defp display_row( row, grid ) do
for x <- [1, 4, 7], do: display_row_group( x, row, grid )
display_row_nl( row )
end
defp display_row_group( start, row, grid ) do
Enum.each(start..start+2, &IO.write " #{Map.get( grid, {&1, row}, ".")}")
IO.write " "
end
defp display_row_nl( n ) when n in [3,6,9], do: IO.puts "\n"
defp display_row_nl( _n ), do: IO.puts ""
defp is_all_correct( grid ), do: map_size( grid ) == 81
defp is_not_allowed( grid ) do
is_not_allowed_rows( grid ) or is_not_allowed_columns( grid ) or is_not_allowed_groups( grid )
end
defp is_not_allowed_columns( grid ), do: values_all_columns(grid) |> Enum.any?(&is_not_allowed_values/1)
defp is_not_allowed_groups( grid ), do: values_all_groups(grid) |> Enum.any?(&is_not_allowed_values/1)
defp is_not_allowed_rows( grid ), do: values_all_rows(grid) |> Enum.any?(&is_not_allowed_values/1)
defp is_not_allowed_values( values ), do: length( values ) != length( Enum.uniq(values) )
defp group_positions( {x, y} ) do
for colum <- group_positions_close(x), row <- group_positions_close(y), do: {colum, row}
end
defp group_positions_close( n ) when n < 4, do: [1,2,3]
defp group_positions_close( n ) when n < 7, do: [4,5,6]
defp group_positions_close( _n ) , do: [7,8,9]
defp positions_not_in_grid( grid ) do
keys = Map.keys( grid )
for x <- 1..9, y <- 1..9, not {x, y} in keys, do: {x, y}
end
defp potentials_one_position( grid ) do
Enum.min_by( potentials( grid ), fn {_position, values} -> length(values) end )
end
defp potentials( grid ), do: List.flatten( for x <- positions_not_in_grid(grid), do: potentials(x, grid) )
defp potentials( position, grid ) do
useds = potentials_used_values( position, grid )
{position, Enum.to_list(1..9) -- useds }
end
defp potentials_used_values( {x, y}, grid ) do
row_values = (for row <- 1..9, row != x, do: {row, y}) |> potentials_values( grid )
column_values = (for column <- 1..9, column != y, do: {x, column}) |> potentials_values( grid )
group_values = group_positions({x, y}) -- [ {x, y} ] |> potentials_values( grid )
row_values ++ column_values ++ group_values
end
defp potentials_values( keys, grid ) do
for x <- keys, val = grid[x], do: val
end
defp values_all_columns( grid ) do
for x <- 1..9, do:
( for y <- 1..9, do: {x, y} ) |> potentials_values( grid )
end
defp values_all_groups( grid ) do
[[g1,g2,g3], [g4,g5,g6], [g7,g8,g9]] = for x <- [1,4,7], do: values_all_groups(x, grid)
[g1,g2,g3,g4,g5,g6,g7,g8,g9]
end
defp values_all_groups( x, grid ) do
for x_offset <- x..x+2, do: values_all_groups(x, x_offset, grid)
end
defp values_all_groups( _x, x_offset, grid ) do
( for y_offset <- group_positions_close(x_offset), do: {x_offset, y_offset} )
|> potentials_values( grid )
end
defp values_all_rows( grid ) do
for y <- 1..9, do:
( for x <- 1..9, do: {x, y} ) |> potentials_values( grid )
end
defp solve_all_sure( grid ), do: solve_all_sure( solve_all_sure_values(grid), grid )
defp solve_all_sure( [], grid ), do: grid
defp solve_all_sure( sures, grid ) do
solve_all_sure( Enum.reduce(sures, grid, &solve_all_sure_store/2) )
end
defp solve_all_sure_values( grid ), do: (for{position, [value]} <- potentials(grid), do: {position, value} )
defp solve_all_sure_store( {position, value}, acc ), do: Map.put( acc, position, value )
defp solve_unsure( [], grid ), do: grid
defp solve_unsure( _potentials, grid ) do
try do
bt( grid )
catch
{:ok, board} -> board
end
end
end
simple = [{{1, 1}, 3}, {{2, 1}, 9}, {{3, 1},4}, {{6, 1}, 2}, {{7, 1}, 6}, {{8, 1}, 7},
{{4, 2}, 3}, {{7, 2}, 4},
{{1, 3}, 5}, {{4, 3}, 6}, {{5, 3}, 9}, {{8, 3}, 2},
{{2, 4}, 4}, {{3, 4}, 5}, {{7, 4}, 9},
{{1, 5}, 6}, {{9, 5}, 7},
{{3, 6}, 7}, {{7, 6}, 5}, {{8, 6}, 8},
{{2, 7}, 1}, {{5, 7}, 6}, {{6, 7}, 7}, {{9, 7}, 8},
{{3, 8}, 9}, {{6, 8}, 8},
{{2, 9}, 2}, {{3, 9}, 6}, {{4, 9}, 4}, {{7, 9}, 7}, {{8, 9}, 3}, {{9, 9}, 5}]
Sudoku.task( simple )
difficult = [{{6, 2}, 3}, {{8, 2}, 8}, {{9, 2}, 5},
{{3, 3}, 1}, {{5, 3}, 2},
{{4, 4}, 5}, {{6, 4}, 7},
{{3, 5}, 4}, {{7, 5}, 1},
{{2, 6}, 9},
{{1, 7}, 5}, {{8, 7}, 7}, {{9, 7}, 3},
{{3, 8}, 2}, {{5, 8}, 1},
{{5, 9}, 4}, {{9, 9}, 9}]
Sudoku.task( difficult )</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>
F#
Backtracking
<lang fsharp>module SudokuBacktrack
//Helpers let tuple2 a b = a,b let flip f a b = f b a let (>>=) f g = Option.bind g f
/// "A1" to "I9" squares as key in values dictionary let key a b = $"{a}{b}"
/// Cross product of elements in ax and elements in bx let cross ax bx = [| for a in ax do for b in bx do key a b |]
// constants let valid = "1234567890.," let rows = "ABCDEFGHI" let cols = "123456789" let squares = cross rows cols
// List of all row, cols and boxes: aka units let unitList =
[for c in cols do cross rows (string c) ]@ // row units [for r in rows do cross (string r) cols ]@ // col units [for rs in ["ABC";"DEF";"GHI"] do for cs in ["123";"456";"789"] do cross rs cs ] // box units
/// Dictionary of units for each square let units =
[for s in squares do s, [| for u in unitList do if u |> Array.contains s then u |] ] |> Map.ofSeq
/// Dictionary of all peer squares in the relevant units wrt square in question let peers =
[for s in squares do units[s] |> Array.concat |> Array.distinct |> Array.except [s] |> tuple2 s] |> Map.ofSeq
/// Should parse grid in many input formats or return None let parseGrid grid =
let ints = [for c in grid do if valid |> Seq.contains c then if ",." |> Seq.contains c then 0 else (c |> string |> int)] if Seq.length ints = 81 then ints |> Seq.zip squares |> Map.ofSeq |> Some else None
/// Outputs single line puzzle with 0 as empty squares let asString = function
| Some values -> values |> Map.toSeq |> Seq.map (snd>>string) |> String.concat "" | _ -> "No solution or Parse Failure"
/// Outputs puzzle in 2D format with 0 as empty squares let prettyPrint = function
| Some (values:Map<_,_>) ->
[for r in rows do [for c in cols do (values[key r c] |> string) ] |> String.concat " " ] |> String.concat "\n"
| _ -> "No solution or Parse Failure"
/// Is digit allowed in the square in question? !!! hot path !!!! /// Array/Array2D no faster and they need explicit copy since not immutable let constraints (values:Map<_,_>) s d = peers[s] |> Seq.map (fun p -> values[p]) |> Seq.exists ((=) d) |> not
/// Move to next square or None if out of bounds let next s = squares |> Array.tryFindIndex ((=)s) |> function Some i when i + 1 < 81 -> Some squares[i + 1] | _ -> None
/// Backtrack recursively and immutably from index let rec backtracker (values:Map<_,_>) = function
| None -> Some values // solved!
| Some s when values[s] > 0 -> backtracker values (next s) // square not empty
| Some s ->
let rec tracker = function
| [] -> None
| d::dx ->
values
|> Map.change s (Option.map (fun _ -> d))
|> flip backtracker (next s)
|> function
| None -> tracker dx
| success -> success
[for d in 1..9 do if constraints values s d then d] |> tracker
/// solve sudoku using simple backtracking let solve grid = grid |> parseGrid >>= flip backtracker (Some "A1")</lang> Usage: <lang fsharp>open System open SudokuBacktrack
[<EntryPoint>] let main argv =
let puzzle = "000028000800010000000000700000600403200004000100700000030400500000000010060000000"
puzzle |> printfn "Puzzle:\n%s"
puzzle |> parseGrid |> prettyPrint |> printfn "Formatted:\n%s"
puzzle |> solve |> prettyPrint |> printfn "Solution:\n%s"
printfn "Press any key to exit"
Console.ReadKey() |> ignore
0</lang>
- Output:
Puzzle: 000028000800010000000000700000600403200004000100700000030400500000000010060000000 Formatted: 0 0 0 0 2 8 0 0 0 8 0 0 0 1 0 0 0 0 0 0 0 0 0 0 7 0 0 0 0 0 6 0 0 4 0 3 2 0 0 0 0 4 0 0 0 1 0 0 7 0 0 0 0 0 0 3 0 4 0 0 5 0 0 0 0 0 0 0 0 0 1 0 0 6 0 0 0 0 0 0 0 Solution: 6 1 7 3 2 8 9 4 5 8 9 4 5 1 7 2 3 6 3 2 5 9 4 6 7 8 1 9 7 8 6 5 1 4 2 3 2 5 6 8 3 4 1 7 9 1 4 3 7 9 2 6 5 8 7 3 1 4 8 9 5 6 2 4 8 9 2 6 5 3 1 7 5 6 2 1 7 3 8 9 4 Press any key to exit
Constraint Satisfaction (Norvig)
<lang fsharp>// https://norvig.com/sudoku.html module SudokuCPS
// Throughout this program we have:
// r is a row, e.g. "A"
// c is a column, e.g. "3"
// s is a square, e.g. "A3"
// d is a digit, e.g. 9
// u is a unit, e.g. ["A1","B1","C1","D1","E1","F1","G1","H1","I1"]
// g is a grid, e.g. 81 non-blank chars, e.g. starting with ".18...7...
// values is a dict of possible values, e.g. {"A1":seq{1;2;3;4;8;9}, "A2":seq{8}, ...}
open System // helpers let tuple2 a b = a,b let (>>=) f g = Result.bind g f
/// folds folder returning Some on compleition or returns None if not let all folder state = Seq.fold (fun acc cur -> acc >>= (fun st -> folder st cur)) state
/// "A1" to "I9" squares as key in values dictionary let key a b = $"{a}{b}"
/// Cross product of elements in ax and elements in bx let cross ax bx = [| for a in ax do for b in bx do key a b |]
// constants let digits = [|1..9|] let rows = "ABCDEFGHI" let cols = "123456789" let empty = "0,." let valid = cols+empty let squares = cross rows cols
// List of all row, cols and boxes: aka units let unitlist =
[for c in cols do cross rows (string c) ]@ [for r in rows do cross (string r) cols ]@ [for rs in ["ABC";"DEF";"GHI"] do for cs in ["123";"456";"789"] do cross rs cs ]
/// Dictionary of units for each square let units =
[for s in squares do s, [| for u in unitlist do if u |> Array.contains s then u |] ] |> Map.ofList
/// Dictionary of all peer squares in the relevant units wrt square in question let peers =
[for s in squares do units.[s] |> Array.concat |> Array.distinct |> Array.except [s] |> tuple2 s] |> Map.ofList
/// Eliminate d from values[s] and propagate when values = 1. /// Return Some values, except return None if a contradiction is detected. let rec eliminate (values:Map<_,int[]>) (s:string) d =
let peerElim (vx:Map<_,_>) =
match Seq.length vx[s] with
| 0 -> Error "peer contradiction" // removed last value
| 1 -> peers[s] |> all (fun st s2 -> eliminate st s2 (vx[s] |> Seq.head) ) (Ok vx)
| _ -> Ok vx
let unitsElim vx =
units[s]
|> all (fun (st:Map<_,_>) u ->
let dKeys = [for s in u do if st[s] |> Seq.contains d then s]
match Seq.length dKeys with
| 0 -> Error "units contradiction"
| 1 -> assign st (Seq.head dKeys) d
| _ -> Ok st) (Ok vx)
match values[s] |> Seq.contains d with
| false -> Ok values // Already eliminated, nothing to do
| true -> // (1) If a square s is reduced to one value d, then eliminate d from the peers.
let values2 = values |> Map.change s (Option.map (fun dx -> dx |> Array.filter ((<>)d)))
peerElim values2
|> function // (2) If a unit u is reduced to only one place for a value d, then put it there.
| Error error -> Error error // from peers
| Ok values3 -> unitsElim values3
/// Eliminate all the other values (except d) from values[s] and propagate. /// Return Some values, except None if a contradiction is detected. and assign (values:Map<_,_>) (s:string) d =
values[s] |> Seq.filter ((<>)d) |> all (fun st d2 -> eliminate st s d2) (Ok values)
/// Convert grid into a Map of {square: char} with "0","."or"," for empties. let parseGrid grid =
let cells = [for c in grid do if valid |> Seq.contains c then if empty |> Seq.contains c then 0 else ((string>>int) c)]
if Seq.length cells = 81 then cells |> Seq.zip squares |> Map.ofSeq |> Ok else Error $"parseGrid length ={Seq.length cells}"
/// Convert grid to a Map of constraint popagated possible values, or return None if a contradiction is detected. let applyCPS grid =
let values = [for s in squares do s, digits]|> Map.ofList
parseGrid grid
>>= (Seq.filter (fun sd -> digits |> Seq.contains sd.Value)
>> all (fun vx sd -> assign vx sd.Key sd.Value) (Ok values) )
/// Calculate string centre for each square - which can contain more than 1 digit when debugging let centre s width =
let n = width - (Seq.length s)
if n <= 0 then s
else
let half = n/2 + ( if (n%2>0 && width%2>0) then 1 else 0)
sprintf "%s%s%s" (String(' ',half)) s (String(' ', n - half))
/// Display these values as a 2-D grid. Used for debugging let prettyPrint (valuesOpt:Result<Map<_,_>,_>) =
let asString = Seq.map string >> String.concat ""
match valuesOpt with
| Error error -> error
| Ok values ->
let width = 1 + ([for s in squares do Seq.length values[s]] |> List.max)
let line = sprintf "%s\n" ((String('-',width*3) |> Seq.replicate 3) |> String.concat "+")
[for r in rows do
for c in cols do
sprintf "%s%s" (centre (asString values[key r c]) width) (if "36".Contains c then "|" else "")
sprintf "\n%s"(if "CF".Contains r then line else "") ]
|> String.concat ""
/// Outputs single line puzzle with 0 as empty squares let asString = function
| Ok values -> values |> Map.toSeq |> Seq.map (snd>>string) |> String.concat "" | Error error -> error
/// Using depth-first search and propagation, try all possible values. let rec search (values:Map<_,_>): Result<Map<_,_>,_> =
let rec some = function
| [] -> Error "No solution"
| s::sx ->
values[s]
|> Seq.tryPick (fun d -> assign values s d >>= search |> function Error _ -> None | Ok seq -> Some seq)
|> function
| Some seqx when Seq.isEmpty seqx -> some sx
| Some seq -> Ok seq
| _ -> Error "No solution"
// Choose the unfilled square(s) s with the fewest possibilities
[for s in squares do if Seq.length values[s] > 1 then Seq.length values[s] ,s]
|> function
| seqx when Seq.isEmpty seqx -> Ok values // solved!
| seqx -> seqx |> List.sortBy fst |> List.map snd|> some
/// Solve Sudoku using Constraint Propagation let solve grid = grid |> applyCPS >>= search |> prettyPrint</lang> Usage<lang fsharp>open System open SudokuCPS
[<EntryPoint>] let main argv =
printfn "Easy board solution automatic with constraint propagation"
let easy = "..3.2.6..9..3.5..1..18.64....81.29..7.......8..67.82....26.95..8..2.3..9..5.1.3.."
easy |> applyCPS |> prettyPrint |> printfn "%s"
printfn "Simple elimination not possible"
let simple = "4.....8.5.3..........7......2.....6.....8.4......1.......6.3.7.5..2.....1.4......"
simple |> parseGrid |> asString |> printfn "%s\n"
simple |> applyCPS |> prettyPrint |> printfn "%s"
printfn "Try again with search:"
simple |> parseGrid |> Result.map (Map.toSeq >> Seq.map (fun (k,v) -> k , string v) >> Map.ofSeq) |> prettyPrint |> printfn "%s"
simple |> solve |> printfn "%s"
let watch = Stopwatch()
let hard = "85...24..72......9..4.........1.7..23.5...9...4...........8..7..17..........36.4."
