A hard Sudoku Puzzle | ||||||||

3 | 6 | |||||||

1 | ||||||||

9 | 7 | 5 | 8 | |||||

9 | 2 | |||||||

8 | 7 | 4 | ||||||

3 | 6 | |||||||

1 | 2 | 8 | 9 | |||||

4 | ||||||||

5 | 1 |

grid'[i] = jwhere grid is the vector of problem variables from the first model and numbers the vector of problem variables from the second model. Note, that the channeling constraints are realized with Reified Constraints among other things.i+ 1 numbers[ j-1],

val inputlist_hard=[(0,5,3),(0,7,6),(1,7,1),(2,1,9), (2,2,7),(2,3,5),(2,7,8),(3,4,9), (3,6,2),(4,2,8),(4,4,7),(4,6,4), (5,2,3),(5,4,6),(6,1,1),(6,5,2), (6,6,8),(6,7,9),(7,1,4),(8,1,5), (8,3,1)] fun sudoku inputlist space = let val grid = Vector.tabulate(9,fn x => FD.rangeVec(space,9,(1,9))) val grid' = Vector.concat(Vector.toList(grid)) val numbers = Vector.tabulate(9,fn x => FS.upperBound(space,#[(1,81)])) val rows = List.tabulate(9,fn x =>(x*9+9-8,x*9+9)) val domainvecs = List.tabulate(9,fn y => Vector.tabulate(9,fn x =>(x*9+1+y,x*9+1+y))) val boxes1 = List.tabulate(3,fn y => Vector.tabulate(3,fn x =>(x*9+1+y*3,x*9+3+y*3))) val boxes2 = List.tabulate(3,fn y => Vector.tabulate(3,fn x =>(x*9+28+y*3,x*9+30+y*3))) val boxes3 = List.tabulate(3,fn y => Vector.tabulate(3,fn x =>(x*9+55+y*3,x*9+57+y*3))) val boxes = List.concat([boxes1,boxes2,boxes3]) fun flatten([])= [] | flatten(x::xs)= x@flatten(xs) fun box(x,y)= flatten(List.tabulate(3,fn k => List.tabulate(3,fn z =>(k+x,z+y)))) fun constr l = List.app(fn y => let val tmp1 = FS.Value.make(space,y) in Vector.app(fn x => let val tmp2 = FS.setvar space in FS.relOp(space,x,FS.INTER,tmp1,FS.SEQ,tmp2); FS.cardRange(space,1,1,tmp2) end)numbers end)l fun fsDisjoint ([]) = () | fsDisjoint (x::xs) = (List.app(fn y => FS.rel(space,y, FS.DISJ,x))xs; fsDisjoint(xs)) in List.app(fn(x,y,z) => FD.relI(space,Vector.sub (Vector.sub(grid,x),y),FD.EQ,z))inputlist; (* distinct values in rows *) Vector.app(fn x => FD.distinct(space,x,FD.DOM))grid; (* distinct values in columns *) Vector.appi(fn(i,y)=> FD.distinct(space,Vector.map(fn x => Vector.sub(x,i))grid,FD.DOM))grid; (* distinct values in 3 x 3 boxes *) Vector.app(fn(k,l)=> let val box' = Vector.map(fn(x,y) => Vector.sub(Vector.sub(grid,x),y)) (Vector.fromList(box(k,l))) in FD.distinct(space,box',FD.DOM) end)(#[(0,0),(0,3),(0,6),(3,0),(3,3),(3,6),(6,0),(6,3),(6,6)]); Vector.app(fn x => FS.cardRange(space,9,9,x))numbers; (* the domains of all numbers are pairwise distinct *) fsDisjoint(Vector.toList numbers); (* distinct numbers in rows *) List.app(fn(y,z)=> let val tmp1 = FS.Value.make(space,#[(y,z)]) in Vector.app(fn x => let val tmp2 = FS.setvar space in FS.relOp(space,x,FS.INTER,tmp1,FS.SEQ,tmp2); FS.cardRange(space,1,1,tmp2) end)numbers end)rows; (* distinct numbers in columns *) constr domainvecs; (* distinct numbers in 3 x 3 boxes *) constr boxes; (* channeling constraints: grid'[i] = j <=> i+1 in numbers[j-1] *) List.app(fn i => List.app(fn j => let val bvar = FD.boolvar space in FD.Reified.relI(space,Vector.sub(grid',i-1), FD.EQ,j,bvar); FS.domR(space,Vector.sub(numbers,j-1), FS.SUP,#[(i,i)],bvar) end) (List.tabulate(9,fn x => x+1))) (List.tabulate(81,fn x => x+1)); FS.setvarbranch(space,numbers,FS.FSB_MIN_CARD,FS.FSB_MIN); numbers end

Below, the three realizations of the a hard Sudoku Problem 12.2.1 are compared:

Problem realized with | Choices | Failures | Solutions |

finite domain constraints | 28 | 28 | 1 |

finite set constraints | 12 | 12 | 1 |

combined model | 10 | 10 | 1 |

Andreas Rossberg 2006-08-28