Sudoku
You are encouraged to solve this task according to the task description, using any language you may know.
Solve a partially filled-in normal 9x9 Sudoku grid and display the result in a human-readable format. Algorithmics of sudoku may be help implement this.
AutoHotkey
by infogulch. More info (and similarly formatted puzzles) <lang AutoHotkey>puzzleEasy = ( LTrim 394 @@2 67@ @@@ 3@@ 4@@ 5@@ 69@ @2@
@45 @@@ 9@@ 6@@ @@@ @@7 @@7 @@@ 58@
@1@ @67 @@8 @@9 @@8 @@@ @26 4@@ 735 ) s := A_TickCount MsgBox % "Easy:`n" Sudoku( puzzleEasy ) "`nIterations: " ErrorLevel "`n`nSeconds: " (A_TickCount-s)/1000
- 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
- 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) (!Mod(A_Index, 9) ? "`n" : !Mod(A_Index,3) ? "`t" : "") return r } </lang>