printfn "Hard"
watch.Start()
hard |> applyCPS >>= search |> prettyPrint |> printfn "%s"
watch.Stop()
printfn $"Elapsed milliseconds = {watch.ElapsedMilliseconds } ms"
watch.Reset()
let num = 10000 // http://www.csse.uwa.edu.au/~gordon/sudoku17 printfn $"First {num} puzzles in sudoku17" let puzzles = File.ReadLines(@"sudoku17.txt") |> Seq.take num |> List.ofSeq watch.Start() let result = puzzles |> List.map (fun p -> p |> applyCPS >>= search |> asString) watch.Stop() let total = watch.ElapsedMilliseconds printf $"Puzzles:{num}, Total:{total} ms, Average:{((float total) /(float num))} ms" printfn " (i7500U @2.7GHz CPU, 16GB Ram)"
Console.ReadKey() |> ignore
0
</lang>
- Output:
Easy board solution automatic with constraint propagation4 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
Simple elimination not possible 400000805030000000000700000020000060000080400000010000000603070500200000104000000
4 1679 12679 | 139 2369 269 | 8 1239 5 26789 3 1256789 | 14589 24569 245689 | 12679 1249 124679 2689 15689 125689 | 7 234569 245689 | 12369 12349 123469 ------------------------+------------------------+------------------------ 3789 2 15789 | 3459 34579 4579 | 13579 6 13789 3679 15679 15679 | 359 8 25679 | 4 12359 12379 36789 4 56789 | 359 1 25679 | 23579 23589 23789 ------------------------+------------------------+------------------------ 289 89 289 | 6 459 3 | 1259 7 12489 5 6789 3 | 2 479 1 | 69 489 4689 1 6789 4 | 589 579 5789 | 23569 23589 23689Try again with search: 4 0 0 |0 0 0 |8 0 5 0 3 0 |0 0 0 |0 0 0 0 0 0 |7 0 0 |0 0 0 ------+------+------ 0 2 0 |0 0 0 |0 6 0 0 0 0 |0 8 0 |4 0 0 0 0 0 |0 1 0 |0 0 0 ------+------+------ 0 0 0 |6 0 3 |0 7 0 5 0 0 |2 0 0 |0 0 0 1 0 4 |0 0 0 |0 0 0
4 1 7 |3 6 9 |8 2 5 6 3 2 |1 5 8 |9 4 7 9 5 8 |7 2 4 |3 1 6 ------+------+------ 8 2 5 |4 3 7 |1 6 9 7 9 1 |5 8 6 |4 3 2 3 4 6 |9 1 2 |7 5 8 ------+------+------ 2 8 9 |6 4 3 |5 7 1 5 7 3 |2 9 1 |6 8 4 1 6 4 |8 7 5 |2 9 3
Hard 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
Elapsed milliseconds = 8 ms 10000 puzzles in sudoku17
Puzzles:10000, Total:46090 ms, Average:4.609 ms (i7500U @2.7GHz CPU, 16GB Ram)
SLPsolve
<lang fsharp> // Solve Sudoku Like Puzzles. Nigel Galloway: September 6th., 2018 let fN y n g=let _q n' g'=[for n in n*n'..n*n'+n-1 do for g in g*g'..g*g'+g-1 do yield (n,g)]
let N=[|for n in 0..(y/n)-1 do for g in 0..(y/g)-1 do yield _q n g|]
(fun n' g'->N.[((n'/n)*n+g'/g)])
let fI n=let _q g=[for n in 0..n-1 do yield (g,n)]
let N=[|for n in 0..n-1 do yield _q n|]
(fun g (_:int)->N.[g])
let fG n=let _q g=[for n in 0..n-1 do yield (n,g)]
let N=[|for n in 0..n-1 do yield _q n|]
(fun (_:int) n->N.[n])
let fE v z n g fn=let N,G,B,fn=fI z,fG z,fN z n g,readCSV ',' fn|>List.ofSeq
let fG n g mask=List.except (N n g@G n g@B n g) mask
let b=List.except(List.map(fun n->(n.row,n.col))fn)[for n in 0..z-1 do for g in 0..z-1 do yield (n,g)]
let q=Map.ofList[for v' in v do yield ((v'),List.choose(fun n->if n.value=v' then Some(n.row,n.col) else None)fn|>List.fold(fun z (n,g)->(n,g)::fG n g z)b)]
(fG,(fun n->Map.find n q),z,v)
let SLPsolve (N,G,z,l)=
let rec nQueens col mask res=[
if col=z then yield res else
yield! List.filter(fun(n,_)->n=col)mask|>List.collect(fun(n,g)->nQueens (col+1) (N n g mask) ((n,g)::res))]
let rec sudoku l res=seq{
match l with
|h::t->let n=nQueens 0 (List.except (List.concat res) (G h)) []
yield! n|>Seq.collect(fun n->sudoku t (n::res))
|_->yield res}
sudoku l []
let printSLP n g=
List.map2(fun n g->List.map(fun(n',g')->((n',g'),n))g) (List.rev n) g|>List.concat|>List.sortBy (fun ((_,n),_)->n)|>List.groupBy(fun ((n,_),_)->n)|>List.sortBy(fun(n,_)->n) |>List.iter(fun (_,n)->n|>Seq.fold(fun z ((_,g),v)->[z..g-1]|>Seq.iter(fun _->printf " |");printf "%s|" v; g+1 ) 0 |>ignore;printfn "")
</lang> Usage: Given sud1.csv:
7,1,,4,,6,,2 ,,,,7,,,,3 4,,,,,1,,8 ,,,,1,3,,,9 ,,1,,,,7 2,,,8,6 ,2,,1,,,,,4 9,,,,8 ,7,,6,,4,,5,2
then <lang fsharp> let n=SLPsolve (fE ([1..9]|>List.map(string)) 9 3 3 "sud1.csv") printSLP ([1..9]|>List.map(string)) (Seq.item 0 n) </lang>
- Output:
7|1|8|4|3|6|9|2|5| 5|6|2|9|7|8|4|1|3| 4|3|9|5|2|1|6|8|7| 6|8|5|7|1|3|2|4|9| 3|9|1|2|4|5|7|6|8| 2|4|7|8|6|9|5|3|1| 8|2|6|1|5|7|3|9|4| 9|5|4|3|8|2|1|7|6| 1|7|3|6|9|4|8|5|2|
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|+-----+-----+-----+
FreeBASIC
<lang freebasic>Dim Shared As Integer cuadricula(9, 9), cuadriculaResuelta(9, 9)
Function isSafe(i As Integer, j As Integer, n As Integer) As Boolean
Dim As Integer iMin, jMin, f, c
If cuadricula(i, j) <> 0 Then Return (cuadricula(i, j) = n)
'cuadricula(i, j) es una celda vacía. Compruebe si n está OK
'primero revisa la fila i
For f = 1 To 9
If cuadricula(i, f) = n Then Return False
Next f
'ahora comprueba la columna j
For c = 1 To 9
If cuadricula(c, j) = n Then Return False
Next c
'finalmente, compruebe el subcuadrado de 3x3 que contiene cuadricula(i,j)
iMin = 1 + 3 * Int((i - 1) / 3)
jMin = 1 + 3 * Int((j - 1) / 3)
For c = iMin To iMin + 2
For f = jMin To jMin + 2
If cuadricula(c, f) = n Then Return False
Next f
Next c
'todas las pruebas estuvieron OK
Return True
End Function
Sub Resolver(i As Integer, j As Integer)
Dim As Integer f, c, n, temp
If i > 9 Then
'salir con cuadriculaResuelta = cuadricula
For c = 1 To 9
For f = 1 To 9
cuadriculaResuelta(c, f) = cuadricula(c, f)
Next f
Next c
Exit Sub
End If
For n = 1 To 9
If isSafe(i, j, n) Then
temp = cuadricula(i, j)
cuadricula(i, j) = n
If j = 9 Then
Resolver i + 1, 1
Else
Resolver i, j + 1
End If
cuadricula(i, j) = temp
End If
Next n
End Sub
Dim As String s(9) 'inicializar la cuadrícula usando 9 cadenas, una por fila 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"
Dim As Integer i, j For i = 1 To 9
For j = 1 To 9
cuadricula(i, j) = Int(Val(Mid(s(i), j, 1)))
Next j
Next i
Resolver 1, 1 Print "Solucion:" Color 12: Print "---------+---------+---------" For i = 1 To 9
For j = 1 To 9
Color 7: Print cuadriculaResuelta(i, j); " ";
Color 12
If (j Mod 3 = 0) And (j <> 9) Then Color 12: Print "|";
Next j
If (i Mod 3 = 0) Then Print !"\n---------+---------+---------" Else Print
Next i Sleep</lang>
- Output:
Solucion: ---------+---------+--------- 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 ---------+---------+---------
FutureBasic
First is a short version: <lang futurebasic> include "ConsoleWindow" include "NSLog.incl" include "Util_Containers.incl"
begin globals dim as container gC end globals
BeginCDeclaration short solve_sudoku(short i); short check_sudoku(short r, short c); CFMutableStringRef print_sudoku(); EndC
BeginCFunction short sudoku[9][9] = {
{3,0,0,0,0,1,4,0,9},
{7,0,0,0,0,4,2,0,0},
{0,5,0,2,0,0,0,1,0},
{5,7,0,0,4,3,0,6,0},
{0,9,0,0,0,0,0,3,0},
{0,6,0,7,9,0,0,8,5},
{0,8,0,0,0,5,0,4,0},
{0,0,6,4,0,0,0,0,7},
{9,0,5,6,0,0,0,0,3},
};
short check_sudoku( short r, short c )
{
short i; short rr, cc;
for (i = 0; i < 9; i++)
{
if (i != c && sudoku[r][i] == sudoku[r][c]) return 0;
if (i != r && sudoku[i][c] == sudoku[r][c]) return 0;
rr = r/3 * 3 + i/3;
cc = c/3 * 3 + i%3;
if ((rr != r || cc != c) && sudoku[rr][cc] == sudoku[r][c]) return 0;
}
return -1;
}
short solve_sudoku( short i )
{
short r, c;
if (i < 0) return 0; else if (i >= 81) return -1;
r = i / 9; c = i % 9;
if (sudoku[r][c])
return check_sudoku(r, c) && solve_sudoku(i + 1);
else
for (sudoku[r][c] = 9; sudoku[r][c] > 0; sudoku[r][c]--)
{
if ( solve_sudoku(i) ) return -1;
}
return 0;
}
CFMutableStringRef print_sudoku()
{
short i, j; CFMutableStringRef mutStr; mutStr = CFStringCreateMutable( kCFAllocatorDefault, 0 );
for (i = 0; i < 9; i++)
{
for (j = 0; j < 9; j++)
{
CFStringAppendFormat( mutStr, NULL, (CFStringRef)@" %d", sudoku[i][j] );
}
CFStringAppendFormat( mutStr, NULL, (CFStringRef)@"\r" );
}
return( mutStr );
} EndC
toolbox fn solve_sudoku( short i ) = short toolbox fn check_sudoku( short r, short c ) = short toolbox fn print_sudoku() = CFMutableStringRef
dim as short solution dim as CFMutableStringRef cfRef
gC = " " cfRef = fn print_sudoku() fn ContainerCreateWithCFString( cfRef, gC ) print : print "Sudoku challenge:" : print : print gC
solution = fn solve_sudoku(0)
print : print "Sudoku solved:" : print if ( solution ) gC = " " cfRef = fn print_sudoku() fn ContainerCreateWithCFString( cfRef, gC ) print gC else print "No solution found" end if </lang>
Output:
Sudoku challenge: 3 0 0 0 0 1 4 0 9 7 0 0 0 0 4 2 0 0 0 5 0 2 0 0 0 1 0 5 7 0 0 4 3 0 6 0 0 9 0 0 0 0 0 3 0 0 6 0 7 9 0 0 8 5 0 8 0 0 0 5 0 4 0 0 0 6 4 0 0 0 0 7 9 0 5 6 0 0 0 0 3 Sudoku solved: 3 2 8 5 6 1 4 7 9 7 1 9 3 8 4 2 5 6 6 5 4 2 7 9 3 1 8 5 7 1 8 4 3 9 6 2 8 9 2 1 5 6 7 3 4 4 6 3 7 9 2 1 8 5 2 8 7 9 3 5 6 4 1 1 3 6 4 2 8 5 9 7 9 4 5 6 1 7 8 2 3
More code in this one, but faster execution:
include "ConsoleWindow" include "Tlbx Timer.incl" begin globals _digits = 9 _setH = 3 _setV = 3 _nSetH = 3 _nSetV = 3 begin record Board dim as boolean f(_digits,_digits,_digits) dim as char match(_digits,_digits) dim as pointer previousBoard // singly-linked list used to discover repetitions dim && end record dim quiz as board dim as long t dim as double sProgStartTime end globals // 'ordinary' timer used for playing local fn Milliseconds as long // time in ms since prog start '~'1 dim as UnsignedWide us long if ( sProgStartTime == 0.0 ) Microseconds( @us ) sProgStartTime = 4294967296.0*us.hi + us.lo end if Microseconds( @us ) end fn = (4294967296.0*us.hi + us.lo - sProgStartTime)'*1e-3 local fn InitMilliseconds '~'1 sProgStartTime = 0.0 fn Milliseconds end fn local mode local fn CopyBoard( source as ^Board, dest as ^Board ) '~'1 BlockMoveData( source, dest, sizeof( Board ) ) dest.previousBoard = source // linked list end fn local fn prepare( b as ^Board ) '~'1 dim as short i, j, n for i = 1 to _digits for j = 1 to _digits for n = 1 to _digits b.match[i, j] = 0 b.f[i, j, n] = _true next n next j next i end fn local fn printBoard( b as ^Board ) '~'1 dim as short i, j for i = 1 to _digits for j = 1 to _digits Print b.match[i, j]; next j print next i end fn local fn verifica( b as ^Board ) '~'1 dim as short i, j, n, first, x, y, ii dim as boolean check check = _true for i = 1 to _digits for j = 1 to _digits long if ( b.match[i, j] == 0 ) check = _false for n = 1 to _digits long if ( b.f[i, j, n] != _false ) check = _true end if next n if ( check == _false ) then exit fn end if next j next i check = _true for j = 1 to _digits for n = 1 to _digits first = 0 for i = 1 to _digits long if ( b.match[i, j] == n ) long if ( first == 0 ) first = i xelse check = _false exit fn end if end if next i next n next j for i = 1 to _digits for n = 1 to _digits first = 0 for j = 1 to _digits long if ( b.match[i, j] == n ) long if ( first == 0 ) first = j xelse check = _false exit fn end if end if next j next n next i for x = 0 to ( _nSetH - 1 ) for y = 0 to ( _nSetV - 1 ) first = 0 for ii = 0 to ( _digits - 1 ) i = x * _setH + ii mod _setH + 1 j = y * _setV + ii / _setH + 1 long if ( b.match[i, j] == n ) long if ( first == 0 ) first = j xelse check = _false exit fn end if end if next ii next y next x end fn = check local fn setCell( b as ^Board, x as short, y as short, n as short) as boolean dim as short i, j, rx, ry dim as boolean check b.match[x, y] = n for i = 1 to _digits b.f[x, i, n] = _false b.f[i, y, n] = _false next i rx = (x - 1) / _setH ry = (y - 1) / _setV for i = 1 to _setH for j = 1 to _setV b.f[ rx * _setH + i, ry * _setV + j, n ] = _false next j next i check = fn verifica( #b ) if ( check == _false ) then exit fn end fn = check local fn solve( b as ^Board ) dim as short i, j, n, first, x, y, ii, ppi, ppj dim as boolean check check = _true for i = 1 to _digits for j = 1 to _digits long if ( b.match[i, j] == 0 ) first = 0 for n = 1 to _digits long if ( b.f[i, j, n] != _false ) long if ( first == 0 ) first = n xelse first = -1 exit for end if end if next n long if ( first > 0 ) check = fn setCell( #b, i, j, first ) if ( check == _false ) then exit fn check = fn solve(#b) if ( check == _false ) then exit fn end if end if next j next i for i = 1 to _digits for n = 1 to _digits first = 0 for j = 1 to _digits if ( b.match[i, j] == n ) then exit for long if ( b.f[i, j, n] != _false ) and ( b.match[i, j] == 0 ) long if ( first == 0 ) first = j xelse first = -1 exit for end if end if next j long if ( first > 0 ) check = fn setCell( #b, i, first, n ) if ( check == _false ) then exit fn check = fn solve(#b) if ( check == _false ) then exit fn end if next n next i for j = 1 to _digits for n = 1 to _digits first = 0 for i = 1 to _digits if ( b.match[i, j] == n ) then exit for long if ( b.f[i, j, n] != _false ) and ( b.match[i, j] == 0 ) long if ( first == 0 ) first = i xelse first = -1 exit for end if end if next i long if ( first > 0 ) check = fn setCell( #b, first, j, n ) if ( check == _false ) then exit fn check = fn solve(#b) if ( check == _false ) then exit fn end if next n next j for x = 0 to ( _nSetH - 1 ) for y = 0 to ( _nSetV - 1 ) for n = 1 to _digits first = 0 for ii = 0 to ( _digits - 1 ) i = x * _setH + ii mod _setH + 1 j = y * _setV + ii / _setH + 1 if ( b.match[i, j] == n ) then exit for long if ( b.f[i, j, n] != _false ) and ( b.match[i, j] == 0 ) long if ( first == 0 ) first = n ppi = i ppj = j xelse first = -1 exit for end if end if next ii long if ( first > 0 ) check = fn setCell( #b, ppi, ppj, n ) if ( check == _false ) then exit fn check = fn solve(#b) if ( check == _false ) then exit fn end if next n next y next x end fn = check local fn resolve( b as ^Board ) dim as boolean check, daFinire dim as long i, j, n dim as board localBoard check = fn solve(b) long if ( check == _false ) exit fn end if daFinire = _false for i = 1 to _digits for j = 1 to _digits long if ( b.match[i, j] == 0 ) daFinire = _true for n = 1 to _digits long if ( b.f[i, j, n] != _false ) fn CopyBoard( b, @localBoard ) check = fn setCell(@localBoard, i, j, n) long if ( check != _false ) check = fn resolve( @localBoard ) long if ( check == -1 ) fn CopyBoard( @localBoard, b ) exit fn end if end if end if next n end if next j next i long if daFinire xelse check = -1 end if end fn = check fn InitMilliseconds fn prepare( @quiz ) DATA 0,0,0,0,2,9,0,8,7 DATA 0,9,7,3,0,0,0,0,0 DATA 0,0,2,0,0,0,4,0,9 DATA 0,0,3,9,0,1,0,0,6 DATA 0,4,0,0,0,0,0,9,0 DATA 9,0,0,7,0,3,1,0,0 DATA 0,0,9,0,0,0,6,0,0 DATA 0,0,0,0,0,5,8,2,0 DATA 2,8,0,1,3,0,0,0,0 dim as short i, j, d for i = 1 to _digits for j = 1 to _digits read d fn setCell(@quiz, j, i, d) next j next i Print : print "quiz:" fn printBoard( @quiz ) print : print "-------------------" : print dim as boolean check t = fn Milliseconds check = fn resolve(@quiz) t = fn Milliseconds - t if ( check ) print "solution:"; str$( t/1000.0 ) + " ms" else print "No solution found" end if fn printBoard( @quiz )
Output:
quiz: 0 0 0 0 0 9 0 0 2 0 9 0 0 4 0 0 0 8 0 7 2 3 0 0 9 0 0 0 3 0 9 0 7 0 0 1 2 0 0 0 0 0 0 0 3 9 0 0 1 0 3 0 5 0 0 0 4 0 0 1 6 8 0 8 0 0 0 9 0 0 2 0 7 0 9 6 0 0 0 0 0 ------------------- solution: 6.956 ms 3 8 6 5 7 9 4 1 2 1 9 5 2 4 6 3 7 8 4 7 2 3 1 8 9 6 5 6 3 8 9 5 7 2 4 1 2 5 1 8 6 4 7 9 3 9 4 7 1 2 3 8 5 6 5 2 4 7 3 1 6 8 9 8 6 3 4 9 5 1 2 7 7 1 9 6 8 2 5 3 4
Go
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
Golfscript
Código sacado de http://www.golfscript.com/
Imprime todas las soluciones posibles, sale con un error, pero funciona. <lang golfscript> 'Solution:'
- '2 8 4 3 7 5 1 6 9
0 0 9 2 0 0 0 0 7 0 0 1 0 0 4 0 0 2 0 5 0 0 0 0 8 0 0 0 0 8 0 0 0 9 0 0 0 0 6 0 0 0 0 4 0 9 0 0 1 0 0 5 0 0 8 0 0 0 0 7 6 0 4 4 2 5 6 8 9 7 3 1' {9/[n]*puts}:p; #optional formatting
~]{:@0?:^~!{@p}*10,@9/^9/=-@^9%>9%-@3/^9%3/>3%3/^27/={+}*-{@^<\+@1^+>+}/1}do </lang>
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.
I consider this a "brute force" solution of sorts, in that it is the same method I use when solving Sudokus manually. <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 (former) 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 Algol 68: ~ 0.1 seconds '394..267....3..4..5..69..2..45...9..6.......7..7...58..1..67..8..9..8....264..735', //Used in C# solution: ~ 0.2 seconds '97.3...6..6.75.........8.5.......67.....3.....539..2..7...25.....2.1...8.4...73..', //Used in Oz solution: ~ 0.2 seconds '4......6.5...8.9..3....1....2.7....1.9.....4.8....3.5....2....7..6.5...8.1......6', //Used in many other solutions, notably C++: ~ 0.3 seconds '85...24..72......9..4.........1.7..23.5...9...4...........8..7..17..........36.4.', //Used in VBA solution: ~ 0.3 seconds '..1..5.7.92.6.......8...6...9..2.4.1.........3.4.8..9...7...3.......7.69.1.8..7..', //Used in Forth solution: ~ 0.8 seconds '.9...4..7.....79..8........4.58.....3.......2.....97.6........4..35.....2..6...8.', //3rd "exceptionally difficult" example in Wikipedia: ~ 2.3 seconds '12.3....435....1....4........54..2..6...7.........8.9...31..5.......9.7.....6...8', //Used in Curry solution: ~ 2.4 seconds '9..2..5...4..6..3...3.....6...9..2......5..8...7..4..37.....1...5..2..4...1..6..9', //"AL Escargot", so-called "hardest sudoku" (HA!): ~ 3.0 seconds '1....7.9..3..2...8..96..5....53..9...1..8...26....4...3......1..4......7..7...3..', //1st "exceptionally difficult" example in Wikipedia: ~ 6.5 seconds '12.4..3..3...1..5...6...1..7...9.....4.6.3.....3..2...5...8.7....7.....5.......98', //Used in Bracmat and Scala solutions: ~ 6.7 seconds '..............3.85..1.2.......5.7.....4...1...9.......5......73..2.1........4...9', //2nd "exceptionally difficult" example in Wikipedia: ~ 8.8 seconds '.......39.....1..5..3.5.8....8.9...6.7...2...1..4.......9.8..5..2....6..4..7.....', //Used in MATLAB solution: ~15 seconds '....839..1......3...4....7..42.3....6.......4....7..1..2........8...92.....25...6', //4th "exceptionally difficult" example in Wikipedia: ~29 seconds '..3......4...8..36..8...1...4..6..73...9..........2..5..4.7..686........7..6..5..']
sudokus.each { sudoku ->
def grid = string2grid(sudoku)
println '\nPUZZLE'
grid.each { println it }
println '\nSOLUTION'
def start = System.currentTimeMillis()
def solution = solve(grid)
def elapsed = (System.currentTimeMillis() - start)/1000
solution.each { println it }
println "\nELAPSED: ${elapsed} seconds"
}</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);
initSubsets();
}
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(" -----------------------");
}
public static void main(String[] args) {
int[][] board = {
{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}
};
Sudoku s = new Sudoku(board);
System.out.print("Starting grid:\n");
s.print();
if (s.solve()) {
System.out.print("\nSolution:\n");
s.print();
} else {
System.out.println("\nUnsolvable!");
}
}
}</lang>
- Output:
Starting grid: ----------------------- | 8 5 - | - - 2 | 4 - - | | 7 2 - | - - - | - - 9 | | - - 4 | - - - | - - - | ----------------------- | - - - | 1 - 7 | - - 2 | | 3 - 5 | - - - | 9 - - | | - 4 - | - - - | - - - | ----------------------- | - - - | - 8 - | - 7 - | | - 1 7 | - - - | - - - | | - - - | - 3 6 | - 4 - | ----------------------- Solution: ----------------------- | 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 | -----------------------
JavaScript
ES6
<lang JavaScript>//-------------------------------------------[ Dancing Links and Algorithm X ]-- /**
* The doubly-doubly circularly linked data object. * Data object X */
class DoX {
/**
* @param {string} V
* @param {!DoX=} H
*/
constructor(V, H) {
this.V = V;
this.L = this;
this.R = this;
this.U = this;
this.D = this;
this.S = 1;
this.H = H || this;
H && (H.S += 1);
}
}
/**
* Helper function to help build a horizontal doubly linked list.
* @param {!DoX} e An existing node in the list.
* @param {!DoX} n A new node to add to the right of the existing node.
* @return {!DoX}
*/
const addRight = (e, n) => {
n.R = e.R; n.L = e; e.R.L = n; return e.R = n;
};
/**
* Helper function to help build a vertical doubly linked list.
* @param {!DoX} e An existing node in the list.
* @param {!DoX} n A new node to add below the existing node.
*/
const addBelow = (e, n) => {
n.D = e.D; n.U = e; e.D.U = n; return e.D = n;
};
/**
* Verbatim copy of DK's search algorithm. The meat of the DLX algorithm.
* @param {!DoX} h The root node.
* @param {!Array<!DoX>} s The solution array.
*/
const search = function(h, s) {
if (h.R == h) {
printSol(s);
} else {
let c = chooseColumn(h);
cover(c);
for (let r = c.D; r != c; r = r.D) {
s.push(r);
for (let j = r.R; r !=j; j = j.R) {
cover(j.H);
}
search(h, s);
r = s.pop();
for (let j = r.R; j != r; j = j.R) {
uncover(j.H);
}
}
uncover(c);
}
};
/**
* Verbatim copy of DK's algorithm for choosing the next column object.
* @param {!DoX} h
* @return {!DoX}
*/
const chooseColumn = h => {
let s = Number.POSITIVE_INFINITY;
let c = h;
for(let j = h.R; j != h; j = j.R) {
if (j.S < s) {
c = j;
s = j.S;
}
}
return c;
};
/**
* Verbatim copy of DK's cover algorithm
* @param {!DoX} c
*/
const cover = c => {
c.L.R = c.R;
c.R.L = c.L;
for (let i = c.D; i != c; i = i.D) {
for (let j = i.R; j != i; j = j.R) {
j.U.D = j.D;
j.D.U = j.U;
j.H.S = j.H.S - 1;
}
}
};
/**
* Verbatim copy of DK's cover algorithm
* @param {!DoX} c
*/
const uncover = c => {
for (let i = c.U; i != c; i = i.U) {
for (let j = i.L; i != j; j = j.L) {
j.H.S = j.H.S + 1;
j.U.D = j;
j.D.U = j;
}
}
c.L.R = c;
c.R.L = c;
};
//-----------------------------------------------------------[ Print Helpers ]-- /**
* Given the standard string format of a grid, print a formatted view of it.
* @param {!string|!Array} a
*/
const printGrid = function(a) {
const getChar = c => {
let r = Number(c);
if (isNaN(r)) { return c }
let o = 48;
if (r > 9 && r < 36) { o = 55 }
if (r >= 36) { o = 61 }
return String.fromCharCode(r + o)
};
a = 'string' == typeof a ? a.split() : a;
let U = Math.sqrt(a.length);
let N = Math.sqrt(U);
let line = new Array(N).fill('+').reduce((p, c) => {
p.push(... Array.from(new Array(1 + N*2).fill('-')));
p.push(c);
return p;
}, ['\n+']).join() + '\n';
a = a.reduce(function(p, c, i) {
let d = i && !(i % U), G = i && !(i % N);
i = !(i % (U * N));
d && !i && (p += '|\n| ');
d && i && (p += '|');
i && (p = + p + line + '| ');
return + p + (G && !d ? '| ' : ) + getChar(c) + ' ';
}, ) + '|' + line;
console.log(a);
};
/**
* Given a search solution, print the resultant grid.
* @param {!Array<!DoX>} a An array of data objects
*/
const printSol = a => {
printGrid(a.reduce((p, c) => {
let [i, v] = c.V.split(':');
p[i * 1] = v;
return p;
}, new Array(a.length).fill('.')));
};
//----------------------------------------------[ Grid to Exact cover Matrix ]-- /**
* Helper to get some meta about the grid.
* @param {!string} s The standard string representation of a grid.
* @return {!Array}
*/
const gridMeta = s => {
const g = s.split(); const cellCount = g.length; const tokenCount = Math.sqrt(cellCount); const N = Math.sqrt(tokenCount); const g2D = g.map(e => isNaN(e * 1) ? new Array(tokenCount).fill(1).map((_, i) => i + 1) : [e * 1]); return [cellCount, N, tokenCount, g2D];
};
/**
* Given a cell grid index, return the row, column and box indexes.
* @param {!number} n The n-value of the grid. 3 for a 9x9 sudoku.
* @return {!function(!number): !Array<!number>}
*/
const indexesN = n => i => {
let c = Math.floor(i / (n * n)); i %= n * n; return [c, i, Math.floor(c / n) * n + Math.floor(i / n)];
};
/**
* Given a puzzle string, reduce it to an exact-cover matrix and use * Donald Knuth's DLX algorithm to solve it. * @param puzString */
const reduceGrid = puzString => {
printGrid(puzString);
const [
numCells, // The total number of cells in a grid (81 for a 9x9 grid)
N, // the 'n' value of the grid. (3 for a 9x9 grid)
U, // The total number of unique tokens to be placed.
g2D // A 2D array representation of the grid, with each element
// being an array of candidates for a cell. Known cells are
// single element arrays.
] = gridMeta(puzString);
const getIndex = indexesN(N);
/**
* The DLX Header row.
* Its length is 4 times the grid's size. This is to be able to encode
* each of the 4 Sudoku constrains, onto each of the cells of the grid.
* The array is initialised with unlinked DoX nodes, but in the next step
* those nodes are all linked.
* @type {!Array.<!DoX>}
*/
const headRow = new Array(4 * numCells)
.fill()
.map((_, i) => new DoX(`H${i}`));
/**
* The header row root object. This is circularly linked to be to the left
* of the first header object in the header row array.
* It is used as the entry point into the DLX algorithm.
* @type {!DoX}
*/
let H = new DoX('ROOT');
headRow.reduce((p, c) => addRight(p, c), H);
/**
* Transposed the sudoku puzzle into a exact cover matrix, so it can be passed
* to the DLX algorithm to solve.
*/
for (let i = 0; i < numCells; i++) {
const [ri, ci, bi] = getIndex(i);
g2D[i].forEach(num => {
let id = `${i}:${num}`;
let candIdx = num - 1;
// The 4 columns that we will populate.
const A = headRow[i];
const B = headRow[numCells + candIdx + (ri * U)];
const C = headRow[(numCells * 2) + candIdx + (ci * U)];
const D = headRow[(numCells * 3) + candIdx + (bi * U)];
// The Row-Column Constraint
let rcc = addBelow(A.U, new DoX(id, A));
// The Row-Number Constraint
let rnc = addBelow(B.U, addRight(rcc, new DoX(id, B)));
// The Column-Number Constraint
let cnc = addBelow(C.U, addRight(rnc, new DoX(id, C)));
// The Block-Number Constraint
addBelow(D.U, addRight(cnc, new DoX(id, D)));
});
}
search(H, []);
}; </lang>
<lang JavaScript>[
'819..5.....2...75..371.4.6.4..59.1..7..3.8..2..3.62..7.5.7.921..64...9.....2..438', '53..247....2...8..1..7.39.2..8.72.49.2.98..7.79.....8.....3.5.696..1.3...5.69..1.', '..3.2.6..9..3.5..1..18.64....81.29..7.......8..67.82....26.95..8..2.3..9..5.1.3..', '394..267....3..4..5..69..2..45...9..6.......7..7...58..1..67..8..9..8....264..735', '97.3...6..6.75.........8.5.......67.....3.....539..2..7...25.....2.1...8.4...73..', '4......6.5...8.9..3....1....2.7....1.9.....4.8....3.5....2....7..6.5...8.1......6', '85...24..72......9..4.........1.7..23.5...9...4...........8..7..17..........36.4.', '..1..5.7.92.6.......8...6...9..2.4.1.........3.4.8..9...7...3.......7.69.1.8..7..', '.9...4..7.....79..8........4.58.....3.......2.....97.6........4..35.....2..6...8.', '12.3....435....1....4........54..2..6...7.........8.9...31..5.......9.7.....6...8', '9..2..5...4..6..3...3.....6...9..2......5..8...7..4..37.....1...5..2..4...1..6..9', '1....7.9..3..2...8..96..5....53..9...1..8...26....4...3......1..4......7..7...3..', '12.4..3..3...1..5...6...1..7...9.....4.6.3.....3..2...5...8.7....7.....5.......98', '..............3.85..1.2.......5.7.....4...1...9.......5......73..2.1........4...9', '.......39.....1..5..3.5.8....8.9...6.7...2...1..4.......9.8..5..2....6..4..7.....', '....839..1......3...4....7..42.3....6.......4....7..1..2........8...92.....25...6', '..3......4...8..36..8...1...4..6..73...9..........2..5..4.7..686........7..6..5..'
].forEach(reduceGrid);
// Or of you want to create all the grids of a particular n-size. // I run out of stack space at n = 9 let n = 2; let s = new Array(Math.pow(n, 4)).fill('.').join(); reduceGrid(s); </lang>
+-------+-------+-------+ | . . 3 | . . . | . . . | | 4 . . | . 8 . | . 3 6 | | . . 8 | . . . | 1 . . | +-------+-------+-------+ | . 4 . | . 6 . | . 7 3 | | . . . | 9 . . | . . . | | . . . | . . 2 | . . 5 | +-------+-------+-------+ | . . 4 | . 7 . | . 6 8 | | 6 . . | . . . | . . . | | 7 . . | 6 . . | 5 . . | +-------+-------+-------+ +-------+-------+-------+ | 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 | +-------+-------+-------+
Julia
<lang julia>function check(i, j)
id, im = div(i, 9), mod(i, 9) jd, jm = div(j, 9), mod(j, 9)
jd == id && return true jm == im && return true
div(id, 3) == div(jd, 3) && div(jm, 3) == div(im, 3)
end
const lookup = zeros(Bool, 81, 81)
for i in 1:81
for j in 1:81
lookup[i,j] = check(i-1, j-1)
end
end
function solve_sudoku(callback::Function, grid::Array{Int64})
(function solve()
for i in 1:81
if grid[i] == 0
t = Dict{Int64, Void}()
for j in 1:81
if lookup[i,j]
t[grid[j]] = nothing
end
end
for k in 1:9
if !haskey(t, k)
grid[i] = k
solve()
end
end
grid[i] = 0
return
end
end
callback(grid) end)()
end
function display(grid)
for i in 1:length(grid)
print(grid[i], " ")
i % 3 == 0 && print(" ")
i % 9 == 0 && print("\n")
i % 27 == 0 && print("\n")
end
end
grid = Int64[5, 3, 0, 0, 2, 4, 7, 0, 0,
0, 0, 2, 0, 0, 0, 8, 0, 0,
1, 0, 0, 7, 0, 3, 9, 0, 2,
0, 0, 8, 0, 7, 2, 0, 4, 9,
0, 2, 0, 9, 8, 0, 0, 7, 0,
7, 9, 0, 0, 0, 0, 0, 8, 0,
0, 0, 0, 0, 3, 0, 5, 0, 6,
9, 6, 0, 0, 1, 0, 3, 0, 0,
0, 5, 0, 6, 9, 0, 0, 1, 0]
solve_sudoku(display, grid)</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
Kotlin
<lang scala>// version 1.2.10
class Sudoku(rows: List<String>) {
private val grid = IntArray(81) private var solved = false
init {
require(rows.size == 9 && rows.all { it.length == 9 }) {
"Grid must be 9 x 9"
}
for (i in 0..8) {
for (j in 0..8 ) grid[9 * i + j] = rows[i][j] - '0'
}
}
fun solve() {
println("Starting grid:\n\n$this")
placeNumber(0)
println(if (solved) "Solution:\n\n$this" else "Unsolvable!")
}
private fun placeNumber(pos: Int) {
if (solved) return
if (pos == 81) {
solved = true
return
}
if (grid[pos] > 0) {
placeNumber(pos + 1)
return
}
for (n in 1..9) {
if (checkValidity(n, pos % 9, pos / 9)) {
grid[pos] = n
placeNumber(pos + 1)
if (solved) return
grid[pos] = 0
}
}
}
private fun checkValidity(v: Int, x: Int, y: Int): Boolean {
for (i in 0..8) {
if (grid[y * 9 + i] == v || grid[i * 9 + x] == v) return false
}
val startX = (x / 3) * 3
val startY = (y / 3) * 3
for (i in startY until startY + 3) {
for (j in startX until startX + 3) {
if (grid[i * 9 + j] == v) return false
}
}
return true
}
override fun toString(): String {
val sb = StringBuilder()
for (i in 0..8) {
for (j in 0..8) {
sb.append(grid[i * 9 + j])
sb.append(" ")
if (j == 2 || j == 5) sb.append("| ")
}
sb.append("\n")
if (i == 2 || i == 5) sb.append("------+-------+------\n")
}
return sb.toString()
}
}
fun main(args: Array<String>) {
val rows = listOf(
"850002400",
"720000009",
"004000000",
"000107002",
"305000900",
"040000000",
"000080070",
"017000000",
"000036040"
)
Sudoku(rows).solve()
}</lang>
- Output:
Starting grid: 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 Solution: 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
Lua
without FFI, slow
<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
with FFI, fast
<lang lua>#!/usr/bin/env luajit ffi=require"ffi" local printf=function(fmt, ...) io.write(string.format(fmt, ...)) end local band, bor, lshift, rshift=bit.band, bit.bor, bit.lshift, bit.rshift local function show(x) for i=0,8 do if i%3==0 then print() end for j=0,8 do printf(j%3~=0 and "%2d" or "%3d", x[j+9*i]) end print() end end local function trycell(x, pos) local row=math.floor(pos/9) local col=pos%9 local used=0 if pos==81 then return true end if x[pos]~=0 then return trycell(x, pos+1) end for i=0,8 do used=bor(used,lshift(1,x[i*9+col]-1)) end for j=0,8 do used=bor(used,lshift(1,x[row*9+j]-1)) end row=math.floor(row/3)*3 col=math.floor(col/3)*3 for i=row,row+2 do for j=col,col+2 do used=bor(used, lshift(1, x[i*9+j]-1)) end end x[pos]=1 while x[pos]<=9 do if band(used,1)==0 and trycell(x, pos+1) then return true end used=rshift(used,1) x[pos]=x[pos]+1 end x[pos]=0 return false end local function solve(str) local x=ffi.new("char[?]", 81) str=str:gsub("[%c%s]","") for i=0,81 do x[i]=tonumber(str:sub(i+1, i+1)) or 0 end if trycell(x, 0) then show(x) else print("no solution") end end
do -- MAIN solve([[ 5.. .7. ... 6.. 195 ... .98 ... .6.
8.. .6. ..3 4.. 8.3 ..1 7.. .2. ..6
.6. ... 28. ... 419 ..5 ... .8. .79 ]]) end</lang>
- Output:
> time ./sudoku_fast.lua 5 3 4 6 7 8 9 1 2 6 7 2 1 9 5 3 4 8 1 9 8 3 4 2 5 6 7 8 5 9 7 6 1 4 2 3 4 2 6 8 5 3 7 9 1 7 1 3 9 2 4 8 5 6 9 6 1 5 3 7 2 8 4 2 8 7 4 1 9 6 3 5 3 4 5 2 8 6 1 7 9 ./sudoku_fast.lua 0,01s user 0,00s system 90% cpu 0,007 total
Speed is about the speed of unoptimized C, half as fast as optimized C (C normal=0.007 C opt=0.004)
Mathematica/Wolfram Language
<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>
Nim
<lang nim>{.this: self.}
type
Sudoku = ref object grid : array[81, int] solved : bool
proc `$`(self: Sudoku): string =
var sb: string = ""
for i in 0..8:
for j in 0..8:
sb &= $grid[i * 9 + j]
sb &= " "
if j == 2 or j == 5:
sb &= "| "
sb &= "\n"
if i == 2 or i == 5:
sb &= "------+------+------\n"
sb
proc init(self: Sudoku, rows: array[9, string]) =
for i in 0..8:
for j in 0..8:
grid[9 * i + j] = rows[i][j].int - '0'.int
proc checkValidity(self: Sudoku, v, x, y: int): bool =
for i in 0..8:
if grid[y * 9 + i] == v or grid[i * 9 + x] == v:
return false
var startX = (x div 3) * 3
var startY = (y div 3) * 3
for i in startY..startY + 2:
for j in startX..startX + 2:
if grid[i * 9 + j] == v:
return false
result = true
proc placeNumber(self: Sudoku, pos: int) =
if solved:
return
if pos == 81:
solved = true
return
if grid[pos] > 0:
placeNumber(pos + 1)
return
for n in 1..9:
if checkValidity(n, pos mod 9, pos div 9):
grid[pos] = n
placeNumber(pos + 1)
if solved:
return
grid[pos] = 0
proc solve(self: Sudoku) =
echo "Starting grid:\n\n", $self placeNumber(0) if solved: echo "Solution:\n\n", $self else: echo "Unsolvable!"
var rows = ["850002400",
"720000009",
"004000000",
"000107002",
"305000900",
"040000000",
"000080070",
"017000000",
"000036040"]
var puzzle = Sudoku() puzzle.init(rows) puzzle.solve()</lang>
- Output:
Starting grid: 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 Solution: 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
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>
PARI/GP
Build plugin for PARI's function interface from C code: sudoku.c <lang C>#include <pari/pari.h>
typedef int SUDOKU [9][9];
static inline int check_num(SUDOKU s, int row, int col, int num) {
int i, r = (row/3)*3, c = (col/3)*3;
for (i = 0; i < 9; i++)
if (s[row][i] == num || s[i][col] == num || s[i%3 + r][i/3 + c] == num)
return 0;
return 1;
}
static int sudoku_solve(SUDOKU s, int row, int col) {
int num;
if (row < 9 && col < 9) {
if (s[row][col]) {
if (col < 8)
return sudoku_solve(s, row, col+1);
if (row < 8)
return sudoku_solve(s, row+1, 0);
return 1;
}
else
for (num = 1; num < 10; num++)
if (check_num(s, row, col, num)) { s[row][col] = num; if (sudoku_solve(s, row, col)) return 1; else s[row][col] = 0; }
return 0; } return 1;
}
GEN plug_sudoku(GEN M) {
SUDOKU s; GEN S; int i, k;
if (typ(M) != t_MAT) pari_err(e_MISC, "parameter not matrix");
S = matsize(M);
if (itos(gel(S, 1)) < 9 || itos(gel(S, 2)) < 9) pari_err(e_MISC, "parameter not 9x9 matrix");
for (i = 0; i < 9; i++)
for (k = 0; k < 9; k++)
s[i][k] = itos(gcoeff(M, i+1, k+1)); /* get sudoku */
if (sudoku_solve(s, 0, 0)) { /* solve sudoku */
S = cgetg(10, t_MAT);
for (k = 0; k < 9; k++) { /* create 9x9 matrix */
gel(S, k+1) = cgetg(10, t_COL);
for (i = 0; i < 9; i++)
gcoeff(S, i+1, k+1) = stoi(s[i][k]); /* fill in elements */
} return S; } return gen_0; /* no solution */
} </lang> Compile plugin: gcc -O2 -Wall -fPIC -shared sudoku.c -o libsudoku.so -lpari
Install plugin from home directory and play: <lang parigp>install("plug_sudoku", "G", "sudoku", "~/libsudoku.so")</lang>
Output:
gp > S=[5,3,0,0,7,0,0,0,0;6,0,0,1,9,5,0,0,0;0,9,8,0,0,0,0,6,0;8,0,0,0,6,0,0,0,3;4,0,0,8,0,3,0,0,1;7,0,0,0,2,0,0,0,6;0,6,0,0,0,0,2,8,0;0,0,0,4,1,9,0,0,5;0,0,0,0,8,0,0,7,9] [5 3 0 0 7 0 0 0 0] [6 0 0 1 9 5 0 0 0] [0 9 8 0 0 0 0 6 0] [8 0 0 0 6 0 0 0 3] [4 0 0 8 0 3 0 0 1] [7 0 0 0 2 0 0 0 6] [0 6 0 0 0 0 2 8 0] [0 0 0 4 1 9 0 0 5] [0 0 0 0 8 0 0 7 9] gp > sudoku(S) [5 3 4 6 7 8 9 1 2] [6 7 2 1 9 5 3 4 8] [1 9 8 3 4 2 5 6 7] [8 5 9 7 6 1 4 2 3] [4 2 6 8 5 3 7 9 1] [7 1 3 9 2 4 8 5 6] [9 6 1 5 3 7 2 8 4] [2 8 7 4 1 9 6 3 5] [3 4 5 2 8 6 1 7 9]
Pascal
Simple backtracking implimentation, therefor it must be fast to be competetive.With doReverse = true same sequence for trycell. nearly 5 times faster than C-Version. <lang pascal>Program soduko; {$IFDEF FPC}
{$CODEALIGN proc=16,loop=8}
{$ENDIF} uses
sysutils,crt;
const
carreeSize = 3; maxCoor = carreeSize*carreeSize; maxValue = maxCoor; maxMask = 1 shl (maxCoor+1)-1;
type
tLimit = 0..maxCoor-1;
tValue = 0..maxCoor;
tSteps = 0..maxCoor*maxCoor;
tValField = array[tLimit,tLimit] of NativeInt;//tValue;
tBitrepr = 0..maxMask;
tcol = array[tLimit] of NativeInt;// tBitrepr;
trow = array[tLimit] of NativeInt;// tBitrepr;
tcar = array[tLimit] of NativeInt;// tBitrepr;
tpValue = ^NativeInt;//^tValue;
tpLimit = ^tLimit;
tpBitrepr= ^NativeInt;//^tBitrepr;
tchgVal = record
cvCol,
cvRow,
cvCar : tpBitrepr;
cvVal : tpValue;
end;
tpChgVal = ^tchgVal;
tchgList = array[tSteps] of tchgVal;
tField = record
fdChgList: tchgList;
fdCol : tcol;
fdRow : trow;
fdcar : tcar;
fdVal : tValField;
fdChgIdx : tSteps;
end;
const
Expl0:tValField = ((9,0,7,0,0,0,3,0,0),
(0,0,0,1,0,0,2,0,0),
(6,0,0,0,0,8,0,0,0),
(0,0,5,0,3,0,0,0,0),
(0,0,0,0,0,0,0,8,4),
(0,0,0,0,0,0,0,6,0),
(0,0,0,2,7,0,0,0,0),
(8,4,0,0,0,0,0,0,0),
(0,6,0,0,0,0,0,0,0));
Expl1:tValField=((0,0,0,1,0,0,0,3,8),
(2,0,0,0,0,5,0,0,0),
(0,0,0,0,0,0,0,0,0),
(0,5,0,0,0,0,4,0,0),
(4,0,0,0,3,0,0,0,0),
(0,0,0,7,0,0,0,0,6),
(0,0,1,0,0,0,0,5,0),
(0,0,0,0,6,0,2,0,0),
(0,6,0,0,0,4,0,0,0));
var
F, solF : TField; solCnt, callCnt: NativeUint; solFound : Boolean;
procedure OutField(const F:tField); var
rw,cl : tLimit; rowS: AnsiString;
Begin
GotoXy(1,1);
For rw := low(tLimit) to High(tLimit) do
Begin
rowS := ' ';
For cl := low(tLimit) to High(tLimit) do
RowS :=RowS+IntToStr(F.fdVal[rw,cl]);
writeln(RowS);
end;
end;
function CarIdx(rw,cl: NativeInt):NativeInt; begin
CarIdx:= (rw DIV carreeSize)*carreeSize +cl DIV carreeSize;
end; function InsertTest(const F:tField;rw,cl:tLimit;value:tValue):boolean; var
msk: tBitrepr;
Begin
result := (Value = 0); IF result then EXIT; msk := 1 shl (value-1); with F do Begin result := fdRow[rw] AND msk = 0; result := result AND (fdCol[cl] AND msk = 0); rw :=CarIdx(rw,cl); result := result AND (fdCar[rw] AND msk = 0); end;
end;
function InitField(var F:tField;const InFd:tValField;DoReverse:boolean):boolean; var
TmpchgVal:tchgVal; rw,cl, value, msk : NativeInt; leftSteps:tSteps;
Begin
Fillchar(F,SizeOf(F),#0);
leftSteps := High(tSteps)-1;
//unknown fields inserted from end
For rw := low(tLimit) to High(tLimit) do
For cl := low(tLimit) to High(tLimit) do
Begin
value := InFd[rw,cl];
IF InsertTest(F,rw,cl,value) then
Begin
with F do
Begin
if value > 0 then
Begin
msk := 1 shl (value-1);
//given state
//use pointer to the relevant places and mark as occupied
with fdChgList[fdChgIdx] do
begin
cvCol := @fdCol[cl];
cvCol^ +=Msk;
cvRow := @fdRow[rw];
cvRow^ +=Msk;
cvCar := @fdCar[CarIdx(rw,cl)];
cvCar^ +=Msk;
cvVal := @fdVal[rw,cl];
cvVal^ := value;
end;
inc(fdChgIdx);
end
else
Begin
//use pointer to the relevant places
with fdChgList[leftSteps] do
begin
cvCol := @fdCol[cl];
cvRow := @fdRow[rw];
cvCar := @fdCar[CarIdx(rw,cl)];
cvVal := @fdVal[rw,cl];
end;
dec(leftSteps);
end;
end
end
else
Begin
writeln(rw:10,cl:10,value:10);
Writeln(' not solvable SuDoKu ');
delay(2000);
result := false;
EXIT;
end;
end;
//reverse direction of left over
IF DoReverse then
Begin
leftSteps := High(tSteps)-1;
rw := F.fdChgIdx;
repeat
TmpchgVal:= F.fdChgList[leftSteps];
F.fdChgList[leftSteps]:= F.fdChgList[rw];
F.fdChgList[rw] :=TmpchgVal;
dec(leftSteps);
inc(rw);
until rw>=leftSteps;
end;
//OutField(F);
solFound := false;
result := true;
end; procedure SolIsFound; begin
solF := F; inc(solCnt); solFound := True;
end;
procedure TryCell(var ChgVal:tpchgVal); var
value :NativeInt; poss,msk: NativeInt;
Begin
IF solFound then EXIT; with ChgVal^ do poss:= (cvRow^ OR cvCol^ OR cvCar^) XOR maxMask; IF Poss = 0 then EXIT;
value := 1; msk := 1;
repeat
IF Poss AND MSK <>0 then
Begin
inc(callCnt);
//insert test value
with ChgVal^ do
Begin
cvCol^ := cvCol^ OR msk;
cvRow^ := cvRow^ OR msk;
cvCar^ := cvCar^ OR msk;
cvVAl^ := value;
end;
//try next in list, if beyond last
inc(ChgVal);
IF ChgVal^.cvCol <> NIL then
TryCell(ChgVal)
else
SolIsFound;
//remove test value
dec(ChgVal);
with ChgVal^ do
Begin
cvCol^ := cvCol^ XOR msk;
cvRow^ := cvRow^ XOR msk;
cvCar^ := cvCar^ XOR msk;
cvVAl^ := 0;
end;
end;
inc(msk,msk);
inc(value);
until value> maxValue;
end;
var
ChangeBegin : tpChgVal; k : NativeInt; T1,T0: TDateTime;
begin
randomize; ClrScr; solCnt := 0; callCnt:= 0; T0 := time; k := 0; repeat InitField(F,Expl1,FALSE); ChangeBegin := @F.fdChgList[F.fdChgIdx]; TryCell(ChangeBegin); inc(k); until k >= 5; T1 := time; Outfield(solF); writeln(86400*1000*(T1-T0)/k:10:3,' ms Test calls :',callCnt/k:8:0);
end.</lang>
- Output:
Expl0
927465318
458193276
613728459
185634792
376912584
294587163
539276841
841359627
762841935
// InitField doReverse = true
9.850 ms Test calls : 532466
// InitField doReverse = false
2609.000 ms Test calls :135196346
...
Expl1
594126738
237895164
618473925
859612473
476539812
123748596
341287659
985361247
762954381
// InitField doReverse = true
857.600 ms Test calls :40980572
// InitField doReverse = false
21.400 ms Test calls : 1089986
Perl
<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
Phix
Simple brute force solution. Generally quite good but will struggle on some puzzles (eg see "the beast" below) <lang Phix>sequence board = split(""" .......39 .....1..5 ..3.5.8.. ..8.9...6 .7...2... 1..4..... ..9.8..5. .2....6.. 4..7.....""",'\n')
function valid_move(integer y, integer x, integer ch)
for i=1 to 9 do
if ch=board[i][x] then return 0 end if
if ch=board[y][i] then return 0 end if
end for
y -= mod(y-1,3)
x -= mod(x-1,3)
for ys=y to y+2 do
for xs=x to x+2 do
if ch=board[ys][xs] then return 0 end if
end for
end for
return 1
end function
sequence solution = {}
procedure brute_solve()
for y=1 to 9 do
for x=1 to 9 do
if board[y][x]<='0' then
for ch='1' to '9' do
if valid_move(y,x,ch) then
board[y][x] = ch
brute_solve()
board[y][x] = ' '
if length(solution) then return end if
end if
end for
return
end if
end for
end for
solution = board -- (already solved case)
end procedure
atom t0 = time() brute_solve() printf(1,"%s\n(solved in %3.2fs)\n",{join(solution,"\n"),time()-t0})</lang>
- Output:
751846239 892371465 643259871 238197546 974562318 165438927 319684752 527913684 486725193 (solved in 0.95s)
OTT solution. Implements line/col and set exclusion, and x-wings. Blisteringly fast
The included program demo\rosetta\Sudoku.exw is an extended version of this that performs extended validation,
contains 339 puzzles, can be run as a command-line or gui program, check for multiple solutions, and produce
a more readable single-puzzle output (example below).
<lang Phix>-- Working directly on 81-character strings ultimately proves easier: Originally I
-- just wanted to simplify the final display, but later I realised that a 9x9 grid
-- encourages laborious indexing/looping everwhere whereas using a flat 81-element
-- approach encourages precomputation of index sets, and once you commit to that,
-- the rest of the code starts to get a whole lot cleaner. Below we create 27+18
-- sets and 5 tables of lookup indexes to locate them quickly.
sequence nines = {}, -- will be 27 in total
cols = repeat(0,9*9), -- remainder(i-1,9)+1
rows = repeat(0,9*9), -- floor((i-1)/9)+10
squares = repeat(0,9*9),
sixes = {}, -- will be 18 in total
dotcol = repeat(0,9*9), -- same col, diff square
dotrow = repeat(0,9*9) -- same row, diff square
procedure set_nines() sequence nine, six integer idx, ndx
for x=0 to 8 do -- columns
nine = {}
ndx = length(nines)+1
for y=1 to 81 by 9 do
idx = y+x
nine = append(nine,idx)
cols[idx] = ndx
end for
nines = append(nines,nine)
end for
for y=1 to 81 by 9 do -- rows
nine = {}
ndx = length(nines)+1
for x=0 to 8 do
idx = y+x
nine = append(nine,idx)
rows[idx] = ndx
end for
nines = append(nines,nine)
end for
if length(nines)!=18 then ?9/0 end if
for y=0 to 8 by 3 do -- small squares [19..27]
for x=0 to 8 by 3 do
nine = {}
ndx = length(nines)+1
for sy=y*9 to y*9+18 by 9 do
for sx=x to x+2 do
idx = sy+sx+1
nine = append(nine,idx)
squares[idx] = ndx
end for
end for
nines = append(nines,nine)
end for
end for
if length(nines)!=27 then ?9/0 end if
for i=1 to 9*9 do
six = {}
nine = nines[cols[i]] -- dotcol
for j=1 to length(nine) do
if squares[i]!=squares[nine[j]] then
six = append(six,nine[j])
end if
end for
ndx = find(six,sixes)
if ndx=0 then
sixes = append(sixes,six)
ndx = length(sixes)
end if
dotcol[i] = ndx
six = {}
nine = nines[rows[i]] -- dotrow
for j=1 to length(nine) do
if squares[i]!=squares[nine[j]] then
six = append(six,nine[j])
end if
end for
ndx = find(six,sixes)
if ndx=0 then
sixes = append(sixes,six)
ndx = length(sixes)
end if
dotrow[i] = ndx
end for
end procedure set_nines()
integer improved = 0
function eliminate_in(sequence valid, sequence set, integer ch)
for i=1 to length(set) do
integer idx = set[i]
if string(valid[idx]) then
integer k = find(ch,valid[idx])
if k!=0 then
valid[idx][k..k] = ""
improved = 1
end if
end if
end for
return valid
end function
function test_comb(sequence chosen, sequence pool, sequence valid) -- -- (see deep_logic()/set elimination) -- chosen is a sequence of length 2..4 of integers 1..9: ordered elements of pool. -- pool is a set of elements of the sequence valid, each of which is a sequence. -- (note that elements of valid in pool not in chosen are not necessarily sequences) -- sequence contains = repeat(0,9) integer ccount = 0, ch object set
for i=1 to length(chosen) do
set = valid[pool[chosen[i]]]
for j=1 to length(set) do
ch = set[j]-'0'
if contains[ch]=0 then
contains[ch] = 1
ccount += 1
end if
end for
end for
if ccount=length(chosen) then
for i=1 to length(pool) do
if find(i,chosen)=0 then
set = valid[pool[i]]
if sequence(set) then
-- (reverse order so deletions don't foul indexes)
for j=length(set) to 1 by -1 do
ch = set[j]-'0'
if contains[ch] then
valid[pool[i]][j..j] = ""
improved = 1
end if
end for
end if
end if
end for
end if
return valid
end function
-- from Combinations -- from http://rosettacode.org/wiki/Combinations#Phix function comb(sequence pool, valid, integer needed, done=0, sequence chosen={}) -- (used by deep_logic()/set elimination)
if needed=0 then -- got a full set
return test_comb(chosen,pool,valid)
end if
if done+needed>length(pool) then return valid end if -- cannot fulfil
-- get all combinations with and without the next item:
done += 1
if sequence(valid[pool[done]]) then
valid = comb(pool,valid,needed-1,done,append(chosen,done))
end if
return comb(pool,valid,needed,done,chosen)
end function
function deep_logic(string board, sequence valid) -- -- Create a grid of valid moves. Note this does not modify board, but instead creates -- sets of permitted values for each cell, which can also be and are used for hints. -- Apply standard eliminations of known cells, then try some more advanced tactics: -- -- 1) row/col elimination -- If in any of the 9 small squares a number can only occur in one row or column, -- then that number cannot occur in that row or column in two other corresponding -- small squares. Example (this one with significant practical benefit): -- 000|000|036 -- 840|000|000 -- 000|000|020 -- ---+---+--- -- 000|203|000 -- 010|000|700 -- 000|600|400 -- ---+---+--- -- 000|410|050 -- 003|000|200 -- 600|000|000 <-- 3 -- ^-- 3 -- Naively, the br can contain a 3 in the four corners, but looking at mid-right and -- mid-bottom leads us to eliminating 3s in column 9 and row 9, leaving 7,7 as the -- only square in the br that can be a 3. Uses dotcol and dotrow. -- Without this, brute force on the above takes ~8s, but with it ~0s -- -- 2) set elimination -- If in any 9-set there is a set of n blank squares that can only contain n digits, -- then no other squares can contain those digits. Example (with some benefit): -- 75.|.9.|.46 -- 961|...|352 -- 4..|...|79. -- ---+---+--- -- 2..|6.1|..7 -- .8.|...|.2. -- 1..|328|.65 -- ---+---+--- -- ...|...|... <-- [7,8] is {1,3,8}, [7,9] is {1,3,8} -- 3.9|...|2.4 <-- [8,8] is {1,8} -- 84.|.3.|.79 -- The three cells above the br 479 can only contain {1,3,8}, so the .. of the .2. -- in column 7 of that square are {5,6} (not 1) and hence [9,4] must be a 1. -- (Relies on plain_logic to spot that "must be a 1", and serves as a clear example -- of why this routine should not bother to attempt updating the board itself - as -- it spends almost all of its time looking in a completely different place.) -- (One could argue that [7,7] and [9,7] are the only places that can hold {5,6} and -- therefore we should eliminate all non-{5,6} from those squares, as an alternative -- strategy. However I think that would be harder to code and cannot imagine a case -- said complementary logic covers, that the above does not, cmiiw.) -- -- 3) x-wings -- If a pair of rows or columns can only contain a given number in two matching places, -- then once filled they will occupy opposite diagonal corners, hence that said number -- cannot occur elsewhere in those two columns/rows. Example (with a benefit): -- .43|98.|25. <-- 6 in [1,{6,9}] -- 6..|425|... -- 2..|..1|.94 -- ---+---+--- -- 9..|..4|.7. <-- hence 6 not in [4,9] -- 3..|6.8|... -- 41.|2.9|..3 -- ---+---+--- -- 82.|5..|... <-- hence 6 not in [7,6],[7,9] -- ...|.4.|..5 <-- hence 6 not in [8,6] -- 534|89.|71. <-- 6 in [9,{6,9}] -- A 6 must be in [1,6] or [1,9] and [9,6] or [9,9], hence [7,9] is not 6 and must be 9. -- (we also eliminate 6 from [4,9], [7,6] and [8,6] to no great use) -- In practice this offers little benefit over a single trial-and-error step, as -- obviously trying either 6 in row 1 or 9 immediately pinpoints that 9 anyway. -- -- 4) swordfish (not attempted) -- There is an extension to x-wings known as swordfish: three (or more) pairs form -- a staggered pair (or more) of rectangles that exhibit similar properties, eg: -- 8-1|-5-|-3- -- 953|-68|--- -- -4-|-*3|5*8 -- ---+---+--- -- 6--|9-2|--- -- -8-|-3-|-4- -- 3*-|5-1|-*7 <-- hence [6,3] is not 9, must be 4 -- ---+---+--- -- 5*2|-*-|-8- -- --8|37-|--9 -- -3-|82-|1-- -- ^---^---^-- 3 pairs of 9s (marked with *) on 3 rows (only) -- It is not a swordfish if the 3 pairs are on >3 rows, I trust that is obvious. -- Logically you can extend this to N pairs on N rows, however I cannot imagine a -- case where this is not immediately solved by a single trial-step being invalid. -- (eg above if you try [3,5]:=9 it is quickly proved to be invalid, and the same -- goes for [6,8]:=9 and [7,2]:=9, since they are all entirely inter-dependent.) -- Obviously where I have said rows, the same concept can be applied to columns. -- Likewise there are "Alternate Pairs" and "Hook or X-Y wing" strategies, which -- are easily solved with a single trial-and-error step, and of course the brute -- force algorithm is going to select pairs first anyway. [Erm, no it doesn't, -- it selects shortest - I've noted the possible improvement below.] -- integer col, row sequence c, r sequence nine, prevsets, set object vj integer ch, k, idx, sx, sy, count
if length(valid)=0 then
-- initialise/start again from scratch
valid = repeat("123456789",9*9)
end if
--
-- First perform standard eliminations of any known cells:
-- (repeated every time so plain_logic() does not have to worry about it)
--
for i=1 to 9*9 do
ch = board[i]
if ch>'0'
and string(valid[i]) then
valid[i] = ch
valid = eliminate_in(valid,nines[cols[i]],ch)
valid = eliminate_in(valid,nines[rows[i]],ch)
valid = eliminate_in(valid,nines[squares[i]],ch)
end if
end for
--
-- 1) row/col elimination
--
for s=19 to 27 do
c = repeat(0,9) -- 0 = none seen, 1..9 this col only, -1: >1 col
r = repeat(0,9) -- "" row row
nine = nines[s]
for n=1 to 9 do
k = nine[n]
vj = valid[k]
if string(vj) then
for i=1 to length(vj) do
ch = vj[i]-'0'
col = dotcol[k]
row = dotrow[k]
c[ch] = iff(find(c[ch],{0,col})!=0?col:-1)
r[ch] = iff(find(r[ch],{0,row})!=0?row:-1)
end for
end if
end for
for i=1 to 9 do
ch = i+'0'
col = c[i]
if col>0 then
valid = eliminate_in(valid,sixes[col],ch)
end if
row = r[i]
if row>0 then
valid = eliminate_in(valid,sixes[row],ch)
end if
end for
end for
--
-- 2) set elimination
--
for i=1 to length(nines) do
--
-- Practical note: Meticulously counting empties to eliminate larger set sizes
-- would at best reduce 6642 tests to 972, not deemed worth it.
--
for set_size=2 to 4 do
--if floor(count_empties(nines[i])/2)>=set_size then -- (untested)
valid = comb(nines[i],valid,set_size)
--end if
end for
end for
--
-- 3) x-wings
--
for ch='1' to '9' do
prevsets = repeat(0,9)
for x=1 to 9 do
count = 0
set = repeat(0,9)
for y=0 to 8 do
idx = y*9+x
if sequence(valid[idx]) and find(ch,valid[idx]) then
set[y+1] = 1
count += 1
end if
end for
if count=2 then
k = find(set,prevsets)
if k!=0 then
for y=0 to 8 do
if set[y+1]=1 then
for sx=1 to 9 do
if sx!=k and sx!=x then
valid = eliminate_in(valid,{y*9+sx},ch)
end if
end for
end if
end for
else
prevsets[x] = set
end if
end if
end for
prevsets = repeat(0,9)
for y=0 to 8 do
count = 0
set = repeat(0,9)
for x=1 to 9 do
idx = y*9+x
if sequence(valid[idx]) and find(ch,valid[idx]) then
set[x] = 1
count += 1
end if
end for
if count=2 then
k = find(set,prevsets)
if k!=0 then
for x=1 to 9 do
if set[x]=1 then
for sy=0 to 8 do
if sy+1!=k and sy!=y then
valid = eliminate_in(valid,{sy*9+x},ch)
end if
end for
end if
end for
else
prevsets[y+1] = set
end if
end if
end for
end for
return valid
end function
function permitted_in(string board, sequence sets, sequence valid, integer ch) sequence set integer pos, idx, bch
for i=1 to 9 do
set = nines[sets[i]]
pos = 0
for j=1 to 9 do
idx = set[j]
bch = board[idx]
if bch>'0' then
if bch=ch then pos = -1 exit end if
elsif find(ch,valid[idx]) then
if pos!=0 then pos = -1 exit end if
pos = idx
end if
end for
if pos>0 then
board[pos] = ch
improved = 1
end if
end for
return board
end function
enum INVALID = -1, INCOMPLETE = 0, SOLVED = 1, MULTIPLE = 2, BRUTE = 3
function plain_logic(string board) -- -- Responsible for: -- 1) cells with only one option -- 2) numbers with only one home -- integer solved sequence valid = {} object vi
while 1 do
solved = SOLVED
improved = 0
valid = deep_logic(board,valid)
-- 1) cells with only one option:
for i=1 to length(valid) do
vi = valid[i]
if string(vi) then
if length(vi)=0 then return {board,{},INVALID} end if
if length(vi)=1 then
board[i] = vi[1]
improved = 1
end if
end if
if board[i]<='0' then
solved = INCOMPLETE
end if
end for
if solved=SOLVED then return {board,{},SOLVED} end if
-- 2) numbers with only one home
for ch='1' to '9' do
board = permitted_in(board,cols,valid,ch)
board = permitted_in(board,rows,valid,ch)
board = permitted_in(board,squares,valid,ch)
end for
if not improved then exit end if
end while
return {board,valid,solved}
end function
function validate(string board) -- (sum9 should be sufficient - if you want, get rid of nine/nines) integer ch, sum9 sequence nine, nines = tagset(9)
for x=0 to 8 do -- columns
sum9 = 0
nine = repeat(0,9)
for y=1 to 81 by 9 do
ch = board[y+x]-'0'
if ch<1 or ch>9 then return 0 end if
sum9 += ch
nine[ch] = ch
end for
if sum9!=45 then return 0 end if
if nine!=nines then return 0 end if
end for
for y=1 to 81 by 9 do -- rows
sum9 = 0
nine = repeat(0,9)
for x=0 to 8 do
ch = board[y+x]-'0'
sum9 += ch
nine[ch] = ch
end for
if sum9!=45 then return 0 end if
if nine!=nines then return 0 end if
end for
for y=0 to 8 by 3 do -- small squares
for x=0 to 8 by 3 do
sum9 = 0
nine = repeat(0,9)
for sy=y*9 to y*9+18 by 9 do
for sx=x to x+2 do
ch = board[sy+sx+1]-'0'
sum9 += ch
nine[ch] = ch
end for
end for
if sum9!=45 then return 0 end if
if nine!=nines then return 0 end if
end for
end for
return 1
end function
function solve(string board, sequence valid={}) sequence solution, solutions integer solved integer minopt, mindx object vi
{solution,valid,solved} = plain_logic(board)
if solved=INVALID then return {{},INVALID} end if
if solved=SOLVED then return {{solution},SOLVED} end if
if solved=BRUTE then return {{solution},BRUTE} end if
if solved!=INCOMPLETE then ?9/0 end if
-- find the cell with the fewest options:
-- (a possible improvement here would be to select the shortest
-- with the "most pairs" set, see swordfish etc above.)
minopt = 10
for i=1 to 9*9 do
vi = valid[i]
if string(vi) then
if length(vi)<=1 then ?9/0 end if -- should be caught above
if length(vi)<minopt then
minopt = length(vi)
mindx = i
end if
end if
end for
solutions = {}
for i=1 to minopt do
board[mindx] = valid[mindx][i]
{solution,solved} = solve(board,valid)
if solved=MULTIPLE then
return {solution,MULTIPLE}
elsif solved=SOLVED
or solved=BRUTE then
if not find(solution[1],solutions)
and validate(solution[1]) then
solutions = append(solutions,solution[1])
end if
if length(solutions)>1 then
return {solutions,MULTIPLE}
elsif length(solutions) then
return {solutions,BRUTE}
end if
end if
end for
if length(solutions)=1 then
return {solutions,BRUTE}
end if
return {{},INVALID}
end function
function test_one(string board) sequence solutions string solution, desc integer solved
{solutions,solved} = solve(board)
if solved=SOLVED then
desc = "(logic)"
elsif solved=BRUTE then
desc = "(brute force)"
else
desc = "???" -- INVALID/INCOMPLETE/MULTIPLE
end if
if length(solutions)=0 then
solution = board
desc = "*** NO SOLUTIONS ***"
elsif length(solutions)=1 then
solution = solutions[1]
if not validate(solution) then
desc = "*** ERROR ***" -- (should never happen)
end if
else
solution = board
desc = "*** MULTIPLE SOLUTIONS ***"
end if
return {solution,desc}
end function
--NB Blank cells can be represented by any character <'1'. Spaces are not recommended since -- they can all too easily be converted to tabs by copy/paste/save. In particular, ? and -- _ are NOT valid characters for representing a blank square. Use any of .0-* instead.
constant tests = {
"..............3.85..1.2.......5.7.....4...1...9.......5......73..2.1........4...9", -- (0.01s, (logic)) -- row/col elimination (was 8s w/o logic first) "000000036840000000000000020000203000010000700000600400000410050003000200600000000", -- (0.04s, (brute force)) ".......39.....1..5..3.5.8....8.9...6.7...2...1..4.......9.8..5..2....6..4..7.....", -- (1.12s, (brute force)) "000037600000600090008000004090000001600000009300000040700000800010009000002540000", -- (0.00s, (logic)) "....839..1......3...4....7..42.3....6.......4....7..1..2........8...92.....25...6", -- (0.04s, (brute force)) "..1..5.7.92.6.......8...6...9..2.4.1.........3.4.8..9...7...3.......7.69.1.8..7..", -- (0.00s, (logic)) -- (the following takes ~8s when checking for multiple solutions) "--3------4---8--36--8---1---4--6--73---9----------2--5--4-7--686--------7--6--5--", -- (0.01s, (brute force)) "..3.2.6..9..3.5..1..18.64....81.29..7.......8..67.82....26.95..8..2.3..9..5.1.3..", -- (0.00s, (logic)) "--4-5--6--6-1--8-93----7----8----5-----4-3-----6----7----2----61-5--4-3--2--7-1--", -- (0.00s, (logic)) -- x-wings ".4398.25.6..425...2....1.949....4.7.3..6.8...41.2.9..382.5.........4...553489.71.", -- (0.00s, (logic)) ".9...4..7.....79..8........4.58.....3.......2.....97.6........4..35.....2..6...8.", -- (0.00s, (logic)) -- "AL Escargot", so-called "hardest sudoku" "1....7.9..3..2...8..96..5....53..9...1..8...26....4...3......1..4......7..7...3..", -- (0.26s, (brute force)) "12.3....435....1....4........54..2..6...7.........8.9...31..5.......9.7.....6...8", -- (0.48s, (brute force)) "12.4..3..3...1..5...6...1..7...9.....4.6.3.....3..2...5...8.7....7.....5.......98", -- (1.07s, (brute force)) "394..267....3..4..5..69..2..45...9..6.......7..7...58..1..67..8..9..8....264..735", -- (0.00s, (logic)) "4......6.5...8.9..3....1....2.7....1.9.....4.8....3.5....2....7..6.5...8.1......6", -- (0.01s, (brute force)) "5...7....6..195....98....6.8...6...34..8.3..17...2...6.6....28....419..5....8..79", -- (0.00s, (logic)) "503600009010002600900000080000700005006804100200003000030000008004300050800006702", -- (0.00s, (logic)) "53..247....2...8..1..7.39.2..8.72.49.2.98..7.79.....8.....3.5.696..1.3...5.69..1.", -- (0.00s, (logic)) "530070000600195000098000060800060003400803001700020006060000280000419005000080079", -- (0.00s, (logic)) -- set exclusion "75..9..46961...3524.....79.2..6.1..7.8.....2.1..328.65.........3.9...2.484..3..79", -- (0.00s, (logic)) -- Worlds hardest sudoku: "800000000003600000070090200050007000000045700000100030001000068008500010090000400", -- (0.21s, (brute force)) "819--5-----2---75--371-4-6-4--59-1--7--3-8--2--3-62--7-5-7-921--64---9-----2--438", -- (0.00s, (logic)) "85...24..72......9..4.........1.7..23.5...9...4...........8..7..17..........36.4.", -- (0.01s, (logic)) "9..2..5...4..6..3...3.....6...9..2......5..8...7..4..37.....1...5..2..4...1..6..9", -- (0.17s, (brute force)) "97.3...6..6.75.........8.5.......67.....3.....539..2..7...25.....2.1...8.4...73..", -- (0.00s, (logic)) -- "the beast" (an earlier algorithm took 318s (5min 18s) on this): "000060080020000000001000000070000102500030000000000400004201000300700600000000050", -- (0.03s, (brute force)) $},
lt = length(tests), run_one_test = 0
constant l = " x x x | x x x | x x x ",
s = "-------+-------+-------",
l3 = join({l,l,l},"\n"),
fmt = substitute(join({l3,s,l3,s,l3},"\n"),"x","%c")&"\n"
procedure print_board(string board)
printf(1,fmt,board)
end procedure
procedure test() string board -- (81 characters) string solution, desc atom t0 = time()
if run_one_test then
board = tests[run_one_test]
print_board(board)
{solution,desc} = test_one(board)
if length(solution)!=0 then
printf(1,"solution:\n")
print_board(solution)
end if
printf(1,"%s, %3.2fs\n",{desc,time()-t0})
else
for i=1 to lt do
atom t1 = time()
board = tests[i]
{solution,desc} = test_one(board)
printf(1," \"%s\", -- (%3.2fs, %s)\n",{board,time()-t1,desc})
-- printf(1," \"%s\", -- (%3.2fs, %s)\n",{solution,time()-t1,desc})
end for
t0 = time()-t0
printf(1,"%d puzzles solved in %3.2fs (av %3.2fs)\n",{lt,t0,t0/lt})
end if
end procedure test()</lang>
- Output:
"..............3.85..1.2.......5.7.....4...1...9.......5......73..2.1........4...9", -- (0.02s, (logic))
"000000036840000000000000020000203000010000700000600400000410050003000200600000000", -- (0.03s, (brute force))
".......39.....1..5..3.5.8....8.9...6.7...2...1..4.......9.8..5..2....6..4..7.....", -- (1.31s, (brute force))
"000037600000600090008000004090000001600000009300000040700000800010009000002540000", -- (0.00s, (logic))
"....839..1......3...4....7..42.3....6.......4....7..1..2........8...92.....25...6", -- (0.03s, (brute force))
"..1..5.7.92.6.......8...6...9..2.4.1.........3.4.8..9...7...3.......7.69.1.8..7..", -- (0.00s, (logic))
"--3------4---8--36--8---1---4--6--73---9----------2--5--4-7--686--------7--6--5--", -- (0.03s, (brute force))
"..3.2.6..9..3.5..1..18.64....81.29..7.......8..67.82....26.95..8..2.3..9..5.1.3..", -- (0.00s, (logic))
"--4-5--6--6-1--8-93----7----8----5-----4-3-----6----7----2----61-5--4-3--2--7-1--", -- (0.01s, (logic))
".4398.25.6..425...2....1.949....4.7.3..6.8...41.2.9..382.5.........4...553489.71.", -- (0.00s, (logic))
".9...4..7.....79..8........4.58.....3.......2.....97.6........4..35.....2..6...8.", -- (0.00s, (logic))
"1....7.9..3..2...8..96..5....53..9...1..8...26....4...3......1..4......7..7...3..", -- (0.26s, (brute force))
"12.3....435....1....4........54..2..6...7.........8.9...31..5.......9.7.....6...8", -- (0.40s, (brute force))
"12.4..3..3...1..5...6...1..7...9.....4.6.3.....3..2...5...8.7....7.....5.......98", -- (1.12s, (brute force))
"394..267....3..4..5..69..2..45...9..6.......7..7...58..1..67..8..9..8....264..735", -- (0.00s, (logic))
"4......6.5...8.9..3....1....2.7....1.9.....4.8....3.5....2....7..6.5...8.1......6", -- (0.03s, (brute force))
"5...7....6..195....98....6.8...6...34..8.3..17...2...6.6....28....419..5....8..79", -- (0.00s, (logic))
"503600009010002600900000080000700005006804100200003000030000008004300050800006702", -- (0.01s, (logic))
"53..247....2...8..1..7.39.2..8.72.49.2.98..7.79.....8.....3.5.696..1.3...5.69..1.", -- (0.00s, (logic))
"530070000600195000098000060800060003400803001700020006060000280000419005000080079", -- (0.00s, (logic))
"75..9..46961...3524.....79.2..6.1..7.8.....2.1..328.65.........3.9...2.484..3..79", -- (0.01s, (logic))
"800000000003600000070090200050007000000045700000100030001000068008500010090000400", -- (0.21s, (brute force))
"819--5-----2---75--371-4-6-4--59-1--7--3-8--2--3-62--7-5-7-921--64---9-----2--438", -- (0.00s, (logic))
"85...24..72......9..4.........1.7..23.5...9...4...........8..7..17..........36.4.", -- (0.01s, (logic))
"9..2..5...4..6..3...3.....6...9..2......5..8...7..4..37.....1...5..2..4...1..6..9", -- (0.18s, (brute force))
"97.3...6..6.75.........8.5.......67.....3.....539..2..7...25.....2.1...8.4...73..", -- (0.00s, (logic))
"000060080020000000001000000070000102500030000000000400004201000300700600000000050", -- (0.01s, (brute force))
27 puzzles solved in 3.74s (av 0.14s)
Running the fuller version mentioned above:
"008002000000600040064000092017005004200000008800100730470000910080001000000900200", -- (0.05s, *** MULTIPLE SOLUTIONS ***) 338 puzzles solved in 36.20s (av 0.11s) --or 338 puzzles solved in 16.46s (av 0.05s) (w/o check for multiple solutions)
Running a single puzzle (run_one_test set non-zero) produces:
. . . | . . . | . . . . . . | . . 3 | . 8 5 . . 1 | . 2 . | . . . -------+-------+------- . . . | 5 . 7 | . . . . . 4 | . . . | 1 . . . 9 . | . . . | . . . -------+-------+------- 5 . . | . . . | . 7 3 . . 2 | . 1 . | . . . . . . | . 4 . | . . 9 solution: 9 8 7 | 6 5 4 | 3 2 1 2 4 6 | 1 7 3 | 9 8 5 3 5 1 | 9 2 8 | 7 4 6 -------+-------+------- 1 2 8 | 5 3 7 | 6 9 4 6 3 4 | 8 9 2 | 1 5 7 7 9 5 | 4 6 1 | 8 3 2 -------+-------+------- 5 1 9 | 2 8 6 | 4 7 3 4 7 2 | 3 1 9 | 5 6 8 8 6 3 | 7 4 5 | 2 1 9 (logic), 0.02s
PHP
<lang php> class SudokuSolver { protected $grid = []; protected $emptySymbol; public static function parseString($str, $emptySymbol = '0') { $grid = str_split($str); foreach($grid as &$v) { if($v == $emptySymbol) { $v = 0; } else { $v = (int)$v; } } return $grid; }
public function __construct($str, $emptySymbol = '0') { if(strlen($str) !== 81) { throw new \Exception('Error sudoku'); } $this->grid = static::parseString($str, $emptySymbol); $this->emptySymbol = $emptySymbol; }
public function solve() { try { $this->placeNumber(0); return false; } catch(\Exception $e) { return true; } }
protected function placeNumber($pos) { if($pos == 81) { throw new \Exception('Finish'); } if($this->grid[$pos] > 0) { $this->placeNumber($pos+1); return; } for($n = 1; $n <= 9; $n++) { if($this->checkValidity($n, $pos%9, floor($pos/9))) { $this->grid[$pos] = $n; $this->placeNumber($pos+1); $this->grid[$pos] = 0; } } }
protected function checkValidity($val, $x, $y) { for($i = 0; $i < 9; $i++) { if(($this->grid[$y*9+$i] == $val) || ($this->grid[$i*9+$x] == $val)) { return false; } } $startX = (int) ((int)($x/3)*3); $startY = (int) ((int)($y/3)*3);
for($i = $startY; $i<$startY+3;$i++) { for($j = $startX; $j<$startX+3;$j++) { if($this->grid[$i*9+$j] == $val) { return false; } } } return true; }
public function display() { $str = ; for($i = 0; $i<9; $i++) { for($j = 0; $j<9;$j++) { $str .= $this->grid[$i*9+$j]; $str .= " "; if($j == 2 || $j == 5) { $str .= "| "; } } $str .= PHP_EOL; if($i == 2 || $i == 5) { $str .= "------+-------+------".PHP_EOL; } } echo $str; }
public function __toString() { foreach ($this->grid as &$item) { if($item == 0) { $item = $this->emptySymbol; } } return implode(, $this->grid); } } $solver = new SudokuSolver('009170000020600001800200000200006053000051009005040080040000700006000320700003900'); $solver->solve(); $solver->display();</lang>
- Output:
3 6 9 | 1 7 5 | 8 4 2 4 2 7 | 6 8 9 | 5 3 1 8 5 1 | 2 3 4 | 6 9 7 ------+-------+------ 2 1 8 | 7 9 6 | 4 5 3 6 3 4 | 8 5 1 | 2 7 9 9 7 5 | 3 4 2 | 1 8 6 ------+-------+------ 1 4 3 | 9 2 8 | 7 6 5 5 9 6 | 4 1 7 | 3 2 8 7 8 2 | 5 6 3 | 9 1 4 (solved in 0.027s)
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
Raku
(formerly Perl 6)
Brute Force
<lang perl6>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
Finesse It
This is an alternative solution that uses a more ellaborate set of choices instead of brute-forcing it.
<lang perl6>#
- 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 (flat 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].raku;
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 False;
}
# 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 False;
}
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 False;
} else {
$sudoku[$x][$y] = @r;
}
}
} while $resolved; # repeat if there was any change
return True;
}
sub solution-complexity-factor($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 (flat 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 (flat 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, $message; # un-comment for verbose logging
}
sub clone-sudoku($sudoku) {
my $clone;
for (flat 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:
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
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.lines
9.times.collect { |i| 9.times.collect { |j| lines[i][j].to_i } }
end
def permissible(matrix, i, j)
ok = [nil, *1..9]
check = ->(x,y) { ok[matrix[x][y]] = nil if matrix[x][y].nonzero? }
# Same as another in the column isn't permissible...
9.times { |x| check[x, j] }
# Same as another in the row isn't permissible...
9.times { |y| check[i, y] }
# Same as another in the 3x3 block isn't permissible...
xary = [ *(x = (i / 3) * 3) .. x + 2 ] #=> [0,1,2], [3,4,5] or [6,7,8]
yary = [ *(y = (j / 3) * 3) .. y + 2 ]
xary.product(yary).each { |x, y| check[x, y] }
# Gathering only permitted one
ok.compact
end
def deep_copy_sudoku(matrix)
matrix.collect { |row| row.dup }
end
def solve_sudoku(matrix)
loop do
options = []
9.times do |i|
9.times do |j|
next if matrix[i][j].nonzero?
p = permissible(matrix, i, j)
# If nothing is permissible, there is no solution at this level.
return if p.empty? # return nil
options << [i, j, p]
end
end
# If the matrix is complete, we have a solution...
return matrix if options.empty?
i, j, permissible = options.min_by { |x| x.last.length }
# If there is an option with only one solution, set it and re-check permissibility
if permissible.length == 1
matrix[i][j] = permissible[0]
next
end
# We have two or more choices. We need to search both...
permissible.each do |v|
mtmp = deep_copy_sudoku(matrix)
mtmp[i][j] = v
ret = solve_sudoku(mtmp)
return ret if ret
end
# We did an exhaustive search on this branch and nothing worked out.
return
end
end
def print_matrix(matrix)
puts "Impossible" or return unless matrix
border = "+-----+-----+-----+"
9.times do |i|
puts border if i%3 == 0
9.times do |j|
print j%3 == 0 ? "|" : " "
print matrix[i][j] == 0 ? "." : matrix[i][j]
end
puts "|"
end
puts border
end
data = <<EOS 394__267_ ___3__4__ 5__69__2_ _45___9__ 6_______7 __7___58_ _1__67__8 __9__8___ _264__735 EOS
matrix = read_matrix(data) print_matrix(matrix) puts print_matrix(solve_sudoku(matrix))</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| +-----+-----+-----+
Rust
<lang rust>type Sudoku = [u8; 81];
fn is_valid(val: u8, x: usize, y: usize, sudoku_ar: &mut Sudoku) -> bool {
(0..9).all(|i| sudoku_ar[y * 9 + i] != val && sudoku_ar[i * 9 + x] != val) && {
let (start_x, start_y) = ((x / 3) * 3, (y / 3) * 3);
(start_y..start_y + 3).all(|i| (start_x..start_x + 3).all(|j| sudoku_ar[i * 9 + j] != val))
}
}
fn place_number(pos: usize, sudoku_ar: &mut Sudoku) -> bool {
(pos..81).find(|&p| sudoku_ar[p] == 0).map_or(true, |pos| {
let (x, y) = (pos % 9, pos / 9);
for n in 1..10 {
if is_valid(n, x, y, sudoku_ar) {
sudoku_ar[pos] = n;
if place_number(pos + 1, sudoku_ar) {
return true;
}
sudoku_ar[pos] = 0;
}
}
false
})
}
fn pretty_print(sudoku_ar: Sudoku) {
let line_sep = "------+-------+------";
println!("{}", line_sep);
for (i, e) in sudoku_ar.iter().enumerate() {
print!("{} ", e);
if (i + 1) % 3 == 0 && (i + 1) % 9 != 0 {
print!("| ");
}
if (i + 1) % 9 == 0 {
println!(" ");
}
if (i + 1) % 27 == 0 {
println!("{}", line_sep);
}
}
}
fn solve(sudoku_ar: &mut Sudoku) -> bool {
place_number(0, sudoku_ar)
}
fn main() {
let mut sudoku_ar: Sudoku = [
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
];
if solve(&mut sudoku_ar) {
pretty_print(sudoku_ar);
} else {
println!("Unsolvable");
}
}</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 ------+-------+------
SAS
Use CLP solver in SAS/OR:
<lang sas>/* define SAS data set */ data Indata;
input C1-C9; datalines;
. . 5 . . 7 . . 1 . 7 . . 9 . . 3 . . . . 6 . . . . . . . 3 . . 1 . . 5 . 9 . . 8 . . 2 . 1 . . 2 . . 4 . . . . 2 . . 6 . . 9 . . . . 4 . . 8 . 8 . . 1 . . 5 . .
/* call OPTMODEL procedure in SAS/OR */ proc optmodel;
/* declare variables */
set ROWS = 1..9;
set COLS = ROWS;
var X {ROWS, COLS} >= 1 <= 9 integer;
/* declare nine row constraints */
con RowCon {i in ROWS}:
alldiff({j in COLS} X[i,j]);
/* declare nine column constraints */
con ColCon {j in COLS}:
alldiff({i in ROWS} X[i,j]);
/* declare nine 3x3 block constraints */
con BlockCon {s in 0..2, t in 0..2}:
alldiff({i in 3*s+1..3*s+3, j in 3*t+1..3*t+3} X[i,j]);
/* fix variables to cell values */
/* X[i,j] = c[i,j] if c[i,j] is not missing */
num c {ROWS, COLS};
read data indata into [_N_] {j in COLS} <c[_N_,j]=col('C'||j)>;
for {i in ROWS, j in COLS: c[i,j] ne .}
fix X[i,j] = c[i,j];
/* call CLP solver */ solve;
/* print solution */ print X;
quit;</lang>
Output:
X 1 2 3 4 5 6 7 8 9 1 9 8 5 3 2 7 6 4 1 2 6 7 1 5 9 4 2 3 8 3 3 2 4 6 1 8 9 5 7 4 2 4 3 7 6 1 8 9 5 5 5 9 7 4 8 3 1 2 6 6 1 6 8 2 5 9 4 7 3 7 4 5 2 8 3 6 7 1 9 8 7 1 6 9 4 5 3 8 2 9 8 3 9 1 7 2 5 6 4
Scala
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!!!
Scilab
The grid should be input in Init_board as a 9x9 matrix. The blanks should be represented by 0. A rule that the initial game should have at least 17 givens is enforced, for it guarantees a unique solution. It is also possible to set a maximum number of steps to the solver using break_point. If it is set to 0, there will be no break until it finds the solution.
<lang>Init_board=[... 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];
break_point=1.0d5; //if 0 there will be no break
function []=disp_board(board)
StringBoard=string(board);
for i=1:9
for j=1:9
if board(i,j)==0 then
StringBoard(i,j)='×';
end
end
end
StringBoard=[StringBoard, string(zeros(9,2))];
StringBoard=[StringBoard; string(zeros(2,11))];
for i=1:9
StringBoard(i,:)=[StringBoard(i,1:3), '|', StringBoard(i,4:6), '|', StringBoard(i,7:9)]
end
StringBoard(9:11,:)=StringBoard(7:9,:);
StringBoard(8,:)=strsplit('-----------');
StringBoard(5:7,:)=StringBoard(4:6,:);
StringBoard(4,:)=strsplit('-----------');
disp(StringBoard)
endfunction
function varargout=validate_input(input,position,board)
row=board(position(1),:);
column=board(:,position(2));
block=zeros(3,3);
if position(1)>=1 & position(1)<=3 then
i=0;
elseif position(1)>=4 & position(1)<=6 then
i=3;
else
i=6;
end
if position(2)>=1 & position(2)<=3 then
j=0;
elseif position(2)>=4 & position(2)<=6 then
j=3;
else
j=6;
end
block=board(i+1:i+3,j+1:j+3)
valid_input=%F;
valid_row=%F;
valid_col=%F;
valid_block=%F;
if find(input==row)==[] then
valid_row=%T;
end
if find(input==column)==[] then
valid_col=%T;
end
if find(input==block)==[] then
valid_block=%T;
end
if valid_row & valid_col & valid_block then
valid_input=%T;
end
varargout=list(valid_input,valid_row,valid_col,valid_block)
endfunction
function varargout=validate_board(board)
valid_flag1=%T;
for i=1:9
for j=1:9
if board(i,j)~= 0 then
check_board=Init_board;
check_board(i,j)=0;
valid_flag1=validate_input(board(i,j),[i j],check_board);
if ~valid_flag1 then
break
end
end
end
if ~valid_flag1 then
break
end
end
valid_flag2 = (length( find(board) ) >= 17); //enforces rule of minimum of 17 givens
//set it to always %T to ignore this rule
valid_board = (valid_flag1 & valid_flag2);
varargout=list(valid_board)
endfunction
disp('Initial board:'); disp_board(Init_board);
valid_init_board=validate_board(Init_board);
if ~valid_init_board then
error('Invalid initial board. Should follow sudoku rules and have at least 17 clues.');
end
blank=[]; for i=1:9
for j=1:9
if Init_board(i,j)== 0 then
blank=[blank; i j];
end
end
end
Solved_board=Init_board;
tic(); i=0; counter=0; breaked=%F; while i<size(blank,'r')
i=i+1;
counter=counter+1;
pos=blank(i,:);
value=Solved_board(pos(1),pos(2));
valid_value=%F;
while valid_value==%F
value=value+1;
if value>=10
break
else
valid_value=validate_input(value,pos,Solved_board);
end
end
if valid_value & value<10 then
Solved_board(pos(1),pos(2))=value
else
Solved_board(pos(1),pos(2))=0;
i=i-2;
end
if counter==break_point
breaked=%T;
break
end
end
valid_solved_board=validate_board(Solved_board); t2=toc();
if valid_solved_board & ~breaked then
disp('Solved!');
disp('Solution:');
disp_board(Solved_board);
disp('Time: '+string(t2)+'s.');
disp('Steps: '+string(counter)+'.');
elseif breaked
disp('Break point reached.');
disp('Time: '+string(t2)+'s.');
disp_board(Solved_board);
elseif ~valid_solved_board & ~breaked
disp('Invalid solution found.');
disp_board(Solved_board);
end</lang>
- Output:
Initial board: !5 3 × | × 7 × | × × × ! ! ! !6 × × | 1 9 5 | × × × ! ! ! !× 9 8 | × × × | × 6 × ! ! ! !- - - - - - - - - - - ! ! ! !8 × × | × 6 × | × × 3 ! ! ! !4 × × | 8 × 3 | × × 1 ! ! ! !7 × × | × 2 × | × × 6 ! ! ! !- - - - - - - - - - - ! ! ! !× 6 × | × × × | 2 8 × ! ! ! !× × × | 4 1 9 | × × 5 ! ! ! !× × × | × 8 × | × 7 9 ! Solved! Solution: !5 3 4 | 6 7 8 | 9 1 2 ! ! ! !6 7 2 | 1 9 5 | 3 4 8 ! ! ! !1 9 8 | 3 4 2 | 5 6 7 ! ! ! !- - - - - - - - - - - ! ! ! !8 5 9 | 7 6 1 | 4 2 3 ! ! ! !4 2 6 | 8 5 3 | 7 9 1 ! ! ! !7 1 3 | 9 2 4 | 8 5 6 ! ! ! !- - - - - - - - - - - ! ! ! !9 6 1 | 5 3 7 | 2 8 4 ! ! ! !2 8 7 | 4 1 9 | 6 3 5 ! ! ! !3 4 5 | 2 8 6 | 1 7 9 ! Time: 1.7142502s. Steps: 8365.
Sidef
<lang ruby>func check(i, j) is cached {
var (id, im) = i.divmod(9) var (jd, jm) = j.divmod(9)
jd == id && return true jm == im && return true
(id//3 == jd//3) && (jm//3 == im//3)
}
func solve(grid) {
for i in ^grid {
grid[i] && next
var t = [grid[{|j| check(i, j) }.grep(^grid)]].freq
{ |k|
t.has_key(k) && next
grid[i] = k
solve(grid)
} << 1..9
grid[i] = 0
return nil
}
for i in ^grid {
print "#{grid[i]} "
print " " if (3 -> divides(i+1))
print "\n" if (9 -> divides(i+1))
print "\n" if (27 -> divides(i+1))
}
}
var grid = %i(
5 3 0 0 2 4 7 0 0 0 0 2 0 0 0 8 0 0 1 0 0 7 0 3 9 0 2
0 0 8 0 7 2 0 4 9 0 2 0 9 8 0 0 7 0 7 9 0 0 0 0 0 8 0
0 0 0 0 3 0 5 0 6 9 6 0 0 1 0 3 0 0 0 5 0 6 9 0 0 1 0
)
solve(grid)</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
Shale
<lang shale>
- !/usr/local/bin/shale
time library
// This solves a sudoku with: // row/column/3x3box constraints (standard sudoku) // row/column/irregular, or jigsaw, region constraints // optionally with a Chess Knight's move constraint. // It is based on the python code from the Computerphile video // https://www.youtube.com/watch?v=G_UYXzGuqvM // The sudoku example from this video is used below, along with // a couple of Cracking The Cryptic examples and one from Andrew Stuart.
// The sudoku grid is stored in a multi-dimensional array under the grid:: namespace. // The row and column of each cell is represented by: // // row column:: grid:: // // A 0 value represents an empty cell, and a 1 to 9 value represents a cell value.
// You can specify which of the sudokus to solve by specifying s={n} on the command line. // If it is not specified then you get a text explaining how to specify an option // and what the options are and the default (s=1) option is solved.
startTime dup var now time::() =
whichSudoku var
s initialised {
whichSudoku s =
} {
"You can choose to solve one of several sudokus by adding s=n to the command line," println "where n is" println " 1: standard sudoku from Computerphile (the default)" println " 2: standard sudoku from Cracking The Cryptic" println " 3: standard sudoku from Cracking The Cryptic, with Knight's move constraint" println " 4: standard sudoku from Cracking The Cryptic" println " 5: irregular sudoku from Cracking The Cryptic" println " 6: irregular sudoku from Andrew Stuart" println "" println "You can also enable colour output by adding colour=true or color=true to the command line." println "" println "For example," println file arg:: shale:: " %s s=3 colour=true\n" printf "will solve the CTC sudoku with the Knight's move constraint" println "" println whichSudoku 1 =
} if
// Prints the sudoku grid. printGrid dup var {
r var c var doColour var irregular var region var
colour initialised {
doColour colour =
} {
color initialised {
doColour color =
} {
doColour false =
} if
} if
irregular 0 0:: regions:: initialised =
r 0 =
{ r 9 < } {
c 0 =
{ c 9 < } {
doColour {
irregular {
region r.value c.value:: regions:: =
region.value colour:: 0x1b "%c[1;%dm" printf
} {
r 3 / c 3 / + 1 + 2 % 3 * 31 + 0x1b "%c[1;%dm" printf
} if
} ifthen
r.value c.value:: grid:: dup 0 == { pop " ." } { " %d" } if printf
doColour {
0x1b "%c[0m" printf
} ifthen
c++
} while
"" println
r++
} while
} =
// This sets up the colour map for irregular sudokus. 1 colour:: dup var 30 = 2 colour:: dup var 31 = 3 colour:: dup var 32 = 4 colour:: dup var 33 = 5 colour:: dup var 34 = 6 colour:: dup var 35 = 7 colour:: dup var 36 = 8 colour:: dup var 30 = // skip 37 (too light) and reuse 30 and 31 (with luck they won't be close to regions 1 and 2). 9 colour:: dup var 31 =
// Assign the cell values to one row of the grid. setRow dup var {
8$ dup var swap = // cell 9 of the row 7$ dup var swap = // cell 8 of the row 6$ dup var swap = // ... 5$ dup var swap = 4$ dup var swap = 3$ dup var swap = 2$ dup var swap = 1$ dup var swap = // ... 0$ dup var swap = // cell 1 of the row r dup var swap = // the row number ns dup var swap &= // the namespace to use i var // loop counter
r 0 < r 8 > or {
r "Illegal row %d\n" printf
1 exit
} ifthen
i 0 =
{ i 9 < } {
i.value$ 0 < i.value$ 9 > or {
i r i.value$ "Illegal value %d specified for r%dc%d\n" printf
1 exit
} ifthen
r.value i.value:: ns->:: defined not { // define this grid cell if not already defined
r.value i.value:: ns->:: var
} ifthen
r.value i.value:: ns->:: i.value$ = // assign the value to the grid cell
i++
} while
} =
// Standard 3x3 box region checker. // This only works when called from within possible(). standardRegions dup var {
r0 var c0 var
r0 r 3 / 3 * =
c0 c 3 / 3 * =
i 0 =
{ i 3 < } {
j 0 =
{ j 3 < } {
r0 i + c0 j +:: grid:: n == {
false return
} ifthen
j++
} while
i++
} while
} =
// Irregular region checker. // This only works when called from within possible(). irregularRegions dup var {
region var i var
region r.value c.value:: regions:: = // The region we are in.
i 0 =
{ i 9 < } {
i.value r:: region.value:: region:: value i.value c:: region.value:: region:: value:: grid:: n == {
false return
} ifthen
i++
} while
} =
// Convert the regions:: namespace into something a little more cpu-time friendly. // Only optimise if there are irregular regions defined. optimiseRegions dup var {
0 0:: regions:: initialised {
region var
r var
c var
i var
region 1 =
{ region 10 < } {
i 0 =
r 0 =
{ r 9 < } {
c 0 =
{ c 9 < } {
r.value c.value:: regions:: region == {
i.value r:: region.value:: region:: dup var r =
i.value c:: region.value:: region:: dup var c =
i++
} ifthen
c++
} while
r++
} while
i 9 != {
i region "Region %d contains the wrong number of cells (%d)\n" printf
1 exit
} ifthen
region++
} while
} ifthen
} =
// Is it possible to place a digit in a given cell? possible dup var { function
n dup var swap = c dup var swap = r dup var swap = i var j var
// Check the column doesn't already contain this value.
i 0 =
{ i 9 < } {
r.value i.value:: grid:: n == {
false return
} ifthen
i++
} while
// Check the row doesn't already contain this value.
i 0 =
{ i 9 < } {
i.value c.value:: grid:: n == {
false return
} ifthen
i++
} while
// Check that the region doesn't already contain this value. regionChecker()
// Check for any other constraint.
constraint initialised { r c n constraint() not } and {
false return
} ifthen
true
} =
// This is a Knight's move constraint. knightsMoveConstraint dup var { function
0$ dup var swap = // n 1$ dup var swap = // column 2$ dup var swap = // row nr var nc var
// Check Knight's move 2 left and 1 up and down.
nc 1$ 2 - =
nc 0 >= {
nr 2$ 1 - =
nr 0 >= {
nr.value nc.value:: grid:: 0$ == {
false return
} ifthen
} ifthen
nr 2$ 1 + =
nr 9 < {
nr.value nc.value:: grid:: 0$ == {
false return
} ifthen
} ifthen
} ifthen
// Check Knight's move 1 left and 2 up and down.
nc 1$ 1 - =
nc 0 >= {
nr 2$ 2 - =
nr 0 >= {
nr.value nc.value:: grid:: 0$ == {
false return
} ifthen
} ifthen
nr 2$ 2 + =
nr 9 < {
nr.value nc.value:: grid:: 0$ == {
false return
} ifthen
} ifthen
} ifthen
// Check Knight's move 1 right and 2 up and down.
nc 1$ 1 + =
nc 9 < {
nr 2$ 2 - =
nr 0 >= {
nr.value nc.value:: grid:: 0$ == {
false return
} ifthen
} ifthen
nr 2$ 2 + =
nr 9 < {
nr.value nc.value:: grid:: 0$ == {
false return
} ifthen
} ifthen
} ifthen
// Check Knight's move 2 right and 1 up and down.
nc 1$ 2 + =
nc 9 < {
nr 2$ 1 - =
nr 0 >= {
nr.value nc.value:: grid:: 0$ == {
false return
} ifthen
} ifthen
nr 2$ 1 + =
nr 9 < {
nr.value nc.value:: grid:: 0$ == {
false return
} ifthen
} ifthen
} ifthen
true
} =
// Set this to standardRegions for the usual 3x3 boxes, // or irregularRegions to handle irregular, or jigsaw, regions. regionChecker var
// Leave this undefined to get the standard sudoku rules, or set this // to constraint code such as knightsMoveConstraint defined above. constraint var
solve dup var { function
r var c var n var
r 0 =
{ r 9 < } {
c 0 =
{ c 9 < } {
r.value c.value:: grid:: dup 0 == { // dup r.value c.value:: grid:: so we don't have to recalculate it in the inner loop.
n 1 =
{ n 10 < } {
r.value c.value n.value possible() {
dup n = // picking up r.value c.value:: grid:: again.
solve()
dup 0 = // picking up r.value c.value:: grid:: again.
} ifthen
n++
} while
pop // get rid of r.value c.value:: grid::
return
} ifthen
pop // get rid of r.value c.value:: grid::
c++
} while
r++
} while
"" println printGrid() now time::() startTime - 1000.0 / "Solution in %0.3f seconds\n" printf
} =
found var found false =
// As mentioned above, this is the example taken from the Computerphile video. // Raspberry Pi3 Model A time: 17s to the solution, 19s to finish. // The time spent between the solution and the finish is the script searching // of other, non-existant, soultion. whichSudoku 1 == {
found true = "From Computerphile: https://www.youtube.com/watch?v=G_UYXzGuqvM" println regionChecker standardRegions = grid 0 5 3 0 0 7 0 0 0 0 setRow() grid 1 6 0 0 1 9 5 0 0 0 setRow() grid 2 0 9 8 0 0 0 0 6 0 setRow() grid 3 8 0 0 0 6 0 0 0 3 setRow() grid 4 4 0 0 8 0 3 0 0 1 setRow() grid 5 7 0 0 0 2 0 0 0 6 setRow() grid 6 0 6 0 0 0 0 2 8 0 setRow() grid 7 0 0 0 4 1 9 0 0 5 setRow() grid 8 0 0 0 0 8 0 0 7 9 setRow()
} ifthen
// From https://www.youtube.com/watch?v=MXUgYxHmKq4&t=0s // This sudoku was featured on Cracking The Cryptic, and takes considerably longer than the one above. // Pi3 Model A time: 3m 23s to the solution, 15m 11s to finish. whichSudoku 2 == {
found true = "From Cracking The Cryptic: https://www.youtube.com/watch?v=MXUgYxHmKq4&t=0s" println regionChecker standardRegions = grid 0 0 6 8 0 0 0 0 1 3 setRow() grid 1 0 0 0 9 0 1 0 0 0 setRow() grid 2 0 0 0 0 0 8 0 0 4 setRow() grid 3 0 1 0 0 4 0 5 0 0 setRow() grid 4 0 3 0 0 0 9 0 0 0 setRow() grid 5 0 8 5 0 0 0 0 7 0 setRow() grid 6 0 2 0 0 0 7 3 0 0 setRow() grid 7 0 0 0 0 9 4 0 0 6 setRow() grid 8 4 0 0 0 6 0 0 0 0 setRow()
} ifthen
// From https://www.youtube.com/watch?v=rQHV-gIAG_0 // Another Cracking The Cryptic video, this one with a Knight's move constraint. // Pi3 Model A time: 48s to the solution, 6m 10s to finish. whichSudoku 3 == {
found true = "From Cracking The Cryptic: https://www.youtube.com/watch?v=rQHV-gIAG_0" println "This includes a Chess Knight's move constraint." println regionChecker standardRegions = constraint knightsMoveConstraint = grid 0 0 0 0 0 0 6 0 0 0 setRow() grid 1 0 0 3 0 0 0 0 0 7 setRow() grid 2 2 0 0 3 0 0 4 9 0 setRow() grid 3 6 0 0 0 0 0 0 4 5 setRow() grid 4 0 0 2 0 0 0 8 0 0 setRow() grid 5 0 0 0 1 0 0 0 0 0 setRow() grid 6 3 0 0 0 0 0 0 0 0 setRow() grid 7 7 0 0 0 0 1 0 0 9 setRow() grid 8 0 0 0 0 0 0 5 0 0 setRow()
} ifthen
// Another CTC sudoku: https://www.youtube.com/watch?v=vH-JooV8RA4&t=0s // Pi3 Model A time: 1m 6s to the solution, 48m 35s to finish. whichSudoku 4 == {
found true = "From Cracking The Cryptic: https://www.youtube.com/watch?v=vH-JooV8RA4&t=0s" println regionChecker standardRegions = grid 0 0 0 0 0 0 0 0 0 0 setRow() grid 1 0 0 9 8 0 0 0 0 7 setRow() grid 2 0 8 0 0 6 0 0 5 0 setRow() grid 3 0 5 0 0 4 0 0 3 0 setRow() grid 4 0 0 7 9 0 0 0 0 2 setRow() grid 5 0 0 0 0 0 0 0 0 0 setRow() grid 6 0 0 2 7 0 0 0 0 9 setRow() grid 7 0 4 0 0 5 0 0 6 0 setRow() grid 8 3 0 0 0 0 6 2 0 0 setRow()
} ifthen
// Another Cracking The Cryptic sudoku: https://www.youtube.com/watch?v=eJIu8w3ZXo8 // This is an irregular (jigsaw) sudoku. // Pi3 Model A time: 2m 5s to the solution, 7m 5s to finish. whichSudoku 5 == {
found true = "From Cracking The Cryptic: https://www.youtube.com/watch?v=eJIu8w3ZXo8" println "An irregular sudoku." println regionChecker irregularRegions = regions 0 1 1 1 1 1 2 2 2 2 setRow() regions 1 1 3 3 3 6 6 2 2 2 setRow() regions 2 1 3 3 3 6 7 7 7 2 setRow() regions 3 1 3 3 3 6 7 7 7 2 setRow() regions 4 1 6 6 6 6 7 7 7 8 setRow() regions 5 9 6 4 4 4 8 8 8 8 setRow() regions 6 9 9 4 4 4 8 5 5 5 setRow() regions 7 9 9 4 4 4 8 5 5 5 setRow() regions 8 9 9 9 9 8 8 5 5 5 setRow() grid 0 3 0 0 0 0 0 0 0 1 setRow() grid 1 0 9 0 1 7 2 0 0 0 setRow() grid 2 0 0 3 0 0 0 9 0 0 setRow() grid 3 0 7 0 0 0 0 0 4 0 setRow() grid 4 0 4 0 0 3 0 0 6 0 setRow() grid 5 0 5 0 0 0 0 0 9 0 setRow() grid 6 0 0 6 0 0 0 5 0 0 setRow() grid 7 0 0 0 8 5 6 0 2 0 setRow() grid 8 7 0 0 0 0 0 0 0 8 setRow()
} ifthen
// This is taken from Andrew Stuart's web site https://www.sudokuwiki.org/Daily_Jigsaw_Sudoku // No. 4294, dated 13 Nov 2020. It is rated 5-star out of 6: Diabolical. // The archived version is here: https://www.sudokuwiki.org/Print_Daily_Jigsaw.asp?day=13/11/2020 // At the time of writing, Andrew only archives sudoku's for 31 days, then they are deleted. // The region layout is called "Andrew Stuart 24" and the numbering is taken directly // from Andrew's solver page. A big thanks goes to Andrew for giving me permission to include this sudoku. // The region numbers must be between 1 and 9 inclusive. // Pi3 Model A time: 26m 23s to the solution, 1h 15m 20s to finish. whichSudoku 6 == {
found true = "From Andrew Stuart's web page: https://www.sudokuwiki.org/Daily_Jigsaw_Sudoku" println "No. 4294, dated 13 Nov 2020" println regionChecker irregularRegions = regions 0 1 1 2 2 2 2 3 3 4 setRow() regions 1 1 1 2 2 2 3 3 3 4 setRow() regions 2 1 1 1 2 3 3 3 4 4 setRow() regions 3 1 1 5 2 3 7 4 4 4 setRow() regions 4 5 5 5 6 6 7 7 4 4 setRow() regions 5 5 5 5 6 6 7 7 7 9 setRow() regions 6 8 5 5 6 6 7 7 7 9 setRow() regions 7 8 8 6 6 6 9 9 9 9 setRow() regions 8 8 8 8 8 8 8 9 9 9 setRow() grid 0 0 0 0 0 0 0 0 9 0 setRow() grid 1 5 0 0 0 8 1 0 0 0 setRow() grid 2 0 0 0 9 0 4 5 0 0 setRow() grid 3 0 0 0 0 0 0 0 0 5 setRow() grid 4 0 1 0 0 0 8 0 0 0 setRow() grid 5 7 0 6 0 0 0 0 0 0 setRow() grid 6 0 0 0 1 0 0 6 0 0 setRow() grid 7 1 0 0 7 2 0 0 0 0 setRow() grid 8 0 9 0 0 0 0 0 0 0 setRow()
} ifthen
// If you want to add your own sudoku, add it here. whichSudoku 7 == {
found true = // add it here // regionChecker standardRegions = // Include a region checker // regionChecker irregularRegions = // constraint knightsMoveConstraint = // Include any extra contraints, as appropriate
} ifthen
// Don't change anything from here to the end.
found {
optimiseRegions()
"" println "Initial grid" println printGrid()
solve()
now time::() startTime - 1000.0 / "Total runtime was %0.3f seconds\n" printf
} {
whichSudoku "Sudoku %d not found\n" printf
} if
// I'd like to see an irregular sudoku with a knight's move constraint. Any takers... </lang>
Output
From Cracking The Cryptic: https://www.youtube.com/watch?v=rQHV-gIAG_0 This includes a Chess Knight's move constraint. Initial grid . . . . . 6 . . . . . 3 . . . . . 7 2 . . 3 . . 4 9 . 6 . . . . . . 4 5 . . 2 . . . 8 . . . . . 1 . . . . . 3 . . . . . . . . 7 . . . . 1 . . 9 . . . . . . 5 . . 1 9 7 8 4 6 2 5 3 8 4 3 5 9 2 6 1 7 2 6 5 3 1 7 4 9 8 6 1 9 2 3 8 7 4 5 5 7 2 4 6 9 8 3 1 4 3 8 1 7 5 9 2 6 3 8 6 9 5 4 1 7 2 7 5 4 6 2 1 3 8 9 9 2 1 7 8 3 5 6 4 Solution in 7.241 seconds Total runtime was 53.995 seconds
SQL
The implementation below uses a recursive WITH clause (aka recursive CTE, recursive query, recursive factored subquery). This is supported - with minimal syntactical differences - by some (perhaps many) but not all SQL dialects. The code was written and tested in Oracle SQL; Oracle has supported recursive subqueries since version 11.2.
The code implements a brute force algorithm. It can solve most problems in less than half a second (depending on hardware too), although I found a few that take 2-3 and up to 6 seconds. The output may either be empty (when the problem is impossible), a single completed grid (for a "correct" problem), or all the correct solutions to problems that admit more than one solution.
The input and output are presented as strings of 81 characters, where each character is either a digit or a space (indicating an empty cell in the grid). The translation between grids and such strings is trivial (convert grid to string by concatenating rows, reading left to right and then top to bottom), and not covered in the solution. The input is given as a bind variable, :game.
<lang sql>with
symbols (d) as (select to_char(level) from dual connect by level <= 9)
, board (i) as (select level from dual connect by level <= 81) , neighbors (i, j) as (
select b1.i, b2.i
from board b1 inner join board b2
on b1.i != b2.i
and (
mod(b1.i - b2.i, 9) = 0
or ceil(b1.i / 9) = ceil(b2.i / 9)
or ceil(b1.i / 27) = ceil(b2.i / 27) and trunc(mod(b1.i - 1, 9) / 3) = trunc(mod(b2.i - 1, 9) / 3)
)
)
, r (str, pos) as (
select :game, instr(:game, ' ')
from dual
union all
select substr(r.str, 1, r.pos - 1) || s.d || substr(r.str, r.pos + 1), instr(r.str, ' ', r.pos + 1)
from r inner join symbols s
on r.pos > 0 and not exists (
select *
from neighbors n
where r.pos = n.i and s.d = substr(r.str, n.j, 1)
)
)
select str from r where pos = 0
- </lang>
A better (faster) approach - taking advantage of database-specific features - is to create a table NEIGHBORS (similar to the inline view in the WITH clause) and an index on column I of that table; then the query execution time drops by more than half in most cases.
Stata
In this implementation, a Sudoku is a 9x9 matrix a, with either digits 1-9 or missing values. A cell is any element of a. A cell can be reference with indices i,j, or with a single index n between 1 and 81.
Three functions are defined:
- sudoku(a) will return 0 if there is no solution, 1 if there is at least one solution, and then a is modified in place with the solution found. This function builds several tables before calling the main "solver" function.
- solve(a,s,t,v,w) is made of two parts: first, all cells that can be filled without any assumption are considered known. Then, if not all cells are known, a cell with minimal number of choices is found, and a recursive call to solve is made with all possibilities in turn for this cell.
- push(a,s,t,v,w,n,z) is an auxiliary function, which pushes the value z in the nth cell of matrix a. Here n is a value between 1 and 81.
Fixed tables:
- t(n,.) is a row i,j,k: the cell n (1-81) has row index i, column index j, and square index k, where i, j, k are all in 1-9.
- s(n,.) is a row that contains the indices of the 20 "neighbors" of cell n: these are the cells in the same row, column or square.
Varying tables:
- v(n,z)=1 if the cell n may have the value z, otherwise 0. When a value z is pushed into the matrix, all neighboring cells are updated, as they can't take the value z anymore.
- w(n)=1 if cell n is not yet known, otherwise 0.
- In the initialization step of the sudoku() function, w has another meaning: it stores the column index of the last entry in row n of the array s while it is built. When it's done, every element of w is thus twenty.
The example grid given below is taken from Wikipedia. It does not require any recursive call (it's entirely filled in the first step of solve), as can be seen with additional printf in the code to follow the algorithm.
<lang stata>mata function sudoku(a) { s = J(81,20,.) t = J(81,3,.) v = J(81,9,1) w = J(81,1,0) for (i=1; i<=9; i++) { for (j=1; j<=9; j++) { n = (i-1)*9+j k = floor((i-1)/3)*3+floor((j-1)/3)+1 t[n,.] = i,j,k } } for (i=1; i<=81; i++) { for (j=i+1; j<=81; j++) { if (any(t[i,.]:==t[j,.])) { w[i]=w[i]+1 w[j]=w[j]+1 s[i,w[i]] = j s[j,w[j]] = i } } } w = J(81,1,1) for (i=1; i<=9; i++) { for (j=1; j<=9; j++) { if (a[i,j]<.) { push(a,s,t,v,w,(i-1)*9+j,a[i,j]) } } } return(solve(a,s,t,v,w)) }
function solve(a,s,t,v,w) { for (q=1; q;) { q = 0 for (n=1; n<=81; n++) { if (w[n]) { r = sum(v[n,.]) if (r==0) { return(0) } else if (r==1) { q = 1 push(a,s,t,v,w,n,selectindex(v[n,.])) } } } }
if (all(w:==0)) { return(1) } else { m0 = n0 = . for (n=1; n<=81; n++) { m = sum(v[n,.]) if (w[n] & m>1 & m<m0) { m0 = m n0 = n } } z = selectindex(v[n0,.]) for (i=1; i<=m0; i++) { a2 = a v2 = v w2 = w push(a2,s,t,v2,w2,n0,z[i]) if (solve(a2,s,t,v2,w2)) { a = a2 return(1) } } return(0) } }
function push(a,s,t,v,w,n,z) { w[n] = 0 i = t[n,1] j = t[n,2] a[i,j] = z for (k=1; k<=20; k++) { v[s[n,k],z] = 0 } }
a = 5,3,.,.,7,.,.,.,.\
6,.,.,1,9,5,.,.,.\ .,9,8,.,.,.,.,6,.\ 8,.,.,.,6,.,.,.,3\ 4,.,.,8,.,3,.,.,1\ 7,.,.,.,2,.,.,.,6\ .,6,.,.,.,.,2,8,.\ .,.,.,4,1,9,.,.,5\ .,.,.,.,8,.,.,7,9
sudoku(a) a end</lang>
Output
1
1 2 3 4 5 6 7 8 9
+-------------------------------------+
1 | 5 3 4 6 7 8 9 1 2 |
2 | 6 7 2 1 9 5 3 4 8 |
3 | 1 9 8 3 4 2 5 6 7 |
4 | 8 5 9 7 6 1 4 2 3 |
5 | 4 2 6 8 5 3 7 9 1 |
6 | 7 1 3 9 2 4 8 5 6 |
7 | 9 6 1 5 3 7 2 8 4 |
8 | 2 8 7 4 1 9 6 3 5 |
9 | 3 4 5 2 8 6 1 7 9 |
+-------------------------------------+
Two more examples, from here and there.
<lang stata>a = 7,9,.,.,.,3,.,.,2\
.,6,.,5,1,.,.,.,.\ .,.,.,.,.,2,.,.,6\ .,.,.,8,.,1,.,4,.\ .,.,.,.,.,.,1,.,3\ 5,.,.,.,2,.,.,.,.\ .,.,.,.,.,.,.,7,4\ .,.,1,.,.,.,.,6,5\ .,.,8,.,9,7,.,.,.
sudoku(a) a
1 2 3 4 5 6 7 8 9 +-------------------------------------+ 1 | 7 9 5 6 8 3 4 1 2 | 2 | 2 6 4 5 1 9 7 3 8 | 3 | 1 8 3 7 4 2 5 9 6 | 4 | 9 3 6 8 5 1 2 4 7 | 5 | 8 4 2 9 7 6 1 5 3 | 6 | 5 1 7 3 2 4 6 8 9 | 7 | 3 2 9 1 6 5 8 7 4 | 8 | 4 7 1 2 3 8 9 6 5 | 9 | 6 5 8 4 9 7 3 2 1 | +-------------------------------------+
a = .,.,4,1,.,.,.,.,9\
.,9,.,.,8,.,.,6,.\ 6,.,.,.,.,3,7,.,.\ 5,.,.,.,.,2,8,.,.\ .,4,.,.,3,.,.,2,.\ .,.,1,8,.,.,.,.,5\ .,.,2,5,.,.,.,.,3\ .,8,.,.,7,.,.,1,.\ 7,.,.,.,.,1,4,.,.
sudoku(a) a
1 2 3 4 5 6 7 8 9 +-------------------------------------+ 1 | 2 7 4 1 5 6 3 8 9 | 2 | 1 9 3 7 8 4 5 6 2 | 3 | 6 5 8 2 9 3 7 4 1 | 4 | 5 3 7 6 1 2 8 9 4 | 5 | 8 4 6 9 3 5 1 2 7 | 6 | 9 2 1 8 4 7 6 3 5 | 7 | 4 1 2 5 6 8 9 7 3 | 8 | 3 8 5 4 7 9 2 1 6 | 9 | 7 6 9 3 2 1 4 5 8 | +-------------------------------------+</lang>
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 | -----------------------
<lang Swift> func solving(board: Int) -> Int { var board = board var isSolved = false while !isSolved { for x in 0 ..< 9 { for y in 0 ..< 9 { if board[x][y] == 0 { let known = Set(board.map { $0[y] } + board[x] + subgrid(board, pos: (x, y))) let possible = Set(Array(1...9)).subtracting(known) if possible.count == 1 { board[x][y] = possible.first! } } } isSolved = 45 == board[x].reduce(0, +) } } return board }
func subgrid(_ board: Int, pos: (Int, Int)) -> [Int] { var r = [Int]() var (x, y) = pos x = x / 3 * 3 y = y / 3 * 3 for i in x ..< x + 3 { for j in y ..< y + 3 { r.append(board[i][j]) } } return r }
func print(_ board: Int) { for i in board.indices { if i % 9 == 0 { print(" -------------------") } for j in board.indices { if j % board.count == 0 { print("| ", terminator: "") } let digit = board[i][j] print(digit != 0 ? digit : " ", terminator: "") print(" ", terminator: "") } print("|") } print(" -------------------") }
let puzzle = [ [0,2,0,4,5,0,7,0,9], [0,0,0,1,0,9,0,3,0], [0,0,8,0,0,0,1,0,4], [0,4,0,0,6,1,0,7,0], [5,0,6,0,3,0,0,1,0], [0,3,0,0,0,2,0,9,0], [3,0,4,0,7,5,0,6,8], [0,9,0,0,1,0,3,0,7], [0,0,2,0,0,3,0,0,1] ]
print(solving(board: puzzle)) </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 | -------------------
SystemVerilog
<lang systemverilog>
////////////////////////////////////////////////////////////////////////////// /// SudokuSolver /// /// A class that fills up a sudoku board, the initial board is given /// /// as an array preset_rows, the positions where preset_rows is zero /// /// are to be determined. Three views of the sudoku board are created /// /// and the uniqueness of its elements are defined by on constraint for /// /// each view, one constraint ensures that the values are between 1 and /// /// 9, and two other constraints are used to ensure that the values in /// /// all three views agree to each other. /// /// /// /// /// /// A solution using only the "rows" array would be possible, however /// /// this illustrates better how one can relate different variables in /// /// SystemVerilog Constrained randomization. /// ////////////////////////////////////////////////////////////////////////////// class SudokuSolver;
rand int tiles[0:8][0:8];
rand int rows[0:8][0:8];
rand int cols[0:8][0:8];
int preset_rows[0:8][0:8];
constraint board_input {
foreach(preset_rows[i])foreach(preset_rows[i][j])
if(preset_rows[i][j] != 0) rows[i][j] == preset_rows[i][j];
}
constraint range {
foreach(rows[i]) foreach(rows[i][j])
rows[i][j] inside {[1:9]};
}
////////////////////////////////////////////////
/// Every number in a row is unique ///
////////////////////////////////////////////////
constraint rows_permutation {
foreach(rows[i]) foreach(rows[i][j1])
foreach(rows[i][j2])
if(j1 != j2) rows[i][j1] != rows[i][j2];
}
///////////////////////////////////////////////
/// Every number in a column is unique ///
///////////////////////////////////////////////
constraint cols_permutation {
foreach(cols[i]) foreach(cols[i][j1])
foreach(cols[i][j2])
if(j1 != j2) cols[i][j1] != cols[i][j2];
}
/////////////////////////////////////////////////
/// Every number in a tile (square) is unique ///
/////////////////////////////////////////////////
constraint tiles_permutation {
foreach(tiles[i]) foreach(tiles[i][j1])
foreach(tiles[i][j2])
if(j1 != j2) tiles[i][j1] != tiles[i][j2];
}
///////////////////////////////////////////////////
/// Makes sure that sure that the numbers in ///
/// each view agree with other views ///
///////////////////////////////////////////////////
constraint rows_vs_tiles {
foreach(tiles[i]) foreach(tiles[i][j])
tiles[i][j] == rows[(i/3) * 3 + (j/3)][3*(i%3)+(j%3)];
}
constraint rows_vs_cols {
foreach(cols[i]) foreach(cols[i][j])
cols[i][j] == rows[j][i];
}
///////////////////////////////////////////////////
/// Print the current state of the board in the ///
/// standard output ///
///////////////////////////////////////////////////
function void printBoard;
int i, j;
for(i = 0; i < 9; ++i) begin
if(i % 3 == 0)$display(" -------------");
$write(" ");
for(j = 0; j < 9; ++j) begin
if(j % 3 == 0) $write("|");
$write("%c", "0" + rows[i][j]);
end
$display("|");
end
$display(" -------------");
endfunction
function void printInitial;
int i, j;
for(i = 0; i < 9; ++i) begin
if(i % 3 == 0)$display("-------------");
for(j = 0; j < 9; ++j) begin
if(j % 3 == 0) $write("|");
if(preset_rows[i][j]) begin
$write("%c", "0" + preset_rows[i][j]);
end
else begin
$write(".");
end
end
$display("|");
end
$display("-------------");
endfunction
endclass
////////////////////////////////////////////////////// /// Simple program instantiating the sudoku object /// ////////////////////////////////////////////////////// program SudokuTest;
SudokuSolver board;
initial begin
board = new;
foreach(board.preset_rows[0][i]) board.preset_rows[i][i] = i+1;
$display("Initial Board:");
board.printInitial();
// Generate two different solutions for the board
if(board.randomize())begin
$display("One solution:");
board.printBoard();
end
else begin
$display("ERROR: Failed to generate first solution");
end
if(board.randomize())begin
$display("Another solution:");
board.printBoard();
end
else begin
$display("ERROR: Failed to generate second solution");
end
end endprogram </lang>
It can be seen that SystemVerilog randomization is a very powerfull tool, in this implementation I directly described the game constraints and the randomization engine takes care of producing solutions, and when multiple solutions are possible they will be chosen at random.
Running the above code using Cadence ncverilog I get
> ncverilog +sv sudoku.sv Initial Board: ------------- |1..|...|...| |.2.|...|...| |..3|...|...| ------------- |...|4..|...| |...|.5.|...| |...|..6|...| ------------- |...|...|7..| |...|...|.8.| |...|...|..9| ------------- One solution: ------------- |168|547|392| |925|631|478| |743|928|156| ------------- |832|479|561| |671|852|934| |459|316|827| ------------- |396|284|715| |214|795|683| |587|163|249| ------------- Another solution: ------------- |197|642|358| |425|389|176| |683|715|294| ------------- |956|431|827| |842|957|631| |731|826|945| ------------- |214|598|763| |569|173|482| |378|264|519| -------------
Tailspin
<lang tailspin> templates deduceRemainingDigits
data row <"1">, col <"1"> local
templates findOpenPosition
@:{options: 10};
$ -> \[i;j](when <[](..~$@findOpenPosition.options)> do @findOpenPosition: {row: $i, col: $j, options: $::length}; \) -> !VOID
$@ !
end findOpenPosition
templates selectFirst&{pos:}
def digit: $($pos.row;$pos.col) -> $(1);
$ -> \[i;j](
when <?($i <=$pos.row>)?($j <=$pos.col>)> do $digit !
when <[]?($i <=$pos.row>)
|[]?($j <=$pos.col>)
|[]?(($i-1)~/3 <=($pos.row-1)~/3>)?(($j-1)~/3 <=($pos.col-1)~/3>)> do [$... -> \(when <~=$digit> do $! \)] !
otherwise $ !
\) !
end selectFirst
@: $;
$ -> findOpenPosition -> #
when <{options: <=0>}> do [] !
when <{options: <=10>}> do $@ !
otherwise def next: $;
$@ -> selectFirst&{pos: $next} -> deduceRemainingDigits
-> \(when <~=[]> do @deduceRemainingDigits: $; {options: 10} !
when <=[]> do ^@deduceRemainingDigits($next.row;$next.col;1)
-> { $next..., options: $next.options::raw-1} ! \) -> #
end deduceRemainingDigits
test 'internal solver'
def sample: [ [5,3,4,6,7,8,9,1,2], [6,7,2,1,9,5,3,4,8], [1,9,8,3,4,2,5,6,7], [8,5,9,7,6,1,4,2,3], [4,2,6,8,5,3,7,9,1], [7,1,3,9,2,4,8,5,6], [9,6,1,5,3,7,2,8,4], [2,8,7,4,1,9,6,3,5], [3,4,5,2,8,6,1,7,9] ]; assert $sample -> deduceRemainingDigits <=$sample> 'completed puzzle unchanged' assert [ [[5],3,4,6,7,8,9,1,2], $sample(2..last)...] -> deduceRemainingDigits <=$sample> 'final digit gets placed' assert [ [[],3,4,6,7,8,9,1,2], $sample(2..last)...] -> deduceRemainingDigits <=[]> 'no remaining options returns empty' assert [ [[5],3,4,6,[2,5,7],8,9,1,[2,5]], $sample(2..last)...] -> deduceRemainingDigits <=$sample> 'solves 3 digits on row' assert [ [5,3,4,6,7,8,9,1,2], [[6,7,9],7,2,1,9,5,3,4,8], [1,9,8,3,4,2,5,6,7], [8,5,9,7,6,1,4,2,3], [4,2,6,8,5,3,7,9,1], [[7],1,3,9,2,4,8,5,6], [[7,9],6,1,5,3,7,2,8,4], [2,8,7,4,1,9,6,3,5], [3,4,5,2,8,6,1,7,9] ] -> deduceRemainingDigits <=$sample> 'solves 3 digits on column' assert [ [5,3,[4,6],6,7,8,9,1,2], [[6],7,2,1,9,5,3,4,8], [1,[4,6,9],8,3,4,2,5,6,7], $sample(4..last)... ] -> deduceRemainingDigits <=$sample> 'solves 3 digits in block' // This gives a contradiction if 3 gets chosen out of [3,5] assert [ [[3,5],[3,4,6],[3,4,6],[3,4,6],7,8,9,1,2], $sample(2..last)...] -> deduceRemainingDigits <=$sample> 'contradiction is backtracked'
end 'internal solver'
composer parseSudoku
[<section>=3] rule section: <row>=3 (<'-+'>? <WS>?) rule row: [<triple>=3] (<WS>?) rule triple: <digit|dot>=3 (<'\|'>?) rule digit: [<'\d'>] rule dot: <'\.'> -> [1..9 -> '$;']
end parseSudoku
test 'input sudoku' def parsed: '53.|.7.|...
6..|195|... .98|...|.67 ----------- 8..|.6.|..3 4..|8.3|..1 7..|.2.|..6 ----------- .6.|...|28. ...|419|..5 ...|.8.|.79' -> parseSudoku; assert $parsed <[<[<[]>=9](9)>=9](9)> 'parsed sudoku has 9 rows containing 9 columns of lists' assert $parsed(1;1) <=['5']> 'a digit' assert $parsed(1;3) <=['1','2','3','4','5','6','7','8','9']> 'a dot'
end 'input sudoku'
templates solveSudoku
$ -> parseSudoku -> deduceRemainingDigits -> #
when <=[]> do 'No result found' !
when <> do $ -> \[i](
'$(1..3)...;|$(4..6)...;|$(7..9)...;$#10;' !
$i -> \(when <=3|=6> do '-----------$#10;' !\) !
\) -> '$...;' !
end solveSudoku
test 'sudoku solver'
assert
'53.|.7.|...
6..|195|... .98|...|.67 ----------- 8..|.6.|..3 4..|8.3|..1 7..|.2.|..6 ----------- .6.|...|28. ...|419|..5 ...|.8.|.79' -> solveSudoku <=
'534|678|912 672|195|348 198|342|567
859|761|423 426|853|791 713|924|856
961|537|284 287|419|635 345|286|179 '> 'solves sudoku and outputs pretty solution' end 'sudoku solver' </lang>
Tcl
Adapted from a page on the Tcler's Wiki to use a standard object system.
Note that you can implement more rules if you want. Just make another subclass of Rule and the solver will pick it up and use it automatically.
or
<lang tcl>package require Tcl 8.6 oo::class create Sudoku {
variable idata
method clear {} {
for {set y 0} {$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
Wren
<lang ecmascript>class Sudoku {
construct new(rows) {
if (rows.count != 9 || rows.any { |r| r.count != 9 }) {
Fiber.abort("Grid must be 9 x 9")
}
_grid = List.filled(81, null)
for (i in 0..8) {
for (j in 0..8 ) _grid[9 * i + j] = rows[i][j]
}
_solved = false
}
checkValidity_(v, x, y) {
for (i in 0..8) {
if (_grid[y * 9 + i] == v || _grid[i * 9 + x] == v) return false
}
var startX = (x / 3).floor * 3
var startY = (y / 3).floor * 3
for (i in startY...startY + 3) {
for (j in startX...startX + 3) {
if (_grid[i * 9 + j] == v) return false
}
}
return true
}
placeNumber_(pos) {
if (_solved) return
if (pos == 81) {
_solved = true
return
}
if (_grid[pos].bytes[0] > 48) {
placeNumber_(pos + 1)
return
}
for (n in 1..9) {
if (checkValidity_(n.toString, pos % 9, (pos/9).floor)) {
_grid[pos] = n.toString
placeNumber_(pos + 1)
if (_solved) return
_grid[pos] = "0"
}
}
}
solve() {
System.print("Starting grid:\n\n%(this)")
placeNumber_(0)
System.print(_solved ? "Solution:\n\n%(this)" : "Unsolvable!")
}
toString {
var sb = ""
for (i in 0..8) {
for (j in 0..8) {
sb = sb + _grid[i * 9 + j] + " "
if (j == 2 || j == 5) sb = sb + "| "
}
sb = sb + "\n"
if (i == 2 || i == 5) sb = sb + "------+-------+------\n"
}
return sb
}
}
var rows = [
"850002400", "720000009", "004000000", "000107002", "305000900", "040000000", "000080070", "017000000", "000036040"
] Sudoku.new(rows).solve()</lang>
- Output:
Starting grid: 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 Solution: 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
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
zkl
Note: Unlike in the C solution, 1<<-1 is defined (as 0).
<lang zkl>fcn trycell(sdku,pos=0){
row,col:=pos/9, pos%9;
if(pos==81) return(True);
if(sdku[pos]) return(trycell(sdku, pos + 1));
used:=0;
foreach r in (9){ used=used.bitOr((1).shiftLeft(sdku[r*9 + col] - 1)) }
foreach c in (9){ used=used.bitOr((1).shiftLeft(sdku[row*9 + c] - 1)) }
row,col = row/3*3, col/3*3;
foreach r,c in ([row..row+2], [col..col+2])
{ used=used.bitOr((1).shiftLeft(sdku[r*9 + c] - 1)) }
sdku[pos]=1; while(sdku[pos]<=9){
if(used.isEven and trycell(sdku, pos + 1)) return(True);
sdku[pos]+=1; used/=2;
}
sdku[pos]=0;
return(False);
}</lang> <lang zkl>problem:=
- <<<
" 5 3 0 0 7 0 0 0 0
6 0 0 1 9 5 0 0 0
0 9 8 0 0 0 0 6 0
8 0 0 0 6 0 0 0 3
4 0 0 8 0 3 0 0 1
7 0 0 0 2 0 0 0 6
0 6 0 0 0 0 2 8 0
0 0 0 4 1 9 0 0 5
0 0 0 0 8 0 0 7 9";
- <<<
s:=problem.split().apply("toInt").copy(); // writable list of 81 ints trycell(s).println(); println("+-----+-----+-----+"); foreach n in (3){
s[n*27,27].pump(Console.println,T(Void.Read,8),("| " + "%s%s%s | "*3).fmt); // 3 lines
println("+-----+-----+-----+");
}</lang>
- Output:
True +-----+-----+-----+ | 534 | 678 | 912 | | 672 | 195 | 348 | | 198 | 342 | 567 | +-----+-----+-----+ | 859 | 761 | 423 | | 426 | 853 | 791 | | 713 | 924 | 856 | +-----+-----+-----+ | 961 | 537 | 284 | | 287 | 419 | 635 | | 345 | 286 | 179 | +-----+-----+-----+
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